# Why is van der Waals equation used?

##### 1 Answer

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 = (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,

#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