# Show that b = 4V for a hard sphere potential?

Jul 27, 2017

I assume you mean the second virial coefficient $b \equiv {B}_{2 V} \left(T\right)$, i.e the one seen in the virial equation of state:

$Z = \frac{P \overline{V}}{R T} = 1 + \frac{{B}_{2 V} \left(T\right)}{\overline{V}} + \frac{{B}_{3 V} \left(T\right)}{\overline{V}} ^ 2 + . . .$

where $\overline{V} = \frac{V}{n}$ is the molar volume, $Z$ is the compressibility factor, ${B}_{n V} \left(T\right)$ is the $n$th virial coefficient, and the rest of the variables are known from the ideal gas law.

For example, ${B}_{2 V} \left(T\right) = 0.00064$ at $\text{1 bar}$ of pressure, and $0.00648$ at $\text{10 bar}$ of pressure.

The molar second virial coefficient is given by (Physical Chemistry, McQuarrie & Simon, Ch. 16-5):

${B}_{2 V} \left(T\right) = - 2 \pi {N}_{A} {\int}_{0}^{\infty} \left({e}^{- u \left(r\right) / {k}_{B} T} - 1\right) {r}^{2} \mathrm{dr}$

where:

• $u \left(r\right)$ is the intermolecular pair potential, or the potential energy of two molecules separated by a distance $r$. This has more than one functional form.
• ${k}_{B}$ is the Boltzmann constant.
• $T$ is the temperature in $\text{K}$.
• ${N}_{A} = 6.0221413 \times {10}^{23}$ ${\text{mol}}^{- 1}$ is Avogadro's number. Thus, ${B}_{2 V} \left(T\right)$ is in units of $\text{[ . . . ]/mol}$.

In general, this integral is very hard to evaluate, but for the hard sphere potential (i.e. for molecules that are assumed to have spherical shapes and collide elastically), we can use:

$u \left(r\right) = \left\{\begin{matrix}\infty & \text{ "r < sigma \\ 0 & " } r > \sigma\end{matrix}\right.$

where $\sigma$ is the distance from the center of one molecule to the center of the other (sometimes called the "collision diameter").

This applies well at high temperatures, because intermolecular forces are minimized as collisions get stronger and faster. We can easily obtain ${B}_{2 V} \left(T\right)$ under this approximation.

Under a transformation of integral bounds, $\infty$ is replaced by $\sigma$ so that ${e}^{- u \left(r\right) / {k}_{B} T} \to 0$. This gives:

${B}_{2 V} \left(T\right) = - 2 \pi {N}_{A} {\int}_{0}^{\sigma} \left(0 - 1\right) {r}^{2} \mathrm{dr}$

$= - 2 \pi {N}_{A} \cdot - \left[\frac{{\sigma}^{3}}{3} - \frac{{0}^{3}}{3}\right]$

$= \frac{2}{3} \pi {\sigma}^{3} {N}_{A}$

For one molecule, we then have that ${B}_{2 V} \left(T\right)$ under this approximation is $\frac{2}{3} \pi {\sigma}^{3}$. Now, recall the definition of $\sigma$. For two identical adjacent molecules, $\sigma = 2 r$, where $r$ is the radius of the molecular sphere.

So, we can rewrite this as:

$\frac{{B}_{2 V} \left(T\right)}{{N}_{A}} = \frac{2}{3} \pi {\left(2 r\right)}^{3}$

$= \frac{2}{3} \pi \cdot 8 \cdot {r}^{3} = \frac{16}{3} \pi {r}^{3}$

The volume of a single sphere is $\frac{4}{3} \pi {r}^{3}$. We can consider that the volume of one molecule in this scenario, or its molecular volume $V$.

Thus, the hard sphere potential shows that for a single molecule,

$\textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "(B_(2V)(T))/(N_A) = 4V" }}{|}}}}$,

where $V$ is the volume of one molecular sphere.