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 where half of the gas leaks out of a vessel at constant temperature. We will use the formula for root mean square (RMS) speed of gas molecules.

### Step-by-Step Solution:

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

2. **Initial Conditions**:
We know that the initial RMS speed of the gas is \( v_{rms} = 400 \, \text{m/s} \).

3. **Effect of Gas Leakage**:
When half of the gas leaks out at constant temperature, the number of molecules decreases, but the temperature \( T \) and the molar mass \( M \) of the gas remain unchanged.

4. **RMS Speed Dependence**:
The RMS speed depends only on the temperature and the molar mass of the gas. Since neither of these parameters changes 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:
\[ v_{rms} = 400 \, \text{m/s} \]

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

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