text: Consider one mole of gas composed of non-interacting point particles that satisfy the ideal gas law:(see any standard Physical Chemistry text, op. cit.)
  Next, assume that all particles are hard spheres of the same finite radius r (the Van der Waals radius). The effect of the finite volume of the particles is to decrease the available void space in which the particles are free to move. We must replace V by V − b, where b is called the excluded volume (per mole) or "co-volume". The corrected equation becomes
  The excluded volume  is not just equal to the volume occupied by the solid, finite-sized particles, but actually four times the total molecular volume for one mole of a Van der waals gas. To see this, we must realize that a particle is surrounded by a sphere of radius 2r (two times the original radius) that is forbidden for the centers of the other particles. If the distance between two particle centers were to be smaller than 2r', it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.
  The excluded volume for the two particles (of average diameter d or radius r) is
  which, divided by two (the number of colliding particles), gives the excluded volume per particle:
  So  is four times the proper volume of the particle. It was a point of concern to Van der Waals that the factor four yields an upper bound; empirical values for  are usually lower. Of course, molecules are not infinitely hard, as Van der Waals thought, and are often fairly soft. To obtain the excluded volume per mole we just need to multiply by the number of molecules in a mole, i.e. by the Avogadro number:
  Next, we introduce a (not necessarily pairwise) attractive force between the particles. Van der Waals assumed that, notwithstanding the existence of this force, the density of the fluid is homogeneous; furthermore, he assumed that the range of the attractive force is so small that the great majority of the particles do not feel that the container is of finite size. Given the homogeneity of the fluid, the bulk of the particles do not experience a net force pulling them to the right or to the left. This is different for the particles in surface layers directly adjacent to the walls. They feel a net force from the bulk particles pulling them into the container, because this force is not compensated by particles on the side where the wall is (another assumption here is that there is no interaction between walls and particles, which is not true, as can be seen from the phenomenon of droplet formation; most types of liquid show adhesion). This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density. On considering one mole of gas, the number of particles will be N
  The number of particles in the surface layers is, again by assuming homogeneity, also proportional to the density. In total, the force on the walls is decreased by a factor proportional to the square of the density, and the pressure (force per unit surface) is decreased by: 
  so that:
  Upon writing n for the number of moles and , the equation obtains the second form given above:
  It is of some historical interest to point out that Van der Waals, in his Nobel prize lecture, gave credit to Laplace for the argument that pressure is reduced proportional to the square of the density.