The rms speed of the molecules of a gas in a vessel is `400ms^(-1)`. If half of the gas leaks out at constant temperature, the rms speed of the ramaining molecules will be…………..

Answer

To solve the problem, we need to analyze the situation based on the concepts of kinetic theory of gases and the formula for root mean square (RMS) speed.

### Step-by-Step Solution:

1. **Understanding RMS Speed**:
The root mean square speed (RMS speed) of gas molecules is given by the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Given Information**:
- The initial RMS speed of the gas molecules is \( v_{\text{rms}} = 400 \, \text{m/s} \).
- Half of the gas leaks out at constant temperature.

3. **Effect of Gas Leakage**:
When half of the gas leaks out, the number of moles of gas reduces, but the temperature \( T \) and the molar mass \( M \) remain constant because the process occurs at constant temperature.

4. **RMS Speed After Leakage**:
Since the formula for RMS speed depends only on \( R \), \( T \), and \( M \), and none of these parameters change when half of the gas leaks out, the RMS speed of the remaining gas molecules will remain the same.

5. **Conclusion**:
Therefore, the RMS speed of the remaining molecules after half of the gas has leaked out will still be:
\[ v_{\text{rms}} = 400 \, \text{m/s} \]

### Final Answer:
The RMS speed of the remaining molecules will be \( 400 \, \text{m/s} \).

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