91Ó°ÊÓ

Equation for Mayer's f-function

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

here the factor $\mathrm{β}$=

$\mathrm{β}=\frac{1}{k{T}_c}$

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

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

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

$u\left(r\right)→∞$

The second viral coefficient:

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

Using the spherical coordination, the volume element is,

$d^{3}r=\left(dr\right)\left(rd\mathrm{θ}\right)\left(r\sin \mathrm{θ}d\mathrm{ϕ}\right)$

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

we have,

$B\left(T\right)=-\frac{1}{2}{∫}_0^{∞}f\left(r\right)4\mathrm{π}{r}^{2}dr$

$=-2\mathrm{π}{∫}_0^{∞}{r}^{2}f\left(r\right)dr$

$=-2\mathrm{π}\left[{∫}_0^{r_0}{r}^{2}f\left(r\right)dr\right]$

$=-2\mathrm{π}\left[{∫}_0^{r_0}{r}^2\left(e^{-\frac{u\left(r\right)}{kT}}-1\right)dr\right]-2\mathrm{π}\left[{∫}_{s_0}^{∞}{r}^2\left(e^{\frac{u\left(r\right)}{kT}}-1\right)dr\right]$$B\left(T\right)=-2\mathrm{π}\left[{∫}_0^{r_0}{r}^2\left(e^{-\frac{u\left(r\right)}{kT}}-1\right)dr\right]-2\mathrm{π}\left[{∫}_0^{∞}{r}^2\left(e^{-\frac{u\left(r\right)}{kT}}-1\right)dr\right]$

$=-2\mathrm{π}{∫}_0^{r^2}{e}^{\frac{-∞}{kT}}dr+2\mathrm{π}{∫}_0^{r_0}{r}^{2}dr-2\mathrm{π}{∫}_0^{∞}{r}^2\left(1-\frac{u\left(r\right)}{kT}-1\right)dr$

$=0+\frac{2\mathrm{π}{r}_0^3}{3}-\frac{2\mathrm{π}}{KT}{∫}_{5_0}^{∞}{r}^{2}u\left(r\right)dr$

On comparing,

$B\left(T\right)=\frac{2\mathrm{π}{r}_0^3}{3}+2\mathrm{π}{∫}_{v_0}^{∞}{r}^2\frac{u\left(r\right)}{kT}dr$

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

and,

$\frac{a}{kT}=2\mathrm{π}{∫}_{s_0}^{∞}{r}^2\frac{u\left(r\right)}{kT}dr$

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

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