A cubical box with porous walls containing an equal number of `O_(2)` and `H_(2)` molecules is placed in a larger evacuated chamber. The entire system is maintained at constant temperature `T`. The ratio of `v_(rms)` of `O_(2)` molecules to that of the `v_(rms)` of `H_(2)` molecules, found in the chamber outside the box after a short interval is

Options

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

Answer

To solve the problem, we need to find the ratio of the root mean square (rms) velocities of oxygen (O₂) and hydrogen (H₂) molecules after they are allowed to mix in the evacuated chamber.

### Step-by-Step Solution:

1. **Understanding the Formula for v_rms**:
The root mean square velocity (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

2. **Identifying Molar Masses**:
- The molar mass of oxygen (O₂) is 32 g/mol.
- The molar mass of hydrogen (H₂) is 2 g/mol.

3. **Calculating the Ratio of v_rms**:
We need to find the ratio of the rms velocities of O₂ to H₂:
\[ \frac{v_{rms(O_2)}}{v_{rms(H_2)}} = \frac{\sqrt{\frac{3RT}{M_{O_2}}}}{\sqrt{\frac{3RT}{M_{H_2}}}} \]
Since \( R \) and \( T \) are constants and common to both gases, they will cancel out:
\[ \frac{v_{rms(O_2)}}{v_{rms(H_2)}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]

4. **Substituting the Molar Masses**:
Now substituting the molar masses:
\[ \frac{v_{rms(O_2)}}{v_{rms(H_2)}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

5. **Final Result**:
Thus, the ratio of the rms velocity of O₂ to that of H₂ is:
\[ \frac{v_{rms(O_2)}}{v_{rms(H_2)}} = \frac{1}{4} \]

### Conclusion:
The final answer is:
\[ \text{The ratio of } v_{rms} \text{ of } O_2 \text{ to } H_2 \text{ is } \frac{1}{4}. \]