Two closed vessels A,B are at the same temperature T and contain gases which obey Maxwellian distribution of velocities. Vessel A contains `O_(2)`, and B contain mixture of `H_(2) and O_(2)`. If the average speed of the `O_(2)` molecule in vessel A is `V_(1)`, then average speed of `H_(2)` in container B is

Options

A. Zero
B. `(V_(1))/(4)`
C. `(V_(1))/(2)`
D. `4V_(1)`

Answer

To find the average speed of the \( H_2 \) molecules in vessel B, we can use the relationship between the average speeds of different gases based on their molecular masses, as derived from the Maxwellian distribution of velocities.

### Step-by-Step Solution:

1. **Understand the relationship of average speed and molecular mass**:
The average speed of a gas molecule is given by the formula:
\[ V_{\text{avg}} = \sqrt{\frac{8RT}{\pi m}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( m \) is the molecular mass of the gas.

2. **Identify the gases in the vessels**:
- Vessel A contains \( O_2 \) with an average speed \( V_1 \).
- Vessel B contains a mixture of \( H_2 \) and \( O_2 \).

3. **Relate the average speeds of \( O_2 \) and \( H_2 \)**:
Since both vessels are at the same temperature \( T \), we can use the relation:
\[ \frac{V_{O_2}}{V_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]
where \( M_{O_2} \) and \( M_{H_2} \) are the molecular masses of \( O_2 \) and \( H_2 \) respectively.

4. **Substitute the known values**:
The molecular mass of \( O_2 \) is approximately 32 g/mol, and the molecular mass of \( H_2 \) is approximately 2 g/mol. Thus,
\[ \frac{V_{O_2}}{V_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

5. **Express \( V_{H_2} \) in terms of \( V_1 \)**:
From the above relation, we can rearrange to find \( V_{H_2} \):
\[ V_{H_2} = 4 \cdot V_{O_2} = 4 \cdot V_1 \]

6. **Conclusion**:
Therefore, the average speed of the \( H_2 \) molecules in vessel B is:
\[ V_{H_2} = 4V_1 \]

### Final Answer:
The average speed of \( H_2 \) in container B is \( 4V_1 \).

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