A thermodynamic cycle xyzx is shown on a V-T | vedclass

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(D) In the given $V-T$ diagram:$1$. Process $x \rightarrow y$: The line passes through the origin,so $V \propto T$. Since $PV = nRT$,this implies $P$

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(D) In the given $V-T$ diagram:$1$. Process $x \rightarrow y$: The line passes through the origin,so $V \propto T$. Since $PV = nRT$,this implies $P$ is constant. Thus,$x \rightarrow y$ is an isobaric process.$2$. Process $y \rightarrow z$: The line is horizontal,meaning $V$ is constant. Thus,$y \rightarrow z$ is an isochoric process.$3$. Process $z \rightarrow x$: The line is vertical,meaning $T$ is constant. Thus,$z \rightarrow x$ is an isothermal process.Comparing these with the $P-V$ diagrams:- $x \rightarrow y$ must be a horizontal line (constant $P$).- $y \rightarrow z$ must be a vertical line (constant $V$).- $z \rightarrow x$ must be a hyperbolic curve (isothermal,$P \propto 1/V$).Looking at the options,Option $D$ correctly represents an isobaric process $(x \rightarrow y)$,an isochoric process $(y \rightarrow z)$,and an isothermal process $(z \rightarrow x)$.

$A$ thermodynamic cycle $xyzx$ is shown on a $V-T$ diagram.
The $P-V$ diagram that best describes this cycle is
(Diagrams are schematic and not to scale)

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$A$ thermodynamic cycle $xyzx$ is shown on a $V-T$ diagram.The $P-V$ diagram that best describes this cycle is(Diagrams are schematic and not to scale)