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In the given pressure $(\mathrm{P})$ - absolute temperature $(\mathrm{T})$ graph of an ideal gas, the relation between volumes $V_1, V_2, V_3$ and $\mathrm{V}_4$ is
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Verified Answer
The correct answer is:
$\mathrm{V}_1 < \mathrm{V}_2 < \mathrm{V}_3 < \mathrm{V}_4$
From the ideal gas equation
$$
\begin{aligned}
& \mathrm{PV}=\mathrm{n} \cdot \mathrm{R} T \\
& \frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{C}}{\mathrm{V}} \\
& \mathrm{V}=\frac{\mathrm{C}}{\mathrm{P} / \mathrm{T}} \propto \frac{1}{\text { slop of the graph }}
\end{aligned}
$$
Hence, $\mathrm{V}_1 < \mathrm{V}_2 < \mathrm{V}_3 < \mathrm{V}_4$
$$
\begin{aligned}
& \mathrm{PV}=\mathrm{n} \cdot \mathrm{R} T \\
& \frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{C}}{\mathrm{V}} \\
& \mathrm{V}=\frac{\mathrm{C}}{\mathrm{P} / \mathrm{T}} \propto \frac{1}{\text { slop of the graph }}
\end{aligned}
$$
Hence, $\mathrm{V}_1 < \mathrm{V}_2 < \mathrm{V}_3 < \mathrm{V}_4$
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