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HW 2 Ch 16 18

Due 1159pm on Tuesday September 15 2015

To understand how points are awarded read the Grading Policy for this assignment

Storing Ammonia

Ammonia ( ) is a colorless pungent gas at standard pressure and temperature It is a natural metabolic byproduc

usually getting released in respiration sweat and urine However because ammonia is caustic exposure to large

quantities of it can cause illness and even death Despite its inherent dangers ammonia is environmentally friendly in

small quantities and has many applications in our economy It can be used as a fertilizer as a source of hydrogen gas

for welding as a refrigerant in the production of nitric acid and sodium carbonate in metallugy and in the infamous

smelling salts used to revive unconscious people Ammonia today can be mass produced inexpensively in chemical

refineries

To safely produce and store ammonia its physical and thermodynamic properties must be understood Physically

ammonia is a strong base that reacts with acids and metals The thermodynamic properties describing the phases of

ammonia (solid liquid and gas) and the transitions between the phases are just as important The relationship of thes

phases to pressure and temperature is quantitatively described by ammonias pT phase diagram Note that in this

diagram the pressure axis is not to scale

From the diagram the melting temperature boilingtemperature and other quantities can be determined for any

pressure The pressure and temperature range for each of the

phases is shown by its own unique area of the graph The

lines bounding each of the phases on the diagram represent

the temperatures and pressures at which two states can

coexist

For this problem any section of curve on the diagram can be

named using two letters on the boundary in alphabetical

order Other points not lying on the boundary can also be

used to help identify various thermodynamic processes

Part A

On the phase diagram which section of curve represents the pressure and temperature values at which ammonia

will boil

Express your answer as two letters that lie on a section of the appropriate curve

Hint 1 Describe boiling in terms of phase changes

Boiling is the transition from one phase to another with both phases existing together The phases and

direction of change involved in boiling are __________ to __________

Express your answer as two words separated by commas in the order they appear in the sentence

Choose from the following list gas liquid or solid

ANSWER

N H

3

liquidgasTypesetting math 100

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Correct

ANSWER

Correct

Part B

The line between which two points would describe a process of liquid ammonia boiling completely away

Express the answer as two letters representing the endpoints of the line in order so that going from the fir

letter to the second letter would show a process of boiling Be careful to put the letters in the correct order

ANSWER

Correct

Part C

On the phase diagram which section of curve represents the pressure and temperature values at which ammonia

will sublimate

Express the answer as two letters that lie on a section of the appropriate curve

Hint 1 Describe the process of sublimation

The phases and direction of change for sublimation are __________ to __________

Express your answer as two words separated by a comma in the order they appear in the sentence

Choose from the following list gas liquid or solid

ANSWER

ANSWER

Correct

boiling curve = CE

HG

sublimation curve = BC

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Part D

The line between which two points would describe a process of sublimation for ammonia

Express your answer with two letters ordered in the direction of sublimation Be careful to put the letters i

the correct order

ANSWER

Correct

The heat added to a substance undergoing sublimation must be equal to the heat of fusion plus the heat of

vaporization

Part E

On the phase diagram which section of curve represents the pressure and temperature values at which ammonia

will melt

Express the answer as two letters that lie on a section of the appropriate curve

ANSWER

Correct

Part F

The line between which two points would describe the process of complete melting of ammonia

Express your answer as two letters ordered in the direction of melting Be careful to put the letters in the

correct order

ANSWER

Correct

Part G

One of the most important points on a phase diagram is the triple point where gas liquid and solid phases all ca

exist at once What are the coordinates ( ) of the triple point of ammonia in the diagram

Express your answer as an ordered pair Determine the temperature to the nearest 5 and the pressure to

one significant digit

AF

melting curve = CD

AH

T

t r i p l e

983152

t r i p l e

K

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Hint 1 Determine the letter name for the triple point

Given that the triple point is the pressure and temperature at which all three phases can coexist what is the

letter name for the triple point of ammonia

Express the answer as a single letter

ANSWER

Correct

ANSWER

Correct

Temperature scales were originally based upon the melting and boiling points of a substance but these values

vary with pressure as seen in the phase diagram of ammonia So if one did not control and measure the

pressure precisely the measurement used to calibrate the temperature scale would be inaccurate

To circumvent this problem modern temperature scales are based on the triple point of water The triple point

temperature of water is 001 at 0006 The three phases of water will not coexist at any other

temperature regardless of the pressure so the temperature can be calibrated without an additional pressure

measurement The triple points of mercury and other substances are also used as standards for calibrating

thermometers

Part H

At one atmosphere of pressure and temperatures above ammonia exists as a gas For transportation

ammonia is stored as a liquid under its own vapor pressure This means that the liquid and gas phases exist

simultaneously If a container of ammonia is transported in an temperature-controlled truck that is maintained at n

greater than 330 what maximum pressure must the sides of the container be able to withstand

Express the answer numerically in atmospheres to one significant figure

Hint 1 How to approach the problem

Because the liquid and vapor forms both exist in the container find the point on the graph where the given

temperature intersects the curve representing the phase change of liquid to gas

ANSWER

triple point = C

= 195005T

t r i p l e

983152

t r i p l e

C

∘

a t m

minus C 3 3 3

∘

K 983152

= 8 983152 a t m

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Correct

plusmn Cooling a Soft Drink

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 199

Part A

What is the magnitude of its temperature change 199 in degrees Celsius

Hint 1 How to approach the problem

The equation to convert a temperature from Celsius to Kelvin is To determine how to

convert a temperature change from Kelvin to Celsius imagine increasing by 1 degree by how much

would have to change for the temperature equation to remain correct

ANSWER

Correct

A simple way to figure out how the change in temperature in one scale is related to the change in temperature

in another scale is to start with the equation relating the temperature in the two scales and then find the slope

of the line that represents this equation In the example here so the slope ie the

factor in front of is 1 This implies that In other words one degree Kelvin is equal in size

to one degree Celsius

Part B

What is the magnitude of the temperature change 199 in degrees Fahrenheit

Hint 1 Equation for Fahrenheit temperature conversion

The equation to convert a temperature from Celsius to Fahrenheit is To convert achange in temperature try the slope method explained in the comment following Part A

ANSWER

K

| Δ T

| = K

= + 2 7 3 1 5 T

K

T

C

T

K

T

C

| | = 199Δ T C

∘

= + 2 7 3 1 5 T

K

T

C

T

C

Δ = Δ T

K

T

C

| Δ T

| = K

= + 3 2 T

F

9

5

T

C

| | = 358Δ T

F

∘

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Correct

In converting a temperature change from one temperature scale to the other only a multiplicative factor is

needed and not the additive factor that is also used for temperature conversions

Introduction to the Ideal Gas Law

Learning Goal

To understand the ideal gas law and be able to apply it to a wide variety of situations

The absolute temperature volume and pressure of a gas sample are related by the ideal gas law which state

that

Here is the number of moles in the gas sample and is a gas constant that applies to all gases This empirical law

describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the

verge of condensing

In applying the ideal gas law must be the absolute pressure measured with respect to vacuum and not with respecto atmospheric pressure and must be the absolute temperature measured in kelvins (that is with respect to

absolute zero defined throughout this tutorial as ) If is in pascals and is in cubic meters use

If is in atmospheres and is in liters use instead

Part A

A gas sample enclosed in a rigid metal container at room temperature (200 ) has an absolute pressure The

container is immersed in hot water until it warms to 400 What is the new absolute pressure

Express your answer in terms of

Hint 1 How to approach the problem

To find the final pressure you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

