Question

How can I derive the Van der Waals equation?

Answer 1

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

`\mathbf(P = (RT)/(barV - b) - a/(barV^2))`

where:

• `P` is the pressure in `"bar"`s.
• `R` is the universal gas constant (`0.083145 "L"*"bar/mol"cdot"K"`).
• `T` is the temperature in `"K"`.
• `barV` is the molar volume of the gas in `"L/mol"`.
• `b` is the volume excluded by the real gas in `"L/mol"`.
• `a` is the average attraction between gas particles in `("L"^2"bar")/("mol"^2)`.

`\mathbf(a)` and `\mathbf(b)` 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:

`PbarV = RT`

`P = (RT)/(barV)`

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 `barV` 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 `a/(barV^2)`.

`color(blue)(P = (RT)/(barV - b) - a/(barV^2))`

If we note that the compressibility factor `Z = (PbarV)/(RT)` 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`, `barV` 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`, `barV` 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 `barV`, the magnitude of `a/(barV^2)` increases more than the magnitude of `(RT)/(barV - 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 `barV`, the magnitude of `a/(barV^2)` decreases more than the magnitude of `(RT)/(barV - b)` decreases, thereby increasing the pressure `P` acquired from the van der Waals equation than from the ideal gas law.