A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is [figure]
- 2PV
- 4PV
- 12PV
- PV
Answer
(1) 2PV
Solution
For a cyclic process, = 0, so Q = W.
Work done in the cycle = area enclosed by the cycle on the P–V diagram:
|W| = (2P - P)(3V - V) = 2PV
The cycle A → B → C → D is traversed anticlockwise, so the net work done by the gas is negative:
W = -2PV
Q = -2PV
Heat rejected by the gas = 2PV
Answer: (1)