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

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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,$\Delta Q = \Delta U + \Delta W$.Since $\Delta U = 0$,the net heat supplied is equal to the net work done: $\Delta Q = \Delta W_{net} = 5 \ J$.The net work done in the cycle is $\Delta W_{net} = W_{AB} + W_{BC} + W_{CA}$.Process $AB$ is an isochoric process (constant volume),so $W_{AB} = 0$.Process $BC$ is an isobaric process (constant pressure) at $P = 10 \ N/m^2$ from $V = 1 \ m^3$ to $V = 2 \ m^3$. Thus,$W_{BC} = P \Delta V = 10 	imes (2 - 1) = 10 \ J$.Substituting these values into the net work equation: $5 = 0 + 10 + W_{CA}$.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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