At 400K,the root mean square (rms) speed of a gas X (molecular weight=40) is equal to the most probable speed of gas Y at 60K . The molecular weight of the gas Y is :

Explanation

To solve the problem, we need to find the molecular weight of gas Y given that the root mean square (rms) speed of gas X at 400 K is equal to the most probable speed of gas Y at 60 K.

### Step-by-Step Solution:

1. **Understand the formulas**:
- The formula for the root mean square speed (u_rms) of a gas is given by:
\[ u_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass (molecular weight) of the gas.
- The formula for the most probable speed (u_mp) of a gas is given by:
\[ u_{mp} = \sqrt{\frac{2RT}{M}} \]

2. **Set up the equation**:
- According to the problem, the rms speed of gas X is equal to the most probable speed of gas Y:
\[ u_{rms, X} = u_{mp, Y} \]
- Substituting the formulas:
\[ \sqrt{\frac{3RT_X}{M_X}} = \sqrt{\frac{2RT_Y}{M_Y}} \]

3. **Square both sides**:
- Squaring both sides to eliminate the square roots gives:
\[ \frac{3RT_X}{M_X} = \frac{2RT_Y}{M_Y} \]

4. **Cancel out the gas constant R**:
- Since \( R \) is a constant, we can cancel it out from both sides:
\[ \frac{3T_X}{M_X} = \frac{2T_Y}{M_Y} \]

5. **Rearrange to find M_Y**:
- Rearranging the equation to solve for \( M_Y \):
\[ M_Y = \frac{2T_Y \cdot M_X}{3T_X} \]

6. **Substitute the known values**:
- We know:
- \( M_X = 40 \) (molecular weight of gas X)
- \( T_X = 400 \) K
- \( T_Y = 60 \) K
- Substituting these values into the equation:
\[ M_Y = \frac{2 \cdot 60 \cdot 40}{3 \cdot 400} \]

7. **Calculate**:
- Simplifying the right side:
\[ M_Y = \frac{4800}{1200} = 4 \]

8. **Conclusion**:
- The molecular weight of gas Y is \( 4 \).

### Final Answer:
The molecular weight of gas Y is **4**.