What happens when $V=Nb$ in the Van der Waals equation (divergent)?

Question

The VdW equation: $$\left(P+a\left(\frac{N^2}{V^2}\right) \right)\left(V-Nb\right)=Nk_BT$$

when the intermolecular forces are zero $a=0$, so $P=\frac{Nk_BT}{V-Nb}$ which diverges at $V=Nb$ for fixed temperature.

I'm simulating some hard sphere collisions, with $r_{sphere}=0.5, R_{contianer}=10 $. From some curvefit I got $b \approx 2.2$, so when $N>143$ (or less than 50% of the volume/area),the VdW equation breaks down.

Does that imply a different law holds when the number of particles becomes large? But isn't $b \propto r^3$ anyway, so it should account for the case when the number of particles are large?

Answer 1

If $V=Nb$, then container volume is 4 times total volume of molecules, according to the geometry used in the reasoning of van der Waals real gas model.

Molecules of a gas ( or rather a fluid ) at such a state are not able to freely move anymore, forming for each other not passable obstacles. Therefore the statistical pressure model breaks down. Signs of this progressively occur when $V \gg Nb$ is not true any more and the former is approaching the latter.

Additionally, at such a density, intermolecular interactions are not negligible.

Note that this and similar states are far, unjustified extrapolations of applicability of the van der Waals model. This model is not intended for high pressure/density nor states near the critical state, as the model breaks physically much sooner than mathematically.