chapter 14 (page 440). why is it important to know the volume - mass relationship of gases, the idea...

Chapter 14(page 440)

Why is it important to know the volume - mass relationship of gases, the Idea Gas law, and the stoichiometry of gases??

Kinetic molecular theory Brownian motion Pressure Pascal Barometer Diffusion Boltzmann’s constant

Gases consist of widely separated atoms or molecules in constant, random motion

No interaction between atoms or molecules, except during collisions

Mostly empty spaceStraight trajectory until a collision occurs

Properties of gases

Gases have a unique set of physical properties:

1. Gases are translucent or transparent.

2. Gases have very low densities when

compared to liquids or solids.

3. Gases are highly compressible

compared to liquids and solids.

4. Gases can expand or contract to fill

any container. Mostly empty space

Properties of gases

These can be explained by the kinetic molecular theory

kinetic molecular theory: the theory that explains the observed thermal and physical properties of matter in terms of the average behavior of a collection of atoms and molecules.

Gases consist of atoms or molecules with

a lot of space in between,

that are in constant, random motion

If the liquid and gas are both made from the same molecules (H2O),

you can explain the “disappearance” by assuming that

the molecules are much more spread out in the gas phase.

Properties of gases

Evidence for the atomic / molecular nature of matter:

Brownian motion can be seen by

magnifying diluted milk and observing

tiny fat globules getting knocked around by the

surrounding water molecules

Brownian motion

What Brownian motion tells us:1. Matter consists of discrete particles (molecules or atoms)

2. Molecules (or atoms) are in constant, vigorous motion as a result of temperature

Brownian motion

Brownian motion provides a peek into the microscopic world

of atoms to see details that are normally hidden by the law of averages,

and the enormous number of incredibly small atoms.

In the early 1800’s Joseph Gay-Lussac studied gas volume relationships involving a chemical reaction between hydrogen and oxygen and observed that 2 L of hydrogen would react with 1 L of oxygen to form 2 L of water vapor at constant temperature and pressure

Hydrogen gas + oxygen gas ---- water vapor 2 L 1 L ---- 2 L 2 volumes 1 volume ---- 2 volumes

Take a new sheet of paper and fold it into three sections

Write your name, the title of the chapter and the number

On the first section from the sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

In Gay-Lussac experiment it was found that the reaction took place in a 2:1:2 relationship between hydrogen, oxygen and water vapor

If you had 600 L H2, 300 L O2 you would get 600 L H2O formed

With hydrogen and chlorine combining then:

Hydrogen gas + chlorine gas ---- hydrogen chloride gas 1 L 1 L ---- 2 L 1 volumes 1 volume ---- 2 volumes

The relationship that he found between gas volumes is now known to be a law

Gay-Lussac’s law of combining volumes of gases states – at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

Remember from Dalton’s atomic theory that atoms are indivisible

Also remember that Dalton believed that one atom of one reactant always combined with one atom of the other reactant (which caused questions when forming water vapor)

Gay-Lussac disproved the second theory of Dalton, but a scientist by the name of Avogadro formulated an explanation of the problem

Avogadro reasoned that molecules contained more than one atom and formulated his theory which later became a law

Avogadro’s law: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

Remember that one mole is equal to the atomic mass of that atom

Well, one mole of a molecule or a compound is equal to the combined molecular masses of the atoms

Avogadro suggested that one mole of an element, molecule or an compound contained the same number of particles which is 6.022 x 1023 particles

So 1 mole of H2 (2.015g) contained the same number of particles as 1 mole O2 (32g) at standard pressure (1 atm) and temperature (273 K) STP

Standard volume at STP is 22.414 L

Write a detailed three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)

Turn to page 468 and complete # 1 – 4 then turn them in

Honors chemistry homework Page 468 # 7 - 16

Molar Volume Ideal gas

Gases consist of widely separated atoms or molecules in constant, random motion

Mostly empty space

No interaction between gas atoms or molecules except in collisions

Straight trajectories until collision occurs

Gas pressure is increased by more frequent and/or harder collisions

1. the density

More molecules means more impacts and a higher pressure.

2. the volume of the container

With less space to move around, there are more collisions and a higher pressure.

You can affect the gas pressure by changing:

1. the density

More molecules means more impacts and a higher pressure.

2. the volume of the container

With less space to move around, there are more collisions and a higher pressure.

3. the temperature

With more kinetic energy, the molecules move faster. The collisions are harder and more frequent.

