Chapter 14

Properties of Gases

Section 14.1

The Behavior of Gases

1

The Properties of Gases

• Gas can expand to fill its container

• Gases are easily compressed, or squeezed into a smaller volume.

CompressibilityCompressibility – measure of how much the volume of matter decreases under pressure.

2

Compressibility of Gases

What factors do you think affect the pressure of the air inside the soccer ball?

1. Temperature of the air inside the ball

2. Volume of the ball3

Kinetic Theory & GasesGases are easily compressed because of the space between the particles in a gas.

Under pressure, the particles in a gas are forced closer together, or compressed.

4

Factors Affecting Gas PressurePressure (P) - kPa

Volume (V) - liters

Temperature (T) - Kelvin

Number of moles (n)

The amount of gas, volume, and temperature are factors that affect gas pressure

5

Amount of Gas & Gas Pressure

Collisions of particles with the inside walls of the raft result in the pressure that is exerted by the gas.

By adding gas, you increase the number of particles.

Increasing the number of particles increases the number of collisions, which is why the gas pressure increases.

6

When a gas is put into a closed rigid container, the pressure increases as more particles are added

Because the container is rigid, thevolume of the gas is constant.

Assume the temperature doesn't change

Doubling the number of particles of gas, doubles the pressure.

As gas is removed, the pressure inside the container is reduced.

7

Cause and EffectIf the pressure of the gas in a sealed container is lower than the outside air pressure, air will rush into the container when the container is opened.

When the pressure of the gas in a sealed container is higher than the outside air pressure, the gas will flow out of the container when the container is unsealed.

8

Volume & Gas Pressure

You can raise the pressure exerted by a contained gas by reducing its volume.

The more gas is compressed, the greater is the pressure the gas exerts inside the container.

9

When cylinder has a volume of 1 L, the pressure is 100 kPa

If volume is halved to 0.5 L, the pressure doubles to 200kPa

If volume is doubled to 2.0 L, the pressure of the volume is cut in half to 50 kPa.

10

Temperature & Gas Pressure

A sealed bag of potato chips may bulge at the seams if placed in the sun.

Bag bulges because an increase in the temperature of the gas inside the bag causes an increase in its pressure.

As gas inside bag is heated, the temperature increases, increasing kinetic energy of the particles, and causing more collisions, thus more pressure.

11

Questions1. What effect would tripling the number of particles of a gas in a closed container have on the pressure exerted?

Gas pressure would triple2. What effect would doubling the volume of an enclosed gas have on the pressure?

Gas pressure would decrease by half3. How does the pressure of an enclosed gas change with increasing temperature?

The number and force of collisions increase with temperature, and the pressure increases

12

4. Why is a gas easy to compress?Because of the space between particles in a gas

5. List three factors that can affect gas pressure?

Temperature, volume, & amount of gas

6. Why does a collision with an air bag cause less damage than a collision with a steering wheel?

Gas in the inflated airbag can be compressed, and absorbs some of the energy from the impact.

The solid steering wheel cannot.

13

7. If temperature is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to one quarter of its original value?

The volume would need to increase by a factor of four

8. How does a decrease in temperature affect the pressure of a contained gas

If temperature decreases, the pressure will decrease

14

End of Section 14.115

Section 14.2

The Gas Laws

16

Boyle’s Law (Pressure & Volume)

Conversely, if the temperature is constant, as the pressure of a gas decreases, the volume increases.

Robert Boyle was the first person to study this pressure-volume relationship.

In 1662, Boyle proposed a law to describe the relationship.

If the temperature is constant, as the pressure of a gas increases, the volume decreases.

17

Boyle’s Law (Pressure & Volume)

Boyle’s LawBoyle’s Law – states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure.

P1V1 = P2V2

18

Sample Problem UsingBoyle’s Law

A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume the temperature remains constant) P1 = 103 kPa P2 = 25.0 kPa

V1 = 30.0 L V2 = ? L

P1V1 = P2V2 103 X 30 = 25 X V2

V2 = (30.0 L) (103 kPa) 25.0 KPa

V2 = 124 L 19

Sample Problem UsingBoyle’s Law

A gas with a volume of 4.00 L at a pressure of 205 KPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?

P1 = 205 kPa P2 = ? kPaV1 = 4.00 L V2 = 12.0 L

P1V1 = P2V2 205 X 4 = P2 X 12

P2 = (4.00 L) (205 kPa) 12.0 L

P2 = 68.3 KPa 20

Sample Problem UsingBoyle’s Law

Nitrous oxide (N2O) is used as an anesthetic. The pressure on 2.50 L of N2O changes from 105 KPa to 40.5 KPa. If the temperature does not change, what will the new volume be?

P1 = 105 kPa P2 = 40.5 kPaV1 = 2.50 L V2 = ? L

P1V1 = P2V2 105 X 2.5 = 40.5 X V2

V2 = (2.50 L) (105 kPa) 40.5 KPa

V2 = 6.48 L 21

Sample Problem UsingBoyle’s Law

The volume of a gas at 99.6 kPa and 24ºC is 4.23L. What volume will it occupy at 93.3 kPa and 24ºC?

P1 = 99.6 kPa P2 = 93.3 kPa T1 = 24ºCV1 = 4.23 L V2 = ? L T2 = 24ºC

P1V1 = P2V2 99.6 X 4.23 = 93.3 X V2

V2 = (4.23 L) (99.6 kPa) 93.3 kPa

V2 = 4.52 L22

Charles’s Law Temperature and Volume

In 1787, French physicist Jacques Charles studies the effect of temperature on the volume of a gas at constant pressure.

As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant.

23

Charles’s LawCharles’s Law – states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant.

