When 1.00 mol of a pure liquid is vaporized at a constant pressure of 1.00 atm and at its boiling point of 320.0 K, 28.80 kJ of energy (heat) is absorbed and the volume change is +24.90 L. What is ΔE for this process? (1 L-atm = 101.3 J)

Question

Quizplus · Accepted Answer

The change in internal energy (ΔE) for a process can be calculated using the formula ΔE = q + w, where q is the heat absorbed and w is the work done on the system. Here, q = 28.80 kJ (since the heat is absorbed, it's positive). The work done on the system due to volume change can be calculated as w = -PΔV (since work done by the system is negative), where P is the pressure and ΔV is the volume change. Given P = 1.00 atm and ΔV = +24.90 L, converting P to the same units as the energy (1 L-atm = 101.3 J), we get w = -1.00 atm * 24.90 L * 101.3 J/L-atm = -2520 J = -2.52 kJ. Therefore, ΔE = 28.80 kJ + (-2.52 kJ) = 26.28 kJ.

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