Hint 2 Convert temperatures to kelvins

To apply the ideal gas law all temperatures must be in absolute units (ie in kelvins) What is the initial

temperature in kelvins

ANSWER

T V 983152

983152 V =

983150 R T

983150 R

983152

T

minus C 2 7 3

∘

983152 V

R = 8 3 1 4 5 J ( m o l sdot K ) 983152 V R = 0 0 8 2 0 6 L sdot a t m ( m o l sdot K )

C

∘

983152

1

C

∘

983152

2

983152

1

R

983152

V

983150

T

T

1

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ANSWER

Correct

This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent Note

that it is critical for the temperatures to be converted to absolute units If you had used Celsius temperatures

you would have predicted that the pressure should double which is far greater than the actual increase

Part B

Nitrogen gas is introduced into a large deflated plastic bag No gas is allowed to escape but as more and more

nitrogen is added the bag inflates to accommodate it The pressure of the gas within the bag remains at 100

and its temperature remains at room temperature (200 ) How many moles have been introduced into the ba

by the time its volume reaches 224

Express your answer in moles

Hint 1 How to approach the problem

Rearrange the ideal gas law to isolate Be sure to use the value for in units that are consistent with the

rest of the problem and hence will cancel out to leave moles at the end

ANSWER

Correct

One mole of gas occupies 224 at STP (standard temperature and pressure 0 and 1 ) This fact

may be worth memorizing In this problem the temperature is slightly higher than STP so the gas expands

and 224 can be filled by slightly less than 1 of gas

Part C

Some hydrogen gas is enclosed within a chamber being held at 200 with a volume of 00250 The chambe

is fitted with a movable piston Initially the pressure in the gas is (148 ) The piston is slow

=

0

20

100

273

293

T

1

K

= 983152

2

1 0 6 8 983152

1

a t m

C

∘

983150

L

983150 R

= 0932983150 m o l

L C

∘

a t m

L m o l

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a

6

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

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Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

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You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

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Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

ANSWER

Correct

Part B

The line between which two points would describe a process of liquid ammonia boiling completely away

Express the answer as two letters representing the endpoints of the line in order so that going from the fir

letter to the second letter would show a process of boiling Be careful to put the letters in the correct order

ANSWER

Correct

Part C

On the phase diagram which section of curve represents the pressure and temperature values at which ammonia

will sublimate

Express the answer as two letters that lie on a section of the appropriate curve

Hint 1 Describe the process of sublimation

The phases and direction of change for sublimation are __________ to __________

Express your answer as two words separated by a comma in the order they appear in the sentence

Choose from the following list gas liquid or solid

ANSWER

ANSWER

Correct

boiling curve = CE

HG

sublimation curve = BC

7212019 Mastering Physics HW 2 Ch 16 18

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Part D

The line between which two points would describe a process of sublimation for ammonia

Express your answer with two letters ordered in the direction of sublimation Be careful to put the letters i

the correct order

ANSWER

Correct

The heat added to a substance undergoing sublimation must be equal to the heat of fusion plus the heat of

vaporization

Part E

On the phase diagram which section of curve represents the pressure and temperature values at which ammonia

will melt

Express the answer as two letters that lie on a section of the appropriate curve

ANSWER

Correct

Part F

The line between which two points would describe the process of complete melting of ammonia

Express your answer as two letters ordered in the direction of melting Be careful to put the letters in the

correct order

ANSWER

Correct

Part G

One of the most important points on a phase diagram is the triple point where gas liquid and solid phases all ca

exist at once What are the coordinates ( ) of the triple point of ammonia in the diagram

Express your answer as an ordered pair Determine the temperature to the nearest 5 and the pressure to

one significant digit

AF

melting curve = CD

AH

T

t r i p l e

983152

t r i p l e

K

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Hint 1 Determine the letter name for the triple point

Given that the triple point is the pressure and temperature at which all three phases can coexist what is the

letter name for the triple point of ammonia

Express the answer as a single letter

ANSWER

Correct

ANSWER

Correct

Temperature scales were originally based upon the melting and boiling points of a substance but these values

vary with pressure as seen in the phase diagram of ammonia So if one did not control and measure the

pressure precisely the measurement used to calibrate the temperature scale would be inaccurate

To circumvent this problem modern temperature scales are based on the triple point of water The triple point

temperature of water is 001 at 0006 The three phases of water will not coexist at any other

temperature regardless of the pressure so the temperature can be calibrated without an additional pressure

measurement The triple points of mercury and other substances are also used as standards for calibrating

thermometers

Part H

At one atmosphere of pressure and temperatures above ammonia exists as a gas For transportation

ammonia is stored as a liquid under its own vapor pressure This means that the liquid and gas phases exist

simultaneously If a container of ammonia is transported in an temperature-controlled truck that is maintained at n

greater than 330 what maximum pressure must the sides of the container be able to withstand

Express the answer numerically in atmospheres to one significant figure

Hint 1 How to approach the problem

Because the liquid and vapor forms both exist in the container find the point on the graph where the given

temperature intersects the curve representing the phase change of liquid to gas

ANSWER

triple point = C

= 195005T

t r i p l e

983152

t r i p l e

C

∘

a t m

minus C 3 3 3

∘

K 983152

= 8 983152 a t m

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

plusmn Cooling a Soft Drink

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 199

Part A

What is the magnitude of its temperature change 199 in degrees Celsius

Hint 1 How to approach the problem

The equation to convert a temperature from Celsius to Kelvin is To determine how to

convert a temperature change from Kelvin to Celsius imagine increasing by 1 degree by how much

would have to change for the temperature equation to remain correct

ANSWER

Correct

A simple way to figure out how the change in temperature in one scale is related to the change in temperature

in another scale is to start with the equation relating the temperature in the two scales and then find the slope

of the line that represents this equation In the example here so the slope ie the

factor in front of is 1 This implies that In other words one degree Kelvin is equal in size

to one degree Celsius

Part B

What is the magnitude of the temperature change 199 in degrees Fahrenheit

Hint 1 Equation for Fahrenheit temperature conversion

The equation to convert a temperature from Celsius to Fahrenheit is To convert achange in temperature try the slope method explained in the comment following Part A

ANSWER

K

| Δ T

| = K

= + 2 7 3 1 5 T

K

T

C

T

K

T

C

| | = 199Δ T C

∘

= + 2 7 3 1 5 T

K

T

C

T

C

Δ = Δ T

K

T

C

| Δ T

| = K

= + 3 2 T

F

9

5

T

C

| | = 358Δ T

F

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

In converting a temperature change from one temperature scale to the other only a multiplicative factor is

needed and not the additive factor that is also used for temperature conversions

Introduction to the Ideal Gas Law

Learning Goal

To understand the ideal gas law and be able to apply it to a wide variety of situations

The absolute temperature volume and pressure of a gas sample are related by the ideal gas law which state

that

Here is the number of moles in the gas sample and is a gas constant that applies to all gases This empirical law

describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the

verge of condensing

In applying the ideal gas law must be the absolute pressure measured with respect to vacuum and not with respecto atmospheric pressure and must be the absolute temperature measured in kelvins (that is with respect to

absolute zero defined throughout this tutorial as ) If is in pascals and is in cubic meters use

If is in atmospheres and is in liters use instead

Part A

A gas sample enclosed in a rigid metal container at room temperature (200 ) has an absolute pressure The

container is immersed in hot water until it warms to 400 What is the new absolute pressure

Express your answer in terms of

Hint 1 How to approach the problem

To find the final pressure you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

