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Mechanical Properties Of Fluids

Question
CBSEENPH11020190

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 (h– h1)/ kBT]
Where n2n1 refer to number density at heights h2 and h1respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n2 = n1 exp [-mg NA(ρ - P′) (h2 –h1)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [NA is Avogadro’s number, and 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 (h– h1) / kBT]                    … (i) 

where,
nis the number density at height h1, and 
n2 is the number density at height h

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 = 

Volume of a suspended particle = 

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'

= mg - V ρ' g 
=  mg - (m/ρ)ρ'

mg(1 - (ρ'/ρ) )                                         ...(ii) 

Gas constant, R = kB

kB = R / N                                                  ...(iii) 

Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get: 
n2 = n1 exp [-mg (h– h1) / kBT
    = n1 exp [-mg (1 - (ρ'/ρ) )(h– h1)(N/RT) ] 
    = n1 exp [-mg (ρ - ρ')(h– h1)(N/RTρ) ]