thermal expansion l = l o t l = change in_______ = coefficient of _______expansion l o =...

Thermal Expansion

L = Lo T

L = change in_______ = coefficient of _______expansion Lo = original ________ T = change in oC or___

Thermal Expansion

L = Lo T

L = change in_______ = coefficient of _______expansion Lo = original ________ T = change in oC or___

Thermal Expansion

L = Lo T

L = change in length = coefficient of linear expansion Lo = original length T = change in oC or K

Thermal Expansion

L = Lo T

L = change in length = coefficient of linear expansion Lo = original length T = change in oC or K

Thermal Expansion

L = Lo T

L = change in length = coefficient of linear expansion Lo = original length T = change in oC or K

Thermal Expansion

L = Lo T

L = change in length = coefficient of linear expansion Lo = original length T = change in oC or K

Thermal Expansion

L = Lo T

L = change in length = coefficient of linear expansion Lo = original length T = change in oC or K

Problem Example

A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.

= 12x10-6 m-1K-1

L = Lo T= 12x10-6 m-1K-1(12m)(32K) L = .0046 m

Problem Example

A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.

= 12x10-6 m-1K-1

L = Lo T= 12x10-6 m-1K-1(12m)(32K) L = .0046 m

Problem Example

A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.

= 12x10-6 m-1K-1

L = Lo T= 12x10-6 m-1K-1(12m)(32K) L = .0046 m

Problem Example

A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.

= 12x10-6 m-1K-1

L = Lo T= 12x10-6 m-1K-1(12m)(32K) L = .0046 m

Thermo Equations Rate of heat transfer is equal to the ratio of the product of the

coefficient of thermal conductivity, Area, temperature difference and the Length of the path

-10oC H 25oC 20oC H 25oC

H = kAT l

Higher heat transfer, A is larger, T is larger and the window is not Thick (l)

Lower heat transfer, A is smaller, TIs smaller and window thickness isGreater (l)

Thermo Equations Rate of heat transfer is equal to the ratio of the product of the

coefficient of thermal conductivity, Area, temperature difference and the Length of the path

-10oC H 25oC 20oC H 25oC

H = kAT l

Higher heat transfer, A is larger, T is larger and the window is not Thick (l)

Lower heat transfer, A is smaller, TIs smaller and window thickness isGreater (l)

Thermo Equations Rate of heat transfer is equal to the ratio of the product of the

coefficient of thermal conductivity, Area, temperature difference and the Length of the path

-10oC H 25oC 20oC H 25oC

H = kAT l

Higher heat transfer, A is larger, T is larger and the window is not Thick (l)

Lower heat transfer, A is smaller, TIs smaller and window thickness isGreater (l)

Thermo Equations Rate of heat transfer is equal to the ratio of the product of the

coefficient of thermal conductivity, Area, temperature difference and the Length of the path

-10oC H 25oC 20oC H 25oC

H = kAT l

Higher heat transfer, A is larger, T is larger and the window is not Thick (l)

Lower heat transfer, A is smaller, TIs smaller and window thickness isGreater (l)

Chapter 13 Temperature

1717 Fahrenheit – Instrument maker 0 Lowest temperature he could achieve with

water,ice, and sea _____ 96 body temperature

By chance water froze at ___Fand boiled at____F

Chapter 13 Temperature

1717 Fahrenheit – Instrument maker 0 Lowest temperature he could achieve with

water,ice, and sea salt 96 body temperature

By chance water froze at 32 and boiled at 212.

Chapter 13 Temperature

1742 Anders Celsius used the freezing and boiling points of water for reference points and then divided them into 100 equal parts.

100 = freezing pt of ______ 0 = boiling pt of ______ at standard pressure Later changed to 0 C = Fpt and 100 C as Bpt First known as centigrade scale 1954 recognized as _______scale

Chapter 13 Temperature

1742 Anders Celsius used the freezing and boiling points of water for reference points and then divided them into 100 equal parts.

100 = freezing pt of water 0 = boiling pt of water at standard pressure Later changed to 0 C = Fpt and 100 C as Bpt First known as centigrade scale 1954 recognized as Celsius scale

Gas Laws

Vol

OoC 100oC

Temperature 0C

Gas Laws

Volume is directly related to temperature

Vol V = constant T

OoC 100oC

Temperature 0C

Gas Laws

Vol

-273 oC OoC 100oC

Temperature 0C

Gas Laws

Pressure

OoC 100oC

Temperature 0C

Gas Laws

Pressure is directly related to temperature

Pressure P = constant T

OoC 100oC Temperature 0C

Gas Laws

Pressure

-273 oC OoC 100oC

Temperature 0C

Absolute Zero

The temperature at which the volume of of an ideal gas is zero and the pressure is zero due to the lack of _______of the particles 1848 – William Thomson – latter to be known as Lord Kelvin formalized the concept theoretically.

