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In the $V-T$ diagram shown in adjoining figure, what is the relation between $p_{1}$ and $p_{2}$ ?
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Verified Answer
The correct answer is:
$p_{2} < p_{1}$
In an isobaric process, $\mathrm{p}=$ constant Hence, $\mathrm{V} \propto \mathrm{T}$
i.e., $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$
$\therefore$ V-T graph is astraight linewith slope $\propto \frac{1}{\mathrm{p}}$
(slope) $_{2}>$ (slope) $_{1}$ $\therefore \mathrm{p}_{2} < \mathrm{p}_{1}$
i.e., $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$
$\therefore$ V-T graph is astraight linewith slope $\propto \frac{1}{\mathrm{p}}$
(slope) $_{2}>$ (slope) $_{1}$ $\therefore \mathrm{p}_{2} < \mathrm{p}_{1}$
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