1. Use the ideal gas equation in density form

For an ideal gas, $PV = nRT$ Also, density is $\rho = \frac{m}{V}$ Using molar mass $M$, we have $n = \frac{m}{M}$ So, $PV = \frac{m}{M}RT$ $P = \frac{m}{V}\frac{R}{M}T$ Hence, $P = \rho \frac{R}{M}T$

Therefore, $\frac{P}{T} = \rho \frac{R}{M}$ For the same gas, $R$ and $M$ are constants, so $\rho \propto \frac{P}{T}$

2. Interpret the $P$-$T$ graph

In a $P$ vs $T$ graph, the slope is $\text{slope} = \frac{P}{T}$ Thus, for the same ideal gas, $\rho \propto \text{slope}$ So, the line with greater slope corresponds to greater density.

3. Compare the three lines in the figure

From the given graph, the line corresponding to $\rho_1$ is steeper than the line corresponding to $\rho_2$. Hence, $\rho_1 > \rho_2$

Also, the three slopes are not equal, so $\rho_1 = \rho_2 = \rho_3$ is false.

4. Check options

• A: $\rho_1 > \rho_2$ ✅
• B: $\rho_2 < \rho_3$ — not the correct relation from the graph
• C: $\rho_1 = \rho_2 = \rho_3$ ❌
• D: $\rho_1 < \rho_2$ ❌

5. Final answer

The correct option is: $\boxed{A}$