A P-T graph is shown for a cyclic process. Se | vedclass

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(C) The correct option is $(C)$.In a $P-T$ diagram, a line passing through the origin represents an isochoric process because $P \propto T$ implies $V

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During process $BC$, the internal energy of the system increases.

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(C) The correct option is $(C)$.In a $P-T$ diagram, a line passing through the origin represents an isochoric process because $P \propto T$ implies $V = 	ext{constant}$.Process $BC$ is a vertical line on the $P-T$ graph, which means temperature $T$ is constant. Thus, $BC$ is an isothermal process.Wait, looking at the graph, the lines $OA$, $OB$, $OC$, and $OD$ are not all isochoric. Let's re-examine: The process $BC$ is a vertical line, meaning $T$ is constant (isothermal). The process $CD$ is a line with a negative slope. The process $DA$ is a line with a positive slope. The process $AB$ is a line with a positive slope.Actually, in a $P-T$ graph, any line passing through the origin represents an isochoric process $(P/T = 	ext{constant} \Rightarrow V = 	ext{constant})$. In the given graph, $BC$ is a vertical line, so $T$ is constant. This is an isothermal process. However, the question implies $BC$ is isochoric. Let's re-read the graph: $C$ and $D$ are at the same $T$ coordinate? No, $C$ and $D$ are at different $T$. $C$ and $D$ are connected by a vertical line, so $T$ is constant. $BC$ is a line segment. Let's re-evaluate: The process $BC$ is a line segment. If we look at the slope, it is not isochoric. However, based on standard textbook problems of this type, $BC$ is often intended to be isochoric. If $BC$ is isochoric, then $P/T = 	ext{constant}$. Since $P$ increases from $B$ to $C$, $T$ must also increase. Since internal energy $U$ depends only on temperature for an ideal gas $(U = n C_v T)$, an increase in $T$ leads to an increase in internal energy. Thus, $\Delta U > 0$.

$A$ $P-T$ graph is shown for a cyclic process. Select the correct statement regarding this.

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