Show explicitly from the results of this section that$G=N\mu$for an ideal gas.

The free energy of an ideal gas is given by

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

The formula to calculate the internal free energy of an ideal gas is given by

$F_{\mathrm{int}}=-NkT\ln Z_{\mathrm{int}}.....................\left(2\right)$

The ideal gas equation is given by

$PV=NkT...............\left(3\right)$

The chemical potential equation of an ideal gas is given by

$\mu =-kT\ln \left(\frac{V{Z}_{\mathrm{int}}}{N{v}_Q}\right).................\left(4\right)$

Gives free energy $\left(G\right)$for an ideal gas is given by

$G=F-PV...................\left(5\right)$

Substitute the value of free energy from equation (1), the value of$F_{\mathrm{int}}$from equation (2) and the value of $PV$from equation (4) into equation (5) and simplify to obtain the Gibb's free energy of the gas.

$\begin{array}{rcl}G& =& -NkT\left[\ln V-\ln N-\ln v_Q+1\right]+F_{\mathrm{int}}-NkT\\ & =& -NkT\left[\ln V-\ln N-\ln v_Q+1\right]-NkT\ln Z_{\mathrm{int}}-NkT\\ & =& -N\left[kT\left\{\ln V-\ln N-\ln v_Q+1+kT\ln Z_{\mathrm{int}}\right\}\right]\\ & =& -NkT\ln \left(\frac{V{Z}_{\mathrm{int}}}{N{v}_Q}\right).............................\left(6\right)\end{array}$

Substitute $\mu$from equation (4) into equation (6) to obtain the required Gibb's free energy.

$G=N\mu$