At what temperature, the rate of effusion of `N_2` would be 1.625 times than the rate of `SO_2` at `500^@C` ?

Options

A. `373^@`
B. `620^@`
C. `110^@C`
D. `173^@`

Answer

To solve the problem of finding the temperature at which the rate of effusion of \( N_2 \) is 1.625 times that of \( SO_2 \) at \( 500^\circ C \), we can 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 and directly proportional to its temperature.

### Step-by-Step Solution:

1. **Convert the temperature of \( SO_2 \) from Celsius to Kelvin**:
\[ T_{SO_2} = 500^\circ C + 273 = 773 \, K \]

2. **Write down Graham's law of effusion**:
\[ \frac{R_{N_2}}{R_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{N_2}}} \cdot \frac{T_{N_2}}{T_{SO_2}} \]
Where:
- \( R \) is the rate of effusion
- \( M \) is the molar mass
- \( T \) is the temperature in Kelvin

3. **Substitute the known values**:
- Molar mass of \( SO_2 \) = 64 g/mol
- Molar mass of \( N_2 \) = 28 g/mol
- We know \( R_{N_2} = 1.625 \cdot R_{SO_2} \)

Substituting these into the equation gives:
\[ \frac{1.625 \cdot R_{SO_2}}{R_{SO_2}} = \sqrt{\frac{64}{28}} \cdot \frac{T_{N_2}}{773} \]

4. **Simplify the equation**:
\[ 1.625 = \sqrt{\frac{64}{28}} \cdot \frac{T_{N_2}}{773} \]

5. **Calculate the square root**:
\[ \sqrt{\frac{64}{28}} = \sqrt{2.2857} \approx 1.511 \]

6. **Rearrange the equation to solve for \( T_{N_2} \)**:
\[ T_{N_2} = 1.625 \cdot 773 \cdot \frac{28}{64} \cdot \frac{1}{1.511} \]

7. **Calculate \( T_{N_2} \)**:
\[ T_{N_2} = 1.625 \cdot 773 \cdot \frac{28}{64} \cdot \frac{1}{1.511} \approx 620 \, K \]

8. **Convert \( T_{N_2} \) back to Celsius**:
\[ T_{N_2} = 620 - 273 = 347^\circ C \]

### Final Answer:
The temperature at which the rate of effusion of \( N_2 \) is 1.625 times that of \( SO_2 \) is approximately \( 347^\circ C \).