Question

The entropy of vaporization of a liquid is 58 JK\textsuperscript{--1mol\textsuperscript{--1}. If 100 g of its vapour condenses at its boiling point of 123$^\circ$C, the value of entropy change for the process is (Molar mass of the liquid = 58 g mol\textsuperscript{--1})}

• A. --100 JK\textsuperscript{--1}
• B. 100 JK\textsuperscript{--1}
• C. --123 JK\textsuperscript{--1}
• D. 123 JK\textsuperscript{--1}
• E. 1230 JK\textsuperscript{--1}

Hint:

Phase change problems often contain "distractor" information. Here, the boiling point (123$^\circ$C) is irrelevant because the entropy change per mole is already provided directly!

Answer 1

The Correct Option is A

Concept: Entropy change ($\Delta S$) for a phase transition at constant temperature and pressure is related to the enthalpy change and the number of moles.
• Vaporization vs. Condensation: Entropy of vaporization ($\Delta S_{vap}$) is positive (increasing disorder). Condensation is the exact opposite process, so $\Delta S_{cond} = -\Delta S_{vap}$ per mole.
• Total Entropy Change: $\Delta S_{total} = n \times \Delta S_{process}$, where $n$ is the number of moles.

Step 1: Calculate the number of moles ($n$). Given mass $m = 100$ g and molar mass $M = 58$ g/mol: \[ n = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{100 \text{ g}}{58 \text{ g/mol}} \]

Step 2: Calculate the total entropy change. Since the process is condensation, the entropy change per mole is $-58 \text{ J K}^{-1}\text{mol}^{-1}$. \[ \Delta S = n \times \Delta S_{cond} \] \[ \Delta S = \left( \frac{100}{58} \right) \text{ mol} \times (-58 \text{ J K}^{-1}\text{mol}^{-1}) \] \[ \Delta S = -100 \text{ J K}^{-1} \]