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Chapter 13Gases

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Some Gases in Our Lives

Air:

oxygen O2 nitrogen N2 ozone O3

argon Ar carbon dioxide CO2 water H2O

Noble gases:

helium He neon Ne krypton Kr xenon XeOther gases:

fluorine F2 chlorine Cl2 ammonia NH3

methane CH4 carbon monoxide CO

nitrogen dioxide NO2 sulfur dioxide SO2

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13.1 The Nature of Gases

Gases are compressible

Why can you put more air in a tire, but

can’t add more water to a glass full of

water?

Gases have low densities

Dsolid or liquid = 2 g/mL

Dgas 2 g/L

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Nature of Gases

1. Why does a round balloon

become spherical when filled

with air?

2. Suppose we filled this room halfway with water. Where would pressure be exerted?

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Nature of Gases

Gases fill a container completely and uniformly

Gases exert a uniform pressure on all inner surfaces of their containers

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Barometers

760 mmHg

atm

pressure

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Unit of Pressure

One atmosphere (1 atm)

Is the average pressure of the atmosphere at

sea level

Is the standard of pressure

P = Force

Area

1.00 atm = 760 mm Hg = 760 torr

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Types of Pressure Units

Pressure Used in

760 mm Hg or 760 torr Chemistry

14.7 lb/in.2 U.S. pressure gauges

29.9 in. Hg U.S. weather reports

101.3 kPa (kilopascals) Weather in all countries except U.S.

1.013 bars Physics and astronomy

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Check Yourself

A.The downward pressure of the Hg in a barometer is _____ than (as) the weight of the atmosphere.

1) greater 2) less 3) the same

B.A water barometer has to be 13.6 times taller than Hg barometer (DHg = 13.6 g/mL)

because

1) H2O is less dense 2) Hg is heavier

3) air is more dense than H2O

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Solutions

A.The downward pressure of the Hg in a barometer is 3) the same (as) the weight of the atmosphere.

B.A water barometer has to be 13.6 times taller than Hg barometer (DHg = 13.6 g/mL)

because

1) H2O is less dense

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Check Yourself

A. What is 475 mm Hg expressed in atm?

1) 475 atm 2) 0.625 atm 3) 3.61 x 105 atm

B. The pressure of a tire is measured as 29.4 psi.

What is this pressure in mm Hg?

1) 2.00 mm Hg

2) 1520 mm Hg

3) 22,300 mm Hg

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Solutions

A. What is 475 mm Hg expressed in atm?

485 mm Hg x 1 atm = 0.625 atm (B)

760 mm Hg

B. The pressure of a tire is measured as 29.4 psi.

What is this pressure in mm Hg?

29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg

14.7 psi 1.00 atm (B)

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13.2 P and V Changes – Boyles Law

P1

P2

V1 V2

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Boyle's Law

The pressure of a gas is inversely related to the volume when T does not change

Then the PV product remains constant

P1V1 = P2V2

P1V1= 8.0 atm x 2.0 L = 16 atm L

P2V2= 4.0 atm x 4.0 L = 16 atm L

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PV Problem

Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 1.6 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?

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PV Calculation

Prepare a list of given information:

Initial conditions Final conditions

P1 = 50 mm Hg P2 = 200 mm

Hg

V1 = 1.6 L V2 = ??

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Find New Volume (V2)

Solve for V2: P1V2 = P2V2

V2 = V1 x P1 /P2

V2 = 1.6 L x 50 mm Hg = 0.4 L

200 mm Hg

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Check Yourself

A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) Explain.

1) 3.2 L

2) 6.4 L

3) 12.8 L

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Solution

A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant)

6.4 L x 0.70 atm = 3.2 L (1)

1.40 atm

Volume must decrease to cause an increase in the pressure

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Check Yourself

A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain.

1) 200. mmHg

2) 400. mmHg

3) 1200 mmHg

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Solution

A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain.

600. mm Hg x 12.0 L = 200. mmHg (1)

36.0 L

Pressure decrease when volume increases.

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13.3 Charles’ Law – Volume & Temp.

V = 125 mL V = 250 mL

T = 273 K T = 546 K

Observe the V and T of the balloons. How does volume change with temperature?

