The ratio of average speed of an oxygen molecule to the r.m.s. speed of a `N_(2)` molecule at the same temperature is :

Options

A. `((3pi)/(7))^(1//2)`
B. `((7)/(3pi))^(1//2)`
C. `((3)/(7pi))^(1//2)`
D. `((7pi)/(3))^(1/2)`

Explanation

To find the ratio of the average speed of an oxygen molecule to the root mean square (RMS) speed of a nitrogen molecule at the same temperature, we can follow these steps:

### Step 1: Write down the formulas for average speed and RMS speed.
- The formula for the average speed \( v_{avg} \) of a gas is given by:
\[ v_{avg} = \frac{8RT}{\pi m} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( m \) is the molar mass in kg.

- The formula for the RMS speed \( v_{rms} \) is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( M \) is the molar mass in kg.

### Step 2: Identify the molar masses of oxygen and nitrogen.
- The molar mass of oxygen (O₂) is \( 32 \, \text{g/mol} = 32 \times 10^{-3} \, \text{kg/mol} \).
- The molar mass of nitrogen (N₂) is \( 28 \, \text{g/mol} = 28 \times 10^{-3} \, \text{kg/mol} \).

### Step 3: Substitute the molar masses into the formulas.
- For oxygen:
\[ v_{avg} = \frac{8RT}{\pi (32 \times 10^{-3})} \]

- For nitrogen:
\[ v_{rms} = \sqrt{\frac{3RT}{28 \times 10^{-3}}} \]

### Step 4: Set up the ratio of average speed of oxygen to RMS speed of nitrogen.
- The ratio \( \frac{v_{avg}}{v_{rms}} \) can be expressed as:
\[ \frac{v_{avg}}{v_{rms}} = \frac{\frac{8RT}{\pi (32 \times 10^{-3})}}{\sqrt{\frac{3RT}{28 \times 10^{-3}}}} \]

### Step 5: Simplify the ratio.
- This can be simplified as follows:
\[ \frac{v_{avg}}{v_{rms}} = \frac{8RT}{\pi (32 \times 10^{-3})} \cdot \frac{\sqrt{28 \times 10^{-3}}}{\sqrt{3RT}} \]
- Cancel \( RT \):
\[ = \frac{8}{\pi (32 \times 10^{-3})} \cdot \frac{\sqrt{28 \times 10^{-3}}}{\sqrt{3}} \]
- Further simplifying:
\[ = \frac{8 \cdot \sqrt{28}}{\pi \cdot 32 \cdot \sqrt{3} \cdot 10^{-3/2}} \]

### Step 6: Calculate the final ratio.
- The numerical simplification gives:
\[ = \frac{8 \cdot \sqrt{28}}{\pi \cdot 32 \cdot \sqrt{3}} = \frac{7}{3\sqrt{\pi}} \]

### Final Answer:
The ratio of the average speed of an oxygen molecule to the RMS speed of a nitrogen molecule at the same temperature is:
\[ \frac{7}{3\sqrt{\pi}} \]