## Diffusion and effusion of gases

Scottish physical chemist Thomas Graham in 1948 studies the rate of diffusion and effusion of gases and formulated **Graham’s Law**.

**Diffusion**is the movement of gas molecules from high concentration to low concentration and occurred because of the random movement of the gas molecules. Thus the phenomenon of diffusion is the tendency for any substance to spread uniformly throughout the space available to it.- If passing out of gas through the pinhole in the wall of the container is called an
**effusion**. Thus the rate of diffusion and effusion of the gases passing out used for calculation density, pressure, and temperature of the gases.

### Graham’s law of diffusion formula

At constant temperature and pressure, the rates of diffusion or effusion of different gases vary inversely as the square root of their densities.

Let at constant temperature and pressure, the rate of diffusion or effusion of the gas molecules = r and density = d.

Thus the formula for Graham’s law of diffusion

r = k/√d

where k is a gas constant.

### Rate of diffusion of gases

Let at constant T and P, r_{1} and r_{2} are the rates of diffusion or effusion of two gases having densities d_{1} and d_{2}.

Thus Graham’s law states as,

But the density of the gases proportional to the vapor density.

Density(d) ∝ vapour density (D)

### Graham’s law to find molar mass

Let r_{1} and r_{2} be the rates of diffusion or effusion of two gases having molar mass M_{1} and M_{2} at constant T and P.

But molecular mass = 2 × Vapour density

∴ M = 2D

Problem

At constant T and P, 432 ml and 288 ml be the volume of the two gases passing through the same hole with the time 36 min and 48 min respectively. If the molar mass of one gas is 64 gm mol^{-1}, how to find the molar mass of another gas by Graham’s law?

Solution

From the above problem,

r_{1} = 432 ml/36 min = 12 ml min^{-1},

r_{2} = 288 ml/48 min = 6 ml min^{-1},

M_{2} = 64 gm mol^{-1}

Thus form **Thomas Graham** law

∴ M1 = 64/4 = 16 gm mol^{-1}

### Volume and rate of diffusion

Let at constant temperature and pressure V_{1} and V_{2} be the volume of the two gases passing through the same hole with the time t.

V_{1} = r_{1} × t_{1} and V_{2} = r_{2} × t_{2}

#### Diffusion of ammonia and hydrogen chloride

The molar mass of ammonia = 17 gm mol^{-1} and hydrogen chloride = 36.5 gm mol^{-1}.

Thus from the Grahams law

∴ r_{NH3}=1.46 × r_{HCl}

Hence ammonia will diffuse 1.46 times faster than hydrogen chloride gas.

### Graham law from kinetic gas equation

The rate of diffusion or effusion can be assumed to be directly proportional to the root mean square speed or any other average speed.

where u_{1} and u_{2} are the RMS velocity of the gas.

From the Kinetic theory gases

Again from the ideal gas law

PV = RT and m N = M = molar mass.

This is Thomas Graham’s law of diffusion and effusion, derived from the kinetic theory of gases.

### Application of effusion and diffusion of gases

The application of Graham’s law mainly uses for the partial separation of the components in a gas mixture.

- If the mixture led out a tube made of porous walls, in a given time the lighter components will diffuse out more than the heavier ones. By repeating the process with each sperate fraction from diffusion.

Thus the concentration of one component considerably increases compared with that of the other. This process is called atmolysis. - Argon has been concentrated with nitrogen in this way. The isotopes of neon, chlorine, bromine, etc have been partially separated by this method.
- Graham’s law also used for detecting marsh gas main components of methane from the mines.