Assume molecules to be spheres with radius r.
therefore the volume of molecule is `(4pir^(3))/(3)`
The closest approach between two molecules =`2r`. Since two molecules cannot come closer than distance 2r. The excluded volume per pair of molecules in binary collision is
Volume of sphere of radius `2r=(4)/(3)pi(2r)^(3)=8xx(4pir^(3))/(3)`
Excluded volume per molecules is
`(1)/(2)[8xx(4pir^(3))/(3)]=4xx(4pir^(3))/(3)`
`=4xx`volume of one molecule
