I know that if $a = 0$ for a gas at certain temperature and pressure, it means that the molecules of gas have almost no attractive forces acting between them.

But does that also mean that the molecules have high repulsive forces acting between them?

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    $\begingroup$ High repulsive forces would probably result in a negative a. $\endgroup$ Jan 27, 2022 at 6:15
  • 6
    $\begingroup$ Even the tag you have used should give you the hint the name is not Vanderwall (like some textbooks say), but Johannes Diderik van der Waals $\endgroup$
    – Poutnik
    Jan 27, 2022 at 6:19
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    $\begingroup$ Note that such a question is hypothetical, as real gases have positive $a$ values. $\endgroup$
    – Poutnik
    Jan 27, 2022 at 6:24
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    $\begingroup$ What do you mean by "high repulsive forces"? The parameter "b" also implies repulsion by the way, due to excluded volume. You can regard this excluded volume as being due to a "high repulsive force" active at very short distances (on the order of the vdW radius). $\endgroup$
    – Buck Thorn
    Jan 27, 2022 at 6:26

1 Answer 1


The simple answer is no. An "$a$" of zero simply means the attractive forces are so low that they don't affect the pressure of the gas in a measurable way.

As Ivan Neretin mentioned in the comments, high repulsive forces would result in a negative $a$. More specifically, a positive $a$ is due to electrostatic attraction. Thus a negative $a$ would mean that, on average, the gas molecules are close enough to feel an average electrostatic repulsion. But you will never see a negative $a$, because there will never be a net electrostatic repulsion between gas particles. Here is why:

At sufficiently high densities, real gases exert more pressure than would be predicted if they behaved ideally. This is not due to a repulsive electrostatic interaction, it is due to an excluded volume effect. Indeed, at even the closest average interparticle spacing that atoms and molecules can have and still be gases, their average electrostatic interaction is attractive. For them to experience a net electrostatic repulsive interaction, they would need to be so close that they are, on average, closer than the energy minimum in their interaction potential (typically modeled as a Lennard-Jones potential). And once the particles are that close together, the substance is no longer a gas.

Rather, the positive pressure deviation from ideality at high densities comes from the fact that real gas particles take up space. At sufficiently high densities, there is consequently reduced free volume left for the gas particles to move around (compared to what they would have if they were ideal gas point particles). Hence the term "excluded volume".

This lower free volume significantly reduces the configuration space available to the gas (i.e., the number of possible microstates), and is thus entropically unfavorable. I.e., the positive pressure deviation caused by the excluded volume effect is not energetic in origin, it's entropic.

This can be most easily confirmed by looking at equations of state for the simplest possible real gases, namely noble gases like argon. I did this and found that, at extremely high densities, it's actually energetically favorable to compress them (confirming that their average electrostatic interactions remain attractive), but entropically unfavorable:

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

$$\pd{E}{V}{T} > 0 \text{ (compression is energetically favorable)}$$

$$\pd{S}{V}{T} > 0\text{ (compression is entropically unfavorable)}$$

  • $\begingroup$ So then if we plot a graph of interatomic potential v/s interatomic distance for the molecules of a real gas with "a"=0 what would it look like? $\endgroup$
    – Vp127
    Jan 27, 2022 at 8:40
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    $\begingroup$ @Vp127 A constant potential until they "hit the wall" when they hit other molecules. Then it raises infinitely steep, as the van der Waals model does not consider compresibility of molecules. $\endgroup$
    – Poutnik
    Jan 27, 2022 at 12:55

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