Hint 2 Convert temperatures to kelvins

To apply the ideal gas law all temperatures must be in absolute units (ie in kelvins) What is the initial

temperature in kelvins

ANSWER

T V 983152

983152 V =

983150 R T

983150 R

983152

T

minus C 2 7 3

∘

983152 V

R = 8 3 1 4 5 J ( m o l sdot K ) 983152 V R = 0 0 8 2 0 6 L sdot a t m ( m o l sdot K )

C

∘

983152

1

C

∘

983152

2

983152

1

R

983152

V

983150

T

T

1

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent Note

that it is critical for the temperatures to be converted to absolute units If you had used Celsius temperatures

you would have predicted that the pressure should double which is far greater than the actual increase

Part B

Nitrogen gas is introduced into a large deflated plastic bag No gas is allowed to escape but as more and more

nitrogen is added the bag inflates to accommodate it The pressure of the gas within the bag remains at 100

and its temperature remains at room temperature (200 ) How many moles have been introduced into the ba

by the time its volume reaches 224

Express your answer in moles

Hint 1 How to approach the problem

Rearrange the ideal gas law to isolate Be sure to use the value for in units that are consistent with the

rest of the problem and hence will cancel out to leave moles at the end

ANSWER

Correct

One mole of gas occupies 224 at STP (standard temperature and pressure 0 and 1 ) This fact

may be worth memorizing In this problem the temperature is slightly higher than STP so the gas expands

and 224 can be filled by slightly less than 1 of gas

Part C

Some hydrogen gas is enclosed within a chamber being held at 200 with a volume of 00250 The chambe

is fitted with a movable piston Initially the pressure in the gas is (148 ) The piston is slow

=

0

20

100

273

293

T

1

K

= 983152

2

1 0 6 8 983152

1

a t m

C

∘

983150

L

983150 R

= 0932983150 m o l

L C

∘

a t m

L m o l

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a

6

7212019 Mastering Physics HW 2 Ch 16 18

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

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Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

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You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

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Part D

The line between which two points would describe a process of sublimation for ammonia

Express your answer with two letters ordered in the direction of sublimation Be careful to put the letters i

the correct order

ANSWER

Correct

The heat added to a substance undergoing sublimation must be equal to the heat of fusion plus the heat of

vaporization

Part E

On the phase diagram which section of curve represents the pressure and temperature values at which ammonia

will melt

Express the answer as two letters that lie on a section of the appropriate curve

ANSWER

Correct

Part F

The line between which two points would describe the process of complete melting of ammonia

Express your answer as two letters ordered in the direction of melting Be careful to put the letters in the

correct order

ANSWER

Correct

Part G

One of the most important points on a phase diagram is the triple point where gas liquid and solid phases all ca

exist at once What are the coordinates ( ) of the triple point of ammonia in the diagram

Express your answer as an ordered pair Determine the temperature to the nearest 5 and the pressure to

one significant digit

AF

melting curve = CD

AH

T

t r i p l e

983152

t r i p l e

K

7212019 Mastering Physics HW 2 Ch 16 18

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Hint 1 Determine the letter name for the triple point

Given that the triple point is the pressure and temperature at which all three phases can coexist what is the

letter name for the triple point of ammonia

Express the answer as a single letter

ANSWER

Correct

ANSWER

Correct

Temperature scales were originally based upon the melting and boiling points of a substance but these values

vary with pressure as seen in the phase diagram of ammonia So if one did not control and measure the

pressure precisely the measurement used to calibrate the temperature scale would be inaccurate

To circumvent this problem modern temperature scales are based on the triple point of water The triple point

temperature of water is 001 at 0006 The three phases of water will not coexist at any other

temperature regardless of the pressure so the temperature can be calibrated without an additional pressure

measurement The triple points of mercury and other substances are also used as standards for calibrating

thermometers

Part H

At one atmosphere of pressure and temperatures above ammonia exists as a gas For transportation

ammonia is stored as a liquid under its own vapor pressure This means that the liquid and gas phases exist

simultaneously If a container of ammonia is transported in an temperature-controlled truck that is maintained at n

greater than 330 what maximum pressure must the sides of the container be able to withstand

Express the answer numerically in atmospheres to one significant figure

Hint 1 How to approach the problem

Because the liquid and vapor forms both exist in the container find the point on the graph where the given

temperature intersects the curve representing the phase change of liquid to gas

ANSWER

triple point = C

= 195005T

t r i p l e

983152

t r i p l e

C

∘

a t m

minus C 3 3 3

∘

K 983152

= 8 983152 a t m

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

plusmn Cooling a Soft Drink

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 199

Part A

What is the magnitude of its temperature change 199 in degrees Celsius

Hint 1 How to approach the problem

The equation to convert a temperature from Celsius to Kelvin is To determine how to

convert a temperature change from Kelvin to Celsius imagine increasing by 1 degree by how much

would have to change for the temperature equation to remain correct

ANSWER

Correct

A simple way to figure out how the change in temperature in one scale is related to the change in temperature

in another scale is to start with the equation relating the temperature in the two scales and then find the slope

of the line that represents this equation In the example here so the slope ie the

factor in front of is 1 This implies that In other words one degree Kelvin is equal in size

to one degree Celsius

Part B

What is the magnitude of the temperature change 199 in degrees Fahrenheit

Hint 1 Equation for Fahrenheit temperature conversion

The equation to convert a temperature from Celsius to Fahrenheit is To convert achange in temperature try the slope method explained in the comment following Part A

ANSWER

K

| Δ T

| = K

= + 2 7 3 1 5 T

K

T

C

T

K

T

C

| | = 199Δ T C

∘

= + 2 7 3 1 5 T

K

T

C

T

C

Δ = Δ T

K

T

C

| Δ T

| = K

= + 3 2 T

F

9

5

T

C

| | = 358Δ T

F

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

In converting a temperature change from one temperature scale to the other only a multiplicative factor is

needed and not the additive factor that is also used for temperature conversions

Introduction to the Ideal Gas Law

Learning Goal

To understand the ideal gas law and be able to apply it to a wide variety of situations

The absolute temperature volume and pressure of a gas sample are related by the ideal gas law which state

that

Here is the number of moles in the gas sample and is a gas constant that applies to all gases This empirical law

describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the

verge of condensing

In applying the ideal gas law must be the absolute pressure measured with respect to vacuum and not with respecto atmospheric pressure and must be the absolute temperature measured in kelvins (that is with respect to

absolute zero defined throughout this tutorial as ) If is in pascals and is in cubic meters use

If is in atmospheres and is in liters use instead

Part A

A gas sample enclosed in a rigid metal container at room temperature (200 ) has an absolute pressure The

container is immersed in hot water until it warms to 400 What is the new absolute pressure

Express your answer in terms of

Hint 1 How to approach the problem

To find the final pressure you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

Hint 2 Convert temperatures to kelvins

To apply the ideal gas law all temperatures must be in absolute units (ie in kelvins) What is the initial

temperature in kelvins

ANSWER

T V 983152

983152 V =

983150 R T

983150 R

983152

T

minus C 2 7 3

∘

983152 V

R = 8 3 1 4 5 J ( m o l sdot K ) 983152 V R = 0 0 8 2 0 6 L sdot a t m ( m o l sdot K )

C

∘

983152

1

C

∘

983152

2

983152

1

R

983152

V

983150

T

T

1

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent Note

that it is critical for the temperatures to be converted to absolute units If you had used Celsius temperatures

you would have predicted that the pressure should double which is far greater than the actual increase

