A thermodynamic system undergoes a cycle composed of a series of three processes for which Q1 = +10 kJ, Q2 = +30 kJ, Q3 = -5 kJ. For the first process,DE = +20 kJ, and for the third process, DE = -20 kJ. What is the work in the second process, and the net work output of the cycle?
Q + W =DE
Work done in the first process = 20 - 10 = 10 kJ (i.e., work is done on the system)
Work done in the third process = -20 - (-5) = -15 kJ
For a cyclic process, the overall internal energy change is zero. (i.e., DE = 0)
Therefore, DE in the second process = (0 - (20 - 20)) = 0 kJ
Therefore, work done in the second process = 0 - 30 kJ = -30 kJ.
Total work done during the cycle= 10 + (-15) + (-30) = -35 kJ (i.e., 35 kJ of work is done by the system).
Total work done can also be calculated as follows:
Total Q = Q1 + Q2 + Q3 = 10 + 30 - 5 = 35 kJ
Therefore, net work done during the cycle = 0 - 35 kJ = -35 kJ (the negative sign indicates that work is done by the system).
Last Modified on: 11-Sep-2014
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