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(A) $1$. In any thermodynamic process,the internal energy $E_{	ext{int}}$ is a state function. For a complete cycle,the system returns to its initial

Vedclass · Accepted Answer

$\Delta E_{	ext{int}} = 0, Q < 0$

Vedclass · Answer

(A) $1$. In any thermodynamic process,the internal energy $E_{	ext{int}}$ is a state function. For a complete cycle,the system returns to its initial state,so the change in internal energy is $\Delta E_{	ext{int}} = 0$.$2$. According to the first law of thermodynamics,$\Delta Q = \Delta E_{	ext{int}} + \Delta W$. Since $\Delta E_{	ext{int}} = 0$,we have $\Delta Q = \Delta W$.$3$. In a $P-V$ diagram,the work done $\Delta W$ is equal to the area enclosed by the cycle. If the cycle is traversed in an anti-clockwise direction,the work done by the gas is negative $(\Delta W $4$. Looking at the provided $P-V$ diagram,the cycle is traversed in an anti-clockwise direction. Therefore,$\Delta W < 0$,which implies $\Delta Q < 0$.

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

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