Question

Consider a gas of "hard spheres," which do not interact at all unless their separation distance is less than $r_0$, in which case their interaction energy is infinite. Sketch the Mayer $f$-function for this gas, and compute the second virial coefficient. Discuss the result briefly.

Short answer

The second order virial expansion is,

$P=\left(2\pi r_0^3\right)\frac{N}{3}$

Step-by-step solution

Step 1. Given information

The expression for Mayer's f function is,

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

The factor,

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

Here,

$k=$Boltzmann constant

$T_c$= Critical temperature

Thus, the Mayer's function=

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

$u\left(r\right)$= potential energy due to interaction of any pair of molecules.

Considering the gaseous molecules are "hard hemispheres". If the separation between them $r$is more than the intermolecular separation $r_0$then potential energy is,

$u\left(r\right)=0$

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

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

Step 2. The second virial coefficient is,

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

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

$=-2\pi \left[{\int }_0^{r_0}{r}^2\left(e^{-\frac{m\left(r\right)}{kT}}\right)\right]$

$=-2\pi \left[{\int }_0^{\pi }{r}^2\left(e^{-0}\right)\right]$

$=-2\pi {\int }_0^{r_0}{r}^{2}dr$

$=2\pi \left(\frac{r_0^3}{3}\right)$

From the equation we can say that the second virial coefficient is independent of temperature.

The second order virial expansion term,

$P=\left(\frac{NB\left(T\right)}{V}\right)$

Substituting the $B\left(T\right)=2\pi \left(\frac{r_0^3}{3}\right)$in the equation,

$P=\left(\frac{N}{V}\right)2\pi \left(\frac{r_0^3}{3}\right)$

$=\left(2\pi r_0^3\right)\frac{N}{3}$

Thus, the second order virial expansion is,

$P=\left(2\pi r_0^3\right)\frac{N}{3}$