When 3.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1 degreesC, 101.8 kJ are absorbed and the P(delta)V for the vaporization process is equal to 8.70 kJ then:
a. delta E=101.8kj and deltaH=93.1kj
b. delta E=110.5kJ and deltaH=101.8kj
c. delta E=93.1kJ and deltaH=101.8kj
d.delta E=101.8kJ and deltaH=110.5kj