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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 socalled '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 :
$n_2=n_1 \exp \left[-m_2 N_A(\rho-P)\left(h_2-h_1\right) /(\rho R T)\right]$ where $\rho$ is the density of the suspended particle, and $\rho^{\prime}$ that of surrounding medium. [ $N_A$ is Avogadro's number, and $\mathbf{R}$ the universal gas constant.]
[Hint : Use Archimedes principle to find the apparent weight of the suspended paricle.]
$n_2=n_1 \exp \left[m g\left(h_2-h_1\right) / k_B T\right]$ 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 :
$n_2=n_1 \exp \left[-m_2 N_A(\rho-P)\left(h_2-h_1\right) /(\rho R T)\right]$ where $\rho$ is the density of the suspended particle, and $\rho^{\prime}$ that of surrounding medium. [ $N_A$ is Avogadro's number, and $\mathbf{R}$ the universal gas constant.]
[Hint : Use Archimedes principle to find the apparent weight of the suspended paricle.]
Solution:
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Verified Answer
According to Archimedes' Principle :
Apparent weight of the suspended particle $(m g)$
$=$ Actual $w t-$ (weight of an equal volume of the liquid)
$$
(m g)^{\prime}=m g-\left(\frac{m}{\rho} \rho^{\prime} g\right)
$$
$(\because$ Volume of particle $=m / \rho$ and weight $=$ Vol $\times$ density $\times g)$
$$
\left(m g^{\prime}\right)=m g\left(1-\rho^{\prime} / \rho\right)
$$
Given equation is
$$
n_2=n_1 \exp \left[-\frac{m g}{k T}\left(h_2-h_1\right)\right] \quad \ldots(2)
$$
Replacing $m g$ by the apparent weight $(m g)^{\prime}$
From equations (1) and (2), we get
$$
n_2=n_1 \exp \left[\frac{-m g N_A}{R T}\left(1-\frac{\rho^{\prime}}{\rho}\right)\left(h_2-h_1\right)\right]\left(\because h=R / N_A\right)
$$
This is the required equation.
Apparent weight of the suspended particle $(m g)$
$=$ Actual $w t-$ (weight of an equal volume of the liquid)
$$
(m g)^{\prime}=m g-\left(\frac{m}{\rho} \rho^{\prime} g\right)
$$
$(\because$ Volume of particle $=m / \rho$ and weight $=$ Vol $\times$ density $\times g)$
$$
\left(m g^{\prime}\right)=m g\left(1-\rho^{\prime} / \rho\right)
$$
Given equation is
$$
n_2=n_1 \exp \left[-\frac{m g}{k T}\left(h_2-h_1\right)\right] \quad \ldots(2)
$$
Replacing $m g$ by the apparent weight $(m g)^{\prime}$
From equations (1) and (2), we get
$$
n_2=n_1 \exp \left[\frac{-m g N_A}{R T}\left(1-\frac{\rho^{\prime}}{\rho}\right)\left(h_2-h_1\right)\right]\left(\because h=R / N_A\right)
$$
This is the required equation.
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