When 10.0 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C,339 kJ are absorbed and PΔV for the vaporization process is equal to 29.0 kJ,then
A)ΔE = 310.kJ and ΔH = 339.kJ.
B)ΔE = 368.kJ and ΔH = 339.kJ.
C)ΔE = 339.kJ and ΔH = 310.kJ.
D)ΔE = 339.kJ and ΔH = 368.kJ.

Question

Quizplus · Accepted Answer

The relationship between ΔH (enthalpy change) and ΔE (internal energy change) for a process is given by the equation ΔH = ΔE + PΔV. Given that ΔH = 339 kJ and PΔV = 29.0 kJ, we can find ΔE by rearranging the equation to ΔE = ΔH - PΔV = 339 kJ - 29.0 kJ = 310 kJ. Therefore, ΔE = 310 kJ and ΔH remains 339 kJ as given.

When 100 Mol of Benzene Is Vaporized at a Constant Pressure