Page 62 - 4498

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 mol by V mol  b, where b is called the excluded

volume. The corrected equation becomes as follows:

RT
p  (3.2.4)
V mol  b

The excluded volume is not just equal to the volume occupied by the

solid, finite-sized particles, but actually four times that volume. To see
this, we must realize that a particle is surrounded by a sphere of radius r =
2r (two times the original radius) that is forbidden for the centres of the

other particles. If the distance between two particle centres 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 per particle (of average diameter d or radius r)

is

3 3 3
4 d  4 d   4 r 
 b   8     b  4   (3.2.5)
3  3   3 
   

which was divided by 2 to prevent overcounting. So b′ is four times bigger
than 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 b′
are usually lower. Of course, molecules are not infinitely hard, as Van der
Waals thought, and are often fairly soft.

Next, we introduce a pairwise attractive force between the particles.
Van der Waals assumed that, not withstanding the existence of this force,
the density of the fluid is homogeneous. Further 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 it
can be seen from the phenomenon of droplet formation; most types of

62