Gases have the tendency to spontaneously intermix and form a homogenous mixture without the help of external agency. This is due to the presence of large amount of empty space between the gas molecules that makes their movement rapid into each other. The gases move from a region of higher concentration to a region of lower concentration until the mixture attains uniform concentration.
Graham studied the rate of diffusion of various gases and gave this law. It states that under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square roots of their densities.
Rate of diffusion a
If r_{1} and r_{2} are the rates of diffusion of two gases 'A' and 'B' and r_{1} and r_{2} are their densities, then
r1 a and r2 a
(same T and P)
Molecular mass is twice the vapour density, substituting this in the above equation, we have
where M_{1} and M_{2} are the molecular masses of the two gases. Thus, the rate of diffusion of gases are inversely proportional to the square root of their molecular masses.
Rate of diffusion is also equal to the volume of the gas, which diffused per unit time,
r = V/t
If V_{1} and V_{2} are the volumes of the gases diffusing in time t_{1} and t_{2} respectively, then
r_{1} = (V_{1} / t_{1}) and r_{2} = (V_{2} / t_{2})
Therefore,
If the volume diffused is the same, (V_{1} = V_{2})
Then,
Graham's law is useful in:
Separation of gases having different densities by diffusion.
Determining the densities and molecular masses of unknown gases by comparing their rates of diffusion with known gases.
Separating the isotopes of some of the elements.
Effusion
When the gases contained in a vessel are allowed to escape through a small aperature, it is effusion.
Fig:  Process of diffusion and effusion
Problem
4. An unknown gas diffuses four times as quickly as oxygen. Calculate the molecular mass of the gas.
Solution
Let the rate of diffusion of oxygen be r(O_{2}) = r_{1}
The rate of diffusion of the unknown gas r (x) = 4r
Molecular mass of O, M(O) = 32
Molecular mass of unknown gas M (x) = M
From Graham's Law of diffusion,
or
Squaring on both sides,
(1/16) = (M(x) / 32)
M(x) = (32/16) = 2

