The temperature at which the r.m.s. velocity of carbon dioxide becomes the same as that of nitrogen at `21^(@)C` is

Options

A. `462^(@)C`
B. 273 K
C. `189^(@)C`
D. 546 K

Explanation

To solve the problem of finding the temperature at which the root mean square (r.m.s.) velocity of carbon dioxide (CO₂) becomes the same as that of nitrogen (N₂) at 21°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for r.m.s. Velocity**:
The r.m.s. velocity (V_rms) of a gas 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 of the gas.

2. **Set Up the Equation**:
We need to find the temperature \( T_{CO2} \) for carbon dioxide such that:
\[ V_{rms, CO2} = V_{rms, N2} \]
This leads to the equation:
\[ \sqrt{\frac{3RT_{CO2}}{M_{CO2}}} = \sqrt{\frac{3RT_{N2}}{M_{N2}}} \]

3. **Cancel Common Terms**:
Since \( 3R \) is common in both sides, we can simplify the equation:
\[ \sqrt{\frac{T_{CO2}}{M_{CO2}}} = \sqrt{\frac{T_{N2}}{M_{N2}}} \]

4. **Square Both Sides**:
Squaring both sides gives:
\[ \frac{T_{CO2}}{M_{CO2}} = \frac{T_{N2}}{M_{N2}} \]

5. **Rearranging the Equation**:
Rearranging the equation to solve for \( T_{CO2} \):
\[ T_{CO2} = T_{N2} \times \frac{M_{CO2}}{M_{N2}} \]

6. **Substituting Known Values**:
- The temperature of nitrogen \( T_{N2} \) at 21°C is:
\[ T_{N2} = 21 + 273 = 294 \text{ K} \]
- The molar mass of nitrogen \( M_{N2} = 28 \text{ g/mol} \)
- The molar mass of carbon dioxide \( M_{CO2} = 44 \text{ g/mol} \)

Substituting these values into the equation:
\[ T_{CO2} = 294 \times \frac{44}{28} \]

7. **Calculating \( T_{CO2} \)**:
\[ T_{CO2} = 294 \times 1.5714 \approx 462 \text{ K} \]

8. **Convert Kelvin to Celsius**:
To convert Kelvin back to Celsius:
\[ T_{CO2} = 462 - 273 = 189 \text{ °C} \]

### Final Answer:
The temperature at which the r.m.s. velocity of carbon dioxide becomes the same as that of nitrogen at 21°C is **189 °C**.