The p-V diagram for a fixed mass of an ideal | vedclass

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(A) $1$. Process $AB$ is isothermal $(T = 	ext{constant})$. In a $p-T$ diagram, this is represented by a vertical line where $p$ changes while $T$ re

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$(G)$

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(A) $1$. Process $AB$ is isothermal $(T = 	ext{constant})$. In a $p-T$ diagram, this is represented by a vertical line where $p$ changes while $T$ remains constant.$2$. Process $BC$ is isobaric $(p = 	ext{constant})$. In a $p-T$ diagram, this is represented by a horizontal line where $T$ changes while $p$ remains constant.$3$. Process $CA$ is adiabatic $(pV^{\gamma} = 	ext{constant})$. Using the ideal gas law $pV = nRT$, we have $p(T/p)^{\gamma} = 	ext{constant}$, which implies $p^{1-\gamma}T^{\gamma} = 	ext{constant}$, or $p \propto T^{\gamma/(\gamma-1)}$. Since $\gamma > 1$, this curve is non-linear.$4$. Analyzing the cyclic order $A ightarrow B ightarrow C ightarrow A$ in the $p-V$ diagram:- $A ightarrow B$: $p$ decreases, $V$ increases (isothermal).- $B ightarrow C$: $p$ is constant, $V$ decreases (isobaric).- $C ightarrow A$: $p$ increases, $V$ decreases (adiabatic).$5$. Matching this to the $p-T$ graphs, option $(G)$ correctly shows the vertical line $AB$, horizontal line $BC$, and the adiabatic curve $CA$ in the correct cyclic order.

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