When 100 Mol of a Pure Liquid Is Vaporized at a a Constant

question 56

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

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