1. Rate of Diffusion and Effusion
Graham’s Law
Rate of Diffusion and Effusion

2. 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.

3. 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.

4. 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.”

5. 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 = ½mBvB2

6. Introduction ½mAvA2 = ½mBvB2 mAvA2 = mBvB2 vA2 mB = vB2 mA vA2 mB =
The ½’s cancel out.
mAvA2 = mBvB2
Get all speed and mass terms together.
vA mB
vB mA =
Take the square root of both sides.
vA mB
vB mA =
Simplify.
vA mB
vB mA =
The speed of an individual gas molecule is inversely proportional to its mass.

7. Introduction vA mB vB mA = rateA mB rateB mA = rateA MB rateB MA =➙ ➙
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.”

8. Application This is how we apply Graham’s law.
rateA MB
rateB MA =
This is how we apply Graham’s law.
We compare the rates of effusion of different gases.

9. Example 1 rateH2 MO2 rateO2 MH2 = 32.00 g/mol 2.00 g/mol = = 16.00
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 = g/mol
rateH MO2
rateO MH2 =
g/mol
2.00 g/mol = =
= 4.00
Hydrogen gas effuses at a rate 4 times faster than oxygen.

10. Example 2 rateHe MA rateA MHe = (rateHe)2 MA (rateA)2 MHe =
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
rateHe MA
rateA MHe =
(rateHe)2 MA
(rateA) MHe =
(MHe)(rateHe )2
(rateA)2
MA = ➙ ➙
(4.00 g/mol)(6.04 )2
(1.00)2
MA = ➙
= 146 g/mol

11. 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 =