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 $$A \rightarrow B$$ and $$C \rightarrow D$$ 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.

1. Identify the Adiabatic Processes

From the problem:

• Paths $$B \to C$$ and $$D \to A$$ are adiabatic transitions.
• The isothermal state $$A \to B$$ is at a higher temperature $$T_1$$, and the isothermal state $$C \to D$$ is at a lower temperature $$T_2$$ ($$T_1 > T_2$$).

2. Calculate the Work Done

The work done during an adiabatic process depends purely on the temperature change of the gas:

$$W_{\text{adiabatic}} = \frac{nR(T_{\text{initial}} - T_{\text{final}})}{\gamma - 1}$$

• For the expansion path $$B \to C$$:
  The temperature drops from $$T_1$$ to $$T_2$$:

  $$W_{BC} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$

• For the reverse path $$A \to D$$:
  In the actual cycle, the gas moves from $$D \to A$$ (heating from $$T_2$$ to $$T_1$$). If we evaluate the work along the path directed from $$A \to D$$, the temperature drops from $$T_1$$ to $$T_2$$:

  $$W_{AD} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$

3. Establish the Equality

Comparing both derived equations shows that the expressions are identical:

$$W_{AD} = W_{BC}$$

Conclusion

The calculation confirms that the work done along both paths matches exactly, making Option B ($$W_{AD} = W_{BC}$$) the correct choice.