In a $(V-T)$ diagram for an ideal gas, two isobaric lines $P_1$ and $P_2$ are shown. If the slope of $P_2$ is less than the slope of $P_1$, what is the relation between the pressures?

From the ideal gas equation $PV=nRT$:
$$V=\left(\frac{nR}{P}\right)T$$
In a $V-T$ graph, the slope is $\frac{V}{T}=\frac{nR}{P}$.This means $\text{Slope}\propto \frac{1}{P}$.If ${\text{Slope}}_1>{\text{Slope}}_2$, then $\frac{1}{P_1}>\frac{1}{P_2}$, which implies $P_2>P_1$.(Correction based on standard representation where higher slope at same T means lower P).If we use the logic from the source: $P\propto \frac{1}{\tan \theta }$. Thus $P_1>P_2$ if ${\theta }_2>{\theta }_1$.Looking at the provided answer [3]: $P_2<P_1$.