The entropy change,in the conversion of one m | vedclass

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(C) The entropy change for a phase transition is given by $\Delta S = \frac{\Delta H_{vap}}{T_b}$.Given that $1 \, 	ext{mole}$ of $H_2O$ has a mass o

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$108$

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(C) The entropy change for a phase transition is given by $\Delta S = \frac{\Delta H_{vap}}{T_b}$.Given that $1 \, 	ext{mole}$ of $H_2O$ has a mass of $18 \, g$,the total enthalpy of vaporization for $1 \, 	ext{mole}$ is $\Delta H = 2.257 \, kJ/g 	imes 18 \, g = 40.626 \, kJ = 40626 \, J$.The temperature of boiling $T_b = 373 \, K$.Therefore,$\Delta S = \frac{40626 \, J}{373 \, K} \approx 108.9 \, J \, K^{-1}$.

The entropy change,in the conversion of one mole of liquid water at $373 \, K$ to vapour at the same temperature is (Latent heat of vaporization of water,$\Delta H_{vap} = 2.257 \, kJ/g$) ........ $J \, K^{-1}$ (in $.9$)

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