# How can I derive the Van der Waals equation?

Feb 28, 2016

First off, the van der Waals equation looks like this:

$\setminus m a t h b f \left(P = \frac{R T}{\overline{V} - b} - \frac{a}{{\overline{V}}^{2}}\right)$

where:

• $P$ is the pressure in $\text{bar}$s.
• $R$ is the universal gas constant ($0.083145 \text{L"*"bar/mol"cdot"K}$).
• $T$ is the temperature in $\text{K}$.
• $\overline{V}$ is the molar volume of the gas in $\text{L/mol}$.
• $b$ is the volume excluded by the real gas in $\text{L/mol}$.
• $a$ is the average attraction between gas particles in $\left({\text{L"^2"bar")/("mol}}^{2}\right)$.

$\setminus m a t h b f \left(a\right)$ and $\setminus m a t h b f \left(b\right)$ depend on the gas itself .

For example, $a = 13.888$ and $b = 0.11641$ for butane (Physical Chemistry: A Molecular Approach, McQuarrie).

We can derive it by starting from the ideal gas law:

$P \overline{V} = R T$

$P = \frac{R T}{\overline{V}}$

But the actual derivation is fairly involved, so we won't do it the complete way, but in more of a conceptual way.

By assuming the gas is a hard sphere, we say that it takes up the space it has available to move, i.e. it decreases $\overline{V}$ by $b$. Again, this depends on the exact gas.

Therefore, the first half becomes:

P = (RT)/(barV - b) pm ?

Now we examine $a$; $a$ essentially accounts for the different extent of the attractive intermolecular forces present in each gas. Basically, when accounting for that we subtract $\frac{a}{{\overline{V}}^{2}}$.

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

If we note that the compressibility factor $Z = \frac{P \overline{V}}{R T}$ is $1$ for an ideal gas, it is less for a real gas if it is easily compressible and greater for a real gas if it is hardly compressible. As such, we have these two relationships:

• When $Z > 1$, $\overline{V}$ is greater than ideal, which means the gas is harder to compress. It corresponds to a higher pressure required to compress the gas.
• When $Z < 1$, $\overline{V}$ is smaller than ideal, which means the gas is easier to compress. It corresponds to a lower pressure required to compress the gas.

From this we should see that :

• For a gas that is easier to compress than the ideal form, with a smaller $\overline{V}$, the magnitude of $\frac{a}{{\overline{V}}^{2}}$ increases more than the magnitude of $\frac{R T}{\overline{V} - b}$ increases, thereby decreasing the pressure $P$ acquired from the van der Waals equation than from the ideal gas law.
• For a gas that is harder to compress than the ideal form, with a larger $\overline{V}$, the magnitude of $\frac{a}{{\overline{V}}^{2}}$ decreases more than the magnitude of $\frac{R T}{\overline{V} - b}$ decreases, thereby increasing the pressure $P$ acquired from the van der Waals equation than from the ideal gas law.