• I will use the kinetic-molecular theory to explain the physical properties of gases, liquids and solids

• I will compare types of intermolecular forces

• I will explain how kinetic energy and intermolecular forces combine to determine the state of a substance

• I will describe the role of energy in phase changes

• I will use the kinetic-molecular theory to explain the physical properties of gases, liquids and solids

• I will compare types of intermolecular forces

• I will explain how kinetic energy and intermolecular forces combine to determine the state of a substance

• I will describe the role of energy in phase changes

States of MatterStates of Matter

13.1 Gases13.1 Gases

• I will use the Kinetic-molecular theory to explain the behavior of gases

• I will describe how mass affects the rates of diffusion and effusion

• I will explain how gas pressure is measured and calculate the partial pressure of a gas

Vocabulary kinetic-molecular theory elastic collisiontemperature Graham’s Law of effusiondiffusionPressure barometer pascal

atmosphereDalton’s law of partial pressures

GasesGases

Substances that are gases at room temperature usually display similar physical preperties despitetheir different compositions.

Why is there so little variation in behavior among gases?

Kinetic-Molecular

Theory

Kinetic-Molecular

Theory

Kinetic-Molecular TheoryKinetic-Molecular TheoryDescribes the behavior of gases in terms ofparticles in motion

Makes several assumptions about gas particles:• Size• Motion• Energy

Gas AssumptionsGas Assumptions

Particle Size• Separated by empty space• Volume of particle = small• Volume of empty space = large• NO significant attractive or repulsive forces

Particle Motion• Constant random motion• Move in a STRAIGHT line• Collide with walls or other particles• Collisions are elastic (NO kinetic energy is lost, just

transferred)

Gas Assumptions Cont...Gas Assumptions Cont...Particle Energy• Affected by mass and velocity• KE = ½ mv2

• Single gas• Particles have same mass• Particles have different velocity• Particles have different KE

• Temperature = measure of the average KE of the particles in a sample of matter• At a given temperature, ALL gases have the SAME average KE

Gas Behavior

Gas Behavior

Gas BehaviorGas Behavior

• Low Density• Compressible• Expandable• Diffuse• Effuse

Gas BehaviorGas Behavior

Low Density

Density = mass/volume

Large space between gas particles

Fewer gas molecules than solid or liquid molecules in the same volume

Gas BehaviorGas BehaviorCompression & Expansion

Large amount of empty space between gas particles

Allows particles to be squished into a smaller volume

Stop squishing; random motion of particles fills the available space, expands to original volume

Gas BehaviorGas Behavior

DiffusionThe movement of one

material through another material

Gas particles have no significant forces of attraction

Particles can slide past each other

Mix until evenly distributed

Particles diffuse from areas of

high concentration to areas of low concentration

Rate of diffusion• Depends on

mass of particles• Lighter = faster• Heavier = slower

Gas BehaviorGas Behavior

Gas BehaviorGas Behavior

Related to diffusion

When a gas escapes through a tiny opening

Ex. Puncture a balloon or tire

Inverse relationship between effusion rates and molar mass

Effusion

Gas BehaviorGas BehaviorGraham’s Law of Effusion

The rate of effusion for a gas is inversely proportional to the square root of its molar mass

Also applies to rate of diffusion

Set up a proportion to compare the diffusion rates of 2 gases

Finding a Ratio of Diffusion

Rates

Finding a Ratio of Diffusion

Rates

Let’s Try!Let’s Try!

Practice ProblemsPractice Problems1. RN2/RNe =

0.8492. RCO/RCO2 =

1.253. 2.5mol/min

1. RN2/RNe = 0.849

2. RCO/RCO2 = 1.25

3. 2.5mol/min

1. Calculate the ratio of effusion rates for nitrogen (N2) and Neon (Ne).

2. Calculate the ratio of diffusion rates for carbon monoxide (CO) and carbon dioxide (CO2)

3. What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.6 mol/min?

1. Calculate the ratio of effusion rates for nitrogen (N2) and Neon (Ne).

2. Calculate the ratio of diffusion rates for carbon monoxide (CO) and carbon dioxide (CO2)

3. What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.6 mol/min?

Use tofind ratios

Gas Pressure

Gas Pressure

Gas PressureGas Pressure

Force per unit area

Gas particles exert pressure when they collide with:

• The walls of their container

• Each other

Exert pressure in ALL directions (because particles move in all directions)

Individual particles can only exert little pressure

Many particles colliding can exert substantial pressure

pressure of gases is what keeps our tires inflated, makes our basketballs bounce, makes hairspray come out of the can, etc.

Pressure

Gas PressureGas Pressure

Pressure increases when temperature increases because the molecules are moving with greater speed and colliding against the sides of their containers more often.

Therefore, the pressure inside that container is greater, because there are more collisions.

