Graham's law expresses the relationship between the rate of effusion or diffusion of a gas and that gas's molar mass. Diffusion describes the spreading of a gas throughout a volume or second gas and effusion describes the movement of a gas through a tiny hole into an open chamber.

In 1829, Scottish chemist Thomas Graham determined through experimentation that a gas's rate of effusion is inversely proportional to the square root of the gas particle's density. In 1848, he showed that the rate of effusion of a gas is also inversely proportional to the square root of its molar mass. Graham's law also shows that the kinetic energies of gases are equal at the same temperature.

## Graham's Law Formula

Graham's law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. See this law in equation form below.

**r ∝ 1/(M) ^{½}**

or

**r(M) ^{½ }= constant**

In these equations, *r* = rate of diffusion or effusion and *M* = molar mass.

Generally, this law is used to compare the difference in diffusion and effusion rates between gases, often denoted as Gas A and Gas B. It assumes that temperature and pressure are constant and equivalent between the two gases. When Graham's law is used for such a comparison, the formula is written as follows:

**r _{Gas A}/r_{Gas B} = (M_{Gas B})^{½}/(M_{Gas A})^{½}**

## Example Problems

One application of Graham's law is to determine how quickly a gas will effuse in relation to another and quantify the difference in rate. For example, if you want to compare the effusion rates of hydrogen (H_{2}) and oxygen gas (O_{2}), you can use their molar masses (hydrogen = 2 and oxygen = 32) and relate them inversely.

Equation for comparing effusion rates: **rate H _{2}/rate O_{2} = 32^{1/2} / 2^{1/2} = 16^{1/2} / 1^{1/2} = 4/1 **

This equation shows that hydrogen molecules effuse four times faster than oxygen molecules.

Another type of Graham's law problem may ask you to find the molecular weight of a gas if you know its identity and the effusion ratio between two different gases.

Equation** **for finding molecular weight: **M _{2} = M_{1}Rate_{1}^{2} / Rate_{2}^{2}**

### Uranium Enrichment

Another practical application of Graham's law is uranium enrichment. Natural uranium consists of a mixture of isotopes with slightly different masses. In gaseous effusion, uranium ore is first made into uranium hexafluoride gas, then repeatedly effused through a porous substance. Through each effusion, the material passing through the pores becomes more concentrated in U-235 (the isotope used to generate nuclear energy) because this isotope diffuses at a faster rate than the heavier U-238.