WBJEE · Chemistry · Thermodynamics (C)
The enthalpy of vaporisation of a certain liquid at its boiling point of \(35^{\circ} \mathrm{C}\) is \(24.64 \mathrm{kJ} \mathrm{mol}^{-1}\). The value of change in entropy for the process is
- A \(704 \mathrm{K}\mathrm{mol}^{-1} \mathrm{mol}^{-1}\)
- B \(80 \mathrm{K}\mathrm{mol}y^{-1} \mathrm{mol}^{-1}\)
- C \(24.64 \mathrm{K}\mathrm{mol}^{-1} \mathrm{mol}^{-1}\)
- D \(7.04 \mathrm{K}\mathrm{mol}^{-1} \mathrm{mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(80 \mathrm{K}\mathrm{mol}y^{-1} \mathrm{mol}^{-1}\)
Step-by-step Solution
Detailed explanation
Entropy of vaporisation ( \(\Delta\) S) \[ =\frac{\text { enthalpy of vaporisation }(\Delta H)}{\text { boiling point (in K)} } \] Given, \(\Delta H=24.64 \mathrm{kJ} \mathrm{mol}^{-1}\) and \(\text { boiling point }=35+273=308 \mathrm{k}\) \(\therefore\)…
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