For a perfect gas,two pressures P_{1} and P_{ | vedclass

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(A) The ideal gas equation is given by $PV = nRT$.Rearranging this for the volume $V$ as a function of temperature $T$,we get $V = (\frac{nR}{P})T$.Th

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$P_{1} > P_{2}$

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(A) The ideal gas equation is given by $PV = nRT$.Rearranging this for the volume $V$ as a function of temperature $T$,we get $V = (\frac{nR}{P})T$.This represents a straight line passing through the origin with a slope $m = \frac{nR}{P}$.From the graph,the slope of the line for $P_{2}$ is greater than the slope of the line for $P_{1}$ (since the line for $P_{2}$ is steeper).Therefore,$\frac{nR}{P_{2}} > \frac{nR}{P_{1}}$.This implies that $\frac{1}{P_{2}} > \frac{1}{P_{1}}$,which further simplifies to $P_{1} > P_{2}$.

For a perfect gas,two pressures $P_{1}$ and $P_{2}$ are shown in the figure. The graph shows:

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