-273.150C is considered absolute ______or 273.16 degrees below the _____point of water (ice, water and water vapor exists at .010C at a pressure of 610 Pa)

Absolute Zero

The temperature at which the volume of of an ideal gas is zero and the pressure is zero due to the lack of motion of the particles 1848 – William Thomson – latter to be known as Lord Kelvin formalized the concept theoretically.

-273.150C is considered absolute zero or 273.16 degrees below the triple point of water (ice water and water vapor exists at .010C at a pressure of 610 Pa)

Absolute Zero

0C to Kelvin = 0C +_____= Kelvin

Kelvin to oC = K - _____= oC

Absolute Zero

0C to Kelvins = 0C + 273 = Kelvins

Kelvins to oC = K - 273 = oC

Absolute Zero

0C to Kelvins = 0C + 273 = Kelvins

Kelvins to oC = K - 273 = oC

Gas laws continued

P

Volume

Gas laws continued

The pressure of a gas sample is inversely

proportional to its volume

P PV = constant

Volume

Gas law problem

A tank having a volume of 1.00 m3 is filled with air at 00C to 20 times atmospheric pressure. How much volume will that gas occupy at 1.00 atm and 200C?

Gas law problem

A tank having a volume of 1.00 m3 is filled with air at 00C to 20 times atmospheric pressure. How much volume will that gas occupy at 1.00 atm and 200C?

P1V1 = P2V2

T1 T2

1.00 m3(20 atm) = V2(1.00 atm)

273 K 293 K

V2 = 21.46 m3

Ideal Gas Equation

PV = nRT

P = 1.013 x 105 Pa V =.0224 m3

n = ________ T = 273 Kelvins R = 8.31 m3 Pa / mol K

Ideal Gas Equation

PV = nRT

P = 1.013 x 105 Pa V =.0224 m3

n = 1.0 moles T = 273 Kelvins R = 8.31 m3 Pa / mol K

Ideal Gas Equation Problem

PV = nRT

What is the volume of 1.00 mole of any gas at 1.013 x105 Pa and 273 K?

V = n R T P

Ideal Gas Equation Problem

PV = nRT

What is the volume of 1.00 mole of any gas at 1.013 x105 Pa and 273 K?

V = n R T = 1.00 mol (8.31 J / mol k) (273 K) P 1.013 x105 Pa

Ideal Gas Equation Problem

PV = nRT

What is the volume of 1.00 mole of any gas at 1.013 x105 Pa and 273 K?

V = n R T = 1.00 mol (8.31 J / mol k) (273 K) P 1.013 x105 Pa V= .0224 m3

Ideal Gas Equation

PV = nRT

n = N NA=6.02x1023

PV= N kB T

kB = 1.38 x 10-23 J/K

Kinetic Theory

How does the microscopic movement of particles determine the macroscopic properties of matter?

James Clerk Maxwell and Ludwig Boltzman

Gas molecules are extremely small, hard, and perfectly _______-

Kinetic Theory

How does the microscopic movement of particles determine the macroscopic properties of matter?

James Clerk Maxwell and Ludwig Boltzman

Gas molecules are extremely small, hard, and perfectly elastic-

Kinetic Theory Assumptions

Ideal gas molecules – have no ______of their own and have ___attraction for other gas molecules.

Theoretical calculations based on the conservation of momentum associated with the _______of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.

K.E.avg = 3/2 KBT T is in _____Kb=____________

Kinetic Theory Assumptions

Ideal gas molecules – have no volume of their own and have NO attraction for other gas molecules.

Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.

K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

Kinetic Theory Assumptions

Ideal gas molecules – have no volume of their own and have NO attraction for other gas molecules.

Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.

K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

Kinetic Theory Assumptions

Ideal gas molecules – have no volume of their own and have NO attraction for other gas molecules.

Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.

K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

Average speed of gas molecules K.E.avg =

½ mv2avg=

(v2)avg=

Root mean square velocity =

Average speed of gas molecules K.E.avg = 3/2 KBT

½ mv2avg=3/2 KBT

(v2)avg= 3kBT m

Root mean square velocity = 3kBT m

Example Problem

What is the root mean square velocity of hydrogen molecules and oxygen molecules at STP? H2 = 2.0g / mole 2.0 amu

O2 = 32.0g / mole 32.0 amu

Root mean square velocity= 3kBT m

vrms= 3 (1.38x10-23 J / k) (273 K) =1839 m/s 2.0 amu(1.67x10-27kg)

Example Problem

What is the root mean square velocity of hydrogen molecules and oxygen molecules at STP?

