An ideal gas goes from state A to state B via | vedclass

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(D) The change in internal energy $(\Delta U)$ is a state function,meaning it is path-independent and depends only on the initial and final states.Sin

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$Q_1 > Q_2 > Q_3$ and $\Delta U_1 = \Delta U_2 = \Delta U_3$

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(D) The change in internal energy $(\Delta U)$ is a state function,meaning it is path-independent and depends only on the initial and final states.Since the initial state $A$ and final state $B$ are the same for all three processes,the change in internal energy is identical for all:$\Delta U_1 = \Delta U_2 = \Delta U_3$The work done $(W)$ by the gas in a $P-V$ diagram is equal to the area under the curve.From the given diagram,the area under curve $1$ is the largest,followed by curve $2$,and then curve $3$ is the smallest.Therefore,$W_1 > W_2 > W_3$.According to the first law of thermodynamics,$Q = \Delta U + W$.Since $\Delta U$ is constant for all processes and $W_1 > W_2 > W_3$,it follows that the heat absorbed must satisfy:$Q_1 > Q_2 > Q_3$

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