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 a 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$\mathbf{n}_{2}=\mathbf{n}_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]$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 $\rho$ ', that of surrounding medium. [ $\mathrm{N}_{\mathrm{A}}$ is Avogadro's number, and $\mathbf{R}$ the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Home » 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 a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmosphereswhere refer to number density at heights and respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: where is the density of the suspended particle, and ‘, that of surrounding medium. [ is Avogadro’s number, and the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

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 a 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
$\mathbf{n}_{2}=\mathbf{n}_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]$
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 $\rho$ ‘, that of surrounding medium. [ $\mathrm{N}_{\mathrm{A}}$ is Avogadro’s number, and $\mathbf{R}$ the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

CBSE | Class 11 | Kinetic Theory | NCERT | Physics

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 a 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
$\mathbf{n}_{2}=\mathbf{n}_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]$
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 $\rho$ ‘, that of surrounding medium. [ $\mathrm{N}_{\mathrm{A}}$ is Avogadro’s number, and $\mathbf{R}$ the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Law of atmosphere states that,

$\mathrm{n}_{2}=\mathrm{n}_{1} \exp \left[-\mathrm{mg}\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right) / \mathrm{k}_{\mathrm{B}} T\right]$

According to Archimedes principle, we have,

Apparent weight $=$ Weight of the water displaced – weight of the suspended particle

$=\mathrm{mg}-\mathrm{m}^{\prime} \mathrm{g}$

$=m g-V \rho^{\prime} g=m g-(m / \rho) \rho^{\prime} g$

$=m g\left(1-\left(\rho^{\prime} / \rho\right)\right)$

$=m g(1-(\rho / \rho))^{-}-(2)$
$\rho^{\prime}$ is the Density of the water
$\rho$ is the Density of the suspended particle
$m^{\prime}$ is the Mass of the suspended particle
$m$ is the Mass of the water displaced
$V$ is the Volume of a suspended particle