Chapter 14Properties of Gases

The Properties of Gases

• Gas can expand to fill its container

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

• Gases occupy far more space than a liquid or a solid

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

Compressibility of Gases

When an airbag triggers due to a sudden stop, a chemical reaction inside the airbag occurs.

One product of the reaction is nitrogen gas, which causes the bag to inflate.

When a person collides with an inflated air bag, the impact forces the molecules of gas in the bag closer together

The compression of the gas absorbs the energy of the impact.

Compressibility of Gases

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

Temperature of the air inside the ballVolume of the ball

Kinetic Theory & Gases

What is kinetic energy

The energy of motion

How are temperature and kinetic energy related?

Temperature is a measure of average kinetic energy.

Kinetic Theory & Gases

Gases 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.

Factors Affecting Gas Pressure

Pressure (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

Amount of Gas and Gas Pressure

When you inflate an air raft, the pressure inside the raft will increase. (this is a container with a volume that can vary. A balloon is another example)

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.

Amount of Gas & Gas Pressure

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.

Cause and Effect

If 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.

Cause and Effect

Gas pressure inside a new spray paint can is greater than the air pressure outside the can.

As the can is used, the pressure inside the can decreases until there is not enough pressure inside the can to force the paint out.

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.

Volume & Gas Pressure

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.

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.

Temperature & Gas Pressure

If volume and amount of gas are constant, when the Kelvin temperature of gas doubles, the gas pressure doubles.

Gas in sealed container may generate enormous pressure when heated.

For that reason, an aerosol can, even an “empty” one, may explode if thrown onto a fire.

As the temperature of an enclosed gas decreases by half, the pressure decreases by half.

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

Gas pressure would triple

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

Gas pressure would decrease by half

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

QuestionsWhy is a gas easy to compress?

Because of the space between particles in a gas

List three factors that can affect gas pressure?

Temperature, pressure, & amount of gas

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.

QuestionsIf 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

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

If temperature decreases, the pressure will decrease

If gas temperature in a container is constant, how could you increase the pressure one hundredfold?

Increase the amount of gas in the container one hundredfold.

End of Section 14.1

Boyle’s Law (Pressure & Volume)

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

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

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

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

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

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)

What do you think will happen to the volume at a higher temperature knowing what you know already about gases?

Sample Problem UsingBoyle’s Law

P1 = 103 kPa P2 = 25.0 kPaV1 = 30.0 L V2 = ? L

P1V1 = P2V2 or P1V1 / P2 = V2

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

V2 = 1.24 x 102 L (3 sig figs)

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. It 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 or P1V1 / P2 = V2

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

V2 = 6.48 L (3 sig figs)

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 or P1V1 / V2 = P2

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

P2 = 68.3 KPa (3 sig figs)

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 or P1V1 / P2 = V2

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

V2 = 4.52 L (3 sig figs)

Charles’s Law Temperature and Volume

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

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

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.

V1 = V2

T1 T2

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?

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

When using gas laws always express the temperatures in kelvins!

T1 = 24ºC + 273 = 297 KT2 = 58ºC + 273 = 331 K

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?

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

V1 = V2 or V1T2 = V2

T1 T2 T1

V2 = (4.00 L) (331 K) = 4.46 L297 K

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

V1 = V2 or V1T2 = V2

T1 T2 T1

V2 = (6.80 L) (298 K) = 3.39 L598 K

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.

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

P1 = P2

T1 T2

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

P1 = P2 or P1T2 = P2

T1 T2 T1

P2 = (6.58 K) (211 K) = 2.58kPa539 K

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

P1 = P2 or P2T1 = T2

T1 T2 P1

T2 = (225 kPa) (300 K) = 341 K198 kPa

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)

Combined Gas Law

There 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 = P2 V2

T1 T2

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 = ?

P1V1 = P2 V2 or P1V1 T2 = V2

T1 T2 T1 P2

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

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 = 1.00 LP2 = ? kPa T2 = 375 K V2 = 7.00 L

P1V1 = P2 V2 or P1V1 T2 = P2

T1 T2 T1 V2

P2 = (107 kPa)(5.00 L)(375 K) = 1.29 x 102kPa (223K)(7.00 L)

Questions

How are the pressure and volume of a gas related at constant temperature?

The volume of a gas decreases as the pressure increases. (Boyle’s Law)

If pressure is constant, how does a change in temperature affect the volume of a gas?

As the temperature increases, the volume increases. (Charles’s Law)

Questions

What is the relationship between the temperature and pressure of a contained gas at constant volume?

