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.
If r1 and r2 are the rates of diffusion of two gases 'A' and 'B' and r1 and r2 are their densities, then
Molecular mass is twice the vapour density, substituting this in the above equation, we have
where M1 and M2 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,
If V1 and V2 are the volumes of the gases diffusing in time t1 and t2 respectively, then
Therefore,
If the volume diffused is the same, (V1 = V2)
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: 2.6 - 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(O2) = r1
The rate of diffusion of the unknown gas r (x) = 4rMolecular mass of O, M(O) = 32
Molecular mass of unknown gas M (x) = MFrom Graham's Law of diffusion,
or
