Chapter 14

The

Behavior of Gases

Section 14.2Section 14.2The Gas LawsThe Gas Laws

l\l\

Converting PressureConverting Pressure

1 atmosphere = 760 mmHg1 atmosphere = 760 mmHg 1 atmosphere = 101 325 Pa 1 atmosphere = 101 325 Pa 1 atmosphere = 101.325 kPa1 atmosphere = 101.325 kPa 1 atmosphere = 760 torrs1 atmosphere = 760 torrs 1 atmosphere = 14.7 psi1 atmosphere = 14.7 psi 1 atmosphere = 29.92 in Hg1 atmosphere = 29.92 in Hg

STPSTP Standard Standard

temperature and temperature and pressurepressure

1atmosphere1atmosphere

0 degrees Celsius0 degrees Celsius

STP is also a motor STP is also a motor oil. That’s cool, if oil. That’s cool, if irrelevantirrelevant..

The Gas Laws are mathematical.

The gas laws will describe HOW gases behave.

Four Variables Describe a Gas

1.pressure (P) in atm

2. volume (V) in Liters

3. temperature (T) in Kelvin

4. amount (n) in moles

•

Four Variables Describe a Gas

1.pressure (P) in atm

2. volume (V) in Liters

3. temperature (T) in Kelvin

4. amount (n) in moles

•

Held constant in Section 14.2

Robert Boyle(1627-1691)

• Boyle was born into an aristocratic Irish family

•

#1. Boyle’s Law - 1662

Pressure x Volume = a constant

Equation: P1V1 = P2V2 (T = constant)

Gas pressure is inversely proportional to the volume, when temperature is held constant.

Graph of Boyle’s Law – page 418

Boyle’s Law says the pressure is inverse to the volume.

Note that when the volume goes up, the pressure goes down

http://www.grc.nasa.gov/WWW/k-12/airplane/Animation/frglab2.html

Boyle’s lawBoyle’s law

Boyle’s lawBoyle’s law

Key mathematical Key mathematical points:points:

The product of corresponding P and V The product of corresponding P and V values is a constant PV = constant values is a constant PV = constant

PP11VV11=P=P22VV22

V is inversely proportional to PV is inversely proportional to P

V is directly proportional to 1/P: V~1/PV is directly proportional to 1/P: V~1/P

A sample problemA sample problemon volume-pressure relationshipon volume-pressure relationship..

A helium balloon contains 30.0 L A helium balloon contains 30.0 L of helium gas sat 103 kPa. What of helium gas sat 103 kPa. What is the volume of the helium when is the volume of the helium when the balloon rises to an altitude the balloon rises to an altitude where the pressure is only 25.0 where the pressure is only 25.0 kPa. Assume the temperature kPa. Assume the temperature remains constant.remains constant.

Answer: 124 L Answer: 124 L

Jacques Charles (1746-1823)• French Physicist

#2. Charles’s Law - 1787The volume of a fixed mass of gas is directly proportional to the Kelvin temperature, when pressure is held constant.

This extrapolates to zero volume at a temperature of zero Kelvin.

VT

VT

P1

1

2

2 ( constant)

Converting Celsius to Kelvin

• Always use the temperature in Kelvin.

Kelvin = C + 273.15

°C = Kelvin – 273.15

Joseph Louis Gay-Lussac (1778 – 1850)

French chemist and physicist

#3. Gay-Lussac’s Law - 1802•The pressure and Kelvin temperature of a gas are directly proportional, provided that the volume remains constant.

2

2

1

1

T

P

T

P

http://www.grc.nasa.gov/WWW/k-12/airplane/Animation/frglab2.html

Charles’s lawCharles’s law

Charles’s lawCharles’s law

A sample problemA sample problemon volume-temperature on volume-temperature

relationshiprelationship..

A balloon inflated in a room at 24 A balloon inflated in a room at 24 ˚C has a volume of 4.00 L. The ˚C has a volume of 4.00 L. The balloon is then heated to a balloon is then heated to a temperature of 58 ˚C . What is the temperature of 58 ˚C . What is the new volume if the pressure new volume if the pressure remains constant?remains constant?

Answer: 4.46 L Answer: 4.46 L

A sample problemA sample problemon pressure-temperature on pressure-temperature

relationshiprelationship..

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

Answer: 2.58 kPa Answer: 2.58 kPa

Given:Given: A 58 L sample of dry air is A 58 L sample of dry air is cooled from 127°C to -23°C while cooled from 127°C to -23°C while the pressure is maintained at 2.85 the pressure is maintained at 2.85 atm. What is the final volume?atm. What is the final volume?

