A gas in equilibrium has uniform density and pressure throughout … - Vidyadip

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 $n_2, n_1$ 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: $\text{n}_2=\text{n}_1\text{exp}\Big[\frac{-\text{mgN}_\text{A}(\rho-\rho')(\text{h}_2-\text{h}_1)}{(\rho\text{RT})}\Big]$ 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]

✓

Answer

Considering the particles and molecules to be spherical, the weight of the particle is $\text{W}=\text{mg}=\frac{4}{3}\pi\text{r}^3\rho\text{g}\ ...(\text{i})$ Where r = radius of the particle and $\rho=$ density of the particle. Its motion under gravity causes buoyant force to act upward which is equal to B = Volume of particle × fensity of the surrounding medium × g $=\frac{4}{3}\pi\text{r}^3\rho'\text{g}\ ...(\text{ii})$ If F be the downward force acting on the particle, then $\text{F}=\text{W}-\text{B}=\frac{4}{3}\pi\text{r}^3(\rho-\rho')\text{g}\ ...(\text{iii})$ Also $\text{n}_2=\text{n}_1\text{exp}\Big[\frac{-\text{mg}}{\text{k}_\text{B}\text{T}}(\text{h}_2-\text{h}_1)\Big]\ ...(\text{iv})$ Where KB = Boltzman constant $n_1$ and $n_2$ are number densities at heights $h_1$ and $h_2$ respectively. Here, mg can be replaced by effective force F given by equation (iii) $\therefore$ From (iii) and (iv) we get $\text{n}_2=\text{n}_1\text{exp}\Big[-\frac{4\pi}{3}\text{r}^3\frac{(\rho-\rho')}{\text{k}_\text{B}\text{T}}\text{g}(\text{h}_2-\text{h}_1)\Big]$ $\text{n}_2=\text{n}_1\text{exp}\Bigg[-\frac{4\pi}{3}\text{r}^3\frac{\rho\text{g}\big(1-\frac{\rho'}{\rho}\big)(\text{h}_2-\text{h}_1)}{\big(\frac{\text{RT}}{\text{N}_\text{A}}\big)}\Bigg]$ $\Big[\because\text{ k}_\text{B}=\frac{\text{R}}{\text{N}_\text{A}}\Big]$ $\text{n}_2=\text{n}_1\text{exp}\Bigg[-\frac{\text{mgN}_\text{A}\big(1-\frac{\rho'}{\rho}\big)(\text{h}_2-\text{h}_1)}{\text{RT}}\Bigg]$ Which is required relation where, $\frac{4}{3}\pi\text{r}^3\rho\text{g}=$ mass of the particle × g = mg

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