A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is [figure]
- P0V0
- 2P0V0
- P0V02
- zero
Answer
(4) zero
Solution
The cycle forms two triangles meeting at the crossing point (at 2P0).
Lower triangle (A → crossing → D → A) is traced clockwise positive work = 12 V0 P0 = P0V02
Upper triangle (crossing → B → C → crossing) is traced anticlockwise negative work = - P0V02
Net work = P0V02 - P0V02 = 0
Answer: (4)