Kinetic Theory NCERT Solutions Class 11 Physics - Solved Exercise Question 13.13

Question 13.13:

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n₂ = n₁ exp [-mg (h₂ – h₁)/ k_BT]
Where n₂, n₁ refer to number density at heights h₂ and h₁ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n₂ = n₁ exp [-mg N_A(ρ - P′) (h₂ –h₁)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [N_A is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Solution:

According to the law of atmospheres, we have:
n₂ = n₁ exp [-mg (h₂ – h₁) / kBT] … (i)
Where,
n₁ is thenumber density at height h₁, and n₂ is the number density at height h₂
mg is the weight of the particle suspended in the gas column
Density of the medium = ρ'
Density of the suspended particle = ρ
Mass of one suspended particle = m'
Mass of the medium displaced = m
Volume of a suspended particle = V
According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:
Weight of the medium displaced – Weight of the suspended particle
= mg – m'g
= mg - V ρ' g  =  mg - (m/ρ)ρ'g
= mg(1 - (ρ'/ρ) )   ....(ii)
Gas constant, R = k_BN
k_B = R / N     ....(iii)
Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:
n₂ = n₁ exp [-mg (h₂ – h₁) / k_BT]
= n₁ exp [-mg (1 - (ρ'/ρ) )(h₂ – h₁)(N/RT) ]
= n₁ exp [-mg (ρ - ρ')(h₂ – h₁)(N/RTρ) ]