An ideal gas is taken through the cycle A to | vedclass

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

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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,$\Delta Q = \Delta U + \Delta W$,so $\Delta Q = \Delta W_{net}$.Given $\Delta Q = 5 \ J$,the total work done in the cycle is $\Delta W_{net} = W_{AB} + W_{BC} + W_{CA} = 5 \ J$.In process $A 	o B$,the volume increases from $1 \ m^3$ to $2 \ m^3$ at a constant pressure of $10 \ N/m^2$. Thus,$W_{AB} = P \Delta V = 10 	imes (2 - 1) = 10 \ J$.In process $B 	o C$,the volume is constant at $2 \ m^3$,so $W_{BC} = 0 \ J$.Substituting these into the net work equation: $10 + 0 + W_{CA} = 5$.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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