In the P-V diagram,I is the initial state and | vedclass

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(C) According to the first law of thermodynamics,the heat absorbed by the system is given by $\Delta Q = \Delta U + \Delta W$.Since the initial state

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greater in $(i)$ than in $(ii)$

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(C) According to the first law of thermodynamics,the heat absorbed by the system is given by $\Delta Q = \Delta U + \Delta W$.Since the initial state $I$ and final state $F$ are the same for all three processes,the change in internal energy $\Delta U$ is the same for all three paths.Therefore,the heat absorbed $\Delta Q$ depends only on the work done $\Delta W$ by the gas,which is equal to the area under the $P-V$ curve.$1$. For path $(i) IAF$,the area under the curve is positive and large.$2$. For path $(ii) IBF$,the area under the curve is zero because it is a vertical line (isochoric process).$3$. For path $(iii) ICF$,the area under the curve is negative (as the volume decreases from $I$ to $C$ and then increases from $C$ to $F$,but the net area is smaller than path $IAF$).Comparing the work done: $\Delta W_{IAF} > \Delta W_{IBF} > \Delta W_{ICF}$.Since $\Delta Q = \Delta U + \Delta W$,it follows that $\Delta Q_{IAF} > \Delta Q_{IBF} > \Delta Q_{ICF}$.Thus,the heat absorbed is greater in $(i)$ than in $(ii)$.

In the $P-V$ diagram,$I$ is the initial state and $F$ is the final state. The gas goes from $I$ to $F$ by $(i) IAF, (ii) IBF, (iii) ICF$. The heat absorbed by the gas is

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