The rms velocity of CO gas molecules at `27^(@)C` is approximately `1000 m//s`. For `N_(2)` molecules at 600 K the rms velocity is approximately

Options

A. `2000 m//s`
B. `1414 m//s`
C. `1000 m//s`
D. `1500 m//s`

Explanation

To find the root mean square (rms) velocity of nitrogen gas (N₂) at 600 K, we can use the formula for rms velocity:

### Step 1: Write the formula 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,
- \( M \) is the molar mass of the gas in kg/mol.

### Step 2: Identify known values for CO
We know the rms velocity of CO at 27°C (300 K) is approximately 1000 m/s. The molar mass of CO (carbon monoxide) is:
- Carbon (C) = 12 g/mol,
- Oxygen (O) = 16 g/mol,
- Therefore, Molar mass of CO = 12 + 16 = 28 g/mol = 0.028 kg/mol.

### Step 3: Set up the ratio of rms velocities
Using the relationship of rms velocities for two different gases, we can write:
\[ \frac{v_{rms, CO}}{v_{rms, N_2}} = \sqrt{\frac{T_{CO}}{T_{N_2}} \cdot \frac{M_{N_2}}{M_{CO}}} \]

### Step 4: Substitute known values
- For CO:
- \( v_{rms, CO} = 1000 \, m/s \)
- \( T_{CO} = 300 \, K \)
- \( M_{CO} = 0.028 \, kg/mol \)

- For N₂:
- \( T_{N_2} = 600 \, K \)
- Molar mass of N₂ = 28 g/mol = 0.028 kg/mol.

### Step 5: Calculate the ratio
Substituting the values into the ratio:
\[ \frac{1000}{v_{rms, N_2}} = \sqrt{\frac{300}{600} \cdot \frac{0.028}{0.028}} \]
\[ \frac{1000}{v_{rms, N_2}} = \sqrt{\frac{1}{2}} \]

### Step 6: Solve for \( v_{rms, N_2} \)
Now, we can rearrange the equation to find \( v_{rms, N_2} \):
\[ v_{rms, N_2} = 1000 \cdot \sqrt{2} \]
\[ v_{rms, N_2} \approx 1000 \cdot 1.414 \]
\[ v_{rms, N_2} \approx 1414 \, m/s \]

### Final Answer
The rms velocity of nitrogen molecules at 600 K is approximately **1414 m/s**.

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