## Graham’s Law of Diffusion and Effusion

**Graham’s Law** of diffusion or effusion defines by Scottish physical chemist Thomas Graham in 1948 studies the rate of diffusion and effusion formula of gases and liquid molecules. Therefore, **diffusion** is the movement of gas molecules from high concentration to low concentration and occurred because of the random movement of the gas molecules. 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**. Therefore, 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. Therefore, according to Graham’s law formula of diffusion, r = k/√d, where k is a gas constant.

### Rate of Diffusion of Gas Molecule

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}. Therefore, Graham’s law states as, r_{1} = 1/√d_{1} and r_{2} = 1/√d_{2}. But the density of the gases proportional to the vapor density or density(d) ∝ vapour density (D). Therefore, r_{1} = 1/√d_{1} and r_{2} = 1/√d_{2}.

### 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 or M = 2D. Therefore, r_{1}/r_{2} = √M_{2}/√M_{1}.

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}. Therefore, form the **Thomas Graham** law, molar mass (M_{1}) = 64/4 = 16 gm mol^{-1}.

### Volume and Graham 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 and V_{2} = r_{2} × t. Therefore, the rate of diffusion, r_{1}/r_{2} = V_{2}/V_{1}.

#### Diffusion of Ammonia and Hydrogen Chloride

The molar mass of ammonia = 17 gm mol^{-1} and hydrogen chloride = 36.5 gm mol^{-1}. Hence from Graham’s law of diffusion or effusion, the rate of diffusion of ammonia molecule (r_{NH3}) =1.46 × r_{HCl}. Therefore, ammonia will diffuse 1.46 times faster than hydrogen chloride gas.

### Graham Law from Kinetic Gas Equation

The rate of diffusion or effusion assumed to be directly proportional to the root mean square speed or any other average speed. From the Kinetic theory gases and the ideal gas law, we can derive the formula of Graham’s law of diffusion or effusion.

### 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, oxygen, etc have been partially separated by this method.
- Graham’s law also used for detecting marsh gas main components of methane from the mines.