Question

Entropy of vaporisation of water at $100^{\circ} C$, if molar heat of vaporisation is $9710 \,cal \,mol ^{-1}$, will be:

• A. $20 \,cal \,mol ^{-1} \,K^{-1}$
• B. $26 \,cal\, mol ^{-1} \,K ^{-1}$
• C. $24\, cal \,mol ^{-1}\, K^{-1}$
• D. $28\, cal\, mol ^{-1} \,K ^{-1}$

Answer 1

Answer: (B)

Use the following formula to solve the problem
$ \Delta S=\frac{\Delta H}{T} $
where $ \Delta S= $
entropy $= ?$
$ \Delta H= $
enthalpy $ =9710\,cal\,mo{{l}^{-1}} $
$T =$ temperature $ ={{100}^{\circ}}C$
$=100+273=373\,K $
$ \therefore \Delta S=\frac{9710}{373} $
$ =26.032\,cal\,mo{{l}^{-1}}{{K}^{-1}} $
$ \approx 26\,cal\,mo{{l}^{-1}}{{K}^{-1}} $