Question

Two vessels separately contain two ideal gases A and B at the same temperature. The pressure of A is twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weights of A and B is:

• A. 2
• B. \( \frac{3}{4} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{2}{3} \)

Hint:

In ideal gases, the pressure, volume, and temperature are related, and the density is inversely proportional to the molecular weight for gases under the same conditions.

Answer 1

The Correct Option is B

Step 1: Ideal Gas Law.
The ideal gas law is given by: \[ PV = nRT \] where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature. The number of moles \( n \) is related to the density \( \rho \) by: \[ n = \frac{\rho V}{M} \] where \( M \) is the molar mass of the gas. Substituting this into the ideal gas law, we get: \[ P = \frac{\rho RT}{M} \] Thus, the ratio of the densities of gases A and B is related to their molecular weights by: \[ \frac{\rho_A}{\rho_B} = \frac{M_A}{M_B} \] Given that \( \rho_A = 1.5 \rho_B \), we find: \[ \frac{M_A}{M_B} = 1.5 \] Thus, the ratio of the molecular weights of A and B is \( \frac{3}{4} \).
Step 2: Final Answer.
Thus, the ratio of the molecular weights of A and B is \( \frac{3}{4} \).