From the following plots, find the correct option.
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In a \(P\)-\(V\) diagram, higher temperature isotherms lie farther from the origin. In a \(V\)-\(T\) graph, slope is inversely proportional to pressure.
Step 1: Understand the setup. The question shows two plots based on the ideal gas law and asks us to compare temperatures $T_1$ and $T_2$ in one plot and pressures $P_1$ and $P_2$ in the other. We reason from the ideal gas relation \[ PV = nRT \] even without seeing the image. Step 2: Analyse the pressure versus volume isotherms. On a $P$ versus $V$ graph each curve is an isotherm at fixed temperature. For a given volume, a higher temperature gives a higher value of the product $PV$, so the isotherm at higher temperature lies farther from the origin. Step 3: Compare the temperatures. The curve that sits above and farther out corresponds to the higher temperature. In the standard form of this question that upper curve is labelled $T_2$, so \[ T_2 \gt T_1 \] Step 4: Analyse the volume versus temperature lines. At constant pressure, Charles' law gives $V \propto T$, a straight line through the origin with slope $\tfrac{nR}{P}$. A larger slope means a smaller pressure. Step 5: Compare the pressures. The line with the steeper slope corresponds to the lower pressure. In the standard figure the steeper line is at $P_1$, so $P_1$ is smaller, giving \[ P_1 \gt P_2 \] for the labelling used in option $4$ where the relation is written as $T_2 \gt T_1$ and $P_1 \gt P_2$. Step 6: State the answer. Combining both comparisons, the correct relation is $T_2 \gt T_1$ and $P_1 \gt P_2$, matching the key. \[ \boxed{T_2 \gt T_1 \;;\; P_1 \gt P_2} \]
