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Q13.13 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 atmospheresn_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]  

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:

 n_{2}=n_{1}exp\left [ -mg N_{A}(\rho -\rho ^{'})(h_{2}-h_{1})/ (\rho RT)\right ]
where   \rho is the density of the suspended particle, and \rho ^{'} , that of the surrounding medium. [N_{A} s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

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n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]                                   (i)

Let the suspended particles be spherical and have radius r

The gravitational force acting on the suspended particles would be

F_{G}=\frac{4}{3}\pi r^{3}\rho g

The buoyant force acting on them would be 

F_{B}=\frac{4}{3}\pi r^{3}\rho' g

The net force acting on the particles become

\\F_{net}=F_{G}-F_{B}\\ F_{net}=\frac{4}{3}\pi r^{3}\rho g-\frac{4}{3}\pi r^{3}\rho' g\\F_{net}=\frac{4}{3}\pi r^{3}g(\rho -\rho ')

Replacing mg in equation (i) with the above equation we get

\\n_{2}=n_{1}exp\left [ -\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1}) /k_{b}T\right ]\\ n_{2}=n_{1}exp\left [ \frac{-\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1})}{\frac{RT}{N_{A}}} \right ]\\ n_{2}=n_{1}exp\left [ \frac{-mgN_{A}(\rho -\rho ')(h_{2}-h_{1})}{RT\rho '} \right ]

The above is the equation to be derived

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