You can affect the gas pressure by changing:

Boyle’s law: V versus P

Robert Boyle’s experiment:

- Mercury (Hg) was poured down a tube shaped like the letter “J.”

- The tube was closed on the lower end.

- The gas inside the tube took up space (volume).

- The temperature and number of gas molecules inside the tube stayed constant.

- Boyle observed the change in pressure in mmHg as a function of volume. He then graphed the relationship between pressure and volume.


Pressure versus volume

• Temperature• Number of moles constant

If a weather balloon is released on the ground with a volume of 3.0 m3

and a pressure of 1.00 atm, how large will it get when it reaches an

altitude of 100,000 ft, where the pressure is 0.0100 atm?


On the second section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.


If a weather balloon is released on the ground with a volume of 3.0 m3 and a pressure of 1.00 atm, how large will it get when it reaches an altitude of 100,000 ft, where the pressure is 0.0100 atm?

Asked: Volume of the balloon when it reaches 100,000 ft

Given:

Relationships:

Solve:

Answer: The balloon will have a volume of 300 m3.

31 1 2

32

3

2

1 1 2 2

3

1.00 3.0 0.0100

1.00 3.0

1.00 , 3.0 , 0.0

0.0010030

0

0

10

atm m atm P

atm m

P atm V m P atm

PV PV

Pat

mm

Charles’s law: V versus T

The volume increases as

the temperature increases


Volume versus temperature

• Pressure• Number of moles constant


Doubling the Kelvin temperature will double the volume of a gas

Kelvin temperatures simplify the V versus T relationship

If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC?

Charles’s law: V vs. T

If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC?

Charles’s law: V vs. T

Asked: Volume of the balloon when the temperature drops to 3oC

Given:

Relationships:

Solve:

1 1 2

1 2

1 2

1

2

22

8.0 8.0 276

296 276 29

8.0 , 23 , 3

273

:

23 273 296

3 27

7.5

2

6

3 76

o o

Kelvin Celsius

o

o

V L T C T C

V Vand T T

T T

Convert temperature to Kelv

LVL L K

so VK K

in

T C

C

K

K

T K

Combined gas law

Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20oC, you’ll need 18 m3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50oC?

Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20oC, you’ll need 18 m3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50oC?

Asked: Volume of the balloon when it reaches 100,000 ft

Given:

Relationships:

Solve:

31 1 1

2 2

1.00 , 18.0 , 20 293 ,

0.0100 , 50 223

o

o

P atm V m T C K

P atm T C K

32

3

2

1

1 2

3

1 2 2

0.01001.00 18.0

293 223

1.00 18.0 223

293 0.01001,370

atm Vatm m

K K

atm m KV

PV P

K atm

T

m

V

T

On the third section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

Avogadro’s law: V versus moles

Two equal volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapor.

Gas volumes act like moles because the same size container has the same number of molecules (at the same temperature and pressure).


Moles versus volume

• Temperature• Pressure constant

Twice the volume,

twice the number of molecules

assuming the temperature and

pressure are constant

Moles versus volume

• Temperature• Pressure constant


Restrictions

The ideal gas law

Combining the previous gas laws, we obtain the ideal gas law

In reality, the ideal gas law is an approximation which is accurate for many gases over a wide range of conditions.

The ideal gas law is not accurate at very high density or at very low temperature.

The ideal gas law

The universal gas constant

R is the only constant

Calculating the universal gas constant using various units

Watch out for the units!

The ideal gas law

On the first section of back side on that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

The ideal gas law

Limitations of the ideal gas law

In an ideal gas, we assume that:

1. individual gas molecules take up

no space

2. gas molecules do not interact

with each other

For very small volumes or very low temperatures, gas atoms and molecules are very close together, and van der Waals attractions are no longer negligible

PV = nRT

What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20oC?

PV = nRT


PV = nRT

Asked: Tank pressure

Given:

Relationships:

Solve:

50.0 , 20 , 1,252 , : 4.003oV L T C mass g He g mole

, 0.08206atm L

PV nRT Rmole K

:

20.0 273

50.0

293

11,25 312.82

4.003

o

Convert to units of R

V

T C

molen g

g

L

K

moles


PV = nRT

Asked: Tank pressure

Given:

Relationships:

Solve:

50.0 , 20 , 1,252 , : 4.003oV L T C mass g He g mole

, 0.08206atm L

PV nRT Rmole K

50.0 , 293 , 312.8V L T K n moles

50.0 312.8 0.08206 293

312.8 0.0820

15

6 29

0

3

50.0

atm LP L mole

atm

s Kmole K

atm Lmoles K

mole KP

L

P

What would be the new volume of a bubble in a bread dough once it goes from a room temperature (20oC) volume of 0.050 cm3 to a 191oC oven?