O2 = 32.0g / mole 32.0 amu

Root mean square velocity= 3RT M

vrms= 3 (1.38x10-23 J / k) (273 K) =460 m/s 32.0 g(1 kg/100g)

Relative Velocities – Graham law of diffusion vA = mB

VB mA

The velocity of particle A is inversely related

to the square root of the mass

1st Law of Thermodynamicschapter 15

The internal energy of a sample is defined as the sum total of the __________ ___________energy of the molecules in a sample

U=N(1/2mv2) U=3/2 NkT N= U=3/2 nRT n=

1st Law of Thermodynamicschapter 15

The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample

U=N(1/2mv2) U=3/2 NkT N= U=3/2 nRT n=

1st Law of Thermodynamicschapter 15

The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample

U=N(1/2mv2) U=3/2 NkT N=# of molecules U=3/2 nRT n=# of moles of molecules

1st Law of Thermodynamicschapter 15

The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample

U=N(1/2mv2) U=3/2 NkT N=# of molecules U=3/2 nRT n=# of moles of molecules U=3/2 PV

1st Law of Thermodynamicschapter 15

The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample

U=N(1/2mv2) U=3/2 NkT N=# of molecules U=3/2 nRT n=# of moles of molecules U=3/2 PV 2/3 U = PV

1st Law of Thermodynamicschapter 15

The internal energy of a system would _________ if heat were added to the system. Q=____

The internal energy of a system would _________ if work was done on the system. A Piston _______ a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compresses a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compresses a gas.

1st Law of Thermodynamicschapter 15

The internal energy of a system would increase if heat were added to the system. Q=+

The internal energy of a system would increase if work was done on the system. A Piston compresses a gas.

1st Law of Thermodynamicschapter 15

First Law of Thermodynamics Change in Internal Energy = Heat

transferred plus work done ON the system Formula

U = Q + W

1st Law of Thermodynamicschapter 15

First Law of Thermodynamics Change in Internal Energy = Heat

transferred plus work done ON the system Formula

U = Q + W

1st Law of Thermodynamicschapter 15

First Law of Thermodynamics Change in Internal Energy = Heat

transferred plus work done ON the system Formula

U = Q + W

1st Law of Thermodynamicschapter 15

First Law of Thermodynamics Change in Internal Energy = Heat

transferred plus work done ON the system Formula

U = Q + W

1st Law of Thermodynamicschapter 15

If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will ______. System is _________than environment Q=___

If work is done BY the system then the internal energy of the system with respect to the work done will ___________.System is _________W= ____

1st Law of Thermodynamicschapter 15

If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative

If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

1st Law of Thermodynamicschapter 15

If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative

If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

1st Law of Thermodynamicschapter 15

If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative

If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

1st Law of Thermodynamicschapter 15

If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative

If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

1st Law of Thermodynamicschapter 15

If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative

If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

1st Law of Thermodynamicschapter 15

If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative

If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

1st Law of Thermodynamicschapter 15

If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will ______. System is ________ then the environment. Q=___

If work is done ON the system then the internal energy of the system with respect to the work done will ___________. System is ________ W=___

1st Law of Thermodynamicschapter 15

If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +

If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

1st Law of Thermodynamicschapter 15

If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +

If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

1st Law of Thermodynamicschapter 15

If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +

If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

1st Law of Thermodynamicschapter 15

If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +

If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

1st Law of Thermodynamicschapter 15

If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +

If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

1st Law of Thermodynamicschapter 15

If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +

If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

1st Law of Thermodynamicschapter 15

Isothermal Process-A process carried out at constant ___________. The heat reservoir is very _________ which essentially keeps the _________ constant.

P

V

1st Law of Thermodynamicschapter 15

Isothermal Process-A process carried out at constant temperature. The heat reservoir is very large which essentially keeps the temperature constant. U = O

P

V

1st Law of Thermodynamicschapter 15

Isothermal Process-A process carried out at constant temperature. The heat reservoir is very large which essentially keeps the temperature constant. U = O

1st Law of Thermodynamicschapter 15

Isothermal Process-A process carried out at constant temperature. The heat reservoir is very large which essentially keeps the temperature constant. U = O

1st Law of Thermodynamicschapter 15

Adiabatic process is one in which no _____ is allowed to flow into or out of the system. Q = ___ These processes usually occur very ________.

