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(A) $1$. Internal energy $(U)$ is a state function. For any complete thermodynamic cycle,the initial and final states are the same,therefore the chang

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$\Delta U = 0; Q < 0$

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(A) $1$. Internal energy $(U)$ is a state function. For any complete thermodynamic cycle,the initial and final states are the same,therefore the change in internal energy is zero: $(\Delta U)_{	ext{cycle}} = 0$.$2$. The work done $(W)$ in a $P-V$ diagram is equal to the area enclosed by the cycle. If the cycle is traversed in an anticlockwise direction,the net work done by the gas is negative $(W $3$. According to the First Law of Thermodynamics,$Q = \Delta U + W$.$4$. Substituting the values,$Q = 0 + W$. Since $W $5$. Therefore,the correct condition is $\Delta U = 0$ and $Q < 0$.

For one complete cycle of a thermodynamic process on a gas as shown in the $P-V$ diagram,which of the following is true?

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