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Graham's Law Example

Gas Diffusion-Effusion Example Problem

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This diagram shows effusion (left) and diffusion (right).

This diagram shows effusion (left) and diffusion (right). Effusion occurs through a hole smaller than the mean path of moving particles while diffusion occurs through a hole large enough allow multiple particles to pass at once.

Astrang13, Creative Common License
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

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.

rY/rX = [(72 g/mol)/(2)]1/2
rY/rX = [36]1/2
rY/rX = 6

Answer:

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

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