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.
Graham's law of diffusion or effusion
    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.

Graham's law of diffusion

    At constant temperature and pressure, the rates of diffusion or effusion of different gases vary inversely as the square root of their densities.
    At constant temperature(T) and pressure(P) if the rate of diffusion or effusion = r and density = d.
r ∝ 1/√d
or, r = k/√d where k is a constant.

Rates of diffusion and densities

    If r₁ and r₂ be the rates of diffusion or effusion of two gases having densities d₁ and d₂ at constant temperature(T) and pressure (P).
∴ r₁ ∝ 1/√d₁ and r₂ ∝ 1/√d₂
Thus r₁/r₂ = √d₂/√d₁
Again density(d) ∝ Vapour density (D)
r₁/r₂ = √D₂/√D₁

Rates of diffusion and molecular weight

    If r₁ and r₂ be the rates of diffusion or effusion of two gases having molecular weight M₁ and M₂ at constant temperature(T) and pressure (P).
We know that Molecular weight(M) = 2 × Vapour density (D)
r₁/r₂ = √D₂/√D₁ = √M₂/√M₁

Rates of effusion and volume

    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.
Then V₁/t = r₁ and V₂/t = r₂
∴ r₁/r₂ = V₂/V₁ = √M₂/√M₁

Graham's law from the 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.
    Thus, r₁/r₂ = √u₁²/√u₂² where u₁ and u₂ are the root mean square speed of the gas molecules.
From the kinetic gas equation
PV = ⅓ m N u²
or, u² = 3PV/m N

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

Thus u² = 3RT/M

∴ r₁/r₂ = √u₁²/√u₂² = √(3RT/M₁)/√(3RT/M₂)
or,  r₁/r₂ = √M₂/√M₁
which is Graham's law of diffusion.

Practical application of Graham's law

    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.
    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 in mines.

Problems solutions

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
According to Graham's law of diffusion,
r₁/r₂ = √M₂/√M₁
    Here, 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.
According to Graham's law of diffusion,
r₁/r₂ = √M₂/√M₁
    Here, rate of diffusion of hydrogen (r₁) = 200 ml/30 min = 20/3 ml min⁻¹,  rate of diffusion of oxygen (r₂) = 50 ml/t min = 50/t ml min⁻¹,  molecular weight of hydrogen (M₁) = 2 gm mol⁻¹ and molecular weight of oxygen (M₂) = 32 gm mol⁻¹
∴ (200/30)/(50/t) = √(32/2)
or, t₂ = 30 min
Problem
    Which of the two gases ammonia and hydrogen chloride diffuse faster and by which factor?
Solution
According to Graham's law of diffusion,
rNH₃/rHCl = √MHCl/√MNH₃
    Here molecular weight of ammonia (NH₃) = 17 gm mol⁻¹ and molecular weight of hydrogen chloride (M₂) = 36.5 gm mol⁻¹
∴ rNH₃ = rHCl × (√MHCl/√MNH₃)
= rHCl × (√36.5/√17)
= 1.46 rHCl
    Thus ammonia will diffuse 1.46 times faseter then hydrogen chloride gas.
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. Determine 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.
    Here, rate of diffusion of mixure (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
According to Graham's law of diffusion,
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/do)
or, (100/21.91) = (86.16/d) + (13.84/16)
or, d = 23.3
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.
Thus according to Graham's law of diffusion,
rO₂/rH₂ = √MH₂√MO₂
or, (nO₂/t)/(nH₂/t) = √2/√32
or, (nO₂/t)/(0.5/t) = 1/4
or, nO₂ = 1/8