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Kinetic Theory

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Question : 13 of 14
Marks: +1, -0
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=n1exp⁡[−mg(h2−h1)kBT]n_{2}=n_{1} \exp \left[ -\frac{m g (h_{2}-h_{1})}{k_{B} T} \right]
where n2,n1n_2, n_1 refer to number density at heights h2h_2 and h1h_1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column : n2=n1exp⁡[−mgNA(ρ−ρ)(h2−h1)ρRT].n_{2}=n_{1} \exp \left[ -\frac{m g N_{A} (\rho - \rho) (h_{2}-h_{1})}{\rho R T} \right] . where r is the density of the suspended particle, and ρ′ that of surrounding medium.
[NAN_A is Avogadro’s number, and R is the universal gas constant.]
Solution:  
According to the law of atmospheres,
n2=n1exp⁡[−mgkBT(h2−h1)]n_{2}=n_{1} \exp \left[ -\frac{m g}{k_{B} T} (h_{2}-h_{1}) \right]... (i)
where n2,n1n_2, n_1 refer to number density of particles at heights h2 and h1h_2 \text{ and } h_1 respectively.
If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of the suspended particles.
Let V = average volume of a suspended particle, r = density of suspended particle,
r′ = density of liquid, m = mass of equal volume of liquid displaced.
According to Archimede’s principle, effective weight of one suspended particle = actual weight – weight of liquid displaced =mg−m′g= mg - m' g
=mg−Vρ′g=mg−(mρ)ρ′g=mg(1−ρ′ρ)= m g - V \rho' g = m g - \left( \frac{m}{\rho} \right) \rho' g = m g \left( 1 - \frac{\rho'}{\rho} \right)
Also, Boltzman constant, kB=RNAk_{B} = \frac{R}{N_{A}}
where, R is gas constant and NAN_A is Avogadro’s number.
Putting, mg(1−ρ′ρ)m g \left( 1 - \frac{\rho'}{\rho} \right) in place of mgm g and value of kBk_{B} in (i), we get
n2=n1exp⁡[−mgNART(1−ρ′ρ)(h2−h1)]n_{2}=n_{1} \exp \left[ -\frac{m g N_{A}}{R T} \left( 1 - \frac{\rho'}{\rho} \right) (h_{2}-h_{1}) \right]
which is the required relation.
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