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:

Updated On: Nov 4, 2025
  • \( \rho_2 < \rho_3 \)
  • \( \rho_1 > \rho_2 \)
  • \( \rho_1 < \rho_2 \)
  • \( \rho_1 = \rho_2 = \rho_3 \)
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The Correct Option is B

Approach Solution - 1

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).

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Approach Solution -2

Step 1: Understanding the given diagram.
The given graph is a P–T (Pressure–Temperature) diagram of an ideal gas for three different densities: \( \rho_1, \rho_2, \rho_3 \).
Each line represents the relationship between pressure and temperature for a particular density of the same gas.

Step 2: Relation between pressure, temperature, and density for an ideal gas.
For an ideal gas, we know:
\[ PV = nRT \] Also, \( n = \frac{m}{M} \) and \( \rho = \frac{m}{V} \), hence:
\[ P = \frac{\rho RT}{M} \] For a given gas (R and M are constants):
\[ P \propto \rho T \] This means that, on a P–T graph, the slope of each line is directly proportional to the density \( \rho \):
\[ \text{slope} = \frac{P}{T} = \frac{\rho R}{M} \]

Step 3: Interpretation of the graph.
Since \( \rho_1, \rho_2, \rho_3 \) correspond to lines with decreasing slopes, and slope \( \propto \rho \):
\[ \rho_1 > \rho_2 > \rho_3 \] Therefore, the line with the steepest slope corresponds to the highest density.

Step 4: Final Answer.
\[ \boxed{\rho_1 > \rho_2} \]
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