Question

The entropy of vaporisation of a liquid is $58J{K}^{-1}mo{l}^{-1}$. If $100g$ of its vapor condenses at its boiling point of ${123}^{\circ }C$, the value of entropy change for the process is
(Molar mass of the liquid $=58{\text{g mol}}^{-1}$)

• A. $-100J{K}^{-1}$
• B. $100J{K}^{-1}$
• C. $-123J{K}^{-1}$
• D. $123J{K}^{-1}$
• E. $1230J{K}^{-1}$

Answer 1

The correct option is A $-100J{K}^{-1}$

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

$\therefore$ Entropy change when $1mole$ of the liquid condenes $=-58J{K}^{-1}$

$(\because$ Condensation and vaporisation are opposite process).

Moles of the liquid used in the change $=\frac{100}{58}=1.724mol$

$\therefore$ Entropy change when $1.724$ mole (or $100g$) of the liquid condenses

$\Delta S=\frac{-58\times 1.724}{1}$

$\Delta S=-100J{K}^{-1}$.