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Chapter 14“The Behavior of Gases”

Pre-AP ChemistryCharles Page High School

Stephen L. CottonMonday, April 8, 13

Section 14.1The Properties of Gases

OBJECTIVES:Explain why gases are easier to compress than solids or liquids are.

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Section 14.1The Properties of Gases

OBJECTIVES:Describe the three factors that affect gas pressure.

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Compressibility

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CompressibilityGases can expand to fill its

container, unlike solids or liquids

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container, unlike solids or liquidsThe reverse is also true:

They are easily compressed, or squeezed into a smaller volume

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container, unlike solids or liquidsThe reverse is also true:

They are easily compressed, or squeezed into a smaller volume

Compressibility is a measure of how much the volume of matter decreases under pressure

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Compressibility

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CompressibilityThis is the idea behind placing “air

bags” in automobiles

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bags” in automobilesIn an accident, the air compresses

more than the steering wheel or dash when you strike it

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bags” in automobilesIn an accident, the air compresses

more than the steering wheel or dash when you strike it

The impact forces the gas particles closer together, because there is a lot of empty space between them

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Variables that describe a Gas

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Variables that describe a Gas The four variables and their common

units:

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units:1. pressure (P) in kilopascals

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units:1. pressure (P) in kilopascals2. volume (V) in Liters

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units:1. pressure (P) in kilopascals2. volume (V) in Liters3. temperature (T) in Kelvin

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units:1. pressure (P) in kilopascals2. volume (V) in Liters3. temperature (T) in Kelvin4. amount (n) in moles

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units:1. pressure (P) in kilopascals2. volume (V) in Liters3. temperature (T) in Kelvin4. amount (n) in moles

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

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1. Amount of Gas

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1. Amount of GasWhen we inflate a balloon, we are

adding gas molecules.

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adding gas molecules. Increasing the number of gas

particles increases the number of collisionsthus, the pressure increases

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adding gas molecules. Increasing the number of gas

particles increases the number of collisionsthus, the pressure increases

If temperature is constant, then doubling the number of particles doubles the pressure

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Pressure and the number of molecules are directly related

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More molecules means more collisions, and…

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Fewer molecules means fewer collisions.

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Fewer molecules means fewer collisions.

Gases naturally move from areas of high pressure to low pressure, because there is empty space to move into – a spray can is example.

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2. Volume of Gas

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2. Volume of Gas In a smaller container, the

molecules have less room to move.

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The particles hit the sides of the container more often.

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The particles hit the sides of the container more often.

As volume decreases, pressure increases. (think of a syringe)Thus, volume and pressure are

inversely related to each other

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3. Temperature of Gas

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3. Temperature of Gas Raising the temperature of a gas increases

the pressure, if the volume is held constant. (Temp. and Pres. are directly related)The molecules hit the walls harder, and

more frequently!

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more frequently! Should you throw an aerosol can into a

fire? What could happen?

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more frequently! Should you throw an aerosol can into a

fire? What could happen? When should your automobile tire pressure

be checked?

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Section 14.2The Gas Laws

OBJECTIVES:Describe the relationships among the temperature, pressure, and volume of a gas.

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Section 14.2The Gas Laws

OBJECTIVES:Use the combined gas law to solve problems.

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The Gas Laws are mathematical

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The Gas Laws are mathematicalThe gas laws will describe HOW

gases behave.Gas behavior can be predicted by

the theory.

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the theory.The amount of change can be

calculated with mathematical equations.

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the theory.The amount of change can be

calculated with mathematical equations.

You need to know both of these: the theory, and the math

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Robert Boyle(1627-1691)

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• Boyle was born into an aristocratic Irish family

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• Became interested in medicine and the new science of Galileo and studied chemistry.

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• A founder and an influential fellow of the Royal Society of London

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• A founder and an influential fellow of the Royal Society of London

• Wrote extensively on science, philosophy, and theology.

