Question

Why is van der Waals equation used?

Answer 1

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

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

`P = (RT)/(barV - b) - a/(barV^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, `barV -= V/n`. Upon solving for the cubic equation in terms of the molar volume,

`barul|stackrel(" ")(" "barV^3 - (b + (RT)/P)barV^2 + a/PbarV - (ab)/P = 0" ")|`

For this, we need

• specified pressure `P` in `"bar"`,
• temperature `T` in `"K"`,
• `R = "0.083145 L"cdot"bar/mol"cdot"K"`,
• vdW constants `a` in `"L"^2"bar/mol"^2` and `b` in `"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 `barV` is of the liquid.
• One `barV` is of the gas.
• One `barV` is a so-called spurious (i.e. UNPHYSICAL) solution.

To know what you have just gotten, compare with the other `barV` to see if you have found the largest one. If you did not maximize `barV`, try a different guess until you do.