When 1.00 mol of a pure liquid is vaporized at a constant pressure of 1.06 atm and at its boiling point of 332.6 K, 32.16 kJ of energy (heat) is absorbed and the volume change is 27.44 L. What is ΔH for this process? (1 L-atm = 101.3 J)
A) -35.11 kJ
B) 35.11 kJ
C) 29.21 kJ
D) -29.21 kJ
E) 32.16 kJ

Question

Quizplus · Accepted Answer

The Answer of When 1.00 mol of a pure liquid is vaporized at a constant pressure of 1.06 atm and at its boiling point of 332.6 K, 32.16 kJ of energy (heat) is absorbed and the volume change is 27.44 L. What is ΔH for this process? (1 L-atm = 101.3 J)
A) -35.11 kJ
B) 35.11 kJ
C) 29.21 kJ
D) -29.21 kJ
E) 32.16 kJ

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