Question

Consider a monatomic ideal gas that lives at a height z above sea level, so each molecule has potential energy $mgz$in addition to its kinetic energy.

(a) Show that the chemical potential is the same as if the gas were at sea level, plus an additional term $mgz$:

$\mu \left(z\right)=-kT\ln \left[\frac{V}{N}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right]+mgz.$

(You can derive this result from either the definition $\mu =-T(\partial S/\partial N{)}_{U,V}$or the formula $\left.\mu =(\partial U/\partial N{)}_{S,V.}\right)$

(b) Suppose you have two chunks of helium gas, one at sea level and one at height z, each having the same temperature and volume. Assuming that they are in diffusive equilibrium, show that the number of molecules in the higher chunk is

$N\left(z\right)=N\left(0\right)e^{-mgz/kT}$

in agreement with the result of Problem 1.16.

Short answer

(a) $\mu =-kT\left[\ln \left(\frac{V}{N}{\left(\frac{2\pi mkT}{k^2}\right)}^{\frac{3}{2}}\right)\right]+mgz$

(b)$N\left(z\right)=N\left(0\right)e^{-mgz/kT}$

Step-by-step solution

Part (a) Step 1 : Given Information

Monoatomic ideal gas that lives at a height z above sea level, so each molecules has potential energy $mgz$in addition to its kinetic energy.

Part (a) Step 2 : Formula used

Expression for entropy with multiplicity $\Omega$is given by

$S=k\ln \left(\Omega \right)\dots ..\left(1\right)$

Here, $k$is Boltzmann constant, $N$is number of atoms and $\Omega$is multiplicity.

Chemical potential is defined in terms of entropy as

$\mu =-T{\left(\frac{\partial S}{\partial N}\right)}_{U,V}\cdots \dots \left(2\right)$

Here, $T$is temperature applied.

One of the thermodynamic identities of internal energy is

$dU=TdS-PdV+\mu dN\dots ..\text{(3)}$

Here, $V$is volume of the system.

Expression for entropy from the Sackur-Tetrode equation

$S=Nk\left[\ln \left(\frac{V}{N}{\left(\frac{4\pi mU}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]\dots \dots ..\left(4\right)$

Here, $V$is the volume of the system.

Part (a) Step 3 : Calculation

When the monoatomic ideal gas is at height $z$above the sea level then the total energy is the summation of kinetic energy and potential energy.

$U=U_k+mgz$

Where $U_k$is the kinetic energy and $mgz$is the potential energy.

Thus, using the total energy expression, the entropy is,

$\begin{array}{r}S=Nk\left[\ln \left(\frac{V}{N}{\left(\frac{4\pi m{U}_k}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]\\ S=Nk\left[\ln \left(\frac{V}{N}{\left(\frac{4\pi m(U-mgz)}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]\\ S=Nk\left[\ln \left(V{\left(\frac{4\pi m(U-mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)-\ln \left(N^{5/2}\right)+\frac{5}{2}\right]\end{array}$

Part (a) Step 4 : Chemical Potential

Now, calculating for chemical potential:

$\begin{array}{r}\mu =-T{\left(\frac{\mathrm{\partial }S}{\mathrm{\partial }N}\right)}_{U,V}\\ \mu =-kT\frac{\mathrm{\partial }}{\mathrm{\partial }N}\left\{N\left[\ln \left(V{\left(\frac{4\pi m(U-mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)-\ln \left(N^{5/2}\right)+\frac{5}{2}\right]\right\}\end{array}$

$\begin{array}{c}\mu =\\ -kT\left[\frac{\mathrm{\partial }}{\mathrm{\partial }N}N\cdot \ln \left(V{\left(\frac{4\pi m(U-mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)+N\frac{\mathrm{\partial }}{\mathrm{\partial }N}\ln \left(V{\left(\frac{4\pi m(U-mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)-\frac{\mathrm{\partial }}{\mathrm{\partial }N}N\right.\\ \left.\cdot \ln \left(N^{5/2}\right)-N\frac{\mathrm{\partial }}{\mathrm{\partial }N}\ln \left(N^{5/2}\right)+\frac{\mathrm{\partial }}{\mathrm{\partial }N}\frac{5}{2}N\right]\\ \mu =-kT\left[\ln \left(V{\left(\frac{4\pi m(U-mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)-\ln \left(N^{5/2}\right)-N\frac{5}{2N}+\frac{5}{2}\right]\\ \mu =-kT\left[\ln \left(V{\left(\frac{4\pi m(U-mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)-\ln \left(N^{5/2}\right)\right]\\ \mu \approx -kT\left[\ln \left(V{\left(\frac{4\pi m(U-mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)-\ln \left(N^{5/2}\right)\right]+NkT\frac{3mgz}{2(U-mgz)}\\ \mu =-kT\left[\ln \left(V{\left(\frac{4\pi m}{3h^2}\right)}^{\frac{3}{2}}{\left(\frac{3}{2}NkT\right)}^{\frac{3}{2}}\right)-\ln \left(N^{5/2}\right)\right]+NkT\frac{3mgz}{2\left(\frac{3}{2}NkT\right)}\\ \mu =-kT\left[\ln \left(\frac{V}{N}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]+mgz\end{array}$

Part (a) Step 5 : Conclusion

The required entropy is,

$\mu =-kT\left[\ln \left(\frac{V}{N}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]+mgz$

Part (b) Step 1 : Given Information

Two chunk of helium one at sea level $z=0$, and other at height $z$, each having the same temperature and volume.

They are in diffusive equilibrium.

Part (b) Step 2 : Calculation

At diffusive equilibrium, the chemical potentials of the ideal gases are equal.

$\begin{array}{r}\mu \left(z\right)=\mu \left(0\right)\\ -kT\left[\ln \left(\frac{V}{N\left(z\right)}{\left(\frac{2mmkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]+mgz=-kT\left[\ln \left(\frac{V}{N\left(0\right)}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]\\ \left[\ln \left(\frac{V}{N\left(z\right)}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]-\frac{mgz}{kT}=\left[\ln \left(\frac{V}{N\left(0\right)}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]\end{array}$

Taking exponential both sides:

$\begin{array}{r}\left(\frac{V}{N\left(z\right)}{\left(\frac{2\mathrm{sink}}{h^2}\right)}^{\frac{3}{2}}\right)e^{\frac{-mgz}{NT}}=\frac{V}{N\left(0\right)}{\left(\frac{2\mathrm{smk}T}{h^2}\right)}^{\frac{3}{2}}\\ \frac{1}{N\left(z\right)}{e}^{\frac{-mg}{NT}}=\frac{1}{N\left(0\right)}\\ N\left(z\right)=N\left(0\right)e^{-mgz/kT}\end{array}$

Part (b) Step 3 : Conclusion

Thus, the number of molecules in the higher chunk of ideal gas is$N\left(z\right)=N\left(0\right)e^{-mgz/kT}$.