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

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$-15$

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(C) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta W_{net}$.Given $\Delta Q = 5\,J$,therefore the net work done in the cycle $\Delta W_{net} = 5\,J$.The total work done is the sum of work done in each process: $\Delta W_{net} = \Delta W_{AB} + \Delta W_{BC} + \Delta W_{CA}$.Process $A 	o B$: This is an isochoric process (volume is constant at $V = 1\,m^3$),so $\Delta W_{AB} = 0$.Process $B 	o C$: This is an isobaric process at $p = 10\,N/m^2$. The volume changes from $V_B = 2\,m^3$ to $V_C = 1\,m^3$. Thus,$\Delta W_{BC} = p(V_C - V_B) = 10(1 - 2) = -10\,J$.Now,$\Delta W_{net} = \Delta W_{AB} + \Delta W_{BC} + \Delta W_{CA} \implies 5 = 0 + (-10) + \Delta W_{CA}$.Solving for $\Delta W_{CA}$,we get $\Delta W_{CA} = 5 + 10 = 15\,J$.

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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