Gases

What do we know?1. Gases have mass.2. Gases are easily compressed.3. Gases uniformly and completely fill their

containers.4. Different gases move through each other quite

rapidly.5. Gases exert pressure.6. The pressure of a gas depends on its

temperature.

Measuring Gases• Amount of gas (moles = n)

n = mass (g) / molar mass (g/mol)

• Volume (V)Vgas = Vcontainer

Measured in liters

Measuring Gases• Temperature (T)

Always measured in Kelvin (K)T(K) = T(oC) + 273

• Pressure (P)Gases exert pressure when particles collide with

walls and the force is spread over the area of the container

STP = 1 atm pressure, 0oC (273 K)

Measuring Pressure• Pressure of air is Pressure of air is

measured with a measured with a barometerbarometer (developed (developed by Torricelli in 1643)by Torricelli in 1643)

• Hg rises in tube until Hg rises in tube until force of Hg (down) force of Hg (down) balances the force of balances the force of atmosphere (pushing atmosphere (pushing up). up).

Measuring Pressure1.000 atm (standard atmosphere) =

760.0 mm Hg (millimeters of mercury) =760 torr (named after Torricelli) =14.69 psi (pounds per square inch) =101,325 Pa (pascals)

The Pascal is the SI unit for pressure, but it is used sparingly because it is so small.

Just use the equalities as conversion factors to get from unit to another.

The Gas Laws• Boyle’s Law

(pressure-volume relationship)• Charles’s Law

(volume-temperature relationship)• Gay-Lussac’s Law

(pressure-temperature relationship)• Combined Gas Law

(relates pressure, volume, and temperature)• Dalton’s Law of Partial Pressures• The Ideal Gas Law

Boyle’s LawThe pressure of a given sample of a gas is

inversely proportional to the volume of the gas at constant temperature.

If we know the volume of a gas at a given pressure, we can predict the new volume if the pressure is

changed, as long as the temperature and the amount of the gas remain the same.

P1V1 = P2V2

Boyle’s LawFreon-12 (the common name for the compound

CCl2F2) was once widely used in refrigeration systems, but has now been replaced by other

compounds that do not lead to the breakdown of the protective ozone in the upper atmosphere. Consider a 1.5-L sample of gaseous CCl2F2 at a pressure of 56 torr. If the pressure is changed to

150 torr at a constant temperature, will the volume of the gas increase or decrease? What

will be the new volume of the gas?

Boyle’s LawGiven:• P1 = 56 torr

• V1 = 1.5 L

• P2 = 150 torr• Temperature is constant

Pressure increases so volume decreases.

Boyle’s Law: P1V1 = P2V2

(56 torr)(1.5 L) = (150 torr) V2

V2 = (56 torr)(1.5 L) / (150 torr)

V2 = 0.56 L

Unknown:• V2 = ? L

Boyle’s LawIn an automobile engine the gaseous fuel-air

mixture enters the cylinder and is compressed by a moving piston before it is ignited. In a certain

engine the initial cylinder volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The

fuel-air mixture initially has a pressure of 1.00 atm. Calculate the pressure of the compressed

fuel-air mixture, assuming that both the temperature and the amount of gas remain

constant.

Boyle’s LawGiven:• P1 = 1.00 atm

• V1 = 0.725 L

• V2 = 0.075 L• Temperature and amount of gas are constant

Volume decreases so pressure increases.

Boyle’s Law: P1V1 = P2V2

(1.00 atm)(0.725 L) = P2(0.075 L)

P2 = (1.00 atm)(0.725 L) / (0.075 L)

P2 = 9.7 atm

Unknown:• P2 = ? atm

Charles’s LawThe volume of a given sample of a gas is directly proportional to the temperature of

the gas at constant pressure.

Initial values can also be used to find final values

V1/T1 = V2/T2

Charles’s Law A 2.0-L sample of air is collected at 298 K and

then cooled to 278 K. The pressure is held constant at 1.0 atm. Does the volume increase or decrease? Calculate the volume of the air at

278 K.

Charles’s LawGiven:• T1 = 298 K

• V1 = 2.0 L

• T2 = 278 K• Pressure is constant

Temperature decreases so volume decreases.

Unknown:• V2 = ? L

298 K

Charles’s Law

2.0 L

298 K 278 K

V2 =

V1/T1 = V2/T2

V2 = 2.0 L x 278 K

V2 = 1.9 L

(2.0 L)(278 K) = V2 (298K)

Charles’s LawA sample of gas at 15oC (at 1 atm) has a volume of

2.58 L. The temperature is then raised to 38oC (at 1 atm). Does the volume of the gas increase

or decrease? Calculate the new volume.

Charles’s LawGiven:• T1 = 15oC

• V1 = 2.58 L

• T2 = 38oC• Pressure is constant

Temperature increases so volume increases

Unknown:• V2 = ? L+ 273 = 288 K

+ 273 = 311 K

288 K

Charles’s Law

2.58 L

288 K 311 K

V2 =

V1/T1 = V2/T2

V2 = 2.58 L x 311 K

V2 = 2.79 L

(2.58 L)(311 K) = V2 (288K)

Equal volumes of gases at the same temperature and pressure contain an

equal number of particles.

