The entropy change for vaporisation of water to steam at $100^{o}$ C in ${J K}^{- 1} {m o l}^{- 1}$ is :

[Given that the heat of vaporisation is ${40.8 k J m o l}^{- 1}$ .]

109.38

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100.38

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110.38

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120.38

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Solution

The correct option is B 109.38
$Δ S = \frac{{Δ H}_{v}}{T}$

$Δ H_{v} = 40.8 k J m o l^{- 1}$

$T = 373 K$

$∴ Δ S = \frac{40.8 \times 10^{3}}{373} = {109.38 J K}^{- 1} {m o l}^{- 1}$

Hence, option A is correct.

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