In the given P-V diagram,the initial state of | vedclass

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(C) For all paths,the initial and final states are the same. Therefore,the change in internal energy $\Delta U$ is the same for all paths.According to

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Greater in process $(i)$ than in process $(iii)$

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(C) For all paths,the initial and final states are the same. Therefore,the change in internal energy $\Delta U$ is the same for all paths.According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Since $\Delta U$ is constant,$\Delta Q$ depends on the work done $\Delta W$ by the gas.Work done $\Delta W$ is equal to the area under the $P-V$ curve.For path $(i) abe$,the area under the curve is the largest.For path $(ii) ace$,the area under the curve is the smallest.For path $(iii) ade$,the process is isochoric (constant volume),so $\Delta W = 0$.Since the area under the curve for path $(i)$ is greater than the area for path $(iii)$,the heat absorbed $\Delta Q$ is greater in process $(i)$ than in process $(iii)$.

In the given $P-V$ diagram,the initial state of an ideal gas is represented by point $a$ and the final state by point $e$. The gas moves from state $a$ to state $e$ along paths $(i) abe, (ii) ace, (iii) ade$. The heat absorbed by the gas will be:

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