(i) For vaporization of diethyl ether,
∆vapHΘ = 26 kJ mol−1 = 26 × 103 J mol−1
T = 35 + 273= 308 K
\(\therefore\) ΔvapSΘ = \(\frac{\Delta_{vap}H^{\ominus}}{T}\)
= \(\frac{26\times10^3\,J\,mol^{-1}}{35K}\)
= 84.4 JK-1 mol-1
(ii) The conversion of vapour into liquid is condensation. The enthalpy of condensation is negative of enthalpy of vaporization.
\(\therefore\) ∆condHΘ = −26 kJ mol−1
= −26 × 103 J mol−1
\(\therefore\) ∆condSΘ = \(\frac{\Delta_{cond}H^{\ominus}}{T}\)
= \(\frac{-26\times10^3\,J\,mol^{-1}}{308K}\)
= -84.4 J K-1 mol-1