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(D) For the process $A \rightarrow B$:Heat absorbed,$q_{AB} = +5\, J$Work done by the gas,$w_{AB} = -8\, J$ (since work is done by the gas,it is negat

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$6\, J$ of work will be done by the surrounding on the gas.

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(D) For the process $A \rightarrow B$:Heat absorbed,$q_{AB} = +5\, J$Work done by the gas,$w_{AB} = -8\, J$ (since work is done by the gas,it is negative in the system's perspective for $\Delta U = q + w$)Change in internal energy,$\Delta U_{AB} = q_{AB} + w_{AB} = 5 + (-8) = -3\, J$For the reverse process $B \rightarrow A$:Since internal energy is a state function,$\Delta U_{BA} = -\Delta U_{AB} = -(-3) = +3\, J$Heat evolved,$q_{BA} = -3\, J$Using the first law of thermodynamics: $\Delta U_{BA} = q_{BA} + w_{BA}$$3 = -3 + w_{BA}$$w_{BA} = +6\, J$Since the work done $w_{BA}$ is positive,it means work is done by the surrounding on the gas.

$A$ gas undergoes a change from state $A$ to state $B$. In this process,the heat absorbed and work done by the gas are $5\, J$ and $8\, J$,respectively. Now,the gas is brought back to $A$ by another process during which $3\, J$ of heat is evolved. In this reverse process of $B$ to $A$:

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