Consider a gas of molecules whose interaction energy $u\left(r\right)$u is infinite for $r<r_0$and negative for $r>r_0$, with a minimum value of $-u_0$. Suppose further that $kT\gg u_0$, so you can approximate the Boltzmann factor for$r>r_0$using $e^x\approx 1+x$. Show that under these conditions the second virial coefficient has the form $B\left(T\right)=b-(a/kT)$, the same as what you found for a van der Waals gas in Problem 1.17. Write the van der Waals constants $a$and $b$ in terms of $r_0$and $u\left(r\right)$, and discuss the results briefly.

Equation for Mayer's f-function

$f\left(r\right)=e^{-\beta u\left(r\right)}-1$

here the factor $\beta$=

$\beta =\frac{1}{k{T}_c}$

Here,

$T_c$= Critical temperature

$k=$Bolztmann constant

Rewriting the Mayer's equation:

$f\left(r\right)=e^{\frac{M\left(r\right)}{kT}}-1$

$$\begin{gathered} \text{Here,} \\ u\left(r\right)=\text{potential energy due to interaction of any pair of molecules.} \end{gathered}$$

Considering the gaseous molecules as hard hemispheres.

If the separation between them is $r$is more than the intermolecular separation$r_0$then potential is

$r<r_0\text{for}r>r_0$

$\text{If}r<r_0\text{, then the potential energy is,}$

$u\left(r\right)\to \infty$

The second viral coefficient:

$B\left(T\right)=-\frac{1}{2}\int d^{3}rf\left(r\right)$

Using the spherical coordination, the volume element is,

$d^{3}r=\left(dr\right)\left(rd\theta \right)\left(r\sin \theta d\varphi \right)$

$\text{The integral}f\left(r\right)\text{is independent of the angles}\theta \&\varphi \text{. so}{d}^{3}r=4\pi r^{2}dr$

we have,

$B\left(T\right)=-\frac{1}{2}{\int }_0^{\infty }f\left(r\right)4\pi r^{2}dr$

$=-2\pi {\int }_0^{\infty }{r}^{2}f\left(r\right)dr$

$=-2\pi \left[{\int }_0^{r_0}{r}^{2}f\left(r\right)dr\right]$

$=-2\pi \left[{\int }_0^{r_0}{r}^2\left(e^{-\frac{u\left(r\right)}{kT}}-1\right)dr\right]-2\pi \left[{\int }_{s_0}^{\infty }{r}^2\left(e^{\frac{u\left(r\right)}{kT}}-1\right)dr\right]$$B\left(T\right)=-2\pi \left[{\int }_0^{r_0}{r}^2\left(e^{-\frac{u\left(r\right)}{kT}}-1\right)dr\right]-2\pi \left[{\int }_0^{\infty }{r}^2\left(e^{-\frac{u\left(r\right)}{kT}}-1\right)dr\right]$

$=-2\pi {\int }_0^{r^2}{e}^{\frac{-\infty }{kT}}dr+2\pi {\int }_0^{r_0}{r}^{2}dr-2\pi {\int }_0^{\infty }{r}^2\left(1-\frac{u\left(r\right)}{kT}-1\right)dr$

$=0+\frac{2\pi r_0^3}{3}-\frac{2\pi }{KT}{\int }_{5_0}^{\infty }{r}^{2}u\left(r\right)dr$

On comparing,

$B\left(T\right)=\frac{2\pi r_0^3}{3}+2\pi {\int }_{v_0}^{\infty }{r}^2\frac{u\left(r\right)}{kT}dr$

$=b-\frac{a}{kT}$

and,

$\frac{a}{kT}=2\pi {\int }_{s_0}^{\infty }{r}^2\frac{u\left(r\right)}{kT}dr$

The second and third viral coefficient from van der waal model is,

$B=b-\frac{a}{RT}\&C=b^2$