Multiple Choice
When 1.00 mol of a pure liquid is vaporized at a constant pressure of 1.00 atm and at its boiling point of 320.0 K, 28.80 kJ of energy (heat) is absorbed and the volume change is +24.90 L. What is ΔE for this process? (1 L-atm = 101.3 J)
A) 26.28 kJ
B) 31.32 kJ
C) -31.32 kJ
D) -2.49 × 103 kJ
E) -26.28 kJ
Correct Answer:

Verified
Correct Answer:
Verified
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