Graham's Law Example

Graham's law is a gas law which relates the rate of diffusion or effusion of a gas to its molar mass. Diffusion is the process of slowly mixing two gases together. Effusion is the process that occurs when a gas is permitted to escape its container through a small opening.

Graham's law states the rate at which a gas will effuse or diffuse is inversely proportional to the square root of the molar masses of the gas. This means light gasses effuse/diffuse quickly and heavier gases effuse/diffuse slowly.

This example problem uses Graham's law to find how much faster one gas effuses than another.

Problem:

Gas X has a molar mass of 72 g/mol and Gas Y has a molar mass of 2 g/mol. How much faster or slower does Gas Y effuse from a small opening than Gas X at the same temperature?

Solution:

Graham's Law can be expressed as

r_X(MM_X)^1/2 = r_Y(MM_Y)^1/2

where
r_X = rate of effusion/diffusion of Gas X
MM_X = molar mass of Gas X
r_Y = rate of effusion/diffusion of Gas Y
MM_Y = molar mass of Gas Y

We want to know how much faster or slower Gas Y effuses compared to Gas X. To get this value, we need the ratio of the rates of Gas Y to Gas X. Solve the equation for r_Y/r_X.

r_Y/r_X = (MM_X)^1/2/(MM_Y)^1/2

r_Y/r_X = [(MM_X)/(MM_Y)]^1/2

Use the given values for molar masses.

r_Y/r_X = [(72 g/mol)/(2)]^1/2
r_Y/r_X = [36]^1/2
r_Y/r_X = 6

Answer:

Gas Y will effuse six times faster than the heavier Gas X.