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Chemistry · THERMODYNAMICS
हिन्दीA film of 36.0 g of water evaporates at 298 K. Given that the standard enthalpy of vaporisation of water at 298 K is \( \Delta_{\text{vap}} H^\ominus = 44.01\text{ kJ mol}^{-1} \) and \( R = 8.314\text{ J mol}^{-1}\text{ K}^{-1} \), the change in internal energy (\( \Delta_{\text{vap}} U \)) for the evaporation of this water film is: (Assume steam behaves as an ideal gas)
Correct answer · Option 2
83.07 kJ
Explanation
\( n = 36.0 / 18.0 = 2.0\text{ mol} \). \( \Delta H = 2 \times 44.01 = 88.02\text{ kJ} \). \( \Delta n_g RT = 2 \times (8.314 \times 298 \times 10^{-3}) = 4.955\text{ kJ} \). \( \Delta U = 88.02 - 4.955 = 83.065\text{ kJ} \approx 83.07\text{ kJ} \).
