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(B) For a cyclic process $ABCA$,the change in internal energy is $\Delta U_{ABCA} = 0$.From the first law of thermodynamics,$Q_{ABCA} = W_{ABCA}$.This

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can only be adiabatic

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(B) For a cyclic process $ABCA$,the change in internal energy is $\Delta U_{ABCA} = 0$.From the first law of thermodynamics,$Q_{ABCA} = W_{ABCA}$.This can be written as: $Q_{A 	o B} + Q_{B 	o C} + Q_{C 	o A} = W_{A 	o B} + W_{B 	o C} + W_{C 	o A} \quad (1)$.Given that the heat interaction along path $B 	o C 	o A$ is equal to the work done along path $A 	o B 	o C$,we have:$Q_{B 	o C} + Q_{C 	o A} = W_{A 	o B} + W_{B 	o C} \quad (2)$.Subtracting equation $(2)$ from equation $(1)$:$Q_{A 	o B} = W_{C 	o A}$.From the $P-V$ diagram,the process $C 	o A$ is an isochoric process (constant volume),so the work done $W_{C 	o A} = 0$.Therefore,$Q_{A 	o B} = 0$.$A$ process in which the heat interaction is zero is an adiabatic process. Thus,the process $A 	o B$ is adiabatic.

$A$ fixed amount of a gas undergoes a thermodynamic process as shown in the figure,such that the heat interaction along path $B \to C \to A$ is equal to the work done by the gas along path $A \to B \to C$. Then the process $A \to B$ is:

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