1st Law of Thermodynamicschapter 15

Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . U = + if compressed

U = - if expands

1st Law of Thermodynamicschapter 15

Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . U = + if compressed

U = - if expands

1st Law of Thermodynamicschapter 15

Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . U = + if compressed

U = - if expands

1st Law of Thermodynamicschapter 15

Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . U = + if compressed

U = - if expands

1st Law of Thermodynamicschapter 15

Isobaric Processes-one in which the _______ is keep ________.

1st Law of Thermodynamicschapter 15

Isobaric Processes-one in which the pressure is keep constant. Work = -P V = + if compressed

Work =-P V = - if expanded

1st Law of Thermodynamicschapter 15

Isobaric Processes-one in which the pressure is keep constant. Work = -P V = + if compressed

Work = -P V = - if expanded

1st Law of Thermodynamicschapter 15

Isobaric Processes-one in which the pressure is keep constant. Work = -P V = + if compressed

Work = -P V = - if expanded

1st Law of Thermodynamicschapter 15

Isobaric Processes-one in which the pressure is keep constant. Work = -P V = + if compressed

Work =-P V = - if expanded

1st Law of Thermodynamicschapter 15

Isochoric-A process in which there is no change in _________

1st Law of Thermodynamicschapter 15

Isochoric-A process in which there is no change in volume. No work is done

W=O

1st Law of Thermodynamicschapter 15

Isochoric-A process in which there is no change in volume. No work is done

W=O

Work and volume changes

Work = _______ * __________

Work = _________ * ___ ___________ Work= If the systems volume increases (it ________) then the work is

done ___ the system and Work has a ___ sign. The change in volume(__ V is ___)

If the systems volume decreases (it ________) then the work is done ___ the system and Work has a ___ sign. The change in volume(__V is ____)

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

Work and volume changes

Work = force *distance Work = Force *Volume ___ Area Work= - Pressure x Change in Volume = - P V If the systems volume increases (it expands) then the work is

done by the system and Work has a - sign. The change in volume(__ V is + )

If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

2nd Law of Thermodynamics

The 1st law of thermodynamics states that ________ is conserved.

Many processes which could obey the 1st law of thermodynamics do not occur ________________.

The second law of thermodynamics is a statement of which processes_______ in nature and which ________.

The second law of thermodynamics - the total ______of a system plus that of the environment _________ as the result of any natural process.

Natural processes tend to move toward a state of ________.

2nd Law of Thermodynamics

The 1st law of thermodynamics states that energy is conserved. Many processes which could obey the 1st law of

thermodynamics do not occur _spontaneously _. The second law of thermodynamics is a statement of which

processes occur in nature and which do not. The second law of thermodynamics - the total entropy of a

system plus that of the environment increases_ as the result of any natural process.

Natural processes tend to move toward a state of disorder / choas/entropy________.

2nd Law of Thermodynamics

The 1st law of thermodynamics states that energy is conserved. Many processes which could obey the 1st law of

thermodynamics do not occur _spontaneously _. The second law of thermodynamics is a statement of which

processes occur in nature and which do not. The second law of thermodynamics - the total entropy of a

system plus that of the environment increases_ as the result of any natural process.

Natural processes tend to move toward a state of disorder / choas/entropy________.

2nd Law of Thermodynamics

The 1st law of thermodynamics states that energy is conserved. Many processes which could obey the 1st law of

thermodynamics do not occur _spontaneously _. The second law of thermodynamics is a statement of which

processes occur in nature and which do not. The second law of thermodynamics - the total entropy of a

system plus that of the environment increases_ as the result of any natural process.

Natural processes tend to move toward a state of disorder / choas/entropy________.

2nd Law of Thermodynamics

The 1st law of thermodynamics states that energy is conserved. Many processes which could obey the 1st law of

thermodynamics do not occur _spontaneously _. The second law of thermodynamics is a statement of which

processes occur in nature and which do not. The second law of thermodynamics - the total entropy of a

system plus that of the environment increases_ as the result of any natural process.

Natural processes tend to move toward a state of disorder / choas/entropy________.

2nd Law of Thermodynamics

The 1st law of thermodynamics states that energy is conserved. Many processes which could obey the 1st law of

thermodynamics do not occur _spontaneously _. The second law of thermodynamics is a statement of which

processes occur in nature and which do not. The second law of thermodynamics - the total entropy of a

system plus that of the environment increases_ as the result of any natural process.

Natural processes tend to move toward a state of disorder / choas/entropy________.

2nd Law of Thermodynamics

The 1st law of thermodynamics states that energy is conserved. Many processes which could obey the 1st law of

thermodynamics do not occur _spontaneously _. The second law of thermodynamics is a statement of which

processes occur in nature and which do not. The second law of thermodynamics - the total entropy of a

system plus that of the environment increases_ as the result of any natural process.

