Question

A thermodynamic system is taken through the cycle $ABCD$ as shown in the figure. Heat rejected by the gas during the cycle is:

• A. 2PV
• B. 4PV
• C. 1/2PV
• D. PV

Answer 1

Answer: (D)

PV

Answer 2

For a cyclic process, the change in internal energy is zero ($\Delta U_{cycle} = 0$). According to the first law of thermodynamics, $Q_{cycle} = W_{cycle}$. The heat rejected is $-Q_{cycle}$.

Work done in each segment: $W_{AB} = P(2V - V) = PV$ (isobaric expansion) $W_{BC} = 0$ (isochoric process) $W_{CD} = 2P(V - 2V) = -2PV$ (isobaric compression) $W_{DA} = 0$ (isochoric process)

Total work done: $W_{cycle} = PV + 0 - 2PV + 0 = -PV$. Heat absorbed: $Q_{cycle} = -PV$. Heat rejected: $-Q_{cycle} = -(-PV) = PV$.