To the desired approximation, [ begin{equation*} frac{P}{k T} equiv frac{1}{V} ln mathscr{Q}=frac{1}{lambda^{3}}left(z-a_{2} z^{2} ight), quad n=frac{N}{V}=frac{1}{lambda^{3}}left(z-2 a_{2}

Question:

To the desired approximation,

\[
\begin{equation*}
\frac{P}{k T} \equiv \frac{1}{V} \ln \mathscr{Q}=\frac{1}{\lambda^{3}}\left(z-a_{2} z^{2}\right), \quad n=\frac{N}{V}=\frac{1}{\lambda^{3}}\left(z-2 a_{2} z^{2}\right) \tag{1a,b}
\end{equation*}
\]

where \(a_{2}\) is the second virial coefficient of the gas. It follows that

\[
\begin{equation*}
z=n \lambda^{3}\left(1+2 a_{2} \cdot n \lambda^{3}\right) \text {, whence } P=n k T\left(1+a_{2} \cdot n \lambda^{3}\right) \text {. } \tag{2a,b}
\end{equation*}
\]

Next

\[
\begin{aligned}
A & =N k T \ln z-P V=N k T\left\{\ln \left(n \lambda^{3}\right)-1+a_{2} \cdot n \lambda^{3}\right\} \\
G & =N k T \ln z=N k T\left\{\ln \left(n \lambda^{3}\right)+2 a_{2} \cdot n \lambda^{3}\right\} \\
S & =-\left(\frac{\partial A}{\partial T}\right)_{N, \mathrm{~V}}=N k\left\{\frac{5}{2}-\ln \left(n \lambda^{3}\right)-n \frac{\partial}{\partial T}\left(T a_{2} \lambda^{3}\right)\right\}
\end{aligned}
\]

remember that the coefficient \(a_{2}\) is a function of \(T\). Furthermore,

\[
\begin{aligned}
U & =A+T S=N k T\left\{\frac{3}{2}-n T \frac{\partial}{\partial T}\left(a_{2} \lambda^{3}\right)\right\}, \\
H & =U+P V=N k T\left\{\frac{5}{2}-n T^{2} \frac{\partial}{\partial T}\left(\frac{a_{2} \lambda^{3}}{T}\right)\right\}, \\
C_{\mathrm{V}} & =\left(\frac{\partial U}{\partial T}\right)_{N, \mathrm{~V}}=N k\left\{\frac{3}{2}-n \frac{\partial}{\partial T}\left(T^{2} \frac{\partial}{\partial T}\left(a_{2} \lambda^{3}\right)\right)\right\}, \text { and } \\
C_{P}-C_{\mathrm{V}} & =-T \frac{(\partial P / \partial T)_{N, \mathrm{~V}}^{2}}{(\partial P / \partial V)_{N, T}}=N k\left\{1+2 n T \frac{\partial}{\partial T}\left(a_{2} \lambda^{3}\right)\right\} .
\end{aligned}
\]

For the second part, use the expression for \(a_{2} \lambda^{3}\) derived in Problem 10.5 and examine the temperature dependence of the various thermodynamic quantities.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  answer-question
Question Posted: