At what temperature the molecule of nitrogen will have same rms velocity as the molecule of oxygen at `127^(@)C` ?

Options

A. `457^(@)C`
B. `273^(@)C`
C. `350^(@)C`
D. `77^(@)C`

Explanation

To find the temperature at which nitrogen molecules will have the same root mean square (rms) velocity as oxygen molecules at 127°C, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the Temperature of Oxygen to Kelvin**:
\[ T_O = 127°C + 273 = 400 \, K \]

2. **Identify Molecular Masses**:
- Molecular mass of Oxygen (O2), \( M_O = 32 \, g/mol \)
- Molecular mass of Nitrogen (N2), \( M_N = 28 \, g/mol \)

3. **Write the Equation for RMS Velocity**:
The rms velocity \( v_{rms} \) is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass.

4. **Set the RMS Velocities Equal**:
Since we want the rms velocities of nitrogen and oxygen to be equal:
\[ v_{rms, O} = v_{rms, N} \]
This leads to:
\[ \sqrt{\frac{3RT_O}{M_O}} = \sqrt{\frac{3RT_N}{M_N}} \]

5. **Square Both Sides**:
Squaring both sides eliminates the square root:
\[ \frac{3RT_O}{M_O} = \frac{3RT_N}{M_N} \]

6. **Cancel Out Common Terms**:
The \( 3R \) cancels out:
\[ \frac{T_O}{M_O} = \frac{T_N}{M_N} \]

7. **Rearrange to Solve for \( T_N \)**:
Rearranging gives:
\[ T_N = T_O \cdot \frac{M_N}{M_O} \]

8. **Substitute Known Values**:
Substitute \( T_O = 400 \, K \), \( M_N = 28 \, g/mol \), and \( M_O = 32 \, g/mol \):
\[ T_N = 400 \cdot \frac{28}{32} \]

9. **Calculate \( T_N \)**:
\[ T_N = 400 \cdot 0.875 = 350 \, K \]

10. **Convert \( T_N \) to Celsius**:
Finally, convert the temperature back to Celsius:
\[ T_N (°C) = 350 - 273 = 77°C \]

### Final Answer:
The temperature at which nitrogen molecules will have the same rms velocity as oxygen molecules at 127°C is **77°C**.

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