the behavior of gases

The Behavior of Gases

Properties of Gases (Review)

• No definite shape• No definite volume• compressible

Kinetic Molecular Theory

moving molecules well supported ideas

Basic Kinetic Theory of Gases

1. Composed of particles like atoms (ex: He) or molecules like (O2 and CO2)

There are no attractive/repulsive forces.

Lots of empty space!!


2. Particles move in random, constant, straight-line motion.

Move independently of each other.


3. All collisions are elastic meaning that KE is transferred without loss of energy.

No change in kinetic energy.

Gases tend to diffuse towards areas of lower concentration.

Gas Pressure

• Pressure- force exerted on container walls by particles in a gas

• Units used- kPa, atm, Torr, mmHg• STP (Standard Temperature and Pressure)

Table A

273 K or 0°C and 101.3 kPa = 1 atm = 760 Torr (mmHg)

Factors Affecting Pressure

Amount of Amount of Gas (number Gas (number of moles)of moles)

Increasing amount Increasing amount will increase P will increase P (and vice versa)(and vice versa)

Ex: bicycle tires, Ex: bicycle tires, car tirescar tires

TemperatureTemperature Increasing temp. Increasing temp. will increase P will increase P (and vice versa)(and vice versa)

Ex: Tires deflate Ex: Tires deflate in winterin winter

VolumeVolume Decreasing Decreasing volume will volume will increase P, increase P, increasing volume increasing volume decreases Pdecreases P

Ex: press down Ex: press down on a balloon and on a balloon and it popsit pops

• Pressure and volume have an inverse relationship, if temperature remains constant.

• If volume is increased, pressure is decreased by the same factor.

Mathematically, the product of PV is constant or PV = k (where k

is some constant).

Boyle’ Law

P1 V1 = P2 V2 = P3 V3…

• Volume and temperature have a direct relationship, if pressure is held constant.

• If temperature (K) is increased, volume is increased by the same factor.

Mathematically, the relationship of volume divided by Kelvin

temperature is constant or V/T = k.

Charles’ Law

V1 /T1 = V2 /T2 = V3 /T3 …

• Pressure and temperature have a direct relationship, if volume remains constant.

• If temperature (K) is increased, pressure will be increased by the same factor.

Mathematically, the relationship of volume divided by Kelvin

temperature is constant or P/T = k.

Pressure

Gay-Lussac’s Law

P1 /T1 = P2 /T2 = P3 /T3 …

Combined Gas Law Equation

P1 V1 = P2 V2

T1 T2

Combined Gas Law Equation• Steps: • Determine which variable (if any) is kept constant.• Cancel those terms and remove them from the

equation (Ex: If the question says that temperature remains constant the new equation becomes P1V1 = P2V2).

• Plug in values that are given.• Solve for the unknown. • Be sure to always use temperature in Kelvins.

Ideal Gases vs. Real Gases

• “Ideal gases” behave as predicted by Kinetic Molecular Theory.

• Examples: H2 and He

• Gases are most ideal at high temperature and low pressure (also have low mass and low polarity).

• “Real gases” deviate from ideal behavior. • Why? • At low temps, gas particles become

attracted to each other (KMT says they are not).

• Under high pressure, gases occupy a specific volume (KMT says they don’t).

Avogadro’s Law

• Avogadro’s number: 6.02 x 1023

• Simply refers to the quantity of particles found in a

mole.

• At STP, 6.02 x 1023 particles of a gas occupies 22.4 L.

• At STP, 3.01 x 1023 particles of a gas occupies 11.2 L.

kn

VV

n

A. Avogadro’s Principle• Equal volumes of gases contain

equal numbers of moles– at constant temp & pressure– true for any gas

Avogadro’s Law

V1 = V2

n1 n2

Practice with Avogadro’s Law

• Suppose 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

Suppose 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and

pressure)?

Remember your formula!V1 = V2

n1 n2

Plug in your values!

(5.00L) = x

(0.965 mol) (1.80 mol)9 = 0.965xX = 9.33 L

• Avogadro also hypothesized that equal volumes of different gases at the same temperature and pressure contain equal number of particles (or equal moles).

