The compression factor (compressibility factor) for `1 mol` of a van der Waals gas at `0^(@)C` and `100 atm` pressure is found to be `0.5`. Assuming that the volume of a gas molecule is neligible, calculate the van der Waals constant `a`.

Explanation

Using van der Wall's equation of state `:`
`( P+ ( an^(2))/( V^(2))) ( V - nb) = nRT`
Now `: V -nb ~~V` ( given )
`rArr` The equation is reduced to `: ( P+ ( a)/( V^(2))) V =RT`
or `Z=(PV)/( RT) = ( V)/( RT)( (RT)/( V) -(a)/(V^(2)))=1 - ( a)/( RTV)`
`= 0.5= ( 100 xx V)/( 0.0821 xx 273) rArr V = 0.11 2L `
Also, `0.5 = 1- ( a)/( RTV ) `
Substitute the values of V and T `:`
`rArr a = 1.25 litre^(2) mol^(-2)` atm.