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

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?

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.