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{\mbox{Rate}_1 \over \mbox{Rate}_2}=\sqrt{M_2 \over M_1}

-where:
  • ''Rate1'' is the rate of effusion of the first gas.

  • ''Rate2'' is the rate of effusion for the second gas.

  • ''M1'' is the Molar Mass of gas 1

  • ''M2'' is the molar mass of gas 2.


Graham's law works for both effusion and Diffusion .


EXAMPLE

Let gas 1 be H2 and gas 2 be O2.

{\mbox{Rate H}_2 \over \mbox{Rate O}_2}={\sqrt{32} \over \sqrt{2}}={\sqrt{16} \over \sqrt{1}}= rac{4}{ 1}

Therefore, the particles of hydrogen gas diffuse four times as fast as those of oxygen.

Graham's Law can also be used to find the approximate molecular weight of a gas if one gas is a known species, and if there is a specific ratio between the rates of two gases (such as in the previous example). The equation can be solved for either one of the molecular weights provided the subscripts are consistent.

{M_2}={M_1 \mbox{Rate}_1^2 \over \mbox{Rate}_2^2}


SEE ALSO