Graham’s LawRate of Diffusion and Effusion

Introduction

When we first open a container of ammonia, it takes time for the odor to travel from the container to all parts of a room.

This shows the motion of gases through other gases.

In this case, ammonia gas, NH3, moves through air.

This is an example of diffusion and effusion.

Introduction

Diffusion is the tendency of a gas to move toward areas of lower density.

Ammonia moving throughout a room.

Effusion is the escape of a gas from a container from a small hole.

Air escaping from a car tire.

Introduction

In 1831, the Scottish physical chemist, Thomas Graham, first showed the relationship between the mass of a gas molecule and its rate of diffusion or effusion.

This is called Graham’s Law.

“The rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.”

Introduction

The law comes from the relationship between the speed, mass, and kinetic energy of a gas molecule.

At a given temperature, the average kinetic energy of all gas molecules in a mixture is the same value.

If gas A has KEA = ½mAvA2

If gas B has KEB = ½mBvB2

Then KEA = KEB ➙ ½mAvA2 = ½mBvB

2

Introduction

½mAvA2 = ½mBvB

2

mAvA2 = mBvB

2

vA2 mB

vB2 mA

=

vA2 mB

vB2 mA

=

vA mB

vB mA

=

The speed of an individual gas molecule is inversely proportional to its mass.

The ½’s cancel out.

Get all speed and mass terms together.Take the square root of both sides.Simplify.

Introduction

If we extend this to all of the gas,

vA mB

vB mA

=

the speed becomes the rate

rateA mB

rateB mA

=

the mass becomes the molar mass

rateA MB

rateB MA

=

“The rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.”

➙ ➙

Which leads us back to Graham’s Law:

Application

This is how we apply Graham’s law.

We compare the rates of effusion of different gases.

rateA MB

rateB MA

=

Example 1

Compare the rate of effusion of hydrogen gas to the rate of effusion of oxygen gas at a constant temperature.MH2 = 2.00 g/mol

rateH2

MO2rateO2

MH2

=

MO2 = 32.00 g/mol

32.00 g/mol 2.00 g/mol

= = 16.00= 4.00

Hydrogen gas effuses at a rate 4 times faster than oxygen.

Example 2

A sample of helium, He, effuses through a porous container 6.04 times faster than does unknown gas A. What is the molar mass of the unknown gas?MHe = 4.00 g/mol

rateHe

MArateA

MHe

=

MA = A g/mol

rateHe = 6.04

rateA = 1.00

➙ (rateHe)2 MA

(rateA)2

MHe = ➙ (MHe)(rateHe

)2

(rateA)2

MA =

(4.00 g/mol)(6.04 )2 (1.00)2

MA = ➙ = 146 g/mol

Summary

Diffusion is the tendency of a gas to move toward areas of lower density.

Effusion is the escape of a gas from a container from a small hole.

Graham’s Law: the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.rateA

MB

rateB MA

=