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(A) For a cyclic process, the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics, $\Delta Q = \Delta U + W$, so $\

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$5$

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(A) For a cyclic process, the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics, $\Delta Q = \Delta U + W$, so $\Delta Q = W_{net} = 5 \,J$.The net work done in the cycle is $W_{net} = W_{AB} + W_{BC} + W_{CA} = 5 \,J$.From the $V-P$ graph (note: the axes are $V$ on y-axis and $P$ on x-axis):Process $A ightarrow B$: This is a constant pressure process $(P = 10 \,N/m^2)$ where volume increases from $1 \,m^3$ to $2 \,m^3$. Work $W_{AB} = P \Delta V = 10 	imes (2 - 1) = 10 \,J$.Process $B ightarrow C$: This is a constant volume process $(V = 2 \,m^3)$. Work $W_{BC} = 0 \,J$.Substituting these into the net work equation:$10 \,J + 0 \,J + W_{CA} = 5 \,J$$W_{CA} = 5 - 10 = -5 \,J$.The magnitude of work done during the process $C ightarrow A$ is $|W_{CA}| = |-5 \,J| = 5 \,J$.

An ideal gas is taken through the cycle $A \rightarrow B \rightarrow C \rightarrow A$ as shown in the figure. If the net heat supplied to the gas in the cycle is $5 \,J$, the magnitude of work done during the process $C \rightarrow A$ is (in $\,J$)

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