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#1. Boyle’s Law - 1662

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#1. Boyle’s Law - 1662Gas pressure is inversely proportional to the volume, when temperature is held constant.

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Pressure x Volume = a constant

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

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Pressure x Volume = a constant Equation: P1V1 = P2V2 (T = constant)

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

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Graph of Boyle’s Law – page 418

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Graph of Boyle’s Law – page 418Boyle’s Law says the pressure is inverse to the volume.

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

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- Page 419

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Jacques Charles (1746-1823)

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Jacques Charles (1746-1823)• French Physicist

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Jacques Charles (1746-1823)• French Physicist• Part of a scientific

balloon flight on Dec. 1, 1783 – was one of three passengers in the second balloon ascension that carried humans

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• This is how his interest in gases started

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• This is how his interest in gases started

• It was a hydrogen filled balloon – good thing they were careful!

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#2. Charles’s Law - 1787

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

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#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.

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Converting Celsius to Kelvin

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Converting Celsius to Kelvin•Gas law problems involving temperature will always require that the temperature be in Kelvin. (Remember that no degree sign is shown with the kelvin scale.)

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•Reason? There will never be a zero volume, since we have never reached absolute zero.

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Kelvin = °C + 273Monday, April 8, 13



Kelvin = °C + 273 and

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Kelvin = °C + 273 °C = Kelvin - 273and

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- Page 421

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Joseph Louis Gay-Lussac (1778 – 1850)

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Joseph Louis Gay-Lussac (1778 – 1850) French chemist and physicist

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Joseph Louis Gay-Lussac (1778 – 1850) French chemist and physicist Known for his studies on the physical properties of gases.

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Joseph Louis Gay-Lussac (1778 – 1850) French chemist and physicist Known for his studies on the physical properties of gases. In 1804 he made balloon ascensions to study magnetic forces and to observe the composition and temperature of the air at different altitudes.

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#3. Gay-Lussac’s Law - 1802

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

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•How does a pressure cooker affect the time needed to cook food? (Note page 422)

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•How does a pressure cooker affect the time needed to cook food? (Note page 422)

•Sample Problem 14.3, page 423Monday, April 8, 13

#4. The Combined Gas Law

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

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

Sample Problem 14.4, page 424

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The combined gas law contains all the other gas laws!

If the temperature remains constant...

P1 V1

T1

x=

P2 V2

T2

x

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If the temperature remains constant...

P1 V1

T1

x=

P2 V2

T2

x

Boyle’s Law

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If the pressure remains constant...

P1 V1

T1

x=

P2 V2

T2

x

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If the pressure remains constant...

P1 V1

T1

x=

P2 V2

T2

x

Charles’s Law

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u The combined gas law contains all the other gas laws!

u If the volume remains constant...

P1 V1

T1

x=

P2 V2

T2

x

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u The combined gas law contains all the other gas laws!

u If the volume remains constant...

P1 V1

T1

x=

P2 V2

T2

x

Gay-Lussac’s Law

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Section 14.3Ideal Gases

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

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Section 14.3Ideal Gases

OBJECTIVES:Compare and contrast real an ideal gases.

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5. The Ideal Gas Law #1

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5. The Ideal Gas Law #1 Equation: P x V = n x R x T

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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.

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R = 8.31 (L x kPa) / (mol x K)

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

the constant, in order to cancel out.

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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)

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The Ideal Gas Law

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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:

The Ideal Gas Law

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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 =

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Ideal Gases

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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

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

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Particles have no volume? Wrong!

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

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Ideal Gases

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Ideal GasesThere are no gases for which this

is true (acting “ideal”); however,

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

a) high temperature, and b) low pressure.

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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!

