At what temperature will the `rms` velocity of `SO_(2)` be the same as that of `O_(2)` at `303 K`?

Options

A. 273 K
B. 606 K
C. 303 K
D. 403 K

Answer

To find the temperature at which the root mean square (RMS) velocity of sulfur dioxide (SO₂) is the same as that of oxygen (O₂) at 303 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for RMS Velocity**:
The RMS velocity (V_RMS) of a gas is given by the formula:
\[ V_{RMS} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) = gas constant
- \( T \) = temperature in Kelvin
- \( M \) = molar mass of the gas in kg/mol

2. **Identify the Molar Masses**:
- Molar mass of SO₂ (sulfur dioxide) = 64 g/mol = 0.064 kg/mol
- Molar mass of O₂ (oxygen) = 32 g/mol = 0.032 kg/mol

3. **Set Up the Equation**:
According to the problem, we need to find the temperature \( T_1 \) at which the RMS velocity of SO₂ equals the RMS velocity of O₂ at 303 K. Thus, we set:
\[ V_{RMS, SO_2} = V_{RMS, O_2} \]
This leads to:
\[ \sqrt{\frac{3RT_1}{M_{SO_2}}} = \sqrt{\frac{3RT_2}{M_{O_2}}} \]

4. **Square Both Sides**:
Squaring both sides eliminates the square root:
\[ \frac{3RT_1}{M_{SO_2}} = \frac{3RT_2}{M_{O_2}} \]

5. **Cancel Out Common Terms**:
The \( 3R \) terms cancel out:
\[ \frac{T_1}{M_{SO_2}} = \frac{T_2}{M_{O_2}} \]

6. **Rearrange the Equation**:
Rearranging gives us:
\[ T_1 = T_2 \cdot \frac{M_{SO_2}}{M_{O_2}} \]

7. **Substitute Known Values**:
Substitute \( T_2 = 303 \, K \), \( M_{SO_2} = 0.064 \, kg/mol \), and \( M_{O_2} = 0.032 \, kg/mol \):
\[ T_1 = 303 \cdot \frac{0.064}{0.032} \]

8. **Calculate the Ratio**:
The ratio \( \frac{0.064}{0.032} = 2 \):
\[ T_1 = 303 \cdot 2 \]

9. **Final Calculation**:
\[ T_1 = 606 \, K \]

### Conclusion:
The temperature at which the RMS velocity of SO₂ will be the same as that of O₂ at 303 K is **606 K**.