An ideal gas undergoes a thermodynamic cycle | vedclass

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(D) In the given $V-T$ graph:$AB$ is a straight line passing through the origin,so $V \propto T$,which implies $P$ is constant. Thus,$AB$ is an isobar

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(D) In the given $V-T$ graph:$AB$ is a straight line passing through the origin,so $V \propto T$,which implies $P$ is constant. Thus,$AB$ is an isobaric process.$BC$ is a vertical line,meaning $T$ is constant. Thus,$BC$ is an isothermal process.$CA$ is a vertical line,meaning $V$ is constant. Thus,$CA$ is an isochoric process.Now,let's analyze the $P-T$ graphs:For $AB$ (isobaric),$P$ must be constant. In a $P-T$ graph,this is a horizontal line.For $BC$ (isothermal),$T$ must be constant. In a $P-T$ graph,this is a vertical line.For $CA$ (isochoric),$V$ is constant. From the ideal gas equation $PV = nRT$,we have $P = (nR/V)T$. Since $V$ is constant,$P \propto T$,which is a straight line passing through the origin.Comparing these with the options,option $D$ correctly represents the cycle $A \rightarrow B \rightarrow C \rightarrow A$ in the $P-T$ plane.

An ideal gas undergoes a thermodynamic cycle as shown in the figure. Which of the following graphs represents the same cycle?

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