Interacting gas - Aetherwisp

An interacting gas is a generalization of an ideal gas which also includes a two-body interaction term. The Hamiltonian of the system is

H=\sum_{i=1}^{N} \frac{\mathbf{p}_{i}^{2}}{2m}+\sum_{i<j} u_{ij}

The second sum represents the interaction as a sum of interparticle potentials $u_{ij}$ . Normally, the interaction depends on the distance between particles: $u_{ij}\propto |\mathbf{r}_{i}-\mathbf{r}_{j}|$ .

Using cluster integrals#

Our goal is to find an equation of state similar to the ideal gas one, that is $PV=Nk_{B}T$ , so that we can compare the two. We'll do this by using a canonical ensemble, for which

\frac{PV}{Nk_{B}T}=v \frac{ \partial }{ \partial v } \ln Q_{N}\tag{1}

where $v=V/N$ . The issue is, as always, figuring out the canonical partition function $Q_{N}$ . Let's write it out explicitly:

\begin{align} Q_{N}&=\frac{1}{h^{3N}N!}\int d^{3N}p \int e^{-\beta\left( \sum_{i=1}^{N} \frac{\mathbf{p}_{i}^{2}}{2m}+\sum_{i<j} u_{ij} \right)} \ d^{3N}r \\ &=\frac{1}{h^{3N}N!}\int e^{-\beta \sum_{i}\mathbf{p}^{2}/2m} d^{3N}p \int e^{-\beta \sum_{i<j}u_{ij}}\ d^{3N}r \\ &=\ldots \end{align}

The integral on the left is the usual momentum integral that's common to all ideal gases. Including the $h^{3N}$ , it evaluates to $1/\lambda ^{3N}$ , using the de Broglie thermal wavelength $\lambda$ . Using $e^{\sum n}=\prod e^{n}$ , we can write

\begin{align} \\ \ldots&=\frac{1}{\lambda^{3N}N!}\int \prod_{i<j}e^{-\beta u_{ij}} \\ &=\frac{1}{\lambda^{3N}N!}\int \prod_{i<j}(1+e^{-\beta u_{ij}}-1) \\ &=\frac{1}{\lambda^{3N}N!} \int \prod_{i<j}(1+f_{ij}) \ d^{3N}r \\ \end{align}

Where we defined the Mayer function $f_{ij}(r)=e^{-\beta u_{ij}}-1$ . The integral we have left is generally referred to as the configuration integral $Z_{N}$ of the system. For non-interacting particles, the Mayer function is zero and the integral is just the measure of the space occupied by the system to the $N$ -th power, i.e. $V^{N}$ . In that case, we just find our usual $Q_{N}=V^{N}/(\lambda ^{3N}N!)$ . This shows that the ideal gas really is just a specific case of the interacting gas.

When the particles do interact, the Mayer function describes how they do so. Its utility lies in the fact that at high temperatures, it is a very small number and can therefore be expressed in an approximate form with little error. The following is a typical shape for an interparticle potential of effective range $r_{0}$ .

We can modify the partition function as follows:

Q_{N}=\frac{V^{N}}{\lambda^{3N}N!}\int\frac{1}{V^{N}} \prod_{i=1}^{N} \prod_{i<j}(1+f_{ij})d\mathbf{r}=\frac{V^{N}}{\lambda ^{3N}N!}Z_{N}

We multiplied and divided by $V^{N}$ and noticed that the position integrals over $d^{3N}r$ are actually $N$ almost identical integrals over $d^{3}r=d\mathbf{r}$ , each with a different $i$ . We also included $1/V^{N}$ in $Z_{N}$ . Using Stirling's approximation:

\ln Q_{N}=N\ln V-N\ln N-N\ln \lambda ^{3}+\ln Z_{N}

The derivative is

v\frac{ \partial }{ \partial v } \ln Q_{N}=N+v\frac{ \partial }{ \partial v } \ln Z_{N}

If we put this in $(1)$ we get (???)

\frac{PV}{Nk_{B}T}=1+v \frac{ \partial Z_{N} }{ \partial v }

If we solve $Z_{N}$ and take the derivative we are done. As mentioned before, $f_{ij}$ is small for large temperatures, so

\prod_{i<j}(1+f_{ij})\simeq 1+\sum_{i<j}f_{ij}

Using this approximation we find

\begin{align} Z_{N}&=\frac{1}{N}\ln\left[ \frac{1}{V^{N}}\int \prod_{i=1}^{N} \prod_{j<k}(1+f_{jk}) \right] \\ &\simeq\frac{1}{N}\ln\left[ \frac{1}{V^{N}}\int \prod_{i=1}^{N} \left( 1+\sum_{j<k}f_{jk} \right) \right] \\ &=\frac{1}{N}\ln\left[ 1+ \frac{V^{N-2}}{V^{N}}\int f_{12} \frac{N(N-1)}{2}d\mathbf{r}_{1}d\mathbf{r}_{2} \right] \\ &=\frac{1}{N}\ln\left[ 1+ \frac{N(N-1)}{2V}\int f(\mathbf{r})d\mathbf{r} \right] \\ &\simeq \frac{1}{N}\ln\left[ 1+ \frac{N^{2}}{2V}\int f(\mathbf{r})d\mathbf{r} \right] \end{align}

The $v$ derivative is

\frac{ \partial Z }{ \partial v } =- \frac{1}{2v^{2}}\int f(\mathbf{r})d\mathbf{r}=- \frac{1}{2v^{2}}\int_{0}^{\infty} 4\pi r^{2}f(r)dr

Our equation of state therefore becomes

\boxed{\frac{PV}{Nk_{B}T}=1- \frac{1}{2v}\int_{0}^{\infty}4\pi r^{2}f(r)dr}

The negative term is called second virial coefficient.

Virial expansion#

The above result is a special case of a more general case called a virial expansion. In it, the equation of state becomes

\frac{P}{k_{B}T}=\frac{1}{\lambda ^{3}_{T}}\sum_{l=1}^{\infty} b_{l}z^{l}

where each $b_{l}$ are the virial coefficients.