# Why is van der Waals equation used?

Jun 8, 2018

Well, real gases have intermolecular forces, don't they?

And thus, we use the van der Waals equation of state to account for such forces:

$P = \frac{R T}{\overline{V} - b} - \frac{a}{{\overline{V}}^{2}}$

These forces manifest themselves in:

• $a$, a constant that accounts for the average forces of attraction.
• $b$, a constant that accounts for the fact that gases are not always negligible compared to the size of their container.

and these modify the true molar volume, $\overline{V} \equiv \frac{V}{n}$. Upon solving for the cubic equation in terms of the molar volume,

$\overline{\underline{|}} \stackrel{\text{ ")(" "barV^3 - (b + (RT)/P)barV^2 + a/PbarV - (ab)/P = 0" }}{|}$

For this, we need

• specified pressure $P$ in $\text{bar}$,
• temperature $T$ in $\text{K}$,
• $R = \text{0.083145 L"cdot"bar/mol"cdot"K}$,
• vdW constants $a$ in ${\text{L"^2"bar/mol}}^{2}$ and $b$ in $\text{L/mol}$.

Then this can be solved via whatever method you want to solve this cubic. This is gone into more detail here.

Three solutions arise:

• One $\overline{V}$ is of the liquid.
• One $\overline{V}$ is of the gas.
• One $\overline{V}$ is a so-called spurious (i.e. UNPHYSICAL) solution.

To know what you have just gotten, compare with the other $\overline{V}$ to see if you have found the largest one. If you did not maximize $\overline{V}$, try a different guess until you do.