As the temperature increases, the pressure increases. (Gay-Lusaac’s Law)

In what situations is the combined gas law useful?

Allows you to do calculations when the only constant is the amount of gas.

Question

Explain how Charles’s law can be derived from the combined gas law.

When the pressure is constant, P1 = P2, the pressure terms cancel, leaving an equation for Charles’s Law.

P1V1 = P2 V2

T1 T2

Question

The volume of a weather balloon increases as the balloon rises in the atmosphere. Why doesn’t the drop in temperature at higher altitudes cause the volume to decrease?

The outside pressure decreases, causing a greater increase in the balloon’s volume

The higher it rises, the colder the temperature and the lower the volume. At the same time, atmospheric pressure decreases, allowing the gas to expand.

End of Section 14.2

Ideal Gas Law

With 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.

To calculate the number of moles of a contained gas requires an expression that contains the variable n.

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

Ideal Gas LawThe volume occupied by a gas at a specified temperature and pressure also must depend on the number of particles.

So… moles must be directly proportional to volume.

P1V1 = P2 V2

T1 n T2 n

This equation shows that P1V1 is a constant. T1 n

This constant holds for ideal gases – gases that conform to the gas laws.

Ideal Gas LawIf you know the values for P,V, T and n for one set of condition, you can calculate a value for the constant.

1 mole of every gas occupies 22.4 L at STP. (101.3kPa and 273K)

Ideal gas constant uses the symbol R

R = P1V1 R= (101.3kPa)(22.4L)

T1 n (273 K)(1mol)

R = 8.31 L R = 8.31 L · kPa / mole · K or· kPa / mole · K orR = 0.0831 L R = 0.0831 L · atm / mole · K· atm / mole · K

Ideal Gas Law

PV = nRT

pressure volume moles constant temperature(K) 8.31L · kPa / mole · K

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 V = 685 L T = 621 K

PV = nRT or PV / RT = n

n = (1.89 x 103 kPa) (685 L) mol · K(8.31L · kPa) (621K)

n = 251 mol He

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 kPa) (2.20 L) mol · K(8.31L · kPa) (310K)

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

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.

Ideal Gases & Real Gases

At many conditions of temperature and pressure, real gases behave very much like an ideal gas.

Particles of a real gas do have volume and there are attractions between the particles.

Because of these attractions, a gas can condense or solidify when it is compressed or cooled.

Example – if water vapor is cooled below 100ºC at standard atmospheric pressure, it condenses to a liquid.

Ideal Gases & Real Gases

Real gases differ most from an ideal gas at low temperatures and high pressures.

For real gases at high pressures (thus high densities), attractive forces reduce the distance between particles.

As pressures and density increase, the volume of the molecules themselves becomes significant relative to the size of the container.

For real gases below a critical temperature, the attractive forces cause the particles to “stick” together and the gas condenses to become a liquid.

Questions

Under what conditions do real gases deviate most from ideal behavior?

At low temperatures and high pressures.

What is an ideal gas?

A gas that follows the gas laws at all conditions of pressure and temperature.

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 = (0.450mol )(8.31 L · kPa)(298 K)mol · K (0.650 L)

P = 1.71 x 103 kPa

Question

Determine the volume occupied by 0.582 mol of a gas at 15ºC if the pressure is 81.8 kPa?

P = 81.8kPa n = 0.582 mol T = 288 K

PV = nRT or V = nRT / P

V = (0.582mol )(8.31 L · kPa)(288 K)mol · K (81.8 kPa)

V = 17.0 L

Question

Use the kinetic theory of gases to explain this statement: No gas exhibits ideal behavior at all temperatures and pressures.

In real gases, there are attractions between molecules and the molecules have volume.

At low temperatures, attractions between molecules pull them together and reduce the volume.

At high pressures, the volume occupied by the molecules is significant part of the total volume.

End of Section 14.3

Gases: Mixtures & Movements

Gas pressure depends on the number of particles in a given volume and on their average kinetic energy.

Particles in a mixture of gases at the same temperature have the same average kinetic energy.

The kind of gas particle is not important.

Partial pressurePartial pressure – the contribution each gas in a mixture makes to the total pressure

Dalton’s Law of Partial Pressures

In a mixture of gases, the total pressure is the sum of the partial pressures of the gases.

Ptotal = P1 + P2 + P3 + …..

Partial pressurePartial pressure – the contribution each gas in a mixture makes to the total pressure

Dalton’s law of partial pressuresDalton’s law of partial pressures – states that, at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases.

Gases: Mixtures & Movements

If the % composition of a mixture of gases does not change, the fraction of the pressure exerted by a gas does not change as the total pressure changes.

