The enthalpy of vapour is 10 kcal/mole highe | vedclass

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(C) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_{g}RT$.Given:$\Delta

Vedclass · Accepted Answer

$9300$

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(C) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_{g}RT$.Given:$\Delta H = 10 \ kcal/mole = 10000 \ cal/mole$.$T = 350 \ K$.$P = 2 \ atm$.For the phase change $Liquid ightarrow Vapour$,$\Delta n_{g} = 1$ (assuming ideal gas behavior for vapour).Using the ideal gas law $PV = nRT$,we have $\Delta n_{g}RT = P\Delta V = \Delta(PV)$.However,for a phase change,$\Delta H = \Delta U + P\Delta V$.Assuming the volume of liquid is negligible compared to vapour,$P\Delta V \approx P V_{g} = nRT$.Substituting the values:$10000 = \Delta U + (1 	imes 1.987 	imes 350) \approx \Delta U + 700$.Using $R \approx 2 \ cal \ K^{-1} \ mol^{-1}$ as per standard convention for such problems:$10000 = \Delta U + (1 	imes 2 	imes 350)$.$10000 = \Delta U + 700$.$\Delta U = 10000 - 700 = 9300 \ cal$.

The enthalpy of vapour is $10 \ kcal/mole$ higher than liquid at $350 \ K$ and $2 \ atm$. Under given conditions,the difference in internal energy of vapour and liquid will be $...... \ cal$.

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