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35

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36

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37

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38

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39

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40

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#6. Ideal Gas Law 2

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#6. Ideal Gas Law 2 P x V = m x R x T

M

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M Allows LOTS of calculations, and

some new items are:

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some new items are: m = mass, in grams

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some new items are: m = mass, in grams M = molar mass, in g/mol

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some new items are: m = mass, in grams M = molar mass, in g/mol

Molar mass = m R T P V

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42

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Tro's Introductory Chemistry, Chapter 11

111

Molar Mass of a Gas

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111

Molar Mass of a Gas

• One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.

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Density

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Density Density is mass divided by volume

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mD =

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m VD =

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m Vso, m M P V R T

D =

D = =

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84

Avogadro’s Law

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84

Avogadro’s Law• Volume is directly proportional to

the number of gas molecules.V = constant x n.Constant P and T.More gas molecules = larger

volume.

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84



volume.• Count number of gas molecules

by moles, n.

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84



volume.• Count number of gas molecules

by moles, n.• Equal volumes of gases contain

equal numbers of molecules.The gas doesn’t matter.

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85

Avogadro’s Law, Continued

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Example 11.5—A 0.22 Mol Sample of He Has a Volume of 4.8 L. How Many Moles Must Be Added to Give 6.4 L?

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

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V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

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V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol

n2, and added moles

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

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V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol

n2, and added moles

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

V1, V2, n1 n2

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mol added = n2 – n1,


V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol

n2, and added moles

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

V1, V2, n1 n2

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mol added = n2 – n1,


Since n and V are directly proportional, when the volume increases, the moles should increase, and it does.

V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol

n2, and added moles

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

V1, V2, n1 n2

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95

Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?

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Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?, Continued

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

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V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

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V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

V2

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

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V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

V2

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

V1, n1, n2 V2

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Since n and V are directly proportional, when the moles decreases, the volume should decrease, and it does.

V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

V2

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

V1, n1, n2 V2

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Real Gasesand

Ideal Gases

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Ideal Gases don’t exist, because:

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1. Molecules do take up space

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1. Molecules do take up space2. There are attractive forces between

particles- otherwise there would be no liquids formed

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Real Gases behave like Ideal Gases...

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When the molecules are far apart.

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

volume.

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

volume. This is at low pressure

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Real Gases behave like Ideal Gases…

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When molecules are moving fast

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When molecules are moving fastThis is at high temperature

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Collisions are harder and faster.

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

other very long.

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

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

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Section 14.4Gases: Mixtures and Movements OBJECTIVES:

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

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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.

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#7 Dalton’s Law of Partial Pressures

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For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

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PTotal = P1 + P2 + P3 + . . .

•P1 represents the “partial pressure”, or the contribution by that gas.

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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.

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Collecting a gas over water – one of the experiments in Chapter 14 involves this.

Connected to gas generator

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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:

1 2 3 4

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2 atm + 1 atm + 3 atm

1 2 3 4

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2 atm + 1 atm + 3 atm = 6 atm

1 2 3 4

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2 atm + 1 atm + 3 atm = 6 atm

Sample Problem 14.6, page 434

1 2 3 4

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Diffusion is:

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Diffusion is:u Molecules moving from areas of high

concentration to low concentration.

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Diffusion is:u Molecules moving from areas of high

concentration to low concentration.uExample: perfume molecules spreading across the room.

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Diffusion is:

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

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

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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.

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

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

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

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

•Fig. 14.18, p. 435Monday, April 8, 13

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

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

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

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8. Graham’s LawRateA √ MassB

RateB √ MassA

=

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8. Graham’s Law

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

RateA √ MassB

RateB √ MassA

=

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8. Graham’s Law


Derived from: Kinetic energy = 1/2 mv2

RateA √ MassB

RateB √ MassA

=

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8. Graham’s Law


Derived from: Kinetic energy = 1/2 mv2

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

RateA √ MassB

RateB √ MassA

=

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Graham’s Law

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Sample: compare rates of effusion of Helium with Nitrogen – done on p. 436

Graham’s Law

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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

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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

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End of Chapter 14

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