Question

The entropy of vaporization of a liquid is $58 \, JK^{-1} \,mol^{-1}$. 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 = $58 \,g \,mol^{-1})$

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

Answer 1

Answer: (A)

Given entropy of vaporisation $=58 \,J K ^{-1}\, mol ^{-1}$ i.e., entropy change when 1 mole of the liquid vapourises $=58 \,JK ^{-1}$

$\therefore $ Entropy change when $1$ mole of the liguid condenes $=-58 JK ^{-1}\,\, (\because$ Condensation and vaporisation are opposite process).

Moles of the liquid used in the reaction

$=\frac{100}{58}=1.72\, mol$

$\therefore $ Entropy change when $1.72$ mole (or $100\, g$ ) of the liquid condenses

$=\frac{-58 \times 1.72}{1}$

$=-100\, JK ^{-1}$