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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$Q = W_{net}$.Given $Q = 5 \ J$,th

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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$Q = W_{net}$.Given $Q = 5 \ J$,therefore $W_{net} = 5 \ J$.The net work done $W_{net}$ is equal to the area enclosed by the cycle in the $P-V$ diagram.Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (10) 	imes (2 - 1) = 5 \ J$.Since the cycle is in the clockwise direction,the work done is positive: $W_{net} = 5 \ J$.The total work $W_{net} = W_{AB} + W_{BC} + W_{CA}$.$W_{AB}$ is the work done from $A$ to $B$ (isochoric process,$V$ increases from $1$ to $2$ at constant $P=10$): $W_{AB} = P \Delta V = 10 	imes (2 - 1) = 10 \ J$.$W_{BC}$ is the work done from $B$ to $C$ (isobaric process,$V$ is constant at $2$): $W_{BC} = 0 \ J$.Thus,$5 = 10 + 0 + W_{CA}$.$W_{CA} = 5 - 10 = -5 \ J$.

In a cyclic process $A \rightarrow B \rightarrow C \rightarrow A$,the heat given to the gas is $5 \ J$. The work done during the process $C \rightarrow A$ is ............ $J$.

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