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Charles’ Law: V and T

At constant pressure, the volume of a gas is

directly related to its absolute (K) temperature

V1 = V2

T1 T2

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Check Yourself

Use Charles’ Law to complete the statements

below:

1. If final T is higher than initial T, final V

is (greater, or less) than the initial V.

2. If final V is less than initial V, final T is

(higher, or lower) than the initial T.

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Solution

V1 = V2

T1 T2

1. If final T is higher than initial T, final V

is (greater) than the initial V.

2. If final V is less than initial V, final T is (lower) than the initial T.

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V and T Problem

A balloon has a volume of 785 mL on a Fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

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VT Calculation

Complete the following setup:

Initial conditions Final conditions

V1 = 785 mL V2 = ?

T1 = 21°C = 294 K T2 = 0°C = 273 K

V2 = _______ mL x __ K = _______ mL

V1 K

Check your answer: If temperature decreases,

V should decrease.

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Check Yourself

A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL?

1) 443°C 2) 170°C 3) - 82°C

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Solution

A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL?

2) 170°C

T2 = 291 K x 640 mL = 443 K

420 mL

= 443 K - 273 K = 170°C

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Gay-Lussac’s Law: P and T

The pressure exerted by a confined gas

is directly related to the temperature

(Kelvin) at constant volume.

P (mm Hg) T (°C)

936 100

761 25

691 0

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Check Yourself

Use Gay-Lussac’s law to complete the statements below:

1. When temperature decreases, the

pressure of a gas (decreases or increases).

2. When temperature increases, the pressure

of a gas (decreases or increases).

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Solution

1. When temperature decreases, the

pressure of a gas (decreases).

2. When temperature increases, the

pressure of a gas (increases).

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PT Problem

A gas has a pressure at 2.0 atm at 18°C. What will be the new pressure if the temperature rises to 62°C? (V constant)

T = 18°C T = 62°C

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PT Calculation

P1 = 2.0 atm T1 = 18°C + 273 = 291 K

P2 = ? T2 = 62°C + 273 = 335 K

What happens to P when T increases?

P increases (directly related to T)

P2 = P1 x T2

T1

P2 = 2.0 atm x K = atm

K

?

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Check Yourself

Complete with 1) Increases 2) Decreases

3) Does not change

A. Pressure _____, when V decreases

B. When T decreases, V _____.

C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T)

D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)

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Solution

A. Pressure 1) Increases, when V decreases

B. When T decreases, V 2) Decreases

C. Pressure 2) Decreases when V changes

from 12.0 L to 24.0 L (constant n and T)

D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C (constant P and n)

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13.4 Volume and Moles

How does adding more molecules of a gas change the volume of the air in a tire?

If a tire has a leak, how does the loss of air (gas) molecules change the volume?

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Avogadro’s Law

When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas

V1 = V2

n1 n2

initial final

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Combined Gas Law

P1V1 = P2V2

T1 T2

Rearrange the combined gas law to solve for V2

P1V1T2 = P2V2T1

V2 = P1V1T2

P2T1

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Combined Gas Law

P1V1 = P2V2

T1 T2

Isolate V2

P1V1T2 = P2V2T1

V2 = P1V1T2

P2T1

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Check Yourself

Solve the combined gas laws for T2.

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Solution

Solve the combined gas law for T2.

(Hint: cross-multiply first.)

P1V1 = P2V2

T1 T2

P1V1T2 = P2V2T1

T2 = P2V2T1

P1V1

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Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

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Data Table

List given information:

P1 = 0.800 atm V1 = 0.180 L T1 = 302 K

P2 = 3.20 atm V2= 90.0 mL T2 = ????

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Solution

Solve for T2

Enter data

T2 = 302 K x atm x mL = K

atm mL

T2 = K - 273 = °C

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Calculation

Solve for T2

T2 = 302 K x 3.20 atm x 90.0 mL = 604 K

0.800 atm 180.0 mL

T2 = 604 K - 273 = 331 °C

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Check Yourself

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

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Solution

T1 = 308 K T2 = ?

V1 = 675 mL V2 = 0.315 L = 315 mL

P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg

T2 = 308 K x 802 mm Hg x 315 mL

646 mm Hg 675 mL

P inc, T inc V dec, T dec

= 178 K - 273 = - 95°C

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Check Yourself

True (1) or False(2)

1.___The P exerted by a gas at constant V is not affected by the T of the gas.

