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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

vAmB

vA2mB

=

=

=

vBmA

vB2 mA

vB2 mA

Introduction½mAvA2 = 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.

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.”

=

rateBMA

ApplicationThis is how we apply Graham’s law.

We compare the rates of effusion of different gases.

(rateHe)2 MA

rateHeMA

(MHe)(rateHe)2

MA =

MA =

=

=

(rateA)2MHe

rateAMHe

(rateA)2

(1.00)2

Example 2A 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

➙

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