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 → B and C → D are T1 and T2 (T1 > T2) respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic. (a) WAB = WDC (b) WAD = WBC (c) WBC + WDA > 0 (d) WAB < WCD

Answer: (b). For an adiabatic process W = nCV(Tinitial - Tfinal). B to C ( T₁ → T₂): WBC = CV(T₁ - T₂). D to A ( T₂ → T₁): WDA = CV(T₂ - T₁), so WBC + WDA = 0; (c) is wrong. A to D is D to A reversed: WAD = -WDA = CV(T₁ - T₂) = WBC. So the answer is (b) WAD = WBC.

In the reported figure, there is a cyclic process ABCDA on a sample… | JEE Main 2021 (27 Jul, Shift 1)

Q52JEE Main 2021 · 27 JUL · SHIFT 1Medium

P-V Diagrams & Cyclic Processes

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 T1 and T2 (T1 > T2) respectively.

Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

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Answer: (b)

For an adiabatic process $W = nC_V(T_{initial} - T_{final})$ .
B to C ( $T_1 \to T_2$ ): $W_{BC} = C_V(T_1 - T_2)$ . D to A ( $T_2 \to T_1$ ): $W_{DA} = C_V(T_2 - T_1)$ , so $W_{BC} + W_{DA} = 0$ ; (c) is wrong.
A to D is D to A reversed: $W_{AD} = -W_{DA} = C_V(T_1 - T_2) = W_{BC}$ .
So the answer is (b) $W_{AD} = W_{BC}$ .

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