Part B

Nitrogen gas is introduced into a large deflated plastic bag No gas is allowed to escape but as more and more

nitrogen is added the bag inflates to accommodate it The pressure of the gas within the bag remains at 100

and its temperature remains at room temperature (200 ) How many moles have been introduced into the ba

by the time its volume reaches 224

Express your answer in moles

Hint 1 How to approach the problem

Rearrange the ideal gas law to isolate Be sure to use the value for in units that are consistent with the

rest of the problem and hence will cancel out to leave moles at the end

ANSWER

Correct

One mole of gas occupies 224 at STP (standard temperature and pressure 0 and 1 ) This fact

may be worth memorizing In this problem the temperature is slightly higher than STP so the gas expands

and 224 can be filled by slightly less than 1 of gas

Part C

Some hydrogen gas is enclosed within a chamber being held at 200 with a volume of 00250 The chambe

is fitted with a movable piston Initially the pressure in the gas is (148 ) The piston is slow

=

0

20

100

273

293

T

1

K

= 983152

2

1 0 6 8 983152

1

a t m

C

∘

983150

L

983150 R

= 0932983150 m o l

L C

∘

a t m

L m o l

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a

6

7212019 Mastering Physics HW 2 Ch 16 18

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

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Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

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You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

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Hint 1 Determine the letter name for the triple point

Given that the triple point is the pressure and temperature at which all three phases can coexist what is the

letter name for the triple point of ammonia

Express the answer as a single letter

ANSWER

Correct

ANSWER

Correct

Temperature scales were originally based upon the melting and boiling points of a substance but these values

vary with pressure as seen in the phase diagram of ammonia So if one did not control and measure the

pressure precisely the measurement used to calibrate the temperature scale would be inaccurate

To circumvent this problem modern temperature scales are based on the triple point of water The triple point

temperature of water is 001 at 0006 The three phases of water will not coexist at any other

temperature regardless of the pressure so the temperature can be calibrated without an additional pressure

measurement The triple points of mercury and other substances are also used as standards for calibrating

thermometers

Part H

At one atmosphere of pressure and temperatures above ammonia exists as a gas For transportation

ammonia is stored as a liquid under its own vapor pressure This means that the liquid and gas phases exist

simultaneously If a container of ammonia is transported in an temperature-controlled truck that is maintained at n

greater than 330 what maximum pressure must the sides of the container be able to withstand

Express the answer numerically in atmospheres to one significant figure

Hint 1 How to approach the problem

Because the liquid and vapor forms both exist in the container find the point on the graph where the given

temperature intersects the curve representing the phase change of liquid to gas

ANSWER

triple point = C

= 195005T

t r i p l e

983152

t r i p l e

C

∘

a t m

minus C 3 3 3

∘

K 983152

= 8 983152 a t m

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

plusmn Cooling a Soft Drink

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 199

Part A

What is the magnitude of its temperature change 199 in degrees Celsius

Hint 1 How to approach the problem

The equation to convert a temperature from Celsius to Kelvin is To determine how to

convert a temperature change from Kelvin to Celsius imagine increasing by 1 degree by how much

would have to change for the temperature equation to remain correct

ANSWER

Correct

A simple way to figure out how the change in temperature in one scale is related to the change in temperature

in another scale is to start with the equation relating the temperature in the two scales and then find the slope

of the line that represents this equation In the example here so the slope ie the

factor in front of is 1 This implies that In other words one degree Kelvin is equal in size

to one degree Celsius

Part B

What is the magnitude of the temperature change 199 in degrees Fahrenheit

Hint 1 Equation for Fahrenheit temperature conversion

The equation to convert a temperature from Celsius to Fahrenheit is To convert achange in temperature try the slope method explained in the comment following Part A

ANSWER

K

| Δ T

| = K

= + 2 7 3 1 5 T

K

T

C

T

K

T

C

| | = 199Δ T C

∘

= + 2 7 3 1 5 T

K

T

C

T

C

Δ = Δ T

K

T

C

| Δ T

| = K

= + 3 2 T

F

9

5

T

C

| | = 358Δ T

F

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

In converting a temperature change from one temperature scale to the other only a multiplicative factor is

needed and not the additive factor that is also used for temperature conversions

Introduction to the Ideal Gas Law

Learning Goal

To understand the ideal gas law and be able to apply it to a wide variety of situations

The absolute temperature volume and pressure of a gas sample are related by the ideal gas law which state

that

Here is the number of moles in the gas sample and is a gas constant that applies to all gases This empirical law

describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the

verge of condensing

In applying the ideal gas law must be the absolute pressure measured with respect to vacuum and not with respecto atmospheric pressure and must be the absolute temperature measured in kelvins (that is with respect to

absolute zero defined throughout this tutorial as ) If is in pascals and is in cubic meters use

If is in atmospheres and is in liters use instead

Part A

A gas sample enclosed in a rigid metal container at room temperature (200 ) has an absolute pressure The

container is immersed in hot water until it warms to 400 What is the new absolute pressure

Express your answer in terms of

Hint 1 How to approach the problem

To find the final pressure you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

Hint 2 Convert temperatures to kelvins

To apply the ideal gas law all temperatures must be in absolute units (ie in kelvins) What is the initial

temperature in kelvins

ANSWER

T V 983152

983152 V =

983150 R T

983150 R

983152

T

minus C 2 7 3

∘

983152 V

R = 8 3 1 4 5 J ( m o l sdot K ) 983152 V R = 0 0 8 2 0 6 L sdot a t m ( m o l sdot K )

C

∘

983152

1

C

∘

983152

2

983152

1

R

983152

V

983150

T

T

1

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent Note

that it is critical for the temperatures to be converted to absolute units If you had used Celsius temperatures

you would have predicted that the pressure should double which is far greater than the actual increase

Part B

Nitrogen gas is introduced into a large deflated plastic bag No gas is allowed to escape but as more and more

nitrogen is added the bag inflates to accommodate it The pressure of the gas within the bag remains at 100

and its temperature remains at room temperature (200 ) How many moles have been introduced into the ba

by the time its volume reaches 224

Express your answer in moles

Hint 1 How to approach the problem

Rearrange the ideal gas law to isolate Be sure to use the value for in units that are consistent with the

rest of the problem and hence will cancel out to leave moles at the end

ANSWER

Correct

One mole of gas occupies 224 at STP (standard temperature and pressure 0 and 1 ) This fact

may be worth memorizing In this problem the temperature is slightly higher than STP so the gas expands

and 224 can be filled by slightly less than 1 of gas

Part C

Some hydrogen gas is enclosed within a chamber being held at 200 with a volume of 00250 The chambe

is fitted with a movable piston Initially the pressure in the gas is (148 ) The piston is slow

=

0

20

100

273

293

T

1

K

= 983152

2

1 0 6 8 983152

1

a t m

C

∘

983150

L

983150 R

= 0932983150 m o l

L C

∘

a t m

L m o l

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a

6

7212019 Mastering Physics HW 2 Ch 16 18

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

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Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

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You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

plusmn Cooling a Soft Drink

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 199

Part A

What is the magnitude of its temperature change 199 in degrees Celsius

Hint 1 How to approach the problem

The equation to convert a temperature from Celsius to Kelvin is To determine how to

convert a temperature change from Kelvin to Celsius imagine increasing by 1 degree by how much

would have to change for the temperature equation to remain correct

ANSWER

Correct

A simple way to figure out how the change in temperature in one scale is related to the change in temperature

in another scale is to start with the equation relating the temperature in the two scales and then find the slope

of the line that represents this equation In the example here so the slope ie the

factor in front of is 1 This implies that In other words one degree Kelvin is equal in size

to one degree Celsius

Part B

What is the magnitude of the temperature change 199 in degrees Fahrenheit

Hint 1 Equation for Fahrenheit temperature conversion

The equation to convert a temperature from Celsius to Fahrenheit is To convert achange in temperature try the slope method explained in the comment following Part A