Asked: Find the volume of a bubble under changing temperature conditions

Given:

Relationships:

Solve: Pressure and moles stay constant, so they cancel out.

1 1 2 2

1 1 2 2

1 2

1 2

PV PV

nT n

V V

T T

T

3

33

2

2

0.050 4640.079

293

0.050

293 464

cm

Vcm

K K

KV cm

K

31 1 20.050 , 20 293 , 191 464o oV cm T C K T C K

1 1 2 2

1 1 2 2

PV PV

nT n T


Asked: Find the volume of a bubble under changing temperature conditions

Given:

Relationships:

Solve: Pressure and moles stay constant, so they cancel out.

31 1 20.050 , 20 293 , 191 464o oV cm T C K T C K

1 1 2 2

1 1 2 2

PV PV

nT n T

1 1 2 2

1 1 2 2

1 2

1 2

PV PV

nT n T

V V

T T

3

33

2

2

0.050 464

293

0

0.079

.050

293 464

cm

Vcm

K K

cK

VK

m

PV k T and n constant

Vk P and n constant

T

Boyle’s law

Charles’s law

Avogadro’s law

1 1 2 2

1 2

PV P VPVk or

T T T

n constant

Vk T and P constant

n

Combined gas law

1 1 2 2

1 1 2 2

PV P VPV nRT or

nT n T

Ideal gas law

R = universal gas law

Write a three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)

Turn to page 468 and complete # 5 – 6 then turn them in

Honors chemistry Homework: Page 468 # 17 - 26

Molar Volume Ideal gas

1 1 2 2

1 1 2 2

PV P VPV nRT or

nT n T

Ideal gas law

R = universal gas law

:

:

:

:

P pressure

V volume

n moles

T temperature

We can now solve

stoichiometry problems

involving gases

We can now solve

stoichiometry problems

involving gases

and

solids

solutions

other gases

Steps for solving stoichiometry problems

2C20H42 + 61O2 → 40CO2 + 42H2O

If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400oC. Assuming the carbon dioxide produced is at that temperature, what volume of CO2 is produced at a pressure of 790 mmHg? The combustion reaction is:

Asked: Volume of CO2 produced

Given: Paraffin: mass of 125 g

CO2: P = 790 mmHg, T = 1,400oC

Relationships: Molar mass of paraffin = 282.6 g/mole

Mole ratio: 2 moles C20H42 ~ 40 moles CO2

PV = nRT

12.011 20 1.0079

282.6

2

/

4

g mole

If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400oC. Assuming the carbon dioxide produced is at that temperature, what volume of CO2 is produced at a pressure of 790 mmHg? The combustion reaction is:

2C20H42 + 61O2 → 40CO2 + 42H2O


Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC



PV = nRT

Solve:

20 4220 4 20 422

20 42

1125

282.0.442

6mo

mole C Hg C H

g C Hles C H

125 g C20H420.442 moles

C20H42





PV = nRT

Solve:

220 42

202

42

400.442

28.85

moles COmoles C H

molesmol

C Hes CO

0.442 moles C20H42

8.85 moles CO2





PV = nRT

Solve:

8.85 0.08206 1,6731,170

1.04

atm

nRTPV nRT s

Lmoles K

mole KV L

a

oP

tm

V

8.85 moles CO2

1.04

8.85

1,673

P atm

n moles

T K





PV = nRT

Solve:

8.85 moles CO2

1.04

8.85

1,673

P atm

n moles

T K

8.85 0.08206 1,673

1.0,

41170

nRTPV nRT so

Patm L

moles Kmole K

V

V

atmL

1,170 L CO2

A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium:

If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22oC?

Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)

Asked: Pressure of H2 produced

Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L

H2: V = 1.00 L, T = 22oC

Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2

PV = nRT

Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)

A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium:

If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22oC?



