Graham's law, known as Graham's law of effusion, was formulated by Scottish physical chemist Thomas Graham in 1846.

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

Graham's gas law is expressed as:
$\mathrm{rate\; of\; effusion}\propto \frac{1}{\sqrt{\mathrm{molar\; mass}}}\mathrm{at\; constant\; P\; and\; T}$

The ratio of effusion of two gases relates to the ratio of their densities (molar masses) and expressed as:
$\frac{{\mathrm{Rate}}_1}{{\mathrm{Rate}}_2}=\sqrt{\frac{M_2}{M_1}}$
where:
Rate₁ is the rate of effusion of the first gas (volume or number of moles per unit time).
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.

Example 1:
Oxygen has an effusion tate of 10 milliliters per second and weight of 32 amu. What is the molecular weight of a gas which diffuses twice as fast as oxygen?
$\frac{{\mathrm{Rate}}_1}{{\mathrm{Rate}}_2}=\sqrt{\frac{M_2}{M_1}}$
$\frac{{\mathrm{Rate}}_2}{{\mathrm{Rate}}_1}=\sqrt{\frac{M_1}{M_2}}\mathrm{or}{\left(\frac{{\mathrm{Rate}}_2}{{\mathrm{Rate}}_1}\right)}^2=\frac{Mx}{MO2}$
${\left(\frac{10\mathrm{mL\; per\; second}}{20\mathrm{mL\; per\; second}}\right)}^2=\frac{Mx}{32\mathrm{amu}}$
$M_x={\left(\frac{10}{20}\right)}^{2}x32=8\mathrm{amu}$