Natural processes tend to move toward a state of disorder / choas/entropy________.

2nd Law of Thermodynamics

The 1st law of thermodynamics states that energy is conserved. Many processes which could obey the 1st law of

thermodynamics do not occur _spontaneously _. The second law of thermodynamics is a statement of which

processes occur in nature and which do not. The second law of thermodynamics - the total entropy of a

system plus that of the environment increases_ as the result of any natural process.

Natural processes tend to move toward a state of disorder / choas/entropy________.

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain ______ energy from __________ enegy

1700’s ________ Engine TH

TL

QH

Q L

W QH =

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Heat Engines- Devices designed to obtain mechanical energy from thermal enegy

1700’s Steam Engine TH temperature of high temperature reservoir

TL temperature of low temperature reservoir

QH heat released by high temperature reservoir

Q Lheat absorbed by high temperature reservoir

W = work output QH = heat input

2nd Law of Thermodynamics

Efficiency e= e= e=

Ideal carnot efficiency

e ideal =

No device is possible whose sole effect is to transform a given amount of heat completely into _______

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

2nd Law of Thermodynamics

Efficiency e= W/QH

e=QH-QL/QH

e= 1-Ql/QH

Ideal carnot efficiency

e ideal = TH - TL / TH

No device is possible whose sole effect is to transform a given amount of heat completely into work

The greater the difference in temperature the greater the ideal efficiency

Carnot Cycle Isothermal expansion U = O W = neg Q = pos(absorbing heat)

Carnot Cycle Adiabiatic expansion U = - W = neg Q = 0

Carnot Cycle Isothermal compression U = O W = + Q = neg (releases heat)

Carnot Cycle Adiabatic compression U = + W = positive Q = 0

Heat

Heat flows from a sample with a high temperature to one with a lower

temperature

Heat Measurements

calorie = the amount of _______ necessary to raise the ________ of ___gram of a water ____ degree celsius

____ cal = 1 ____ = 1 kcal James Prescott Joule-Mechanical Equivalent

of heat-Experiment 1 cal = ________ joules

Heat Measurements

calorie = the amount of energy necessary to raise the temperature of 1 gram of a water

1 degree celsius 1000cal = 1 kcal James Prescott Joule-Mechanical Equivalent

of heat-Experiment 1 cal =4.18 joules

Heat Definition/Formula

Heat is the energy that is ____________ from one body to another because of a difference in ______________

Q =

Heat Definition/Formula

Heat is the energy that is transferred from one body to another because of a difference in temperature

Q = mc t

Temperature Definition/Formula

A measure of the average ______energy of matter

Formula: K.E. =______

Temperature Definition/Formula

A measure of the average kinetic energy of matter

Formula: K.E. = 3/2 kBT

Rate of heat transfer

Formula Q / t = k A ( T1-T2) / l

Q= t = k= A= T1 - T2= l = Can be used to determine the rate of heat transfer across a

__________

Rate of heat transfer

Formula Q/t = K A (T) l

Q= heat transfer in joules t = time heat transferred k= coefficient of heat transfer A= Area of material T1 - T2= temperature difference between sides l = thickness of material Can be used to determine the rate of heat transfer across a window

Internal Energy-Thermal Energy

The sum of the _________ energy of ___the molecules in a sample

Formula: U=____________ U = Internal energy______ N = __________of particles T =___________

Internal Energy-Thermal Energy

The sum of the kinetic energy of all the molecules in a sample

Formula: U=N 3/2 KBT U = Internal energy change N = number of particles T = temperature

1st Law of Thermodynamicschapter 15

Thermodynamics is the study of the ____________ in which energy is transferred as ______ and as _______

Heat is the energy transferred due to a difference in _____________

Work is the energy transferred that is not due to a _____________ difference.

1st Law of Thermodynamicschapter 15

Thermodynamics is the study of the Processes in which energy is transferred as

heat and as work. Heat is the energy transferred due to a

difference in temperature Work is the energy transferred that is not due

to a temperature difference.

1st Law of Thermodynamicschapter 15

Closed system- an object or set of objects for which no __________ enters or leaves. Energy can be __________ with the environment.

Isolated Closed system - an object or set of objects which no __________ enters or leaves and no _____________ is _______

Open system - _______ and ________ may enter or leave.

1st Law of Thermodynamicschapter 15

Closed system- an object or set of objects for which no matter enters or leaves. Energy can be exchanged with the environment.

Isolated Closed system - an object or set of objects which no matter enters or leaves and no energy is

transferred Open system – matter and energy may

enter or leave.