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(A) For a cyclic process,the change in internal energy $\Delta U = 0$.According to the First Law of Thermodynamics,$\Delta Q = \Delta U + W_{net}$.Sin

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(A) For a cyclic process,the change in internal energy $\Delta U = 0$.According to the First Law of Thermodynamics,$\Delta Q = \Delta U + W_{net}$.Since $\Delta U = 0$,we have $\Delta Q = W_{net} = W_{AB} + W_{BC} + W_{CA}$.Given $\Delta Q = 5\,J$.Process $A 	o B$: This is an isochoric process (volume is constant at $V = 1\,m^3$),so $W_{AB} = 0$.Process $B 	o C$: This is an isobaric process (pressure is constant at $P = 10\,N/m^2$). The work done is $W_{BC} = P(V_C - V_B) = 10(1 - 2) = -10\,J$.Substituting these values into the net work equation:$5 = 0 + (-10) + W_{CA}$$W_{CA} = 5 + 10 = 15\,J$.Wait,let's re-examine the graph. The process $A 	o B$ is vertical,so $P$ is constant at $10\,N/m^2$ and $V$ changes from $1$ to $2$. Thus $W_{AB} = P(V_B - V_A) = 10(2 - 1) = 10\,J$.The process $B 	o C$ is horizontal,so $V$ is constant at $2\,m^3$. Thus $W_{BC} = 0$.The net work $W_{net} = W_{AB} + W_{BC} + W_{CA} = 10 + 0 + W_{CA} = 5$.Therefore,$W_{CA} = 5 - 10 = -5\,J$.

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

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