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A fixed amount of a gas undergoes a thermodynamic process as shown such that 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 process $A \to B$ is :-
can only be isothermal
can only be adiabatic
can be isothermal or adiabatic
none of the above
Solution
As process is cyclic $\Rightarrow \Delta \mathrm{U}_{\mathrm{ABCA}}=0$
$\Rightarrow \mathrm{Q}_{\mathrm{ABCA}}=\mathrm{W}_{\mathrm{ABCA}}$
$\Rightarrow \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}+\mathrm{Q}_{\mathrm{C} \rightarrow \mathrm{A}}$
$=\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{W}_{\mathrm{B} \rightarrow \mathrm{C}}+\mathrm{W}_{\mathrm{C} \rightarrow \mathrm{A}}$ $.(1)$
Given $Q_{B \rightarrow C}+Q_{C \rightarrow A}=W_{A \rightarrow B}+W_{B \rightarrow C}$ $(2)$
Subtracting $( 2)$ from $( 1)$
$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{W}_{\mathrm{C} \rightarrow \mathrm{A}}=0$
(as process $\mathrm{C} \rightarrow \mathrm{A}$ is isochoric)
process $A \rightarrow B$ is adiabatic.
