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

question 15

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 330.8 K, 31.86 kJ of energy (heat) is absorbed and the volume change is +25.32 L. What is E for this process? (1 L-atm = 101.3 J)

• A) 29.30 kJ
• B) 34.42 kJ
• C) -34.42 kJ
• D) -2.53  10³ kJ
• E) -29.30 kJ

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