When one mole of benzene is vaporized at a constant pressure of 1.00 atm and at its boiling point of 353.0 K,30.53 kJ of energy (heat)is absorbed and the volume change is +28.90 L.What is $\Delta$H for this process? (1 L·atm = 101.3 J)
A)27.60 kJ
B)33.46 kJ
C)1.63 kJ
D)30.53 kJ
E)59.43 kJ

Question

Quizplus · Accepted Answer

ΔH = q = 30.53 kJ (heat absorbed)
Δn (change in moles of gas) = 1 (benzene goes from liquid to gas)
ΔV = +28.90 L (positive because volume increases from liquid to gas)
ΔS = ΔSvap = ΔH/T = 30.53 kJ / 353.0 K = 0.0868 kJ/K
ΔG = ΔH - TΔS = (30.53 kJ) - (353.0 K)(0.0868 kJ/K) = 27.60 kJ
Using the equation ΔG = ΔH - TΔS, we can calculate the change in Gibbs free energy for the process, which is the maximum work that can be done by the system. The correct answer is therefore (D) 30.53 kJ.

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

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