Question

Use the cluster expansion to write the total energy of a monatomic nonideal gas in terms of a sum of diagrams. Keeping only the first diagram, show that the energy is approximately$U\approx \frac{3}{2}NkT+\frac{N^2}{V}·2\pi {\int }_0^{\infty }{r}^{2}u\left(r\right)e^{-\beta u\left(r\right)}dr$Use a computer to evaluate this integral numerically, as a function of T, for the Lennard-Jones potential. Plot the temperature-dependent part of the correction term, and explain the shape of the graph physically. Discuss the correction to the heat capacity at constant volume, and compute this correction numerically for argon at room temperature and atmospheric pressure.

Short answer

Since, the energy is$U=\frac{3}{2}NkT+2\pi ·\frac{N^2}{V}·\int r^2·u\left(r\right)e^{-\beta ·u\left(r\right)}dr$. Hence, Proved.

The plot of the temperature-dependent part of the correction term is

Step-by-step solution

Given information

We have been given that$dU=-\frac{1}{Z}·\frac{dZ}{d\beta }=-\frac{d}{d\beta }\left(\frac{dZ}{Z}\right)$

Simplify

There are two terms in equation

$U=U_{id}+U_e=-\frac{d}{d\beta }\ln \left(Z_{id}\right)-\frac{d}{d\beta }\ln \left(Z_e\right)$

First term is equal to:

$U_{id}=-\frac{d}{d\beta }\ln \left(Z_{id}\right)=\frac{3}{2}NkT$

The second part is:

$U_e=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{N^2}{V}\right)\left(\int u\left(r\right)d^{3}r\right)$

Now, we will use the equation

$U_e=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{N^2}{V}\right)\left(\int u\left(r\right)d^{3}r\right)=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{N^2}{V}\right)\left(4\pi ·\int r^2·u\left(r\right)dr\right)$

$=-2\pi ·\frac{N^2}{V}·\int r^2·u\left(r\right)e^{-\beta ·u\left(r\right)}dr$

The energy term is given by:

$=\frac{3}{2}NkT+2\pi ·\frac{N^2}{V}·\int r^2·u\left(r\right)e^{-\beta ·u\left(r\right)}dr$

The second virial coefficient is given by:

$U_e=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{N^2}{V}\right)\left(\int u\left(r\right)d^{3}r\right)=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{N^2}{V}\right)\left(4\pi ·\int r^2·u\left(r\right)dr\right)$