Chapter 6.7, Problem 45P

To determine

The expression for the entropy an chemical potential for a ideal gas.

Explanation of Solution

Given:

The equilibrium temperature of an ideal gas is $T$ .

Formula used:

Write the expression for the Helmholtz free energy of an ideal gas.

  $F=-NkT\left[lnV-lnN-ln{v}_Q+1\right]+F_{\mathrm{int}}$ .......... (1)

Here, $F$ is the Helmholtz free energy, $V$ is the total volume of the ideal gas, $N$ is the total number of molecules, $k$ is the Boltzmann constant, $T$ is the equilibrium temperature, $v_Q$ is the quantum volume of the gas and $F_{\mathrm{int}}$ is the internal free energy of the system.

Write the expression for quantum volume of an ideal gas.

  $v_Q={\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$ .......... (2)

Here, $h$ is the Plank’s constant.

Write the expression for entropy of the system.

  $S=-{\left(\frac{\partial F}{\partial T}\right)}_{N,V}$ .......... (3)

Write the expression for the chemical potential of a system.

  $\mu ={\left(\frac{\partial F}{\partial N}\right)}_{T,V}$ .......... (4)

Write the expression for internal energy of an ideal gas.

  $F=-NkTlnZ$

Here, $\mu$ is the chemical potential of the system.

Calculation:

Apply logarithm to the both side of equation (2).

  $\begin{array}{c}ln{v}_Q=ln{\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}\\ =\frac{3}{2}\left[-ln\left(T\right)+ln\left(\frac{h^2}{2\pi mkT}\right)\right]\end{array}$

Substitute $\left(\frac{3}{2}\left[-ln\left(T\right)+ln\left(\frac{h^2}{2\pi mkT}\right)\right]\right)$ for $ln{v}_Q$ in equation (1).

  $\begin{array}{l}F=-NkT\left[lnV-lnN-\frac{3}{2}\left[-ln\left(T\right)+ln\left(\frac{h^2}{2\pi mkT}\right)\right]+1\right]+F_{\mathrm{int}}\\ F=-NKT\left[lnV-lnN+\frac{3}{2}ln\left(T\right)-\frac{3}{2}ln\left(\frac{h^2}{2\pi mkT}\right)+1\right]+F_{\mathrm{int}}\end{array}$

Substitute $\left(-NKT\left[lnV-lnN+\frac{3}{2}ln\left(T\right)-\frac{3}{2}ln\left(\frac{h^2}{2\pi mkT}\right)+1\right]+F_{\mathrm{int}}\right)$ for $F$ in equation (3).

  $\begin{array}{c}S=\frac{\partial }{\partial T}\left(NKT\left[lnV-lnN+\frac{3}{2}ln\left(T\right)-\frac{3}{2}ln\left(\frac{h^2}{2\pi mkT}\right)+1\right]+F_{\mathrm{int}}\right)\\ =NK\left[lnV-lnN+\frac{3}{2}ln\left(T\right)-\frac{3}{2}ln\left(\frac{h^2}{2\pi mkT}\right)+1\right]-\frac{\partial F_{\mathrm{int}}}{\partial T}+\frac{3}{2}Nk\end{array}$

Substitute $\left(ln{v}_Q\right)$ for $\left(\frac{3}{2}ln\left(T\right)-\frac{3}{2}ln\left(\frac{h^2}{2\pi mkT}\right)\right)$ in the above equation.

  $\begin{array}{l}S=NK\left[lnV-lnN+ln{v}_Q+\frac{5}{2}\right]-\frac{\partial F_{\mathrm{int}}}{\partial T}\\ S=Nk\left[ln\left(\frac{V}{N{v}_Q}\right)+\frac{5}{2}\right]-\frac{\partial F_{\mathrm{int}}}{\partial T}\end{array}$

Substitute $\left(-NkT\left[lnV-lnN-ln{v}_Q+1\right]+F_{\mathrm{int}}\right)$ for $F$ in equation (4).

  $\begin{array}{l}\mu =\frac{\partial }{\partial N}\left(-NkT\left[lnV-lnN-ln{v}_Q+1\right]+F_{\mathrm{int}}\right)\\ \mu =-kT\left[lnV-lnN-ln{v}_Q+1\right]+\frac{\partial F_{\mathrm{int}}}{\partial N}\\ \mu =-kTln\left(\frac{V}{N{v}_Q}\right)+\frac{\partial F_{\mathrm{int}}}{\partial N}\end{array}$

Substitute $\left(-NkTlnZ\right)$ for $F_{\mathrm{int}}$ in the above equation.

  $\begin{array}{l}\mu =-kTln\left(\frac{V}{N{v}_Q}\right)+\frac{\partial }{\partial N}\left(-NkTlnZ\right)\\ \mu =-kTln\left(\frac{VZ}{N{v}_Q}\right)\end{array}$

Conclusion:

Thus, the entropy for an ideal gas is $Nk\left[ln\left(\frac{V}{N{v}_Q}\right)+\frac{5}{2}\right]-\frac{\partial F_{\mathrm{int}}}{\partial T}$ and the chemical potential of the system is $-kTln\left(\frac{VZ}{N{v}_Q}\right)$