Gas!!!Gas!!!It’s It’s

Everywhere!!!!Everywhere!!!!

Kinetic Molecular TheoryKinetic Molecular Theory

1. Most of the volume occupied by a gas is empty space

2. Collisions between gas particles are elastic collisions

• Energy is transferred between the particles

Kinetic Molecular Kinetic Molecular TheoryTheory

3. Gas particles are in constant motion

• Possess kinetic energy (energy of motion)

4. Kinetic Energy in a gas depends on the temperature

• KE = ½ mv2

Kinetic Molecular TheoryKinetic Molecular Theory

• Expansion– No definite shape or volume

• Fluidity– Gases flow like liquids

• Low Density– Most of the volume is empty

space

• Compressibility– Since mostly empty space,

particles can be squeezed into a smaller volume

DiffusionDiffusion• Spontaneous mixing of gas particles

caused by their random motion

EffusionEffusion• A process by which gas particles

pass through a small opening

Diffusion vs. EffusionDiffusion vs. Effusion

Real vs. IdealReal vs. Ideal• Ideal Gas – An imaginary gas

that perfectly fits all of the assumptions of the Kinetic Molecular Theory

• Real Gas – A gas that does not behave completely according to the assumptions of the Kinetic Molecular Theory – A real gas can behave like an

ideal gas under high temperature and low pressure conditions

STPSTP

• STP = Standard Temperature and Pressure

Standard Temperature = 0°C Standard Pressure = 1 atm

PressurePressure

• The force per unit area on a surface

• Barometer – a device used to measure atmospheric pressure

Units of PressureUnits of Pressure

• Millimeters of mercury – mmHg• Torr• Atmosphere – atm• Pascal – Pa• Kilopascal - kPa

Important ConversionImportant Conversion

• 1 atm = 760 mmHg• 1 atm = 760 torr• 1 atm = 101.325 kPa

TemperatureTemperature• Kelvin is the standard temperature

scale used when dealing with gases• Absolute zero = 0 K

– Unreachable!

• Temperature in Kelvin = Temperature in Celsius + 273• Temperature in Celsius = Temperature in Kelvin – 273

0°C = 273 K25°C = 298 K (Room Temp)0 K = -273°C (Absolute Zero)

Boyle’s LawBoyle’s Law• The volume of a gas at

constant temperature varies indirectly with pressure.

ExampleExample• The pressure exerted on a 240

mL sample of hydrogen gas at constant temperature is increased from 0.428 atm to 0.724 atm. What will the final volume of the sample be?

• V1 = 240.0 mL

• V2= ?

• P1 = 0.428 atm

• P2 = 0.724 atm

P1V1 = P2V2

(0.428 atm)(240.0 mL) = (0.724 atm)V2

V2 = 142 mL

Charles’ LawCharles’ Law• The volume of a gas at

constant pressure varies directly with the temperature

ExampleExample• A sample of air has a volume

of 140.0 mL at 67°C. At what temperature will its volume be 50.0 mL at constant pressure?

• V1 = 140.0 mL

• V2= 50.0 mL

• T1 = 67°C + 273 = 340 K

• T2 = ?

V1 V2

T1 T2

=V1

V2 T1T2 = =(50.0 mL)(340K)

140.0 mL= 121 K

Gay-Lussac’s LawGay-Lussac’s Law• The pressure of a gas at

constant volume varies directly with the temperature

Combined Gas LawCombined Gas Law

ExampleExample

• A helium filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10°C?

SolutionSolution

• V1 = 50.0 L

• V2= ????

• P1 = 1.08 atm

• P2 = 0.855 atm

• T1 = 25°C + 273 = 298 K

• T2 = 10°C + 273 = 283 K

Step 1: List what you are given.

Must Convert to Kelvin!!!!

SolutionSolutionStep 2: Solve the Combined Gas

Law for your unknown.

SolutionSolution

Step 3: Solve!

V2 =(1.08 atm)(50.0 L)(283 K)

(0.855 atm)(298 K)

60.0 L HeV2 =

Dalton’s LawDalton’s Law• The total pressure of a

mixture of gases is equal to the sum of the partial pressures of the component gases.

PT = P1 + P2 + ..... + Pn

ExampleExample• What is the total pressure of

a gas containing a mixture of three gases whose partial pressures are 20 kPa, 10 kPa, and 30 kPa?

PT = P1 + P2 + ..... + Pn

PT = 20 kPa + 10 kPa + 30 kPa

PT = 60 kPa

Ideal Gas LawIdeal Gas Law• The mathematical relationship

among pressure, volume, temperature, and the number of moles of a gas.

ExampleExample• Calculate the volume, in liters,

occupied by 2.00 mol of H2 at 300 K and 1.25 atm. (R = 0.0821)

PV = nRT

V =nRT

P

V =(2.00mol)(0.0821)(300K)

1.25 atm

V = 39.4 L

Avogadro’s LawAvogadro’s Law• Equal volumes of gases at

the same temperature and pressure contain equal numbers of molecules.

Gas StoichiometryGas Stoichiometry

2CO + O2 2CO2

How many liters of oxygen are required to produce 2 L of carbon dioxide?

2L CO2

2L CO2

1L O2= 1L O2