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

For `1 mol` of a gas, the van der Waals equation is
`(P+(a)/(V_(m)^(2)))(V_(m)-b)=RT`
Ignoring `b`, we get (given volume of gas molecule is negligible)
`(P+(a)/(V_(m)^(2)))V_(m)=RT` ltbgt or `pV_(m)+(a)/(V_(m))=RT`
or `(pV_(m))/(RT)+(a)/(V_(m)RT)=1`
or `Z=(pV_(m))/(RT)=1-(a)/(V_(m)RT)` (`i`)
It is given that
`Z=(pV_(m))/(RT)=0.5implies V_(m)=(0.5RT)/(P)`
With this, equation (`i`) becomes
`0.5=1-(a)/((0.5RT//p)RT)`
or `a=(0.5)((0.5RT)/(p))RT=0.25(R^(2)T^(2))/(p)`
Substiuting the given values, we get
`a=(0.25)[((0.082L atm K^(-1)mol^(-1))^(2)(273 K)^(2))/((100 atm))]`
`=1.2528 L^(2) atm mol^(-2)`