Gas PressureGas Pressure

Atmospheric Pressure

Air pressure at Earth’s surface

Equal to:Pressure exerted by 1 kg mass on a square centimeter

Varies by elevation

Mountains = lessSea Level = more

Gas PressureGas PressureMeasuring Air Pressure

Barometer• Measures atmospheric

pressure

• Mercury in it is always about 760mm• Exact amount determined

by 2 forces• Gravity- downward force• Air pressure- upward force

(air presses on surface of Hg)

Air pressure varies because of:

• Changes in air temperature

• Changes in humidity

• Increase in air pressure = Hg rises

• Decrease in air pressure = Hg falls

Gas PressureGas PressureMeasuring Enclosed Gas Pressure

Manometer• Measures pressure of

enclosed gas

• The difference in height of the mercury in the 2 arms is used to calculate the pressure of the gas in the flask

Gas PressureGas Pressure

Units of Pressure

Comparison of Pressure Units

Unit Compared with 1 atm

Compared with 1 KPa

Kilopascal (kPa)

1atm = 101.3 kPa

mm Hg 1 atm = 760 mm Hg

1 kPa = 7.501 mm Hg

torr 1 atm = 760 torr 1 kPa = 7.501 torr

psi 1 atm = 14.7 psi 1 kPa = 0.145 psi

atm 1 kPa = 0.009 869 atm

SI Unit = Pascal (Pa)Atmosphere = atm (used to report air

pressure)

Gas LawsGas Laws

Gas LawsGas Laws

Dalton’s Law of Partial Pressures

Gay-Lussac’s Law

Boyle’s Law

Charles’s Law

Ideal Gas Law

Combined Gas Law

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

P total = P1 + P2 + P3 +…Pn

The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture.

Partial pressure = the portion of the total pressure contributed by a single gas

Depends on • # of moles of gas• Size of container• Temperature of the mixture

Does NOT depend on identity of gas (at a given temperature and pressure, the partial pressure of 1 mole of ANY gas is the same)

Dalton’s Law of Partial PressuresDalton’s Law of Partial PressuresPractice Problem

A mixture of oxygen (O2), carbon dioxide (CO2), and nitrogen (N2) has a total pressure of 0.97 atm. What is the partial pressure of O2, if the partial pressure of CO2 is 0.70 atm and the partial pressure of N2 is 0.12 atm?

PO2 = 0.97 atm – 0.70 atm -0.12 atmPO2 = 0.15 atm

Boyle’s LawBoyle’s Law

P1V1 = P2V2

The volume of a given amount of gas held at a constant temperature varies inversely with pressure• As pressure increases, volume decreases

and vice versa

Boyle’s LawBoyle’s Law

Practice Problem

A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0 L volume is 210 kPa, what will the pressure be at 2.5 L?

210 kPa (4.0 L) = P2 (2.5 L)P2 = 340 kPa

Charles’s LawCharles’s Law

V1 V2

T1 = T2

The volume of a given mass of a gas is directly proportional to its Kelvin temperature at constant pressure.• When Kelvin temperature increases, volume

increases and vice versa

Tk = 273 + Tc

Charles’s LawCharles’s Law

Practice Problem

A gas sample at 40 degrees C occupies a volume of 2.32 L. If the temperature is raised to 75.0 degrees C, what will the volume be, assuming the pressure remains constant?

40.0 degrees C + 273 = 313K75 degrees C + 273 = 348K

2.32 L = V2 313 K 348 K

V2 = 2.58 L

Gay-Lussac’s LawGay-Lussac’s Law

P1 P2

T1 = T2

The pressure of a given mass of gas varies directly with the Kelvin temperature when the volume remains constant.• As Kelvin temperature increases, pressure

increases and vice versa

Gay-Lussac’s LawGay-Lussac’s Law

Practice Problem

The pressure of a gas in a tank is 3.20 atm at 22.0 degrees C. If the temperature rises to 60.0 degrees C, what will be the gas pressure in the tank?

22.0 degrees C + 273 = 295 K60.0 degrees C + 273 = 333 K

3.20 atm = P2295 K 333 K

P2 = 3.61 atm

The Combined Gas LawThe Combined Gas LawP1V1 = P2V2 T1 T2

The relationship among pressure, volume, and temperature of a fixed amount of gas.• Pressure increases, volume decreases• Pressure increases, temperature increases• Volume increases, temperature increases

The Combined Gas LawThe Combined Gas LawPractice Problem

A gas at 110 kPa and 30.0 degrees C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 degrees C and the pressure increased to 440 kPa, what is the new volume?

30.0 degrees C + 273 = 303 K80.0 degrees C + 273 = 353 K

110(2.00) = 440(V2) 303 353

V2 = 0.58 L

Ideal GasIdeal Gas

Ideal Gas

One whose particles• Take up NO space• Have NO intermolecular attractive

forces• Follows ideal gas laws under ALL

conditions of temperature and pressure

Real GasReal Gas

NO gas in the real world is truly idealparticles have some volumesome attractive forces

Real gases deviate most from ideal behavior at low temperatures and high pressures

Examples:• Liquid nitrogen is used to store biological

tissues at low temps• Increased pressure allows a larger mass of

propane to fit into a smaller volume for easier transport

Ideal Gas LawIdeal Gas Law

PV = nRT

Describes the physical behavior of an ideal gas in terms of the pressure, volumes, temperature, and number of moles of gas present.

R = ideal gas constant

Numerical Values of the Gas Constant, R

Units of R Numerical R Value

Units of P Units of V Units of T Units of n

L atm/ mol K

0.0821 atm L K mol

L kPa/ mol K

8.314 kPa L K mol

L mm Hg / mol K

62.4 mm Hg L K mol

Ideal Gas LawIdeal Gas Law

Calculate the number of moles of gas contained in a 3.0 L vessel at 3.00 x 102 K with a pressure of 1.50 atm.

V = 3.0 L T = 3.00 x 102 KP = 1.50 atm R = 0.0821 L atm/ mol Kn = ?

1.50 (3.0) = n (0.0821)(3.00 x 102 K)

n = 0.18 mol

Density of a GasDensity of a Gas

D = MP RT

Practice Problem

What is the density of a gas at STP that has a molar mass of 44.0 g/mol?

STP = 273 K, 1 atm

D = 44.0(1) 0.0821(273)

D = 1.96 g/L