Question 13.13:
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 = 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) ]
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′) (h2 –h1)/ (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.]
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′) (h2 –h1)/ (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.]
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 = 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) ]
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