A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is J. [figure]
Answer
300
Solution
From the graph: A (2 m3, 100 Pa), B (5 m3, 300 Pa), C (5 m3, 100 Pa).
WAB = area under AB = 12(100 + 300)(5 - 2) = 600 J
WBC = 0 (isochoric)
WCA = P V = 100 (2 - 5) = -300 J
Wcycle = 600 + 0 - 300 = 300 J
(Same as the area of triangle ABC, 12 3 200 = 300 J, positive because the cycle is clockwise on the P–V diagram.)
Answer: 300