The compression factor (compressibility factor) for one mole of a van der Waals' gas at `0^(@)C and 100` atmosphere pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waal's constant 'a'.

Explanation

`Z=(PV)/(RT) =0.5`
`therefore (100xxV)/(0.082xx273)=0.5`
`therefore V=0.112` litre
Now, using van der Waals. equation
`[P+(a)/(V^(2))][V]=RT`
or `[P+(a)/(V^(2))]=(RT)/(V)" "(therefore` b is negligible)
`therefore [100+(a)/((0.112)^(2))]=(0.082xx273)/(0.112)`
`=199.88`
`therefore (a)/((0.112)^(2))=99.88`
`therefore a=1.253" litre"^(2)" mole"^(-2)` atm.