When 2.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C,67.8 kJ are absorbed and PΔV for the vaporization process is equal to 5.80 kJ then

Question

Quizplus · Accepted Answer

The correct answer is ΔE = 62.0 kJ and ΔH = 67.8 kJ. This is because ΔH represents the enthalpy change, which is the heat absorbed at constant pressure, given as 67.8 kJ. ΔE (change in internal energy) can be calculated using the formula ΔE = ΔH - PΔV, where PΔV = 5.80 kJ. Therefore, ΔE = 67.8 kJ - 5.80 kJ = 62.0 kJ.

When 200 Mol of Benzene Is Vaporized at a Constant Pressure