# Graham's Law Example: Gas Diffusion-Effusion

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 that 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.

### Graham's Law 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:

rX(MMX)1/2 = rY(MMY)1/2

where
rX = rate of effusion/diffusion of Gas X
MMX = molar mass of Gas X
rY = rate of effusion/diffusion of Gas Y
MMY = 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 rY/rX.

rY/rX = (MMX)1/2/(MMY)1/2

rY/rX = (MMX)/(MMY)1/2

Use the given values for molar masses and plug them into the equation:

rY/rX = (72 g/mol)/(2)1/2
rY/rX = 361/2
rY/rX = 6

Note that the answer is a pure number. In other words, the units cancel out. What you get is how many times faster or slower gas Y effuses compared to gas X.