ANSWER

K

| Δ T

| = K

= + 2 7 3 1 5 T

K

T

C

T

K

T

C

| | = 199Δ T C

∘

= + 2 7 3 1 5 T

K

T

C

T

C

Δ = Δ T

K

T

C

| Δ T

| = K

= + 3 2 T

F

9

5

T

C

| | = 358Δ T

F

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

In converting a temperature change from one temperature scale to the other only a multiplicative factor is

needed and not the additive factor that is also used for temperature conversions

Introduction to the Ideal Gas Law

Learning Goal

To understand the ideal gas law and be able to apply it to a wide variety of situations

The absolute temperature volume and pressure of a gas sample are related by the ideal gas law which state

that

Here is the number of moles in the gas sample and is a gas constant that applies to all gases This empirical law

describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the

verge of condensing

In applying the ideal gas law must be the absolute pressure measured with respect to vacuum and not with respecto atmospheric pressure and must be the absolute temperature measured in kelvins (that is with respect to

absolute zero defined throughout this tutorial as ) If is in pascals and is in cubic meters use

If is in atmospheres and is in liters use instead

Part A

A gas sample enclosed in a rigid metal container at room temperature (200 ) has an absolute pressure The

container is immersed in hot water until it warms to 400 What is the new absolute pressure

Express your answer in terms of

Hint 1 How to approach the problem

To find the final pressure you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

Hint 2 Convert temperatures to kelvins

To apply the ideal gas law all temperatures must be in absolute units (ie in kelvins) What is the initial

temperature in kelvins

ANSWER

T V 983152

983152 V =

983150 R T

983150 R

983152

T

minus C 2 7 3

∘

983152 V

R = 8 3 1 4 5 J ( m o l sdot K ) 983152 V R = 0 0 8 2 0 6 L sdot a t m ( m o l sdot K )

C

∘

983152

1

C

∘

983152

2

983152

1

R

983152

V

983150

T

T

1

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent Note

that it is critical for the temperatures to be converted to absolute units If you had used Celsius temperatures

you would have predicted that the pressure should double which is far greater than the actual increase

Part B

Nitrogen gas is introduced into a large deflated plastic bag No gas is allowed to escape but as more and more

nitrogen is added the bag inflates to accommodate it The pressure of the gas within the bag remains at 100

and its temperature remains at room temperature (200 ) How many moles have been introduced into the ba

by the time its volume reaches 224

Express your answer in moles

Hint 1 How to approach the problem

Rearrange the ideal gas law to isolate Be sure to use the value for in units that are consistent with the

rest of the problem and hence will cancel out to leave moles at the end

ANSWER

Correct

One mole of gas occupies 224 at STP (standard temperature and pressure 0 and 1 ) This fact

may be worth memorizing In this problem the temperature is slightly higher than STP so the gas expands

and 224 can be filled by slightly less than 1 of gas

Part C

Some hydrogen gas is enclosed within a chamber being held at 200 with a volume of 00250 The chambe

is fitted with a movable piston Initially the pressure in the gas is (148 ) The piston is slow

=

0

20

100

273

293

T

1

K

= 983152

2

1 0 6 8 983152

1

a t m

C

∘

983150

L

983150 R

= 0932983150 m o l

L C

∘

a t m

L m o l

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a

6

7212019 Mastering Physics HW 2 Ch 16 18

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

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Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

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You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 624

Correct

In converting a temperature change from one temperature scale to the other only a multiplicative factor is

needed and not the additive factor that is also used for temperature conversions

Introduction to the Ideal Gas Law

Learning Goal

To understand the ideal gas law and be able to apply it to a wide variety of situations

The absolute temperature volume and pressure of a gas sample are related by the ideal gas law which state

that

Here is the number of moles in the gas sample and is a gas constant that applies to all gases This empirical law

describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the

verge of condensing

In applying the ideal gas law must be the absolute pressure measured with respect to vacuum and not with respecto atmospheric pressure and must be the absolute temperature measured in kelvins (that is with respect to

absolute zero defined throughout this tutorial as ) If is in pascals and is in cubic meters use

If is in atmospheres and is in liters use instead

Part A

A gas sample enclosed in a rigid metal container at room temperature (200 ) has an absolute pressure The

container is immersed in hot water until it warms to 400 What is the new absolute pressure

Express your answer in terms of

Hint 1 How to approach the problem

To find the final pressure you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

Hint 2 Convert temperatures to kelvins

To apply the ideal gas law all temperatures must be in absolute units (ie in kelvins) What is the initial

temperature in kelvins

ANSWER

T V 983152

983152 V =

983150 R T

983150 R

983152

T

minus C 2 7 3

∘

983152 V

R = 8 3 1 4 5 J ( m o l sdot K ) 983152 V R = 0 0 8 2 0 6 L sdot a t m ( m o l sdot K )

C

∘

983152

1

C

∘

983152

2

983152

1

R

983152

V

983150

T

T

1

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent Note

that it is critical for the temperatures to be converted to absolute units If you had used Celsius temperatures

you would have predicted that the pressure should double which is far greater than the actual increase

Part B

Nitrogen gas is introduced into a large deflated plastic bag No gas is allowed to escape but as more and more

nitrogen is added the bag inflates to accommodate it The pressure of the gas within the bag remains at 100

and its temperature remains at room temperature (200 ) How many moles have been introduced into the ba

by the time its volume reaches 224

Express your answer in moles

Hint 1 How to approach the problem

Rearrange the ideal gas law to isolate Be sure to use the value for in units that are consistent with the

rest of the problem and hence will cancel out to leave moles at the end

ANSWER

Correct

One mole of gas occupies 224 at STP (standard temperature and pressure 0 and 1 ) This fact

may be worth memorizing In this problem the temperature is slightly higher than STP so the gas expands

and 224 can be filled by slightly less than 1 of gas

Part C

Some hydrogen gas is enclosed within a chamber being held at 200 with a volume of 00250 The chambe

is fitted with a movable piston Initially the pressure in the gas is (148 ) The piston is slow

=

0

20

100

273

293

T

1

K

= 983152

2

1 0 6 8 983152

1

a t m

C

∘

983150

L

983150 R

= 0932983150 m o l

L C

∘

a t m

L m o l

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a

6

7212019 Mastering Physics HW 2 Ch 16 18

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

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Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

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You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent Note

that it is critical for the temperatures to be converted to absolute units If you had used Celsius temperatures

you would have predicted that the pressure should double which is far greater than the actual increase

Part B

Nitrogen gas is introduced into a large deflated plastic bag No gas is allowed to escape but as more and more

nitrogen is added the bag inflates to accommodate it The pressure of the gas within the bag remains at 100

and its temperature remains at room temperature (200 ) How many moles have been introduced into the ba

by the time its volume reaches 224

Express your answer in moles

Hint 1 How to approach the problem

Rearrange the ideal gas law to isolate Be sure to use the value for in units that are consistent with the

rest of the problem and hence will cancel out to leave moles at the end

ANSWER

Correct

One mole of gas occupies 224 at STP (standard temperature and pressure 0 and 1 ) This fact

may be worth memorizing In this problem the temperature is slightly higher than STP so the gas expands

and 224 can be filled by slightly less than 1 of gas

Part C

Some hydrogen gas is enclosed within a chamber being held at 200 with a volume of 00250 The chambe

is fitted with a movable piston Initially the pressure in the gas is (148 ) The piston is slow

