### What is Graham's Law of effusion and diffusion?

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

Diffusion is the movement of gas molecules from high concentration to low concentration. Diffusion is occurred on the gases because of the random movement of the gas molecules.
The phenomenon of diffusion may be described as the tendency for any substance to spread uniformly throughout the space available to it.

The passing out of gas through the pinhole in the wall of the container is called effusion.
The rate of diffusion and effusion of the gas passing out depends on the density, pressure, and temperature of the gases.

At constant temperature and pressure, the rates of diffusion or effusion of different gases vary inversely as the square root of their densities.
 Graham's Law in chemistry

#### Mathematical equation

At constant temperature and pressure, the rate of diffusion or effusion of the gas molecules = r and density of the gas = d. According to Graham
r ∝ 1/√d
or, r = k/√d where k is a constant of gas.

#### Rates of diffusion and densities

At constant temperature and pressure, r₁ and r₂ are the rates of diffusion or effusion of two gases having densities d₁ and d₂. Graham's law states as,
r₁ ∝ 1/√d₁ and r₂ ∝ 1/√d₂

∴ r₁/r₂ = √d₂/√d₁
The density of the gas molecules proportional to the vapor density of the gas molecules.
Density(d) ∝ Vapour density (D)

r₁/r₂ = √D₂/√D₁

#### Molecular weight determination

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

Molecular weight = 2 × Vapour density
or M = 2D

∴ r₁/r₂ = √D₂/√D₁ = √M₂/√M₁
Problem

At constant temperature and pressure 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, what is the molecular weight of another gas?

Solution
Graham's law for gas molecule,
r₁/r₂ = √M₂/√M₁.

From the above problem, r₁ = 432 ml/36 min = 12 ml min⁻¹,  r₂ = 288 ml/48 min = 6 ml min⁻¹ and M₂ = 64 gm mol⁻¹.
∴ 12/6 = √64/√M₁
or, M₁ = 64/4 = 16 gm mol⁻¹

Problem

At a certain temperature, the time required for the complete diffusion of 200 ml of hydrogen gas is 30 min. How many times required for the complete diffusion of 50 ml of oxygen gas at the same temperature?

Solution

Let time required for the complete diffusion of 50 ml of oxygen gas is t min.

Graham's law for gas
r₁/r₂ = √M₂/√M₁.

Rate of diffusion for hydrogen (r₁) = 200 ml/30 min
= 20/3 ml min⁻¹

Rate of diffusion for oxygen (r₂) = 50 ml/t min
= 50/t ml min⁻¹

The molecular weight of hydrogen (M₁) = 2 gm mol⁻¹

The molecular weight of oxygen (M₂) = 32 gm mol⁻¹

∴ (200/30)/(50/t) = √(32/2)
or, t₂ = 30 min
Problem

Equal moles of hydrogen and oxygen is placed in a container with a pinhole through which escape. What fraction of oxygen escape in the time required for one half of the hydrogen to escape?

Solution

Let the time required for this escape is t, and the fraction of oxygen and hydrogen escape is nO₂ and nH₂ respectively.
Graham's law for gas molecules
rO₂/rH₂ = √MH₂√MO₂
or, (nO₂/t)/(nH₂/t) = √2/√32
or, (nO₂/t)/(0.5/t) = 1/4
or, nO₂ = 1/8.

#### Volume measurement of gas molecules

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.

V₁/t = r₁ and V₂/t = r₂

∴ r₁/r₂ = V₂/V₁ = √M₂/√M₁
Problem

Which of the two gases ammonia and hydrogen chloride diffuse faster and by which factor?

Solution
Graham's law for ammonia and hydrogen
rNH₃/rHCl = √MHCl/√MNH₃

The molecular weight of ammonia (NH₃)
= 17 gm mol⁻¹
The molecular weight of hydrogen chloride (M₂)
= 36.5 gm mol⁻¹

∴ rNH₃ = rHCl × (√MHCl/√MNH₃)
= rHCl × (√36.5/√17)
= 1.46 rHCl

Ammonia will diffuse 1.46 times faster than hydrogen chloride gas.

#### Diffusion and effusion of gases from kinetic theory

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

∴ r₁/r₂ = √u₁²/√u₂²
where u₁ and u₂ are the RMS value of the gas molecules.

Kinetic gas equation
PV = ⅓ m N u²
or, u² = 3PV/m N.

Ideal gas law for 1-mole gas
PV = RT and m N = M = molar mass of the gas.

∴ u² = 3RT/M

r₁/r₂ = √u₁²/√u₂² = √(3RT/M₁)/√(3RT/M₂)

or,  r₁/r₂ = √M₂/√M₁
which is Graham's law of diffusion.
Problem

Landenberg found that a sample of ozonized oxygen containing 86.16% of ozone by weight required 430 seconds to diffuse under conditions where pure oxygen required 367.5 seconds. What is the vapor density of ozone?

Solution

Let V is the volume diffusing out in each case. Let dm, do and d be the densities of the mixture, pure oxygen, and ozone respectively.

Rate of diffusion of mixture (rm) = V ml/430 seconds
= V/430 ml seconds⁻¹.

Rate of diffusion of oxygen (ro) = V ml/367.5 seconds
= 50/t ml seconds⁻¹.

Graham's law for gas molecule
rm/ro = √do/√dm = √16/√dm
or, dm = (430/367)² × 16
≈ 21.91

Again for 100 gms, the volume = 100/dm
= (86.16/d) + (13.84/d0)

∴ (100/21.91) = (86.16/d) + (13.84/16)
or, d = 23.3

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

1. The application of Graham's law is in the partial separation of the components in a gas mixture. If the mixture is 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, the concentration of one component is considerable increases compared with that of the other. This is called the 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 in mines.