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(B) The initial state $i$ and final state $f$ are the same for both processes $A$ and $B$.Since internal energy is a state function,the change in inte

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$\Delta Q_A > \Delta Q_B; \Delta U_A = \Delta U_B$

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(B) The initial state $i$ and final state $f$ are the same for both processes $A$ and $B$.Since internal energy is a state function,the change in internal energy depends only on the initial and final states.Therefore,$\Delta U_A = \Delta U_B$.The work done $W$ in a $P-V$ diagram is equal to the area under the curve.Since the area under curve $A$ is greater than the area under curve $B$,the work done in process $A$ is greater than the work done in process $B$ $(W_A > W_B)$.According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.Since $\Delta U_A = \Delta U_B$ and $W_A > W_B$,it follows that $\Delta Q_A > \Delta Q_B$.

The following figure shows two processes $A$ and $B$ for a gas. If $\Delta Q_A$ and $\Delta Q_B$ are the amounts of heat absorbed by the system in the two cases,and $\Delta U_A$ and $\Delta U_B$ are the changes in internal energies,respectively,then:

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