=

0

20

100

273

293

T

1

K

= 983152

2

1 0 6 8 983152

1

a t m

C

∘

983150

L

983150 R

= 0932983150 m o l

L C

∘

a t m

L m o l

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a

6

7212019 Mastering Physics HW 2 Ch 16 18

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

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Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

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You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

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extracted until the pressure in the gas falls to What is the final volume of the container

Assume that no gas escapes and that the temperature remains at 200

Enter your answer numerically in cubic meters

Hint 1 How to approach the problem

To find the final volume you must first determine which quantities in the ideal gas law remain constant in

the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

ANSWER

Correct

Notice how is not needed to answer this problem and neither is although you do make use of the fact

that is a constant

Part D

Some hydrogen gas is enclosed within a chamber being held at 200 whose volume is 00250 Initially the

pressure in the gas is (148 ) The chamber is removed from the heat source and allowed to

cool until the pressure in the gas falls to At what temperature does this occur

Enter your answer in degrees Celsius

Hint 1 How to approach the problem

To find the final temperature you must first determine which quantities in the ideal gas law remain constant

in the given situation Note that is always a constant Determine which of the other four quantities are

constant for the process described in this part

Check all that apply

ANSWER

0 9 5 0 times P a 1 0

6

V

2

C

∘

R

983152

V

983150

T

= 395times10minus2 V

2

m

3

983150 T

T

C

∘

m

3

1 5 0 times P a 1 0

6

a t m

0 9 5 0 times P a 1 0

6

T

2

R

7212019 Mastering Physics HW 2 Ch 16 18

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ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

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Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

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Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

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Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 924

ANSWER

Correct

This final temperature happens to be close to room temperature Hydrogen remains a gas to temperatures well

below that but if this question had been about water vapor for example the gas would have condensed to

liquid water at 100 and the ideal gas law would no longer have applied

Understanding pV Diagrams

Learning Goal

To understand the meaning and the basic applications of pV diagrams for an ideal gas

As you know the parameters of an ideal gas are described by the equation

where is the pressure of the gas is the volume of the gas is the number of moles is the universal gas

constant and is the absolute temperature of the gas It follows that for a portion of an ideal gas

One can see that if the amount of gas remains constant it is impossible to change just one parameter of the gas At

least one more parameter would also change For instance if the pressure of the gas is changed we can be sure that

either the volume or the temperature of the gas (or maybe both) would also change

To explore these changes it is often convenient to draw a graph showing one parameter as a function of the other

Although there are many choices of axes the most common one is a plot of pressure as a function of volume a pV

diagram

In this problem you will be asked a series of questions related to different processes shown on a pV diagram Theywill help you become familiar with such diagrams and to understand what information may be obtained from them

983152

V

983150

T

= 266T

2

C

∘

C

∘

983152 V =

983150 R T

983152 V 983150 R

T

= c o n s t a n t

983152 V

T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1024

Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1124

Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

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Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1324

Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1424

Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1524

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1024

Part A

Which of the processes are isobaric

Check all that apply

Hint 1 Definition of isobaric

Isobaric comes from the Greek terms isos meaning equal and baros meaning weight Isobaric refers to a

process in which the pressure does not change

ANSWER

Correct

Isobaric (constant-pressure) processes correspond to the horizontal lines on pV diagrams

Part B

Which of the processes are isochoric

Check all that apply

Hint 1 Definition of isochoric

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1124

Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1224

Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1324

Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1424

Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1524

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1124

Isochoric comes from the Greek words isos meaning equal and chwra meaning space Isochoric refers

to a process in which the volume does not change

ANSWER

Correct

Isochoric (constant-volume) processes correspond to the vertical lines on pV diagrams

Part C

Which of the processes may possibly be isothermal

Check all that apply

Hint 1 Definition of isothermal

Isothermal comes from the Greek words isos meaning equal and therme meaning heat or thermos

meaning hot Isothermal refers to a process in which the temperature does not change

ANSWER

Correct

For isothermal (constant-temperature) processes that is pressure is inversely proportional

to volume and the graph is a hyperbola Curve is the only graph that looks reasonably similar to a

hyperbola

In further questions assume that process is indeed an isothermal one

Part D

In which of the processes is the temperature of the gas increasing

1 rarr 2

1 rarr 3 rarr 6

1 rarr 5

6 rarr 5

1 rarr 4 rarr 6

6 rarr 2

2 rarr 1

6 rarr 3 rarr 1

1 rarr 5

5 rarr 6

6 rarr 4 rarr 1

6 rarr 2

983152 V = c o n s t a n t

1 rarr 4 rarr 6

1 rarr 4 rarr 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1224

Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1324

Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1424

Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1524

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1224

Check all that apply

ANSWER

Correct

If the temperature increases then to keep the ratio constant the product must be increasing as

well This should make sense For instance if the pressure is constant the volume is directly proportional to

temperature if the volume is kept constant the pressure of the heated gas increases directly proportional to

the temperature

Part E

During process the temperature of the gas __________

ANSWER

Correct

During process the pressure of the gas decreases more slowly than it does in the isothermal process

therefore its temperature must be increasing

During process the pressure of the gas decreases more rapidly than it does in the isothermal process

therefore its temperature must be decreasing

Problem 1661

40 of oxygen gas starting at 49 follow the process shown in the figure

2 rarr 1

1 rarr 5

5 rarr 6

6 rarr 2

983152 V

T 983152 V

1 rarr 3 rarr 6

decreases and then increases

increases and then decreases

remains constant

1 rarr 3

1 rarr 4

3 rarr 6

4 rarr 6

g C

∘

1 rarr 2

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1324

Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1424

Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1524

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1324

Part A

What is temperature (in )

Express your answer with the appropriate units

ANSWER

Correct

Problem 1639

A 60- -diameter 10- -long cylinder contains 100 of oxygen at a pressure less than 1 The cap o

one end of the cylinder is held in place only by the pressure of the air One day when the atmospheric pressure is 100

it takes a 184 force to pull the cap off

Part A

What is the temperature of the gas

Express your answer using two significant figures

ANSWER

Correct

T

2

C

∘

= 2630T

2

C

∘

c m c m m g ( ) O

2

a t m

k P a N

= 110T C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1424

Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1524

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1424

Equipartition Theorem and Microscopic Motion

Learning Goal

To understand the Equipartition Theorem and its implications for the mechanical motion of small objects

