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.