Pressure of 1 g ideal gas X at 300 K is 2 atm. When 2 g of another gas Y is introduced in the same vessel at same temperature, the pressure become 3 atm. The correct relationship between molar mass of X and Y is :

Options

A. `M_(Y)=2 M_(X)`
B. `M_(Y)=4 M_(X)`
C. `M_(X)=4 M_(Y)`
D. None of these

Explanation

To solve the problem, we need to analyze the information given about the two gases, X and Y, and derive a relationship between their molar masses.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- We have 1 g of gas X at a pressure of 2 atm.
- The temperature is 300 K.

2. **Calculate Moles of Gas X**:
- Using the formula for number of moles:
\[ n = \frac{mass}{molar\ mass} \]
- Let the molar mass of gas X be \( M_x \).
- Therefore, the number of moles of gas X is:
\[ n_x = \frac{1 \text{ g}}{M_x} \]

3. **Introduce Gas Y**:
- When 2 g of gas Y is introduced, the total pressure in the vessel becomes 3 atm.
- The additional pressure contributed by gas Y can be calculated as:
\[ P_y = P_{total} - P_x = 3 \text{ atm} - 2 \text{ atm} = 1 \text{ atm} \]

4. **Calculate Moles of Gas Y**:
- Let the molar mass of gas Y be \( M_y \).
- The number of moles of gas Y is:
\[ n_y = \frac{2 \text{ g}}{M_y} \]

5. **Use the Ideal Gas Law**:
- According to the ideal gas law, pressure is directly proportional to the number of moles when volume and temperature are constant:
\[ \frac{P_x}{P_y} = \frac{n_x}{n_y} \]
- Substituting the known values:
\[ \frac{2 \text{ atm}}{1 \text{ atm}} = \frac{\frac{1}{M_x}}{\frac{2}{M_y}} \]

6. **Cross-Multiply**:
- Rearranging gives:
\[ 2 \cdot \frac{2}{M_y} = \frac{1}{M_x} \]
- This simplifies to:
\[ \frac{4}{M_y} = \frac{1}{M_x} \]

7. **Derive the Relationship**:
- Cross-multiplying gives:
\[ 4M_x = M_y \]
- Therefore, we can express the relationship as:
\[ M_y = 4M_x \]

### Conclusion:
The correct relationship between the molar masses of gases X and Y is:
\[ M_y = 4M_x \]