The figure shows the P-V plot of an ideal gas | vedclass

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(A) The correct options are $(B)$ and $(D)$.$1$. Analysis of $(A)$: An isothermal process for an ideal gas follows $PV = 	ext{constant}$, which is a

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$(B,D)$

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(A) The correct options are $(B)$ and $(D)$.$1$. Analysis of $(A)$: An isothermal process for an ideal gas follows $PV = 	ext{constant}$, which is a rectangular hyperbola on a $P-V$ diagram, not a circular or elliptical arc. Thus, $(A)$ is incorrect.$2$. Analysis of $(B)$: For the path $B ightarrow C ightarrow D$, the volume decreases ($V$ goes from $3$ to $1$), so the work done by the gas $W = \int P \, dV$ is negative. Also, the temperature $T \propto PV$ decreases from $B$ to $C$ and $C$ to $D$ (as $P$ and $V$ both decrease or stay low), so the change in internal energy $\Delta U$ is negative. From the first law of thermodynamics, $\Delta Q = \Delta U + W$. Since both $\Delta U $3$. Analysis of $(C)$: The work done during $A ightarrow B ightarrow C$ is the area under the curve $ABC$ with the $V$-axis. Since the path $A ightarrow B ightarrow C$ involves expansion followed by compression, the net work is the area under the curve, which is clearly non-zero. Thus, $(C)$ is incorrect.$4$. Analysis of $(D)$: The cycle $ABCDA$ is traversed in a clockwise direction on the $P-V$ diagram. For a clockwise cycle, the net work done by the gas is positive, equal to the area enclosed by the cycle. Thus, $(D)$ is correct.

Column-$I$	Column-$II$
$(A)$ $PV$ vs $V$ (Isothermal expansion)	$(P)$ $W$ > 0
$(B)$ $P$ vs $T$ (Isochoric heating)	$(Q)$ $W$ < 0
$(C)$ $P$ vs $V$ (Isobaric expansion)	$(R)$ $\Delta Q$ > 0
$(D)$ $V$ vs $T$ (Isobaric compression)	$(S)$ $\Delta U$ > 0
	$(T)$ $\Delta U$ < 0

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