 Solution

For the one-mole convention, $b$ is the excluded molar volume: finite molecular size reduces the volume available for translational motion from $V$ to $V-b$. It is an effective excluded volume, not simply the sum of the bare molecular volumes. $a/V^2$ corrects for attractive intermolecular forces, which reduce the pressure exerted on the wall. The attraction per molecule scales with number density, and the number of affected molecules also scales with density, producing the density-squared correction. The Van der Waals equation consequently raises the measured pressure to the corresponding repulsive-gas value before multiplying by the available volume. Both corrections vanish relative to the ideal-gas terms at sufficiently low density.