The enthalpy of vaporisation of benzene at `80^(@)C` (boiling point) is 31`kJ*mol^(-1)`. What will be the change in entropy of rthe transformation of `31.2g` of benzene vapour into liquid benzene at `80^(@)C`?

Explanation

31.2g benzene`=(31.2)/(78)=0.4` mol [Molar mass=78].
Enthalpy of condensation for 1 mol benzene vapour `=(-)xx`entahlpy of vaporisation of 1 mol liquid benzene `=-31kJ`. Therefore enthalpy of condensation of 0.4 mol of benzene`=-31xx0.4=-12.4kJ`
`therefore`The change in entropy for the transformation of 0.4 mol of benzene vapour into liquid benzene at `80^(@)C`,
`DeltaS=(12.4xx10^(3)J)/((273+80)K)=-35.127J*K^(-1)`