The density of a gas a is twice that of gas B. Molecular mass of A is half of the molecular of B. The ratio of the partial pressures of A and B is :

Options

A. `1//4`
B. `1//2`
C. `4//1 `
D. `2//1 `

Answer

To solve the problem, we need to find the ratio of the partial pressures of gases A and B given the relationships between their densities and molecular masses.

### Step-by-Step Solution:

1. **Understanding the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

2. **Expressing Moles in Terms of Mass and Molecular Mass**:
The number of moles \( n \) can be expressed as:
\[ n = \frac{m}{M} \]
where \( m \) is the mass of the gas and \( M \) is its molecular mass.

3. **Substituting Moles into the Ideal Gas Law**:
Substituting \( n \) into the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]
Rearranging this, we find:
\[ P = \frac{mRT}{MV} \]

4. **Relating Mass to Density**:
Since density \( \rho \) is defined as:
\[ \rho = \frac{m}{V} \]
we can express mass \( m \) as:
\[ m = \rho V \]
Substituting this into the pressure equation gives:
\[ P = \frac{\rho V RT}{MV} \]
The volume \( V \) cancels out:
\[ P = \frac{\rho RT}{M} \]

5. **Calculating Partial Pressures for Gases A and B**:
For gas A:
\[ P_A = \frac{\rho_A RT}{M_A} \]
For gas B:
\[ P_B = \frac{\rho_B RT}{M_B} \]

6. **Finding the Ratio of Partial Pressures**:
The ratio of the partial pressures \( \frac{P_A}{P_B} \) is given by:
\[ \frac{P_A}{P_B} = \frac{\rho_A}{M_A} \div \frac{\rho_B}{M_B} = \frac{\rho_A \cdot M_B}{\rho_B \cdot M_A} \]

7. **Substituting Given Values**:
We know:
- The density of gas A is twice that of gas B:
\[ \rho_A = 2\rho_B \]
- The molecular mass of gas A is half that of gas B:
\[ M_A = \frac{1}{2}M_B \]
Substituting these into the ratio:
\[ \frac{P_A}{P_B} = \frac{(2\rho_B) \cdot M_B}{\rho_B \cdot \left(\frac{1}{2}M_B\right)} \]

8. **Simplifying the Expression**:
This simplifies to:
\[ \frac{P_A}{P_B} = \frac{2\rho_B \cdot M_B}{\rho_B \cdot \frac{1}{2}M_B} = \frac{2}{\frac{1}{2}} = 4 \]

### Final Result:
The ratio of the partial pressures of gases A and B is:
\[ \frac{P_A}{P_B} = 4 \]