The boiling point of water at 1 atm is 100^{ | vedclass

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(A) The process of vaporization is non-spontaneous below the boiling point at $1 \ atm$ pressure. For any process,the spontaneity is determined by the

Vedclass · Accepted Answer

$\Delta G_{	ext{vap}}^{\circ} > 0$

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(A) The process of vaporization is non-spontaneous below the boiling point at $1 \ atm$ pressure. For any process,the spontaneity is determined by the Gibbs free energy change,$\Delta G = \Delta H - T \Delta S$. At the boiling point $(T_{b} = 373 \ K)$,the system is at equilibrium,so $\Delta G = 0$,which implies $\Delta S = \Delta H / T_{b}$. For vaporization,$\Delta H > 0$ and $\Delta S > 0$. At $T = 75^{\circ} C = 348 \ K$,since $T Therefore,$\Delta G = \Delta H - T \Delta S > 0$. Since $\Delta G > 0$,the process is non-spontaneous at $75^{\circ} C$.

The boiling point of water at $1 \ atm$ is $100^{\circ} C$. Which among the following is true for vaporisation of water at $75^{\circ} C$?

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