Solution:Solution: V V22=V=V11TT22/T/T11 VV22=58L*250K/400K=36L - Looks =58L*250K/400K=36L - Looks realistic.realistic.

#4. The Combined Gas LawThe combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

2

22

1

11

T

VP

T

VP

The combined gas law contains all the other gas laws!

If the temperature remains constant...

P1 V1

T1

x=

P2 V2

T2

x

Boyle’s Law

The combined gas law contains all the other gas laws!

If the pressure remains constant...

P1 V1

T1

x=

P2 V2

T2

x

Charles’s Law

The combined gas law contains all the other gas laws!

If the volume remains constant...

P1 V1

T1

x=

P2 V2

T2

x

Gay-Lussac’s Law

A sample problemA sample problemon pressure -volume-temperature on pressure -volume-temperature

relationshiprelationship..

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

Answer: 0.342 L Answer: 0.342 L

Section 14.3Ideal Gases

OBJECTIVES:OBJECTIVES:

Compute the value of an unknown using the ideal gas law.

Section 14.3Ideal Gases

OBJECTIVES:OBJECTIVES:

Compare and contrast real an ideal gases.

5. The Ideal Gas Law #1 Equation: P x V = n x R x T Pressure times Volume equals the

number of moles (n) times the Ideal Gas Constant (R) times the Temperature in Kelvin.

R = 8.31 (L x kPa) / (mol x K) The other units must match the value of

the constant, in order to cancel out. The value of R could change, if other

units of measurement are used for the other values (namely pressure changes)

We now have a new way to count moles (the amount of matter), by measuring T, P, and V. We aren’t restricted to only STP conditions:

P x V R x T

The Ideal Gas Law

n =

Ideal Gases We are going to assume the gases

behave “ideally”- in other words, they obey the Gas Laws under all conditions of temperature and pressure

An ideal gas does not really exist, but it makes the math easier and is a close approximation.

Particles have no volume? Wrong! No attractive forces? Wrong!

Ideal GasesThere are no gases for which this

is true (acting “ideal”); however,Real gases behave this way at

a) high temperature, and b) low pressure.Because at these conditions, a gas will stay a gas

Sample Problem

#6. Ideal Gas Law 2

Equation: P x V = m x R x T M

Allows LOTS of calculations, and some new items are:

m = mass, in grams M = molar mass, in g/mol

Molar mass = m R T P V

Density Density is mass divided by volume

m

V

so,

m M P

V R T

D =

D = =

Ideal Gases don’t exist, because:

1. Molecules do take up space

2. There are attractive forces between particles

- otherwise there would be no liquids formed

Real Gases behave like Ideal Gases...

When the molecules are far apart.

The molecules do not take up as big a percentage of the space We can ignore the particle

volume. This is at low pressure

Real Gases behave like Ideal Gases…

When molecules are moving fastThis is at high temperature

Collisions are harder and faster.Molecules are not next to each

other very long.Attractive forces can’t play a role.

Section 14.4Gases: Mixtures and Movements

OBJECTIVES:

Relate the total pressure of a mixture of gases to the partial pressures of the component gases.

Section 14.4Gases: Mixtures and Movements

OBJECTIVES:

Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses.

#7 Dalton’s Law of Partial Pressures

For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

•P1 represents the “partial pressure”, or the contribution by that gas.•Dalton’s Law is particularly useful in calculating the pressure of gases collected over water.

Collecting a gas over water

Connected to gas generator

If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3:

2 atm + 1 atm + 3 atm = 6 atm

1 2 3 4

Diffusion is:

Effusion: Gas escaping through a tiny hole in a container.

Both of these depend on the molar mass of the particle, which determines the speed.

Molecules moving from areas of high concentration to low concentration.Example: perfume molecules spreading across the room.

•Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

•Molecules move from areas of high concentration to low concentration.

Effusion: a gas escapes through a tiny hole in its container

-Think of a nail in your car tire…

Diffusion and effusion are explained by the next gas law: Graham’s

8. Graham’s Law

The rate of effusion and diffusion is inversely proportional to the square root of the molar mass of the molecules.

Derived from: Kinetic energy = 1/2 mv2

m = the molar mass, and v = the velocity.

RateA MassB

RateB MassA

=

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.

Helium effuses and diffuses faster than nitrogen – thus, helium escapes from a balloon quicker than many other gases

Graham’s Law