In the empirical demonstration of the Van der Waals equation at Khan Academy, they made a demonstration of the real volume, for example, it was said that it has to be larger than the ideal volume, since the size of the molecules is not disregarded, hence it would be:

$$ V_{\text{ideal}} = V_{\text{real}} + nb $$

Where, $ b $ was said to be related to the size of the molecule.

But why don't I also include intermolecular interaction? Depending on it, the volume may even be smaller.

  • $\begingroup$ How would interatomic interactions affect the volume? Can you please elaborate on how you think interatomic interactions affect the volume? $\endgroup$ Commented Apr 20, 2021 at 15:31
  • $\begingroup$ Interatomic interactions will affect the pressure and not volume. $\endgroup$ Commented Apr 20, 2021 at 16:07
  • $\begingroup$ @NisargBhavsar Intermolecular interactions can cause attraction or repulsion, changing the volume, right? $\endgroup$ Commented Apr 20, 2021 at 16:13
  • $\begingroup$ Atomic interactions can't change the volume of a gas. Just assume if they did decrease the volume than wouldn't there be vacuum spaces inside a gas. $\endgroup$ Commented Apr 20, 2021 at 16:18
  • 1
    $\begingroup$ $V_{\text{ideal}} = V_{\text{real}} - nb$ is correct. Refer this question $\endgroup$ Commented Apr 20, 2021 at 17:28

1 Answer 1


Intermolecular attraction can reduce the volume, in the same way that increasing the pressure can lower the volume (Boyle's law).

One way to see the effect is to take the derivative of the volume, as given by the van der Waals (vdW) equation, with respect to the attractive parameter $a.$

The vdW equation is


Ignore for simplicity the excluded volume term $b,$ so that


Then the derivative of the volume wrt $a$ is

$$\left(\frac{\partial V}{\partial a} \right)_{p,T} = \frac{\bar{V}}{a-p\bar{V}^2}=-(RT)^{-1}\tag{3}$$

In essence what this result says is that the volume decreases as you increase the parameter $a.$


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