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(C) $1$. Internal energy of an ideal gas depends only on temperature: $\Delta U = nC_v \Delta T$. Since all processes start at point $A$ (temperature

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$E_{AB} = E_{AC}  0, W_{AC} = 0, W_{AD} < 0$

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(C) $1$. Internal energy of an ideal gas depends only on temperature: $\Delta U = nC_v \Delta T$. Since all processes start at point $A$ (temperature $T_2$) and end at different points $B, C, D$ which lie on different isotherms,we must check the final temperatures. Point $B$ lies on isotherm $T_1$. Points $C$ and $D$ lie on an isotherm higher than $T_1$. Therefore,the final temperatures are $T_B = T_1$ and $T_C = T_D > T_1$. Thus,$\Delta U_{AB} $2$. Work done $W$ is the area under the $P-V$ curve. For $A \rightarrow B$,volume increases,so $W_{AB} > 0$. For $A \rightarrow C$,it is an isochoric process (vertical line),so $W_{AC} = 0$. For $A \rightarrow D$,volume decreases,so $W_{AD} $3$. Comparing these,we get $E_{AB}  0, W_{AC} = 0, W_{AD}  0, W_{AC} = 0, W_{AD} < 0$. Option $C$ is the closest match if we interpret the internal energy comparison correctly.

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