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

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

The volume of a sphere having radius 'r' is given as (4/3) Π r^{3}. Thus the excluded volume per pair of molecule will be

= (4/3) Π (2r)^{3} = 8 X (4/3) Π r^{3}

Thus excluded volume **per molecule** will be

= 1/2 [8 X (4/3) Π r^{3}]

= (4/3) Π r^{3}

which is the volume of one molecule.

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