This fact is important for people who must operate at high altitudes.

For example, at the top of Mount Everest, the total atmospheric pressure is 33.73 kPa. About 1/3 of its value at sea level.

Gases: Mixtures & Movements

At Mount Everest, the partial pressure of O is also reduced by one third to 7.06 kPa.

The partial pressure of O must be 10.6 kPa or higher to support respiration in humans.

For a climber, he would need an oxygen mask and a cylinder of compressed oxygen to survive.

Sample Problem Using Dalton’s Law of Partial Pressures

Air contains O, N, CO2, and trace amounts of other gases. What is the partial pressure of O (PO) at 101.30 kPa of total pressure if the partial pressures of N, CO2 and other gases are 79.10 kPa, 0.040 kPa, and 0.94 kPa respectively?

Ptotal = PN2 + PCO2 + PTrace + PO2

101.30kPa = 79.10kPa + 0.040kPa + 0.94kPa + PO

101.30kPa = 80.08 kPa + PO

101.3 kPa – 80.08 kPa = PO

21.22 kPa = PO

Sample Problem Using Dalton’s Law of Partial Pressures

A gas mixture containing o, N, and CO2 has a total pressure of 32.9 kPa. If PN2 = 23.0 kPa and PO2 = 6.6 kPa what is PCO2

Ptotal = PN2 + PCO2 + PO2

PCO2 = Ptotal – (PN2 + PO2)

PCO2 = 32.9 kPa – (23.0 kPa + 6.6 kPa)

PCO2 = 32.9 kPa – 29.6 kPa PCO2 = 3.3 kPa

Diffusion

DiffusionDiffusion – is the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout.

Example - if you spray perfume or have an open bottle of perfume at one corner of a room, at some point you could smell the perfume in the opposite corner of the room.

Effusion

EffusionEffusion – during effusion, a gas escapes through a tiny hole in its container.

With effusion and diffusion, the type of particle is important.

Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass.

Graham’s Law

Scottish chemist Thomas Graham studied rates of effusion during the 1840’s.

Graham’s Law of EffusionGraham’s Law of Effusion – states that the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.

This law can also be applied to the diffusion of gas.

Graham’s Law

KE = ½ mv2 (mass x velocity squared)

For the KE to be constant, any increase in mass must be balanced by a decrease in velocity.

Example – a ball with a mass of 2g must travel at 5 m/s to have the same KE as a ball with a mass of 1g traveling at 7 m/s.

If two objects with different masses have the same KE, the lighter object must move faster.

Comparing Effusion Rates

If you have a helium filled balloon, both the helium atoms and the molecules in air can pass freely through the tiny pores in a balloon.

But a helium-filled balloon will deflate faster than an air-filled balloon.

If the balloons are at the same temperature, the particles in each balloon have the save average kinetic energy.

Helium atoms are less massive than oxygen or nitrogen molecules

Graham’s Law

The molecules in air move more slowly than helium atom with the same kinetic energy.(O & N are larger)

The rate of effusion is related only to a particle’s speed.

Rate A = molar massB

Rate B molar mass A

The rates of effusion of two gases are inversely proportional to the square roots of their molar masses.

Graham’s Law

Use Grahams’ Law to compare the effusion rates of nitrogen (molar mass = 28.0g) and helium (molar mass = 4.0g)

Rate He = 28.0g = 7 = 2.7 Rate N2 4.0g

Helium effuses and diffuses nearly three times faster than nitrogen at the same temperature

Question

The partial pressures of oxygen and hydrogen gases in a container are both 100 kPa. In which sample are there more molecules? The greater average kinetic energy?

Both gases have the same number of molecules and same average kinetic energy.

About how much faster does helium diffuse compared to Oxygen?

About 3 times faster than oxygen.

Questions

In a mixture of gases, how is the total pressure determined?

Total pressure is equal to the sum of the partial pressures of the components.

What is the effect of molar mass on rates of diffusion and effusion?

Gases with lower molar masses diffuse and effuse faster than gases with higher molar masses.

Questions

How is the partial pressure of a gas in a mixture calculated?

Subtract the partial pressures of the other gases from the total pressure

What distinguished effusion from diffusion? How are these processes similar?

During effusion, a gas escapes through a tiny hole in its container. In both cases, the rate depends on the molar mass.

Questions

How can you compare the rates of effusion of two gases in a mixture?

Use Graham’s Law equation

Explain why the rates of diffusion of N gas and CO gas are almost identical at the same temperature.

They have almost identical molar masses.

End of Chapter 14