Prove that the excluded volume 'h' is four times the actual volume of the gas molecules - Chemistry - States of Matter

Question

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

Kirti Gogia · Accepted Answer

The Van der Waal's constant 'b ' is also known as excluded
volume or co-volume Excluded volume of a molecule means that this
volume cannot be made available to the other molecule The numerical
value of 'b' is four times the actual volume occupied by a gaseous
molecule This can be explained as follows :
We know that gaseous molecules are regarded as spheres Let 'r'
be the radius of each sphere If we consider only bimolecular
collisions, then the volume occupied by the sphere of radius 2r
represents the excluded volume per pair of molecules as shown in
the following figure
<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/3268232/2012_10_18_12_42_06/3103435%20png">
The volume of a sphere having radius 'r' is given as (4/3)
Î  r3 Thus the excluded volume per pair of
molecule will be
= (4/3) Î  (2r)3 = 8 X (4/3) Î 
r3
Thus excluded volume per molecule will be
= 1/2 [8 X (4/3) Î  r3]
= (4/3) Î  r3
which is the volume of one molecule

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