If the number of molecules of `SO_(2)` (atomic weight=64) effusing through an orifice of unit area of cross-section in unit time at `0^(@)C` and 1 atm pressure in n. the number of He molecules (atomic weight=4) effusin under similar conditions at `273^(@)C` and 0.25 atm is:

Options

A. `(n)/sqrt(2)`
B. `nsqrt(2)`
C. 2n
D. `(n)/(2)`

Answer

To solve the problem, we will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Additionally, the rate of effusion is directly proportional to the pressure and temperature of the gas.

### Step-by-Step Solution:

1. **Identify Given Data:**
- For \( SO_2 \):
- Molecular weight (\( M_1 \)) = 64 g/mol
- Temperature (\( T_1 \)) = \( 0^\circ C = 273 K \)
- Pressure (\( P_1 \)) = 1 atm
- Number of molecules effusing = \( N \)

- For \( He \):
- Molecular weight (\( M_2 \)) = 4 g/mol
- Temperature (\( T_2 \)) = \( 273^\circ C = 546 K \)
- Pressure (\( P_2 \)) = 0.25 atm
- Number of molecules effusing = \( X \)

2. **Apply Graham's Law of Effusion:**
According to Graham's law:
\[ \frac{\text{Rate of effusion of } SO_2}{\text{Rate of effusion of } He} = \frac{P_1}{P_2} \times \sqrt{\frac{M_2 \cdot T_1}{M_1 \cdot T_2}} \]

3. **Substitute the Values:**
\[ \frac{N}{X} = \frac{1}{0.25} \times \sqrt{\frac{4 \cdot 273}{64 \cdot 546}} \]

4. **Calculate the Pressure Ratio:**
\[ \frac{1}{0.25} = 4 \]

5. **Calculate the Square Root Term:**
\[ \sqrt{\frac{4 \cdot 273}{64 \cdot 546}} = \sqrt{\frac{1092}{34944}} = \sqrt{\frac{1}{32}} = \frac{1}{\sqrt{32}} = \frac{1}{4\sqrt{2}} \]

6. **Combine the Results:**
\[ \frac{N}{X} = 4 \times \frac{1}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \]

7. **Rearranging for \( X \):**
\[ X = N \cdot \sqrt{2} \]

8. **Final Result:**
The number of helium molecules effusing under the given conditions is:
\[ X = N \cdot \sqrt{2} \]

### Conclusion:
Thus, the number of helium molecules effusing under the given conditions is \( \sqrt{2} N \).