At what temperature, the rate of effusion of `N_(2)` would be 1.625 times that of `SO_(2)` at `50^(@)C` ?

Options

A. 110K
B. 173K
C. 373 K
D. 273K

Answer

To solve the problem of determining the temperature at which the rate of effusion of \( N_2 \) is 1.625 times that of \( SO_2 \) at \( 50^\circ C \), we can use Graham's law of effusion. Here's a step-by-step solution:

### Step 1: Understand Graham's Law of Effusion
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:
\[ \frac{\text{Rate of } N_2}{\text{Rate of } SO_2} = \sqrt{\frac{M_{SO_2}}{M_{N_2}}} \]
where \( M \) represents the molar mass of the gases.

### Step 2: Set Up the Equation
According to the problem, the rate of effusion of \( N_2 \) is 1.625 times that of \( SO_2 \):
\[ \frac{\text{Rate of } N_2}{\text{Rate of } SO_2} = 1.625 \]

### Step 3: Relate the Rates to Temperature
The rate of effusion is also related to temperature. We can express the rates in terms of temperature and molar mass:
\[ \frac{\text{Rate of } N_2}{\text{Rate of } SO_2} = \frac{\sqrt{\frac{T_{N_2}}{M_{N_2}}}}{\sqrt{\frac{T_{SO_2}}{M_{SO_2}}}} \]
Squaring both sides gives:
\[ \left(\frac{\text{Rate of } N_2}{\text{Rate of } SO_2}\right)^2 = \frac{T_{N_2}}{M_{N_2}} \cdot \frac{M_{SO_2}}{T_{SO_2}} \]

### Step 4: Substitute Known Values
We know:
- Molar mass of \( N_2 \) (approximately 28 g/mol)
- Molar mass of \( SO_2 \) (approximately 64 g/mol)
- Temperature of \( SO_2 \) is \( 50^\circ C = 323 K \)

Substituting these values into the equation:
\[ (1.625)^2 = \frac{T_{N_2}}{28} \cdot \frac{64}{323} \]

### Step 5: Solve for \( T_{N_2} \)
Calculating \( (1.625)^2 \):
\[ (1.625)^2 = 2.640625 \]
Now substituting this into the equation:
\[ 2.640625 = \frac{T_{N_2} \cdot 64}{28 \cdot 323} \]
Rearranging to solve for \( T_{N_2} \):
\[ T_{N_2} = \frac{2.640625 \cdot 28 \cdot 323}{64} \]

### Step 6: Calculate \( T_{N_2} \)
Calculating the right-hand side:
\[ T_{N_2} = \frac{2.640625 \cdot 28 \cdot 323}{64} \approx 373.15 K \]

### Final Answer
The temperature at which the rate of effusion of \( N_2 \) is 1.625 times that of \( SO_2 \) is approximately **373.15 K**.

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