Q. Theenthalpy change for the transition of liquid water to steam is $\mathrm{\Delta}H_{vap} = 37.3\ kJ\ mol^{- 1}$ at 373 K. The entropy change for the process is
A) $132.5\ J\ mol^{- 1}K^{- 1}$
B) $100\ J\ mol^{- 1}K^{- 1}$
C) $135.3\ J\ mol^{- 1}K^{- 1}$
D) $75.5\ J\ mol^{- 1}K^{- 1}$
Explanation
(b)
$\mathrm{\Delta}S_{vap} = \frac{\mathrm{\Delta}H_{vap}}{T} = \frac{37.3\ kJ\ mol^{- 1}}{373\ K} = 0.1\ kJ\ mol^{- 1}K^{- 1}$
$= 100\ J\ mol^{- 1}K^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.