Pressure of `1g` of an ideal gas `A` at `27^(@)C` is found to be 2 bar when `2g` of another ideal gas `B` is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses .

Explanation

To solve the problem, we will use the ideal gas law and the concept of partial pressures. Let's break down the solution step by step.

### Step 1: Understand the given data
- We have 1 gram of gas A at a pressure of 2 bar.
- When 2 grams of gas B are added, the total pressure becomes 3 bar.

### Step 2: Calculate the partial pressure of gas B
The total pressure after adding gas B is 3 bar, and the pressure of gas A alone is 2 bar. Therefore, the partial pressure of gas B (P_B) can be calculated as:
\[ P_B = P_{\text{total}} - P_A = 3 \, \text{bar} - 2 \, \text{bar} = 1 \, \text{bar} \]

### Step 3: Relate the number of moles to the molecular mass
Using the ideal gas equation, we know that:
\[ P = \frac{nRT}{V} \]
Where:
- \( P \) = pressure
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin
- \( V \) = volume

The number of moles (n) can be expressed as:
\[ n = \frac{m}{M} \]
Where:
- \( m \) = mass of the gas
- \( M \) = molecular mass of the gas

### Step 4: Calculate the number of moles for gases A and B
For gas A:
\[ n_A = \frac{1 \, \text{g}}{M_A} \]
For gas B:
\[ n_B = \frac{2 \, \text{g}}{M_B} \]

### Step 5: Use the ideal gas law for both gases
For gas A:
\[ P_A = \frac{n_A RT}{V} \implies 2 = \frac{\left(\frac{1}{M_A}\right) RT}{V} \]
For gas B:
\[ P_B = \frac{n_B RT}{V} \implies 1 = \frac{\left(\frac{2}{M_B}\right) RT}{V} \]

### Step 6: Set up the equations
From the equations above, we can express them as:
1. \( 2 = \frac{RT}{V} \cdot \frac{1}{M_A} \)
2. \( 1 = \frac{RT}{V} \cdot \frac{2}{M_B} \)

### Step 7: Solve for the ratios of molecular masses
From the first equation, we can express \( \frac{RT}{V} \):
\[ \frac{RT}{V} = 2M_A \]
From the second equation:
\[ \frac{RT}{V} = \frac{M_B}{2} \]

### Step 8: Equate the two expressions
Setting the two expressions for \( \frac{RT}{V} \) equal gives:
\[ 2M_A = \frac{M_B}{2} \]

### Step 9: Rearranging the equation
Rearranging this gives:
\[ M_B = 4M_A \]

### Conclusion
The relationship between the molecular masses of gases A and B is:
\[ M_B = 4M_A \]