Question

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 the 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.]

Answer 1

According to the law of atmospheres, we have:

n₂ = n₁ exp [-mg (h₂ – h₁)/ kBT] … (i)

Where,

n₁ is the number 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 rho'g = mg - (m/rho)rho'g`

`= mg (1-(rho')/rho)` ...(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]

`= n_1 exp[(-mg(h_2 - h_1))/ (k_BT)]`

`= n_1 exp [-mg(rho - rho')(h_2 - h_1) N/(RTrho)]`

Answer 2

Considering the particles and molecules to be spherical, the weight of the particle is

`W = mg = 4/3  pir^3rhog`   ... (i)

Where r = radius of the particle and rho = density of the particle. Its motion under gravity causes buoyant force to act upward which is equal to 

B= Volume of particle x density of the surrounding medium x g

`= 4/3 pir^3rho'g`

if F be the downward force acting on the particle then

`F = W - B = 4/3 pir^3(rho -rho')g` .....(iii)

Also `n_2 = n_1 exp[(-mg)/k_BT(h_2-h_1)]`  .... (iv)

Where `k_B` -  boltzman constant

`n_1` and `n_2` are number densities at height `h_1` and `h_2` respectvely. Here mg can be replaced by effective force F given by the equation (iii)

:. From iii and iv we get

`n_2 = n_1 exp [-(4pi)/3 r^3 ((rho -rho'))/(k_BT) g(h_2 - h_1)]`

`= n_1 exp [-(4pi)/3 r^3 (rhog(1-(rho')/rho)(h_2-h_1))/((RT)/N_A)]`       [∵ `K_B =  R/N_A`]

`n_2= n_1 exp [- (mg N_A(1-(rho')/rho)(h_2-h_1))/(RT)]`

which is required relation where `4/3 pir^3 rhog` = mass of the particle x g = mg