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(D) For the cyclic process $A \rightarrow B \rightarrow C \rightarrow A$,the change in internal energy $\Delta U = 0$. By the first law of thermodynam

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Heat absorbed in part $AC$ is given by $2(p_2 V_2 - p_1 V_1) + \frac{1}{2}(p_1 V_2 - p_2 V_1)$

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(D) For the cyclic process $A ightarrow B ightarrow C ightarrow A$,the change in internal energy $\Delta U = 0$. By the first law of thermodynamics,$\Delta Q = \Delta W$,where $\Delta W$ is the area enclosed by the triangle $ABC$ in the $p-V$ diagram.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} (V_2 - V_1)(p_2 - p_1)$.For process $AB$ (isobaric,$p = p_1$):$\Delta Q_{AB} = n C_p \Delta T = n (\frac{5}{2} R) \Delta T = \frac{5}{2} p_1 (V_1 - V_2)$.For process $BC$ (isochoric,$V = V_2$):$\Delta Q_{BC} = n C_V \Delta T = n (\frac{3}{2} R) \Delta T = \frac{3}{2} V_2 (p_2 - p_1)$.Using $\Delta Q_{total} = \Delta Q_{AB} + \Delta Q_{BC} + \Delta Q_{AC} = \Delta W$,we find:$\Delta Q_{AC} = \Delta W - \Delta Q_{AB} - \Delta Q_{BC}$.Substituting the values:$\Delta Q_{AC} = \frac{1}{2}(V_2 - V_1)(p_2 - p_1) - [\frac{5}{2} p_1 (V_1 - V_2) + \frac{3}{2} V_2 (p_2 - p_1)]$.Simplifying this expression yields:$\Delta Q_{AC} = 2(p_2 V_2 - p_1 V_1) + \frac{1}{2}(p_1 V_2 - p_2 V_1)$.

An ideal gas with constant heat capacity $C_V = \frac{3}{2} n R$ is made to carry out a cycle that is depicted by a triangle in the figure given below. The following statement is true about the cycle.

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