Question

Two vessels separately contain two ideal gases $A$ and $B$ at the same temperature, the pressure of $A$ being 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 weight of $A$ and $B$ is :

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

Answer 1

The Correct Option is C

To solve this problem, we need to apply the ideal gas law and the relations between pressure, volume, density, and molecular weight. The ideal gas law is given by:

PV = nRT

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

We are given that the pressure of gas A is twice that of gas B:

P_A = 2P_B

The density relation is given as the density of gas A is 1.5 times the density of gas B:

\[\rho_A = 1.5 \rho_B\]

Density \(\rho\) can be expressed in terms of molecular weight (M) using the relation:

\[\rho = \frac{PM}{RT}\]

Applying this formula for both gases and using the given conditions:

\[\frac{P_A M_A}{RT} = 1.5 \times \frac{P_B M_B}{RT}\]

Simplifying, since the temperature and gas constant are the same and cancel out:

P_A M_A = 1.5 P_B M_B

Substituting P_A = 2P_B into the equation:

2P_B M_A = 1.5 P_B M_B

Canceling P_B from both sides:

2M_A = 1.5M_B

Solving for the ratio of the molecular weights:

\[\frac{M_A}{M_B} = \frac{1.5}{2} = \frac{3}{4}\]

Therefore, the ratio of the molecular weights of A and B is \[\frac{3}{4}\].