For a perfect gas, two pressures $$P_1$$ and $$P_2$$ are shown in figure. The graph shows

Concept:
For an ideal gas:

$$PV=nRT\ \Rightarrow\ \frac{V}{T}=\frac{nR}{P}\ $$

Analysis:
Slope of V-T graph:

slope=VT=nRP$$\text{slope}=\frac{V}{T}=\frac{nR}{P}$$

So, slope $$\propto\frac{1}{P}$$

From graph:
Line $$P_2$$ is steeper ⇒ larger slope ⇒ smaller pressure

$$\frac{nR}{P_2}>\frac{nR}{P_1}$$

$$\Rightarrow P_2$$

Final Answer:

$$P_2\ <\ P_1$$

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