# How can the pressure be 0 for a gas that obeys the van der Waals equation?

According to Halliday-Resnick, if the intermolecular forces are large enough, then the measured pressure $$p$$ of a gas that obeys the van der Waals equation of state, $$\left(p+\frac{an^2}{V^2}\right)(V-nb) = nRT$$, can be $$0$$.

The explanation given is that the intermolecular forces which cause the pressure correction factor $$\frac{an^2}{V^2}$$ may be enough to prevent particles from bouncing off a wall of the container at all, thus no pressure.

However, if the pressure correction is becoming stronger, why isn't it instead $$p+\frac{an^2}{V^2}$$ that becomes $$0$$ as $$\frac{an^2}{V^2}$$ grows more negative? Do I simply have a conceptual misunderstanding of the van der Waals equation?

Additionally, using the book’s explanation (assuming $$p=0$$), I solved for molar volume and obtained a quadratic with two solutions. What is the physical interpretation of these two molar volumes (ie why isn’t there just 1?)

If I understand your question, then the following might help. Solve the van der Waals equation for p and you get: $$p=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$$ So note that the pressure is reduced from what it would be if the factor $$a$$ was 0.
Van der Waals (vdW) equation of state alone cannot correctly describe the behavior of fluids below its critical point. The reason is that it predicts a metastable and an unstable region in the $$p-V$$ plane, which do not correspond to equilibrium states. The presence of such regions qualitatively agrees with the experimental findings. However, the quantitative agreement is missing (it couldn't be different since, in the vdW equation of state, there is no description of the nucleation mechanism driving the system towards its thermodynamic stable state.
Within vdW theory, it is also correct that you get two solutions. The vdW isotherms are regular (continuous) curves. Since the asymptotic limit for large volume remains the perfect gas isotherms $$pV=nRT$$ if the pressure gets negative at some volume and it is positive for large volumes, somewhere in between must go through zero again. Notice that this second crossing corresponds to the unstable region because $$\left.\frac{\partial{p}}{\partial{V}}\right|_T>0$$, i.e. the isothermal compressibility is negative.