Initial values can also be used to find final values

V1/n1 = V2/n2

Avogadro’s Law

The pressure of a given sample of a gas is directly proportional to the temperature of

the gas at constant volume.

Initial values can also be used to find final values

P1/T1 = P2/T2

Gay-Lussac’s Law

Gay-Lussac’s LawCalculate the final pressure inside a scuba tank after it cools from 1000 °C to 25.0 °C. The initial

pressure in the tank is 130.0 atm.

Gay-Lussac’s LawGiven:• P1 = 130 atm

• T1 = 1000 oC

• T2 = 25 oC • Volume is constant

Temperature decreases so pressure decreases.

Unknown:• P2 = ? atm

+ 273 = 1273 K+ 273 = 298 K

1273 K

Gay-Lussac’s Law

130.0 atm

1273 K 298 K

P2 =

P1/T1 = P2/T2

P2 = 130 atm x 298 K

P2 = 30.4 atm

(130 atm)(298 K) = P2 (1273 K)

Gay-Lussac’s LawIf a gas in a closed container is pressurized from

15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the

final temperature of the gas be?

Gay-Lussac’s LawGiven:• P1 = 15 atm

• P2 = 16 atm

• T1 = 25.0 oC • Volume is constant

Pressure increases so temperature increases.

Unknown:• T2 = ? K

+ 273 = 298 K

15 atm

Gay-Lussac’s Law

15.0 atm

298 K T2

16.0 atm =

P1/T1 = P2/T2

T2 = 16 atm x 298 K

T2 = 318 K

(15 atm)T2 = (16 atm)(298 K)

Summary of Laws

Law Statement Equation Constant

Boyle'sP inversely proportional to

V P1V1 =

P2V2 T, n 

Charles's V directly proportional to T V1/T1 =

V2/T2 P, n

Avogadro's V directly proportional to nV1/n1 =

V2/n2  P, T

Gay-Lussac’s P directly proportional to TP1/T1 =

P2/T2  V, n

Combined Gas Law

Combined Gas Law• Combines Boyle’s, Charles’s, and Gay-Lussac’s

laws to relate the pressure, volume, and temperature of constant amounts of gases.

Initial values can also be used to find final values

P1V1/T1 = P2V2/T2

Combined Gas LawA toy balloon has an internal pressure of 1.05 atm

and a volume of 5.0 L. If the temperature above where the balloon is released is 20oC, what will

happen to the volume when the ballooon rises to an altitude where the pressure is 0.65 atm and

the temperature is -15oC?

Combined Gas LawGiven:• P1 = 1.05 atm

• V1 = 5.0 L

• T1 = 20 oC

• P2 = 0.65 atm

• T2 = -15 oC

Unknown:• V2 = ? L

+ 273 = 293 K

+ 273 = 258 K

(0.65 atm)(293 K)

Combined Gas Law

(1.05 atm)(5.0 L)

293 K 258 K

(0.65 atm)V2 =

P1V1/T1 = P2V2/T2

V2 = (1.05 atm)(5.0 L)(258 K)

V2 = 7.1 L

(1.05 atm)(5.0 L)(258 K) = (0.65 atm)V2(293 K)

Combined Gas LawIf I initially have 4.0 L of gas at a pressure of 1.1 atm and a temperature of 25oC, what will happen to the pressure if I decrease the volume to 3.0 L

and decrease the temperature to 20oC?

Combined Gas LawGiven:• V1 = 4.0 L

• P1 = 1.1 atm

• T1 = 25 oC

• V2 = 3.0 L

• T2 = 20 oC

Unknown:• P2 = ? L

+ 273 = 298 K

+ 273 = 293 K

(3.0 L)(298 K)

Combined Gas Law

(1.1 atm)(4.0 L)

298 K 293 K

P2(3.0 L)=

P1V1/T1 = P2V2/T2

P2 = (1.1 atm)(4.0 L)(293 K)

P2 = 1.4 atm

(1.1 atm)(4.0 L)(293 K) = P2(3.0 L)(298 K)

The Ideal Gas Law

The Ideal Gas LawCombines Boyle’s, Charles’s, and Avogadro’s Laws to

describe the behavior of gases as dependent upon volume, pressure, temperature, and the number of moles

present.

PV = nRTwhere R is the combined constant called the

universal gas constantR = 0.08206 L atm / mol K

The Ideal Gas LawA gas that obeys this equation is said to

behave ideally, thus the name “Ideal Gas Law”

There is no such thing as an ideal gas, but many real gases behave ideally at pressures

of approximately 1 atm and lower and temperatures of approximately 273 K or

higher.

The Ideal Gas LawA sample of hydrogen gas has a volume of 8.56 L

at a temperature of 0oC and a pressure of 1.5 atm. Calculate the number of moles of H2

present in this gas sample. Assume that the gas behaves ideally.

The Ideal Gas LawGiven:• P = 1.5 atm• T = 0oC = 273 K• V = 8.56 L

PV = nRT(1.5 atm)(8.56 L) = n(0.08206 L-atm/mol-K)(273 K)

Unknown:• n = ?

