Why is van der Waals equation used?

Md Ariful Islam

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Md Ariful Islam

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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:

$P = (RT)/(barV - b) - a/(barV^2)$

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.

Three 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.

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