Question

Calculate the entropy change, involved in the conversion of one mole of liquid water at $373K$ to vapour at same temperature is ( Latent heat of vaporization $=2.275KJ/g$ ).

• A. $105.9J{k}^{-1}$
• B. $107.9J{k}^{-1}$
• C. $108.9J{k}^{-1}$
• D. $109.9J{k}^{-1}$

Answer 1

The correct option is C $108.9J{k}^{-1}$

Entropy change $=\frac{\Delta H_{vap}}{T_b}\times m$

$m=18$ $g(1$ $mole$ of $H_{2}O=18$ $g$ of $H_{2}O)$

$\therefore$ Entropy change $=\frac{2.257\times {10}^3}{373}\times 18$

$⟹$ Entropy change $=108.9$ $J{K}^{-1}$