Question

Calculate entropy change for vaporization of 1 mole of liquid water to steam at ${100}^{\circ }C$ if $\Delta H_v$=40.8 kJ $mo{l}^{-1}$.

• A. $\Delta S_v=107.38J{K}^{-1}mo{l}^{-1}$
• B. $\Delta S_v=109.48J{K}^{-1}mo{l}^{-1}$
• C. $\Delta S_v=109.38J{K}^{-1}mo{l}^{-1}$
• D. $\Delta S_v=107.38J{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is C

$\Delta S_v=109.38J{K}^{-1}mo{l}^{-1}$

For entropy change of vaporization can be given as, $\Delta S_v=\Delta \frac{H_v}{T}$
Given values are, $\Delta H_v=40.8\times {10}^{3}Jmo{l}^{-1}$; T=373K
Therefore $\Delta S_v=(40.8\times \frac{{10}^3}{373})$
$\Delta S_v=109.38J{K}^{-1}mo{l}^{-1}$