One mole of water at 100,^{circ}C is converte | vedclass

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(D) The change in entropy for a phase transition at constant temperature is given by $\Delta S = \frac{\Delta H_{vap}}{T}$.Given: Heat of vaporisation

Vedclass · Accepted Answer

$26.06$

Vedclass · Answer

(D) The change in entropy for a phase transition at constant temperature is given by $\Delta S = \frac{\Delta H_{vap}}{T}$.Given: Heat of vaporisation of water $= 540\,cal/g$.Since the molar mass of water $(H_2O)$ is $18\,g/mol$,the molar heat of vaporisation is $\Delta H_{vap} = 540\,cal/g 	imes 18\,g/mol = 9720\,cal/mol$.The temperature $T = 100 + 273 = 373\,K$.Therefore,$\Delta S = \frac{9720\,cal/mol}{373\,K} \approx 26.06\,cal\,mol^{-1}\,K^{-1}$.

One mole of water at $100\,^{\circ}C$ is converted into steam at $100\,^{\circ}C$ at a constant pressure of $1\,atm$. The change in entropy is [heat of vaporisation of water at $100\,^{\circ}C = 540\,cal/g$].

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