If both gases have same temperature so rate of diffusion of `O_2` will be

Options

A. Four times of He
B. Two times of He
C. 0.35 times of He
D. eighth times of He

Answer

To determine the rate of diffusion of \( O_2 \) (oxygen) in comparison to helium gas (\( He \)), we can use Graham's law of diffusion. This law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Understand Graham's Law of Diffusion**:
- Graham's law states that:
\[ \text{Rate of diffusion} \propto \frac{1}{\sqrt{M}} \]
where \( M \) is the molar mass of the gas.

2. **Identify the Molar Masses**:
- The molar mass of helium (\( He \)) is approximately \( 4 \, \text{g/mol} \).
- The molar mass of oxygen (\( O_2 \)) is approximately \( 32 \, \text{g/mol} \).

3. **Set Up the Ratio Using Graham's Law**:
- According to Graham's law, the ratio of the rates of diffusion of \( O_2 \) and \( He \) can be expressed as:
\[ \frac{\text{Rate of diffusion of } O_2}{\text{Rate of diffusion of } He} = \sqrt{\frac{M_{He}}{M_{O_2}}} \]

4. **Substitute the Molar Mass Values**:
- Plugging in the molar masses:
\[ \frac{\text{Rate of diffusion of } O_2}{\text{Rate of diffusion of } He} = \sqrt{\frac{4}{32}} \]

5. **Simplify the Ratio**:
- Simplifying the fraction:
\[ \frac{4}{32} = \frac{1}{8} \]
- Therefore:
\[ \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \approx 0.3536 \]

6. **Conclusion**:
- Thus, the rate of diffusion of \( O_2 \) is approximately \( 0.35 \) times the rate of diffusion of helium.
- The final answer is:
\[ \text{Rate of diffusion of } O_2 \approx 0.35 \times \text{Rate of diffusion of } He \]

### Final Answer:
The rate of diffusion of \( O_2 \) will be \( 0.35 \) times that of helium.