P-V diagram of a cyclic process ABCA is as sh | vedclass

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(D) The correct option is $(d)$.$1$. During process $A 	o B$,both pressure $P$ and volume $V$ are decreasing. Since $T \propto PV$,the temperature $T

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(D) The correct option is $(d)$.$1$. During process $A 	o B$,both pressure $P$ and volume $V$ are decreasing. Since $T \propto PV$,the temperature $T$ decreases,which implies the change in internal energy $\Delta U_{A 	o B}$ is negative. Also,since volume is decreasing,the work done $\Delta W_{A 	o B}$ is negative. By the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,so $\Delta Q_{A 	o B}$ is negative.$2$. In process $B 	o C$,the pressure $P$ is constant while the volume $V$ is increasing. Since $PV = nRT$,an increase in volume at constant pressure leads to an increase in temperature,so $\Delta U_{B 	o C}$ is positive.$3$. During process $C 	o A$,the volume $V$ is constant while the pressure $P$ is increasing. This implies an increase in temperature,so $\Delta U_{C 	o A}$ is positive.$4$. During the process $CAB$ (which is the path $C 	o A 	o B$),the net volume of the gas is decreasing (as the final volume at $B$ is less than the initial volume at $C$). Hence,the total work done by the gas is negative.

$P-V$ diagram of a cyclic process $ABCA$ is as shown in figure. Choose the correct statement.

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