# Graham’s law of effusion and diffusion

## Graham’s law of diffusion and effusion formula

Scottish physical chemist Thomas Graham in 1948 studies the diffusion and effusion of gas molecules and formulated Graham’s Law.

1. 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.
2. If passing out of gas through the pinhole in the wall of the container is called an effusion. Thus the rates of diffusion and effusion of the gas passing out used for calculation density, pressure, and temperature of the gases.

### Thomas Graham law of diffusion of gases

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 according to Thomas Graham

r = k/√d
where k is a gas constant.

#### Rate of diffusion of different gases

Let at constant T and P, r1 and r2 are the rates of diffusion or effusion of two gases having densities d1 and d2.

Thus Graham’s law states as,

$r_{1}=&space;\sqrt{\frac{1}{d_{1}}}&space;\,&space;and\,&space;r_{2}&space;=\sqrt{\frac{1}{d_{2}}}$

$\therefore&space;\frac{r_{1}}{r_{2}}=&space;\sqrt{\frac{d_{2}}{d_{1}}}$

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

Density(d) ∝ vapour density (D)

$\therefore&space;\frac{r_{1}}{r_{2}}&space;=&space;\sqrt{\frac{D_{2}}{D_{1}}}$

### Molecular weight diffusion relationship

Let r1 and r2 be the rates of diffusion or effusion of two gases having molecular weight M1 and M2 at constant T and P.

But molecular weight = 2 × Vapour density
∴ M = 2D

$\therefore&space;\frac{r_{1}}{r_{2}}=&space;\sqrt{\frac{D_{2}}{D_{1}}}=&space;\sqrt{\frac{M_{2}}{M_{1}}}$

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 molecular weight of one gas is 64 gm mol-1, how can we calculate the molecular weight of another gas?

Solution

From the above problem,
r1 = 432 ml/36 min = 12 ml min-1,
r2 = 288 ml/48 min = 6 ml min-1,
M2 = 64 gm mol-1

Thus form Thomas Graham law

$\frac{r_{1}}{r_{2}}&space;=&space;\frac{12}{6}&space;=&space;\sqrt{\frac{64}{M_{1}}}$

∴ M1 = 64/4 = 16 gm mol-1

### Relationship between volume and rate of diffusion

Let at constant temperature and pressure V1 and V2 be the volume of the two gases passing through the same hole with the time t.

$\frac{V_{1}}{t}=&space;r_{1}&space;\,&space;and&space;\,&space;\frac{V_{2}}{t}=&space;r_{2}$

$\therefore&space;\frac{r_{1}}{r_{2}}=&space;\frac{V_{2}}{V_{1}}=&space;\sqrt{\frac{M_{2}}{M_{1}}}$

#### Which gas diffuses faster ammonia or hydrogen?

The molecular weight of ammonia = 17 gm mol-1  and hydrogen chloride = 36.5 gm mol-1.

Thus from the Grahams law

$\frac{r_{NH_{3}}}{r_{HCl}}=&space;\sqrt{\frac{M_{HCl}}{M_{NH_{3}}}}$

$\therefore&space;r_{NH_{3}}=&space;r_{HCl}\times&space;\sqrt{\frac{M_{HCl}}{M_{NH_{3}}}}=&space;r_{HCl}\times&space;\sqrt{\frac{36.5}{17}}$

∴ rNH3=1.46 × rHCl

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.

$\therefore&space;\frac{r_{1}}{r_{2}}=&space;\sqrt{\frac{u_{1}^{2}}{u_{2}^{2}}}$

where u1 and u2 are the RMS velocity of the gas.

From the Kinetic theory gases

$PV&space;=&space;\frac{1}{3}mNu^{2}$

$or,u^{2}&space;=&space;\frac{3PV}{mN}$

Again from the ideal gas law
PV = RT and m N = M = molar mass.

$\therefore&space;\frac{r_{1}}{r_{2}}=&space;\sqrt{\frac{u_{1}^{2}}{u_{2}^{2}}}=&space;\sqrt{\frac{\frac{3RT}{M_{1}}}{\frac{3RT}{M_{2}}}}&space;=&space;\sqrt{\frac{M_{2}}{M_{1}}}$

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

### Application of graham’s law in real life

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

1. 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.
2. Argon has been concentrated with nitrogen in this way. The isotopes of neon, chlorine, bromine, etc have been partially separated by this method.
3. Graham’s law also used for detecting marsh gas main components of methane from the mines.