Question

P-T diagram of an ideal gas having three different densities $\rho_1, \rho_2, \rho_3$ (in three different cases) is shown in the figure. Which of the following is correct:

• A. $\rho_2 < \rho_3$
• B. $\rho_1 > \rho_2$
• C. $\rho_1 < \rho_2$
• D. $\rho_1 = \rho_2 = \rho_3$

Answer 1

Answer: (B)

$\rho_1 > \rho_2$

Answer 2

Ideal Gas Equation and Density Relationship:

For an ideal gas, the equation is given by:

$PV = nRT$

or,

$P = \frac{nRT}{V}$

where $P$ is the pressure, $T$ is the temperature, $R$ is the gas constant, $n$ is the number of moles, and $V$ is the volume.

We can express $P$ in terms of density $\rho$ by substituting $\rho = \frac{m}{V}$, where $m$ is the mass of the gas:

$P = \frac{\rho RT}{M}$

where $M$ is the molar mass of the gas. Rearranging, we get:

$\rho = \frac{PM}{RT}$

Analyze the PT Graph for Different Densities:

Since $\rho = \frac{PM}{RT}$, for a given temperature $T$, the density $\rho$ of the gas is directly proportional to the pressure $P$:

$\rho \propto P$

Therefore, at the same temperature, a higher pressure indicates a higher density.

Interpretation of the PT Diagram:

In the given PT diagram, we observe that:

$P_1 > P_2 > P_3$ for the same temperature $T$

Therefore, based on the proportional relationship $\rho \propto P$ at constant temperature, we have:

$\rho_1 > \rho_2 > \rho_3$

Conclusion:

The correct statement is: $\rho_1 > \rho_2$ which corresponds to Option (2).