In statistical mechanics heat is the random motion of the microscopic world The average kinetic or potential energy o

each degree of freedom of the microscopic world therefore depends on the temperature If heat is added molecules

increase their translational and rotational speeds and the atoms constituting the molecules vibrate with larger amplitud

about their equilibrium positions It is a fact of nature that the energy of each degree of freedom is determined solely b

the temperature The Equipartition Theorem states this quantitatively

The average energy associated with each degree of freedom in a system at absolute temperature is

where is Boltzmanns constant

The average energy of the i th degree of freedom is where the angle brackets represent average

or mean values of the enclosed variable A degree of freedom corresponds to any dynamical variable that appears

quadratically in the energy For instance is the kinetic energy of a gas particle of mass with velocity

component along the x axis

The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically

accessible quantum state of a system has equal probability of being populated which in turn leads to the Boltzmann

distribution for a system in thermal equilibrium From the standpoint of an introductory physics course equipartition is

best regarded as a principle that is justified by observation

In this problem we first investigate the particle model of an ideal gas An ideal gas has no interactions among its

particles and so its internal energy is entirely random kinetic energy If we consider the gas as a system its interna

energy is analogous to the energy stored in a spring If one end of the gas container is fitted with a sliding piston the

pressure of the gas on the piston can do useful work In fact the empirically discovered ideal gas law

enables us to calculate this pressure This rule of nature is remarkable in that the value of the mass does not affect th

energy (or the pressure) of the gas particles motion only the temperature It provides strong evidence for the validity o

the Equipartition Theorem as applied to a particle gas

or

for a particle constrained by a spring whose spring constant is If a molecule has moment of inertia about an axi

and is rotating with angular velocity about that axis with associated rotational kinetic energy that angula

velocity represents another degree of freedom

Part A

Consider a monatomic gas of particles each with mass What is the root mean square (rm

of the x component of velocity of the gas particles if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition for one velocity component

For this case the Equipartition Theorem reduces to

ANSWER

T ( 1 2 )

T 983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

⟨ ⟩ = ( 1 2 ) T U

983145

983147

B

( 1 2 ) M 983158

983160

2

M

983158

983160

983152 V =

N T 983147

B

M ⟨ ⟩ = M ⟨ ⟩ = M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983158

2

983161

1

2

983158

2

983162

1

2

983147

B

1

2

983147

s

983160

2

983147

s

I

ω ( 1 2 ) I ω

2

983160

M = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T

T 983147

B

M

M ⟨ ⟩ = T

1

2

983158

2

983160

1

2

983147

B

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1524

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1524

Correct

Part B

Now consider the same system--a monatomic gas of particles of mass --except in three dimensions Find

the rms speed if the gas is at an absolute temperature

Express your answer in terms of and other given quantities

Hint 1 Equipartition Theorem for three degrees of freedom

What is the internal energy of a monotomic ideal gas with three degrees of freedom

Give your answer in terms of and

ANSWER

ANSWER

Correct

Part C

What is the rms speed of molecules in air at Air is composed mostly of molecules so you may

assume that it has molecules of average mass

Express your answer in meters per second to the nearest integer

ANSWER

Correct

Not surprisingly this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air

= = 983158

983160 r m s

⟨ ⟩ 983158

2

983160

minus minus minus minus

radic

T 983147

B

M

minus minus minus minus

radic

M 983158

r m

T

T 983147

B

M

⟨ U ⟩

983147

B

T

⟨ U ⟩ = M =

1

2

983158

2

r m s

= = 983158

r m s

⟨ ⟩ 983158

2

minus minus minus minus

radic

3 T 983147

B

M

minus minus minus minus minus

radic

983158

0

C 0

∘

N

2

2 8 0 times 1 6 6 1 times k g = 4 6 5 times k g 1 0

minus 2 7

1 0

minus 2 6

= 493983158

0

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1624

Now consider a rigid dumbbell with two masses each of mass spaced a distance apart

Part D

Find the rms angular speed of the dumbbell about a single axis (taken to be the x axis) assuming that

the dumbbell is lined up on the z axis and is in equilibrium at temperature

Express the rms angular speed in terms of and other given quantities

Hint 1 Rotational energy equal to

What is the kinetic energy of rotation that is equal to by the Equipartition Theorem

Express your answer in terms of the x component of the angular velocity and the moment of

inertia about this axis

ANSWER

Hint 2 Moment of inertia of a dumbbell

What is the moment of inertia of the dumbbell

Express in terms of and

Hint 1 Finding of a dumbbell

There are two atoms each with mass but each is only a distance from the center of rotation

(ie the center of mass)

ANSWER

ANSWER

Correct

Part E

What is the typical rotational frequency for a molecule like at room temperature ( ) Assume that

for this molecule is Take the total mass of an molecule to be

983149 983140

⟨ ⟩ ω

2

983160

minus minus minus minus

radic

T

T 983147

B

983149 983140

( 1 2 ) T 983147

B

( 1 2 ) T 983147

B

ω

983160

I

983160

=T

1

2

983147

B

I

983160

I

983160

983149 983140

I

983160

983149 983140 2

=I

983160

=⟨ ⟩ ω

2

983160

minus minus minus minus

radic

2 T 983147

B

983149 983140

2

minus minus minus minus minus

radic

983142

r o t

N

2

2 5 C

∘

983140

1 Aring = m 1 0

minus 1 0

N

2

= 4 6 5 times k g 983149

N

2

1 0

minus 2 6

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1724

You will need to account for rotations around two axes (not just one) to find the correct frequency

Express numerically in hertz to three significant figures

ANSWER

Correct

This frequency corresponds to light of wavelength 022 mm and is in the far-infrared region of the

electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced

with microwave techniques) However because the molecules in the air are homonuclear diatomic molecules

their symmetry prevents them from interacting strongly with radiation of this frequency Only nonhomonuclear

molecules such as water vapor absorb energy at infrared frequencies

Kinetic Energy of Hydrogen and Oxygen Gas Conceptual Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50 is 2000 Note that 10 of

diatomic hydrogen at 50 has a total translational kinetic energy of 4000

Part A

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen The root-mean-square speed for

diatomic oxygen at 50 is

Choose the correct value of

Hint 1 Definition of root-mean-square speed

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

velocity of these particles is zero because each particle is equally likely to have a positive or negative

velocity in any direction However the average of the squared speeds can be determined from the definition

of temperature

This can be rearranged to show that

The square root of this quantity is referred to as the root-mean-square or rms speed

ANSWER

Aring

983142

r o t

= 134times1012 Hz983142

r o t

C

∘

m s m o l

C

∘

J

983158

r m s

C

∘

983158

r m s

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

⟨ ⟩ = 983158

2

3 983147 T

983149

= 983158

r m s

3 983147 T

983149

minus minus minus minus

radic

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1824

Correct

Part B

The total translational kinetic energy of 10 mole of diatomic oxygen at 50 is

Choose the correct total translational kinetic energy

Hint 1 Definition of average and total kinetic energy

Each particle in a gas sample has a different velocity and hence a different kinetic energy The average

kinetic energy of the particles in the gas however is defined to be directly proportional to the temperature

of the gas This temperature must be expressed on the absolute or Kelvin scale

The total kinetic energy is then the product of the average kinetic energy and the number of molecules in

the gas

ANSWER

Correct

Part C

The temperature of the diatomic hydrogen gas sample is increased to 100 The root-mean-square speed

for diatomic hydrogen at 100 is

none of the above

( 1 6 ) ( 2 0 0 0 m s ) = 3 2 0 0 0 m s

( 4 ) ( 2 0 0 0 m s ) = 8 0 0 0 m s

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 5 0 0 m s

1

4

( 983081 ( 2 0 0 0 m s ) = 1 2 5 m s

1

1 6

C

∘

⟨ K ⟩ = 983149 ⟨ ⟩ = 983147 T

1

2

983158

2

3

2

none of the above

( 1 6 ) ( 4 0 0 0 J ) = 6 4 0 0 0 J

( 4 ) ( 4 0 0 0 J ) = 1 6 0 0 0 J

4 0 0 0 J

( 983081 ( 4 0 0 0 J ) = 1 0 0 0 J

1

4

( 983081 ( 4 0 0 0 J ) = 1 5 0 J

1

1 6

C

∘

983158

r m s

C

∘

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 1924

Choose the correct

ANSWER

Correct

The Ideal Gas Law Derived

The ideal gas law discovered experimentally is an equation of state that relates the observable state variables of the

gas--pressure temperature and density (or quantity per volume)

