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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-13 Kinetic Theory



Q.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)/ kBT]
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 NA(ρ – P′) (h2h1)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

 

 

Answer:-



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
= mgmg
= mgV ρ’ g  =  mg – (m/ρ)ρg
= mg(1 – (ρ‘/ρ) )   ….(ii)
Gas constant, R = kBN
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 (h2 h1) / kBT]
= n1 exp [-mg (1 – (ρ’/ρ) )(h2 h1)(N/RT) ]
= n1 exp [-mg (ρ – ρ’)(h2 h1)(N/RTρ) ]