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

When 1.00 mol of a pure liquid is vaporized at a constant pressure of 1.06 atm and at its boiling point of 313.4 K, 30.00 kJ of energy (heat) is absorbed and the volume change is +25.80 L. What is $\Delta$ H for this process? (1 L-atm = 101.3 J)

A) -32.77 kJ
B) 32.77 kJ
C) 27.23 kJ
D) -27.23 kJ
E) 30.00 kJ