B. Ideal Gas Law

UNIVERSAL GAS CONSTANT

R=0.0821 Latm/molKR =8.315 dm3kPa/molK

PV=nRT

B. Ideal Gas Law• Calculate the pressure in atmospheres of 0.412

mol of He at 16°C & occupying 3.25 L.

IDEAL GAS LAW :

PV = nRT

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0821Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

B. Ideal Gas Law• Calculate the pressure in atmospheres of 0.412

mol of He at 16°C & occupying 3.25 L.

IDEAL GAS LAW

B. Ideal Gas Law• Find the volume of 85 g of O2 at 25°C and

104.5 kPa. IDEAL GAS LAW

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/molK

B. Ideal Gas Law• Find the volume of 85 g of O2 at 25°C and

104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol = 2.7 mol

32.00 g

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

IDEAL GAS LAW

A. Dalton’s Law• The total pressure of a mixture of

gases equals the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + ...

Patm = PH2 + PH2O

Dalton’s Law of Partial Pressure

• The total pressure of a mixture of gases equals the sum of the pressures that each gas would exert if it were alone.


• Ptotal = P1 + P2 + P3 + . .

Where P1 and P2 are the partial pressure of gas 1and gas 2 in the mixture. Since each gas behaves independently, the ideal gas law can be used to calculate the pressure of that gas if we know the number of moles of the gas, the total volume of the container, and the temperature of the gas.


• When we apply the ideal gas law to mixtures of gases each component gas will have its own P and n, but all of the component gases will have the same T and V.

Example of Dalton’s Law of Partial Pressure

• What is the total pressure exerted by a mixture of 2.00 g of H2 (MW = 2.016 g/mol), 8.00 g of N2 (MW = 28.01 g/mol) and 12.0 g of Ar (AW = 39.95 g/mol) at 273 K in a 10.0 L vessel?

Dalton’s Law of Partial Pressure Ex 1

• First we need to calculate the number of moles of each gas:

• 2.00 g H2 x [1 mol/2.016 g] = 0.992 mol

• 8.00 g N2 x [1 mol/28.01 g] = 0.286 mol

• 12.0 g Ar x [1 mol/39.95 g] = 0.300 mol


• Now that we have found the moles of each gas, how can we find the pressure of each gas GIVEN the number of moles, the volume, the temperature, and the constant R?

We can use…

• The Ideal Gas Law!

•PV = nRT


• Now we can calculate the partial pressure of each gas:

• P(H2) = nRT/V = [(0.992)(0.0821)(273)]/(10.0) = 2.22 atm

• P(N2) = nRT/V = [(0.286)(0.0821)(273)]/(10.0) = 0.641 atm

• P(Ar) = nRT/V = [(0.300)(0.0821)(273)]/(10.0) = 0.672 atm


• We can now use the law of partial pressures to calculate the total pressure:

• Ptotal = P(H2) + P(N2) + P(Ar) = 2.22 + 0.641 + 0.672 = 3.53 atm

Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4

kPa.

GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

A. Dalton’s Law• Hydrogen gas is collected over water at

22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Look up water-vapor pressure

The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.

Vapor Pressure

• In a sealed container, vapor pressure can be measured above a liquid.

• Evaporation occurs when some particles from the surface of a liquid escape causing pressure to build up above the liquid (not to be confused with boiling).

Factors that Increase the Rate of Evaporation

• Heating a liquid (not to boiling point)

• Increasing surface area • Create air currents (blow across the surface)

Liquid-Vapor Equilibrium

• Some of the gas particles condense and then we find both evaporating and condensing occurs at the same rate.

• Rate of Evaporation = Rate of Condensation

Related to Boiling

• Boiling occurs when the vapor pressure becomes equal to the external pressure.

• At normal atmospheric pressure, we call this normal boiling point.

Boiling and Attractive (Intermolecular Forces)

• Boiling occurs when heat energy overcomes attractive forces between molecules.

• The stronger the intermolecular forces, the higher the boiling point.

• The weaker the intermolecular forces, the lower the boiling point.

Table H

Notice, increasing temperature increases vapor pressure.

Line drawn at 101.3 kPa corresponds to normal boiling point.

the behavior of gases

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