The enthalpy change for the evaporation of a liquid at its boiling point 127^\circ{C} is +40.32 {kJmol}^{-1}. What is the value of internal energy change for the above process at 127 ^\circ{C} ? (R=8.3 {JK}^{-1} {mol}^{-1})

Correct Answer is : +37.0 kJmol−1

KEAM 2021 Physics and Chemistry Solved Paper

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The enthalpy change for the evaporation of a liquid at its boiling point 127°C is +40.32 kJmol−1. What is the value of internal energy change for the above process at 127°C? (R=8.3 JK−1mol−1)

−37.0 kJmol−1
+43.0 kJmol−1
−43.0 kJmol−1
+43.64 kJmol−1
+37.0 kJmol−1

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