# Graham ’ s Law - PowerPoint PPT Presentation

1 / 11

Graham ’ s Law. Rate 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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Graham ’ s Law

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

### Graham’s Law

• Rate 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 kineticenergy of a gas molecule.

• At a given temperature, the average kineticenergy 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 = ½mBvB2

vA2 mB

vAmB

vA2mB

=

=

=

vBmA

vB2 mA

vB2 mA

### Introduction

½mAvA2 = ½mBvB2

The ½’s cancel out.

mAvA2 = mBvB2

Get all speed and mass terms together.

Take the square root of both sides.

Simplify.

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

vAmB

rateAMB

rateAmB

=

=

=

vB mA

rateBmA

rateBMA

### Introduction

If we extend this to all of the gas,

• the speed becomes the rate

• the mass becomes the molar mass

• Which leads us back to Graham’s Law:

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

rateAMB

=

rateBMA

### Application

This is how we apply Graham’s law.

We compare the rates of effusion of different gases.

rateH2MO2

32.00 g/mol

= 16.00

=

=

rateO2MH2

2.00 g/mol

### 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

MO2 = 32.00 g/mol

= 4.00

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

(4.00 g/mol)(6.04)2

(rateHe)2 MA

rateHeMA

(MHe)(rateHe)2

MA =

MA =

=

=

(rateA)2MHe

rateAMHe

(rateA)2

(1.00)2

### 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 = 6.04

MA = A g/mol

rateA = 1.00

= 146 g/mol

rateAMB

=

rateBMA

### 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.