Charles’s Law Temperature and Volume

V1

T1

= V2

T224

Sample Problem UsingCharles’s Law

A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant?When using gas laws always express the temperatures in kelvins!

T1 = 24ºC + 273 = 297 K V1 = 4.00 LT2 = 58ºC + 273 = 331 K V2 = ? L

V1

T1

= V2

T2 4

297=

x

331

V2 = (4.00 L) (331 K) = 4.46 L 297 K

25

Sample Problem UsingCharles’s Law

If a sample of gas occupies 6.80 L at 325ºC, what will its volume be at 25ºC if the pressure does not change?

T1 = 325ºC or 598 K V1 = 6.80 LT2 = 25ºC or 298 K V2 = ? L

V2 = (6.80 L) (298 K) = 3.39 L 598 K

V1

T1

= V2

T2 6.8

598=

x

298

26

Gay-Lussac’s LawPressure and Temperature

As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.

Joseph Gay-Lussac discovered the relationship between the pressure and the temperature of gas in 1802.

27

Gay-Lussac’s Law – states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant.

Gay-Lussac’s LawPressure and Temperature

P1

T1

= P2

T228

Sample Problem UsingGay-Lusaac’s Law

A sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at 211 K?

P1 = 6.58 kPa T1 = 539 KP2 = ? kPa T2 = 211 K

P2 = (6.58 kPa) (211 K) = 2.58 kPa 539 K

P1

T1

= P2

T2 6.58

539=

x

211

29

Sample Problem UsingGay-Lusaac’s Law

The pressure in a car tire is 198 kPa at 27ºC. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant.

P1 = 198 kPa T1 = 300 KP2 = 225 kPa T2 = ? K

T2 = (225 kPa) (300 K) = 341 K 198 kPa

P1

T1

= P2

T2 198

300=

225

x

30

GasesIf the gas is heated (T2 > T1), the new pressure is greater. (volume constant)

If the gas is heated (T2 > T1), the new volume is greater because the gas expands. (pressure constant)

If the gas is cooled (T2 < T1), the new pressure is less. (volume constant)

If the gas is cooled (T2 < T1), the new volume is smaller because the gas contracts. (pressure constant)

31

Combined Gas LawThere is a single expression that combines Boyle’s, Charles’s and Gay-Lusaac’s Law.

The combined gas law describes the relationship among the pressure, temperature, and volume of an enclosed gas.

The combined gas law allows you to do calculation for situations in which only the amount of gas is constant

P1V1

T1

= P2V2

T232

Sample Problem UsingCombined Gas Law

A gas at 155 kPa and 25º C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125º C. What is the new volume?

P1 = 155 kPa T1 = 298 K V1 = 1.00 LP2 = 605 kPa T2 = 398 K V2 = ?

V2 = (155)(1.00)(398) = 0.342 L (298)(605)

P1V1

T1

= P2V2

T2 155 x 1

298=

605 x V2

398

33

Sample Problem UsingCombined Gas Law

A 5.00 L air sample has a pressure of 107 kPa at a temp of -50 º C. If the temperature is raised to 102 º C and the volume expands to 7.00 L, what will the new pressure be?

P1 = 107 kPa T1 = 223 K V1 = 5.00 LP2 = ? kPa T2 = 375 K V2 = 7.00 L

P2 = (107)(5.00)(375) = 128 kPa (223)(7.00)

P1V1

T1

=P2V2

T2 107 x 5

223=

P2 x 7

375

34

End of Section 14.2

35

Section 14.3

Ideal Gases

36

Ideal Gas LawWith the combined gas law, you can solve problems with three variables: temperature, volume & pressure

The combined gas law assumes that the amount of gas does not vary.

The number of moles of gas is directly proportional to the number of particles.

An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces.

37

Ideal Gas LawIdeal Gas Law

P V = n R TP V = n R T

Pressure Volume moles ideal gas Temperature (kPa) (L) (mol) constant (K)

8.31 L · kPa / mol · K

38

Sample Problem Using Ideal Gas Law

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 103 kPa. How many moles of helium does the sphere contain?

P = 1.89 x 103 kPa V = 685 L T = 621 K

PV = nRT n = PV / RT

n = (1.89 x 103 ) (685)

(8.31) (621)

n = 251 mol He39

Sample Problem Using Ideal Gas Law

A child’s lungs can hold 2.20 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a body temperature of 37ºC? Use a molar mass of 29 g for air.

P = 102 kPa V = 2.20 L T = 310 K

PV = nRT or PV / RT = n

n = (102) (2.20)

(8.31) (310)

n = 0.087 mol air 0.087 mol air x 29g air / mol air = 2.5 g air

40

Ideal Gases & Real Gases

Ideal gasIdeal gas – one that follows the gas laws at all conditions of pressure and temperature.

Such a gas would have to conform precisely to the assumptions of kinetic theory.

Its particles could have no volume, and there could be no attraction between particles in the gas.

There is no gas for which these assumptions are true.

41

Question

What pressure is exerted by 0.450 mol of gas at 25 C if the gas is in a 0.650 L container?

V = 0.650 L n = 0.450 mol T = 298 K

PV = nRT or P = nRT / V

P = (0.450 )(8.31)(298)(0.650)

P = 1.71 x 103 kPa42

Question

Determine the volume occupied by 0.582 mol of a gas at 15ºC if the pressure is 0.808 atm?

P = 0.808 x 101.3 n = 0.582 mol T = 288 K = 81.8 kPa

PV = nRT or V = nRT / P

V = (0.582 )(8.31)(288)(81.8)

V = 17.0 L43

End of Chapter 14End of Chapter 1444