(1.5 atm)(8.56 L)

(0.08206 L-atm/mol-K)(273 K)n =

n = 0.57 mol

The Ideal Gas LawWhat volume is occupied by 0.250 mol of carbon

dioxide gas at 25oC and 371 torr?

Because the gas constant is measured in L-atm/mol-K, volume must be measured in liters, pressure in

atmospheres, amount of gas in moles, and temperature in Kelvin.

25oC + 273 = 298 K

371 torr x760 torr1.00 atm

= 0.488 atm

The Ideal Gas LawGiven:• P = 0.488 atm• T = 298 K• n = 0.250 mol CO2

PV = nRT(0.488 atm)V = (0.250 mol)(0.08206 L-atm/mol-K)(298K)

Unknown:• V = ? L

(0.250 mol)

(0.488 atm)

(0.08206 L-atm/mol-K)(298 K)V =

V = 12.5 L

Dalton’s Law of Partial Pressures

Dalton’s Law of Partial PressuresFor a mixture of gases in a container, the total pressure

exerted is the sum of the partial pressures of the gases present

The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container.

Ptotal = P1 + P2 + P3

Each gas is responsible for only part of the total pressure.

Dalton’s Law of Partial PressuresWhat is the atmospheric pressure if the partial pressure

of nitrogen, oxygen, and argon are 604.5 mm Hg, 162.8 mm Hg, and 0.5 mm Hg, respectively?

PT = PN2 + PO2

+ PAr

PT = 604.5 mm Hg + 162.8 mm Hg + 0.5 mm Hg

PT = 767.8 mm Hg

Dalton’s Law of Partial PressuresA gas mixture contains hydrogen, helium, neon, and argon. The total pressure of the mixture is 93.6 kPa. The partial pressures of helium, neon, and argon are 15.4 kPa, 25.7 kPa, and 35.6 kPa, respectively. What

is the pressure exerted by hydrogen?

PT = PH2 + PHe + PNe + PAr

93.6 kPa = PH2 + 15.4 kPa + 25.7 kPa + 35.6 kPa

PH2 = 93.6 – (15.4 + 25.7 +35.6) = 16.9 kPa

Dalton’s Law of Partial PressuresMixtures of helium and oxygen are used in the “air” tanks

of underwater divers for deep dives. For a particular dive, 12 L of O2 at 25oC and 1.0 atm and 46 L of He at

25oC and 1.0 atm were pumped into a 5.0-L tank. Calculate the partial pressure of each gas and the

total pressure in the tank at 25oC.

Dalton’s Law of Partial PressuresO2

• V1 = 12 L

• P1 = 1.0 atm

• T1 = 298 K

He• V1 = 46 L

• P1 = 1.0 atm

• T1 = 298 K

Mixture• V2 = 5.0 L

• T2 = 298 K

• P2 = ?

Since only V and P are changing, use Boyle’s Law to solve for the pressure of each

gas. (Do TWO Boyle’s Law problems to solve for the final pressure of each gas in the

mixture!)

O2

P1V1 = P2V2

(1.0 atm)(12 L) = P2 (5.0 L)

P2 = 2.4 atm(partial pressure of oxygen in the mixture)

HeP1V1 = P2V2

(1.0 atm)(46 L) = P2 (5.0 L)

P2 = 9.2 atm(partial pressure of oxygen in the mixture)

PT = 2.4 atm + 9.2 atm = 11.6 atm

Understanding the Laws

What do we want to know?What are the characteristics of the

individual gas particles that cause a gas to behave like it does?

Remember: Laws tell us what happens, not why. Scientists develop models or theories to explain

why nature behaves the way it does.

The Kinetic Molecular Theory of Gases (KMT)

• A simple model that attempts to explain the behavior of an ideal gas

• Based on the behavior of the individual particles (atoms or molecules) in a gas

Assumptions/Postulates of the KMT of Ideal Gases

1. Gases consist of tiny particles (atoms or molecules).2. These particles are so small, compared with the

distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).

3. The particles are in constant random motion, colliding with the walls of the container. These collisions cause the pressure exerted by the gas.

4. The particles are assumed not to attract or repel one another. In other words, the particles don’t interact.

5. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas.

Postulates True for Real Gases?1. Yes, all gases are made up of atoms or molecules.2. Fair assumption, but not perfect. The particles are in fact

very small, but their volumes are not zero. 3. Yes, all gases have particles in constant, random motion

that produce pressure.4. Depends on the gas. Some gases have stronger

intermolecular forces than others.5. Yes, increased temperature does increase motion of the

particles in a real gas.

The KMT of Gases• Temperature reflects how rapidly the individual

particles of a gas are moving.• High temps = fast movement = lots of collisions with walls

of container• Low temps = slow movement = fewer collisions with walls

of container• Consider a container with an expandable volume.

What will happen if the external pressure remains the same and the temperature of the gas is increased?• Higher temp = faster movement = more collisions with the

walls of the container• Increased pressure on walls increase in volume• Proves Charles’s and Gay-Lussac’s Laws!