2.___ At constant P, the V of a gas is directly proportional to the absolute T

3.___ At constant T, doubling the P will cause the

V of the gas sample to decrease to one-half its

original V.

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Solution

True (1) or False(2)

1. (2)The P exerted by a gas at constant V is not affected by the T of the gas.

2. (1) At constant P, the V of a gas is directly proportional to the absolute T

3. (1) At constant T, doubling the P will cause the

V of the gas sample to decrease to one-half its

original V.

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STP

The volumes of gases can be compared when they have the same temperature and pressure (STP).

Standard temperature 0°C or 273 K

Standard pressure 1 atm (760 mm Hg)

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Check Yourself

A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

P1 = V1 = T1 = K

P2 = V2 = ?? T2 = K

V2 = 15 L x atm x K = 6.8 L

atm K

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Solution

P1 = 1.0 atm V1 = 15 L T1 = 273 K

P2 = 2.0 atm V2 = ?? T2 = 248 K

V2 = 15 L x 1.0 atm x 248 K = 6.8 L

2.0 atm 273 K

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Molar Volume

At STP

4.0 g He 16.0 g CH4 44.0 g CO2

1 mole 1 mole 1mole (STP) (STP) (STP)

V = 22.4 L V = 22.4 L V = 22.4 L

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Molar Volume Factor

1 mole of a gas at STP = 22.4 L

22.4 L and 1 mole

1 mole 22.4 L

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Check Yourself

A.What is the volume at STP of 4.00 g of CH4?

1) 5.60 L 2) 11.2 L 3) 44.8 L

B. How many grams of He are present in 8.0 L

of gas at STP?

1) 25.6 g 2) 0.357 g 3) 1.43 g

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Solution

A.What is the volume at STP of 4.00 g of CH4?

4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L

16.0 g CH4 1 mole CH4

B. How many grams of He are present in 8.0 L of gas at STP?

8.00 L x 1 mole He x 4.00 g He = 1.43 g He

22.4 He 1 mole He

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13.5 Ideal Gas Law PV=nRT

• Uses gas constant R= (.008206 L atm/k mol)

• n= moles

• T = temp in K

• P = pressure

• V = volume

• Describes the ideal behavior of gases

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Example

• A weather balloon contains 1.10x105 mol of He and has a volume of 2.70 x 106L at 1.00 atm. What is the temp. of the He balloon in K?

• P=1.00atm• V=2.70 x 106 L• n=1.10 x 105 mol• PV=nRT (divide both sides by nR)

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Example cont.

• T = PV/nR

• Solution is:

299K

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13.6 Daltons’ Law of Partial Pressures

Partial Pressure

Pressure each gas in a mixture would exert if it were the only gas in the container

Dalton's Law of Partial Pressures

The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.

PT = P1 + P2 + P3 + .....

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Gases in the Air

The % of gases in air Partial pressure (STP)

78.08% N2 593.4 mmHg

20.95% O2 159.2 mmHg

0.94% Ar 7.1 mmHg

0.03% CO2 0.2 mmHg

PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2

Total Pressure 760 mm Hg

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Partial Pressures

The total pressure of a gas mixture depends

on the total number of gas particles, not on

the types of particles.

P = 1.00 atm P = 1.00 atm

0.5 mole O2

+ 0.3 mole He+ 0.2 mole Ar

1 mole H2

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Health Note

When a scuba diver is several hundred feet

under water, the high pressures cause N2 from

the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles

in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep

descents.

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Check Yourself

A 5.00 L scuba tank contains 1.05 mole of O2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

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Solution

P = nRT PT = PO + PHe

V 2

PT = 1.47 mol x 0.0821 L-atm x 298 K

5.00 L (K mol)

= 7.19 atm

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Learning Check C6

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?

1) 35.6 2) 156 3) 760

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?

1) 557 2) 9.14 3) 0.109

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Solution C6

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?

2) 156

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?

1) 557

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13.9 Kinetic Theory of Gases

The particles in gases

• Are very far apart

• Move very fast in straight lines until they

collide

• Have no attraction (or repulsion)

• Move faster at higher temperatures

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13.10 Real Gases