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

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(A) For a complete cyclic process,the change in internal energy is zero,i.e.,$\Delta U = 0$.According to the first law of thermodynamics,the net heat

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$-5$

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(A) For a complete cyclic process,the change in internal energy is zero,i.e.,$\Delta U = 0$.According to the first law of thermodynamics,the net heat supplied $Q$ is equal to the net work done $W$ in the cycle: $Q = W_{AB} + W_{BC} + W_{CA}$.Given $Q = 5 \ J$.Process $A 	o B$ is an isochoric process (constant pressure $P = 10 \ N/m^2$,volume changes from $1 \ m^3$ to $2 \ m^3$ is incorrect,looking at the graph,$A 	o B$ is constant pressure $P=10$ and volume increases from $1$ to $2$ is not right. Let's re-read the graph: $A$ is at $(10, 1)$,$B$ is at $(10, 2)$,$C$ is at $(?, 2)$. Wait,the graph shows $A$ at $P=10, V=1$. $B$ is at $P=10, V=2$. $C$ is at $P=?, V=2$. Actually,$A 	o B$ is constant pressure $P=10$,$W_{AB} = P(V_B - V_A) = 10(2 - 1) = 10 \ J$.Process $B 	o C$ is constant volume $(V=2)$,so $W_{BC} = 0$.Total work $W = W_{AB} + W_{BC} + W_{CA} = 5 \ J$.$10 + 0 + W_{CA} = 5$.$W_{CA} = 5 - 10 = -5 \ J$.

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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