When one mole of benzene is vaporized at a constant pressure of 1.00 atm and at its boiling point of 353.0 K,30.56 kJ of energy (heat)is absorbed and the volume change is +28.90 L.What is $\Delta$E for this process? (1 L·atm = 101.3 J)
A)30.56 kJ
B)59.46 kJ
C)33.49 kJ
D)1.66 kJ
E)27.63 kJ

Question

Quizplus · Accepted Answer

The Answer of When one mole of benzene is vaporized at a constant pressure of 1.00 atm and at its boiling point of 353.0 K,30.56 kJ of energy (heat)is absorbed and the volume change is +28.90 L.What is $\Delta$E for this process? (1 L·atm = 101.3 J)
A)30.56 kJ
B)59.46 kJ
C)33.49 kJ
D)1.66 kJ
E)27.63 kJ

When One Mole of Benzene Is Vaporized at a Constant Δ\DeltaΔ

When One Mole of Benzene Is Vaporized at a Constant $\Delta$