Explain a pure liquid is vaporized at a constant pressure

Reference no: EM13527173

When 1.00 mol of a pure liquid is vaporized at a constant pressure of 1.00 atm and at its boiling point of 324.8 K, 29.30 kJ of energy (heat) is absorbed and the volume change is +24.70 L. What is ?E for this process? (1 L-atm = 101.3 J)

A) -2.47 × 103 kJ
B) -26.80 kJ
C) 26.80 kJ
D) 31.80 kJ
E) -31.80 kJ

Reference no: EM13527173

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