H2: V = 1.00 L, T = 22oC


PV = nRT

Solve:Volume

0.050 L HCl1.25 M HCl

1.250.05 0.0620

15

moles HClL HCl

L Hmole

lCl

Cs H

0.0625 moles HCl



H2: V = 1.00 L, T = 22oC


PV = nRT

Solve:

0.0625 moles HCl

221

0.06252

0.0313mole H

moles HClmoles

molH l

HC

es

0.0313 moles H2



H2: V = 1.00 L, T = 22oC


PV = nRT

Solve:

0.0313 moles H2

1.00

0.0313

22 273

295

o

V L

n mole

T

s

T C

K

0.756 atm H2

0.0313 0.08206 295

1.000.756

nRTPV nRT so

Va

P

a

tm Lmoles K

mole KP

Ltm

What volume of butane gas is needed at room temperature (23oC) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825oC and a pressure of 1.04 atm? The reaction is:

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

What volume of butane gas is needed at room temperature (23oC) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825oC and a pressure of 1.04 atm? The reaction is:

Asked: Volume of C4H10 needed to produce 85.0 L of CO2

Given: C4H10: P = 0.984 atm, T = 23oC

CO2: P = 1.04 atm, T = 825oC, V = 85.0 L

Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO21 1 2 2

1 1 2 2

PV PV

nT n T


Given: C4H10: P = 0.984 atm, T = 23oC

CO2: P = 1.04 atm, T = 825oC, V = 85.0 L

Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2

Solve:

1 1 2 2

1 1 2 2

PV PV

nT n T

C4H10 CO2

4 10

4 10

4 10 4 10

4 1

4 10 2

0

4 10

0.984 1

0.984 1.04 85.0

296 1,0

.04 85.0

296 4 1,098

0.00332 0.0201

6. 5

8

0

9

C H

C H C H

C H

C

C O

H

H

C

H C

at

atm V atm L

n K n K

atm a

m atm L

tm LV

V

K K

V L

n K n K

4 10

2

2 4 10

296

1,098

4

C H

CO

CO C H

T K

T K

n n


Given: C4H10: P = 0.984 atm, T = 23oC

CO2: P = 1.04 atm, T = 825oC, V = 85.0 L


Solve:

1 1 2 2

1 1 2 2

PV PV

nT n T

C4H10 CO2

2 4 10

4 10

2

296

1,0

4

98

CO

C H

C

C

O

H

T K

T

n n

K

4 10

4 10

4 10 4 10

4 1

4 10 2

0

4 10

0.984 1

0.984 1.04 85.0

296 1,0

.04 85.0

296 4 1,098

0.00332 0.0201

6. 5

8

0

9

C H

C H C H

C H

C

C O

H

H

C

H C

at

atm V atm L

n K n K

atm a

m atm L

tm LV

V

K K

V L

n K n K

Use the mole ratio to substitute for nCO2


Given: C4H10: P = 0.984 atm, T = 23oC

CO2: P = 1.04 atm, T = 825oC, V = 85.0 L


Solve:

4 10

4

4 10

4

4 10

4 10

1

10 4

0

1

2

0

0.984 1

0.00332

0.984 1.04 85.0

2

0.0201

6.05

.04 85.0

296 4 1,09

96 1,098

8C H C

C H

C

C H

H

C

C H

C H O

H

atm atm L

n K n

atm V atm L

n K n K

atm atm LV

K K

V

L

K

V

2 4 10

4 10

2

296

1,0

4

98

CO

C H

C

C

O

H

T K

T

n n

K

1 1 2 2

1 1 2 2

PV PV

nT n T


Given: C4H10: P = 0.984 atm, T = 23oC

CO2: P = 1.04 atm, T = 825oC, V = 85.0 L


Solve:

4 10

4 10 2

4 10

4 10

4 10

4 10

4 10

0.984 1.04 85.0

296 1,098

0.984 1.04 85.0

296 4 1,098

0

6.0

.00332 0

5

.0201

C H

C H CO

C H

C H

C H C H

C H

atm V atm L

n K n K

atm V atm L

n K n K

at

V

m atm LV

K K

L

4 10

2

2 4 10

296

1,098

4

C H

CO

CO C H

T K

T K

n n

1 1 2 2

1 1 2 2

PV PV

nT n T

M = m R T

PVm = massM = molar mass

D = M P RT

D = density

Write a three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)

Turn to page 469 and complete # 32 then turn it in.

Honors chemistry Homework: Page 471 # 73 - 81

Homework requirement: Learn all terms and concepts covered on this topic.

Make sure you have all assignments between page 224 and 227 completed and turned in by your test date.

chapter 14 (page 440). why is it important to know the volume - mass relationship of gases, the idea...

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