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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.

Expression for entropy with multiplicity $惟$is given by

$S=k\ln \left(惟\right)鈥�..\left(1\right)$

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

Chemical potential is defined in terms of entropy as

$渭=-T{\left(\frac{鈭�S}{鈭�N}\right)}_{U,V}鈰�鈥�\left(2\right)$

Here, $T$is temperature applied.

One of the thermodynamic identities of internal energy is

$dU=TdS-PdV+渭dN鈥�..\text{(3)}$

Here, $V$is volume of the system.

Expression for entropy from the Sackur-Tetrode equation

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

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

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,

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

Now, calculating for chemical potential:

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

$\begin{array}{c}渭=\\ 鈭�kT\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}N}N鈰�\ln 鈦�\left(V{\left(\frac{4蟺m(U鈭�mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)+N\frac{\mathrm{鈭�}}{\mathrm{鈭�}N}\ln 鈦�\left(V{\left(\frac{4蟺m(U鈭�mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}N}N\right.\\ \left.鈰�\ln 鈦�\left(N^{5/2}\right)鈭�N\frac{\mathrm{鈭�}}{\mathrm{鈭�}N}\ln 鈦�\left(N^{5/2}\right)+\frac{\mathrm{鈭�}}{\mathrm{鈭�}N}\frac{5}{2}N\right]\\ 渭=鈭�kT\left[\ln 鈦�\left(V{\left(\frac{4蟺m(U鈭�mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)鈭�\ln 鈦�\left(N^{5/2}\right)鈭�N\frac{5}{2N}+\frac{5}{2}\right]\\ 渭=鈭�kT\left[\ln 鈦�\left(V{\left(\frac{4蟺m(U鈭�mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)鈭�\ln 鈦�\left(N^{5/2}\right)\right]\\ 渭鈮�鈭�kT\left[\ln 鈦�\left(V{\left(\frac{4蟺m(U鈭�mgz)}{3h^2}\right)}^{\frac{3}{2}}\right)鈭�\ln 鈦�\left(N^{5/2}\right)\right]+NkT\frac{3mgz}{2(U鈭�mgz)}\\ 渭=鈭�kT\left[\ln 鈦�\left(V{\left(\frac{4蟺m}{3h^2}\right)}^{\frac{3}{2}}{\left(\frac{3}{2}NkT\right)}^{\frac{3}{2}}\right)鈭�\ln 鈦�\left(N^{5/2}\right)\right]+NkT\frac{3mgz}{2\left(\frac{3}{2}NkT\right)}\\ 渭=鈭�kT\left[\ln 鈦�\left(\frac{V}{N}{\left(\frac{2蟺mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]+mgz\end{array}$

The required entropy is,

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

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.

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

$\begin{array}{r}渭\left(z\right)=渭\left(0\right)\\ 鈭�kT\left[\ln 鈦�\left(\frac{V}{N\left(z\right)}{\left(\frac{2mmkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]+mgz=鈭�kT\left[\ln 鈦�\left(\frac{V}{N\left(0\right)}{\left(\frac{2蟺mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]\\ \left[\ln 鈦�\left(\frac{V}{N\left(z\right)}{\left(\frac{2蟺mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]鈭�\frac{mgz}{kT}=\left[\ln 鈦�\left(\frac{V}{N\left(0\right)}{\left(\frac{2蟺mkT}{h^2}\right)}^{\frac{3}{2}}\right)\right]\end{array}$

Taking exponential both sides:

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

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