Prove that the excluded volume 'b' is four times the actual volume of the gas molecules



Consider two molecules A and B. Taking the molecules to be spherical, if r is the radius of the molecules then the distance of closest approach between the two molecules = 2 r (as shown in Figure). This is the distance between the centres of their nuclei.

Since the molecules cannot come closer than distance 2r, the excluded volume for a pair of molecules = volume of sphere of radius 2r

= 4/3 $\pi$ ( $2{{r})^{3}}$ = 8 x 4/3 $\pi$ ${{r}^{3}}$

Excluded volume per molecule (b) =
1/2 (8 x 4/3 $\pi$ ${{r}^{3}}$)= 4 x 4/3 $\pi$ ${{r}^{3}}$

But 4/3 $\pi$ ${{r}^{3}}$ = $V_{ m }$

i.e., the actual volume of the gas molecule.
b= 4 $V_{ m }$