The entropy change involved in the conversion | vedclass

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(B) The process is $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$.Entropy change is given by $\Delta S = \frac{\Delta H_{vap}}{T}$.Given $\Delta H_{vap} =

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$0.109$

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(B) The process is $H_2O_{(l)} ightleftharpoons H_2O_{(g)}$.Entropy change is given by $\Delta S = \frac{\Delta H_{vap}}{T}$.Given $\Delta H_{vap} = 2.257 \, kJ/g$.Since the molar mass of water is $18 \, g/mol$,the enthalpy of vaporization per mole is $\Delta H_{vap} = 2.257 	imes 18 \, kJ/mol = 40.626 \, kJ/mol$.At $T = 373 \, K$,$\Delta S = \frac{40.626 \, kJ/mol}{373 \, K} \approx 0.109 \, kJ/mol \cdot K$.

The entropy change involved in the conversion of $1 \, \text{mole}$ of liquid water at $373 \, K$ to vapour at the same temperature will be $[\Delta H_{vap} = 2.257 \, kJ/g]$. (in $, kJ/K$)

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