At what temperature is the rms speed of an atom in a nitrogen gas cylinder equals to the rms speed of an oxygen gas atom at `23^(@)C`. Atomic mass of nitrogen is 14 and that of oxygen is 16.

Explanation

To find the temperature at which the root mean square (rms) speed of nitrogen gas equals the rms speed of oxygen gas at \(23^\circ C\), we can follow these steps:

### Step 1: Convert the temperature of oxygen from Celsius to Kelvin
The temperature in Kelvin is given by:
\[ T_{O_2} = 23 + 273 = 296 \, \text{K} \]

### Step 2: Write the formula for rms speed
The rms speed \(v_{rms}\) 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.

### Step 3: Set up the equation for both gases
For oxygen (\(O_2\)):
\[ v_{rms, O_2} = \sqrt{\frac{3R \cdot T_{O_2}}{M_{O_2}}} \]
For nitrogen (\(N_2\)):
\[ v_{rms, N_2} = \sqrt{\frac{3R \cdot T_{N_2}}{M_{N_2}}} \]

### Step 4: Set the rms speeds equal
Since we want the rms speeds to be equal:
\[ \sqrt{\frac{3R \cdot T_{N_2}}{M_{N_2}}} = \sqrt{\frac{3R \cdot T_{O_2}}{M_{O_2}}} \]

### Step 5: Square both sides to eliminate the square root
\[ \frac{3R \cdot T_{N_2}}{M_{N_2}} = \frac{3R \cdot T_{O_2}}{M_{O_2}} \]

### Step 6: Cancel \(3R\) from both sides
\[ \frac{T_{N_2}}{M_{N_2}} = \frac{T_{O_2}}{M_{O_2}} \]

### Step 7: Rearrange to find \(T_{N_2}\)
\[ T_{N_2} = T_{O_2} \cdot \frac{M_{N_2}}{M_{O_2}} \]

### Step 8: Substitute the known values
- Molar mass of nitrogen \(M_{N_2} = 14 \, \text{g/mol}\)
- Molar mass of oxygen \(M_{O_2} = 16 \, \text{g/mol}\)
- \(T_{O_2} = 296 \, \text{K}\)

Substituting these values gives:
\[ T_{N_2} = 296 \cdot \frac{14}{16} \]

### Step 9: Calculate \(T_{N_2}\)
\[ T_{N_2} = 296 \cdot 0.875 = 259 \, \text{K} \]

### Step 10: Convert \(T_{N_2}\) back to Celsius
\[ T_{N_2} = 259 - 273 = -14 \, \text{°C} \]

### Final Answer
The temperature at which the rms speed of nitrogen gas equals the rms speed of oxygen gas at \(23^\circ C\) is \(-14 \, \text{°C}\).

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