(or )

where is the number of atoms is the number of moles and and are ideal gas constants such that

where is Avogadros number In this problem you should use Boltzmanns constant instead of the

gas constant

Remarkably the pressure does not depend on the mass of the gas particles Why dont heavier gas particles generate

more pressure This puzzle was explained by making a key assumption about the connection between the microscop

world and the macroscopic temperature This assumption is called the Equipartition Theorem

The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at

absolute temperature is where is Boltzmanns const ant A degree of freedom is

a term that appears quadratically in the energy for instance for the kinetic energy of a gas particle of mass

with velocity along the x axis This problem will show how the ideal gas law follows from the Equipartition Theorem

To derive the ideal gas law consider a single gas particle of mass that is moving with speed in a container with

length along the x direction

983158

r m s

none of the above

( 2 ) ( 2 0 0 0 m s ) = 4 0 0 0 m s

( ) ( 2 0 0 0 m s ) = 2 8 0 0 m s 2 radic

2 0 0 0 m s

( 983081 ( 2 0 0 0 m s ) = 1 4 0 0 m s

1

2

radic

( 983081 ( 2 0 0 0 m s ) = 1 0 0 0 m s

1

2

983152 V =

N T 983147

B

983152 V =

983150 R T

N 983150 R 983147

B

R = N

A

983147

B

N

A

R

T

T T

1

2

983147

B

= 1 3 8 times J K 983147

B

1 0

minus 2 3

983149

1

2

983158

2

983160

983149

983158

983160

983149 983158

983160

L

983160

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2024

Part A

Find the magnitude of the average force in the x direction that the particle exerts on the right-hand wall of thcontainer as it bounces back and forth Assume that collisions between the wall and particle are elastic and that

the position of the container is fixed Be careful of the sign of your answer

Express the magnitude of the average force in terms of and

Hint 1 How to approach the problem

From the relationship between applied force and the change in momentum per unit time it

follows that the average force in the x direction exerted by the wall on the particle is

where is the change in the particles momentum upon collision with the wall and is the time

interval between collisions with the wall

You want to find the force exerted by the particle on the wall This is related to the force of the wall on the

particle by Newtons 3rd law

Hint 2 Find the change in momentum

Find the change in momentum of the gas particle when it collides elastically with the right-hand wall

of its container

Express your answer in terms of and

Hint 1 Finding the final momentum

The formula for the momentum of a particle of mass traveling with velocity is What

is the x component of the final momentum of the gas particle (ie after the collision)

Express your answer in terms of and

ANSWER

⟨ ⟩ F

983160

983149 983158

983160

L

983160

= 983140

983140 983156

F

983152

⟨ ⟩ = Δ Δ 983156 F

983160

983152

983160

Δ 983152

983160

Δ 983156

Δ 983152

983160

983149 983158

983160

983152

983149 983158

= 983149 983152

983158

983149 983158

983160

= 983152

f x

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2124

ANSWER

Hint 3 Find the time between collisions

Use kinematics to find the time interval between successive collisions with the right-hand wall of the

container

ANSWER

ANSWER

Correct

Part B

Imagine that the container from the problem introduction is now filled with identical gas particles of mass

The particles each have different x velocities but their average x velocity squared denoted is consistent

with the Equipartition Theorem

Find the pressure on the right-hand wall of the container

Express the pressure in terms of the absolute temperature the volume of the container (where

) and any other given quantities The lengths of the sides of the container should not

appear in your answer

Hint 1 Pressure in terms of average force

The pressure is defined as the force per unit area exerted on the wall by the gas particles The area of the

right-hand wall is Thus if the average force exerted on the wall by the particles is thenthe pressure is given by

Hint 2 Find the pressure in terms of velocity

Find the pressure on the right-hand wall due to a single particle whose squared speed in the x direction

is

Express your answer in terms of and

=Δ 983152

983160

Δ 983156

=Δ 983156

=⟨ ⟩ F

983160

983149

( ) 983158

983160

2

L

983160

N 983149

⟨ ⟩ 983158

2

983160

983152

T V

V = L

983160

L

983161

L

983162

983147

B

A = L

983161

L

983162

⟨ ⟩ F

983160

983152 =

⟨ ⟩ F

983160

L

983161

L

983162

983152

1

983158

2

983160

983158

983160

983149 L

983160

L

983161

L

983162

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2224

ANSWER

Hint 3 Find pressure in terms of temperature

To find the pressure from particles with average squared speed you can use the Equipartition Theorem

Find the pressure due to a single particle

Express the pressure due to a single particle in terms of and any other given

quantities

Hint 1 Relate velocity and temperature

Use the Equipartition Theorem to find an expression for

Express your answer in terms of the gas temperature and given quantities

ANSWER

ANSWER

ANSWER

Correct

Very good You have just derived the ideal gas law generally written (or )

This applet shows a small number of atoms in an ideal gas On the right the path of a specific atom is

followed Look at this for different temperatures to get a feel for how temperature affects the motions of the

atoms in an ideal gas

Part C

Which of the following statements about your derivation of the ideal gas law are true

Check all that apply

ANSWER

= 983152

1

983158

2

983160

983152

1

983147

B

T L

983160

L

983161

L

983162

983149 ⟨ ⟩

983158

2

983160

T 983147

B

=983149 ⟨ ⟩ 983158

2

983160

= 983152

1

= 983152 T

N

V

983147

B

983152 V =

N T 983147

B

983152 V =

983150 R T

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2324

Correct

Part D

If you heat a fixed quantity of gas which of the following statements are true

Check all that apply

ANSWER

Correct

Enhanced EOC Problem 1811

The number density in a container of neon gas is 530times1025 The atoms are moving with an rms speed of 630

You may want to review ( pages 505 - 507)

For help with math skills you may want to review

Rearrangement of Equations Involving Multiplication and Division

Part A

What is the pressure inside the container

Express your answer with the appropriate units

Hint 1 How to approach the problem

How is the pressure of the gas related to the number density mass and root-mean-square (rms) speed of

The Equipartition Theorem implies that

owing to inelastic collisions between the gas molecules

With just one particle in the container the pressure on the wall (at ) is independent of and

With just one particle in the container the average force exerted on the particle by the wall (at )

is independent of and

⟨ ⟩ =

⟨ ⟩ 983158

2

983160

983158

2

983161

⟨ ⟩ = ⟨ ⟩ 983158

2

983160

983158

2

983161

983160 = L

983160

L

983161

L

983162

983160 = L

983160

L

983161

L

983162

The volume will always increase

If the pressure is held constant the volume will increase

The product of volume and pressure will increase

The density of the gas will increase

The quantity of gas will increase

m

minus 3

m s

7212019 Mastering Physics HW 2 Ch 16 18

httpslidepdfcomreaderfullmastering-physics-hw-2-ch-16-18 2424

the atoms

What is the atomic mass number of neon What is the mass of an neon atom

What is the pressure of the gas in the container

ANSWER

Correct

Part B

What is the temperature inside the container

Express your answer with the appropriate units

Hint 1 How to approach the problem

How are the mass of an atom the temperature in the container and the root-mean-square (rms) speed

related

What is the atomic mass number of neon What is the mass of an neon atom

What is the temperature of the gas

ANSWER

Correct

Score Summary

Your score on this assignment is 990

You received 99 out of a possible total of 10 points

= 233times105 983152

P a

= 318T K