Why is the excluded volume 4 times the volume of the gas molecule in van der Waal's Volume Correction Of Ideal Gas Equation? [closed]

Question

Why there is a factor of 4 in van der Waal's Volume Correction?

Answer 1

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

$$p=\frac{RT}{V_{m}}=\frac{RT}{v} $$

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:

$$p=\frac{RT}{V_{m}-b} $$

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:

$$b'_{2}=4\pi\frac{d^{3}}{3}=8\cdot(4\pi r^{3}/3) $$

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

$$b'=b'_{2}/2 \rightarrow b'=4\cdot (4\pi r^{3}/3) $$

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 $\textit{van der Waals}$ thought, and are often fairly soft.

Source: Wikipedia