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 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:
$\text{n}_2=\text{n}_1\text{exp}\Big[\frac{-\text{mg}}{\text{k}_\text{B}\text{T}}(\text{h}_2-\text{h}_1)\Big]$
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:
$\text{n}_2=\text{n}_1\text{exp}\Big[\frac{-\text{mgN}_\text{A}(\rho-\rho')(\text{h}_2-\text{h}_1)}{(\rho\text{RT})}\Big]$
where $\rho$ is the density of the suspended particle, and $\rho'$ 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]
$\text{n}_2=\text{n}_1\text{exp}\Big[\frac{-\text{mg}}{\text{k}_\text{B}\text{T}}(\text{h}_2-\text{h}_1)\Big]$
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:
$\text{n}_2=\text{n}_1\text{exp}\Big[\frac{-\text{mgN}_\text{A}(\rho-\rho')(\text{h}_2-\text{h}_1)}{(\rho\text{RT})}\Big]$
where $\rho$ is the density of the suspended particle, and $\rho'$ 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]
