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.
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½mAvA2 = ½mBvB2
The ½’s cancel out.
mAvA2 = mBvB2
Get all speed and mass terms together.
Take the square root of both sides.
The speed of an individual gas molecule is inversely proportional to its mass.
If we extend this to all of the gas,
This is how we apply Graham’s law.
We compare the rates of effusion of different gases.
2.00 g/molExample 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
Hydrogen gas effuses at a rate 4 times faster than oxygen.
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