At what temperature would `N_(2)` molecules have same average speed as `CO` molecules at 200 K.

Options

A. `-73^(@)C`
B. `200^(@)C`
C. `700^(@)C`
D. none

Answer

To find the temperature at which \( N_2 \) molecules have the same average speed as \( CO \) molecules at 200 K, we can use the formula for average speed of gas molecules:

\[ \mu = \sqrt{\frac{8RT}{\pi m}} \]

where:
- \( \mu \) is the average speed,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin,
- \( m \) is the molar mass of the gas.

### Step-by-Step Solution:

1. **Identify the Molar Masses:**
- For \( CO \): The molar mass \( m_{CO} = 12 + 16 = 28 \, \text{g/mol} \).
- For \( N_2 \): The molar mass \( m_{N_2} = 14 + 14 = 28 \, \text{g/mol} \).

2. **Set the Average Speeds Equal:**
Since we want the average speeds of \( N_2 \) and \( CO \) to be equal, we can write:
\[ \sqrt{\frac{8R T_{N_2}}{\pi m_{N_2}}} = \sqrt{\frac{8R T_{CO}}{\pi m_{CO}}} \]

3. **Square Both Sides:**
Squaring both sides eliminates the square root:
\[ \frac{8R T_{N_2}}{\pi m_{N_2}} = \frac{8R T_{CO}}{\pi m_{CO}} \]

4. **Cancel Common Terms:**
The \( 8R \) and \( \pi \) terms can be canceled from both sides:
\[ \frac{T_{N_2}}{m_{N_2}} = \frac{T_{CO}}{m_{CO}} \]

5. **Substitute Known Values:**
Substitute \( T_{CO} = 200 \, K \) and \( m_{CO} = 28 \, g/mol \):
\[ \frac{T_{N_2}}{28} = \frac{200}{28} \]

6. **Solve for \( T_{N_2} \):**
Multiply both sides by \( 28 \):
\[ T_{N_2} = 200 \, K \]

7. **Convert to Celsius:**
To convert Kelvin to Celsius, use the formula:
\[ T_{C} = T_{K} - 273 \]
Thus,
\[ T_{N_2} = 200 - 273 = -73 \, °C \]

### Final Answer:
The temperature at which \( N_2 \) molecules have the same average speed as \( CO \) molecules at 200 K is \( -73 \, °C \).

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