I have seen a question in my textbook which is little bit confusing,

The term that corrects for attractive forces present in real gas in van der Waal equation is

(i) ${nb}$
(ii) $\frac{an^2}{V^2}$
(iii) $\frac{-an^2}{V^2}$
(iv) ${-nb}$

Since the question is about term "forces" I do not see any of the four options dimensionally correct as (i)&(iv) are related to volume and (ii)&(iii) are related to pressure.

But in my textbook answer the correct answer given is (ii) so how is it dimensionally correct?

  • 3
    $\begingroup$ Kudos for thinking of dimensional analysis. It is a very handy tool to have in your toolbox. $\endgroup$
    – MaxW
    Sep 11, 2018 at 16:12

3 Answers 3


Nothing states that the correction is itself a force, only that the correction is for attractive forces. There is thus no need for the correction to have the dimensions of a force.


I think a better way to understand the term is to look at the Van der Waals equation itself. You start with the ideal gas law


Then add a fudge factor to the pressure to account for the attractive forces between molecules, $\dfrac{an^2}{V^2}$, and another fudge factor to account for the actual volume of the gas molecules, $nb$, themselves.

$$\left(P+\dfrac{an^2}{V^2}\right)(V-nb) = nRT$$

So doing a dimensional analysis on $\dfrac{an^2}{V^2}$ must yield the same pressure unit as $P$.

Doing a dimensional analysis on $nb$ must yield the same volume unit as $V$.


The term $an^2/V^2$ is not a force but a pressure (force/area) and is a term that corrects for the effect of attractive forces between molecules. The constant $a$ has units of $\mathrm{dm^6 bar }$ so $an^2/V^2$ has units of pressure. When the vdW equation is expanded it becomes $an^2/V$ which has units of pressure $\times$ volume ($pV$) or energy (Joules) which is the unit of $RT$. The $1/V$ is there because at a constant number of moles, when the volume is large, the molecules spend less time close to one another and so the effect of intermolecular forces is reduced as one would expect.


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