Question

Show that the excluded volume is four time the actual volume of a gas molecule?

Answer 1

Answer:

Consider one mole of gas composed of non-interacting point particles that satisfy the ideal gas law:

p=RTVm=RTv

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 or "co-volume". The corrected equation becomes:

p=RTVm−b

The excluded volume b 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 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:

b′2=4πd33=8⋅(4πr3/3)

which divided by two (the number of colliding particles) gives the excluded volume per particle:

b′=b′2/2→b′=4⋅(4πr3/3)

So b' 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 b' are usually lower. Of course, molecules are not infinitely hard, as van der Waals thought, and are often fairly soft.

Explanation:

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