In the reported figure, there is a cyclic process $ABCDA$ on a sample of $1$ mol of a diatomic gas. The temperature of the gas during the process $A \to B$ and $C \to D$ are $T_1$ and $T_2$ $(T_1 > T_2)$ respectively.
Choose the correct option out of the following for work done if processes $BC$ and $DA$ are adiabatic.
Answer: (B) $W_{AD} = W_{BC}$
For an adiabatic process, work done by the gas $= -\Delta U = nC_V(T_{\text{initial}} - T_{\text{final}})$.
$W_{BC} = nC_V(T_1 - T_2)$ and $W_{DA} = nC_V(T_2 - T_1)$.
For the reverse path $A \to D$: $W_{AD} = -W_{DA} = nC_V(T_1 - T_2) = W_{BC}$.
Also $W_{BC} + W_{DA} = 0$, so option (3) is wrong; $W_{AB} = RT_1\ln 3.5$ and $W_{DC} = RT_2 \ln\frac{11}{3}$ are not equal; $W_{AB} > 0 > W_{CD}$, so (4) is wrong.
Solution by Sreeraj P, M.Sc Physics