91Ó°ÊÓ

The Helmholtz free energy for an ideal gas is given by

$F=-NkT\left[\ln V-\ln N-\ln v_Q+1\right]+F_{\mathrm{int}}.........................\left(1\right)$

Here, $V,N,v_Q,F_{\mathrm{int}}$are the volume, number of molecules, quantum volume and the internal free energy of the ideal gas.

The quantum volume is given by

$v_Q={\left(\frac{h^2}{2\mathrm{Ï€}mkT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}........................\left(2\right)$

Take logarithm from both sides of equation (2).

$\begin{array}{rcl}\ln v_Q& =& \ln {\left(\frac{h^2}{2\mathrm{Ï€}mkT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& -\frac{3}{2}\ln T+\frac{3}{2}\ln {\left(\frac{h^2}{2\mathrm{Ï€}mk}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}...........................\left(3\right)\end{array}$

The formula calculate the entropy$\left(S\right)$ of the gas is given by

$S=-{\left(\frac{∂F}{∂T}\right)}_{V,N}................................\left(4\right)$

Substitute the value of the free energy from equation (1) into equation (4) and simplify to obtain the required entropy.

localid="1647062285813" $\begin{array}{rcl}S& =& -\frac{∂}{∂T}\left[-NkT\left[\ln V-\ln N-\ln v_Q+1\right]+F_{\mathrm{int}}\right]\\ & =& Nk\left[\ln \left(\frac{V}{N{v}_Q}\right)+1\right]+NkT\left[\frac{3}{2}T\right]-\frac{∂F_{\mathrm{int}}}{∂T}\\ & =& Nk\left[\ln \left(\frac{V}{N{v}_Q}\right)+\frac{5}{2}\right]-\frac{∂F_{\mathrm{int}}}{∂T}\end{array}$

The internal free energy of the gas is given by

$F_{\mathrm{int}}=-NkT\ln Z_{\mathrm{int}}............................\left(5\right)$

The formula to calculate the chemical potential of the gas is given by

$\mathrm{μ}=-{\left(\frac{∂F}{∂N}\right)}_{T,V}.............................\left(6\right)$

Substitute the value of the free energy from equation (1) into equation (6) and simplify to obtain the required chemical potential.

$\begin{array}{rcl}\mathrm{μ}& =& \frac{∂}{∂N}\left[-NkT\left[\ln V-\ln N-\ln v_Q+1\right]+F_{\mathrm{int}}\right]\\ & =& \frac{∂}{∂N}\left[-NkT\left[\ln V-\ln N-\ln v_Q+1\right]-NkT\ln Z_{\mathrm{int}}\right][\mathrm{from}\mathrm{equation}(5\left)\right]\\ & =& -kT\left[\ln \left(\frac{V}{N{v}_Q}\right)+1\right]+NkT\frac{1}{N}-kT\ln Z_{\mathrm{int}}\\ & =& -kT\ln \left(\frac{V}{N{v}_Q}\right)-kT\ln Z_{\mathrm{int}}\\ & =& -kT\ln \left(\frac{V{Z}_{\mathrm{int}}}{N{v}_Q}\right)\end{array}$