$A$ 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:
$n_{2}=n_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]$
where $n_{2}, n_{1}$ refer to number density at heights $h_{2}$ and $h_{1}$ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
$n_{2}=n_{1} \exp \left[-m g N_{A}\left(\rho-\rho^{\prime}\right)\left(h_{2}-h_{1}\right) /(\rho R T)\right]$
where $\rho$ is the density of the suspended particle,and $\rho^{\prime}$ that of the surrounding medium. [$N_{A}$ is Avogadro's number,and $R$ the universal gas constant.]

(N/A) According to the law of atmospheres,the number density is given by:
$n_{2}=n_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]$
For a particle of mass $m$ and density $\rho$ suspended in a medium of density $\rho^{\prime}$,the effective weight $W_{eff}$ is the actual weight minus the buoyant force (Archimedes' principle).
Let $V$ be the volume of the particle. Then $m = V\rho$.
The buoyant force is equal to the weight of the displaced medium: $F_{B} = V\rho^{\prime}g = (m/\rho)\rho^{\prime}g$.
The effective weight is:
$W_{eff} = mg - F_{B} = mg - (m/\rho)\rho^{\prime}g = mg(1 - \rho^{\prime}/\rho) = mg(\rho - \rho^{\prime})/\rho$.
Substituting this effective weight into the law of atmospheres and using $k_{B} = R/N_{A}$:
$n_{2} = n_{1} \exp \left[ -\frac{mg(\rho - \rho^{\prime})}{\rho} \frac{(h_{2} - h_{1})}{k_{B}T} \right]$
Substituting $k_{B} = R/N_{A}$:
$n_{2} = n_{1} \exp \left[ -\frac{mg(\rho - \rho^{\prime})}{\rho} \frac{(h_{2} - h_{1}) N_{A}}{RT} \right]$
Rearranging the terms,we get:
$n_{2} = n_{1} \exp \left[ -\frac{mg N_{A} (\rho - \rho^{\prime}) (h_{2} - h_{1})}{\rho RT} \right]$