When 1.00 mol of a pure liquid is vaporized at a constant pressure of 1.00 atm and at its boiling point of 330.8 K, 31.86 kJ of energy (heat) is absorbed and the volume change is +25.32 L. What is $\Delta$E for this process? (1 L-atm = 101.3 J)

Question

Quizplus · Accepted Answer

The change in internal energy (ΔE) can be calculated using the formula ΔE = q - PΔV. Here, q = 31.86 kJ and PΔV = (25.32 L)(101.3 J/L) = 2.56 kJ. Therefore, ΔE = 31.86 kJ - 2.56 kJ = 29.30 kJ.

When 100 Mol of a Pure Liquid Is Vaporized at a a Constant