OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीIf the molar enthalpy change for vaporisation of \( 1\text{ mol} \) of water at \( 1\text{ bar} \) and \( 100^\circ\text{C} \) is \( 41.00\text{ kJ mol}^{-1} \), assuming water vapour behaves as an ideal gas (\( R = 8.314\text{ J K}^{-1}\text{ mol}^{-1} \)), the change in internal energy (\( \Delta U \)) is approximately:
Correct answer · Option 3
\( 37.90\text{ kJ mol}^{-1} \)
Explanation
For \( \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g) \), \( \Delta n_g = 1 \). \( \Delta U = \Delta H - \Delta n_g RT = 41.00 - (1 \times 8.314 \times 373.15 \times 10^{-3}) \approx 41.00 - 3.10 = 37.90\text{ kJ mol}^{-1} \).
