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

### Step-by-Step Solution:

1. **Identify the formulas:**
- The formula for the root mean square speed (rms) of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
- The formula for the most probable speed (MP) of a gas is given by:
\[ v_{mp} = \sqrt{\frac{2RT}{M}} \]

2. **Set the conditions:**
- We know that the rms speed of gas X at 400 K is equal to the most probable speed of gas Y at 60 K:
\[ v_{rms, X} = v_{mp, Y} \]

3. **Substitute the known values:**
- For gas X (molecular weight = 60, T = 400 K):
\[ v_{rms, X} = \sqrt{\frac{3R \cdot 400}{60}} \]
- For gas Y (T = 60 K, molecular weight = M):
\[ v_{mp, Y} = \sqrt{\frac{2R \cdot 60}{M}} \]

4. **Set the two equations equal:**
\[ \sqrt{\frac{3R \cdot 400}{60}} = \sqrt{\frac{2R \cdot 60}{M}} \]

5. **Square both sides to eliminate the square root:**
\[ \frac{3R \cdot 400}{60} = \frac{2R \cdot 60}{M} \]

6. **Cancel R from both sides:**
\[ \frac{3 \cdot 400}{60} = \frac{2 \cdot 60}{M} \]

7. **Simplify the left side:**
\[ \frac{1200}{60} = 20 \]
So, we have:
\[ 20 = \frac{120}{M} \]

8. **Cross-multiply to solve for M:**
\[ 20M = 120 \]

9. **Divide both sides by 20:**
\[ M = \frac{120}{20} = 6 \]

Thus, the molecular weight of gas Y is **6**.