The temperature at which the r.m.s. velocity of oxygen molecules equal that of nitrogen molecules at `100^(@)C` is nearly.

Options

A. `426.3 K`
B. `456.3 K`
C. `436.3 K`
D. `446.3 K`

Explanation

To find the temperature at which the root mean square (r.m.s.) velocity of oxygen molecules equals that of nitrogen molecules at 100°C, we can follow these steps:

### Step 1: Understand the formula for r.m.s. velocity
The r.m.s. velocity (v_rms) is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Set up the equation for both gases
Since we need the r.m.s. velocity of oxygen (O₂) to equal that of nitrogen (N₂), we can write:
\[ v_{rms, O_2} = v_{rms, N_2} \]

Substituting the formula for each gas:
\[ \sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{N_2}}{M_{N_2}}} \]

### Step 3: Simplify the equation
We can square both sides to eliminate the square root:
\[ \frac{3RT_{O_2}}{M_{O_2}} = \frac{3RT_{N_2}}{M_{N_2}} \]

The \( 3R \) cancels out from both sides:
\[ \frac{T_{O_2}}{M_{O_2}} = \frac{T_{N_2}}{M_{N_2}} \]

### Step 4: Substitute known values
We know:
- Molar mass of oxygen, \( M_{O_2} = 32 \, \text{g/mol} \)
- Molar mass of nitrogen, \( M_{N_2} = 28 \, \text{g/mol} \)
- Temperature of nitrogen, \( T_{N_2} = 100^\circ C = 373 \, \text{K} \)

Now substituting these values into the equation:
\[ \frac{T_{O_2}}{32} = \frac{373}{28} \]

### Step 5: Solve for \( T_{O_2} \)
Cross-multiplying gives:
\[ T_{O_2} = \frac{373 \times 32}{28} \]

Calculating the right side:
\[ T_{O_2} = \frac{11936}{28} \approx 426.3 \, \text{K} \]

### Final Answer
The temperature at which the r.m.s. velocity of oxygen molecules equals that of nitrogen molecules at 100°C is approximately **426.3 K**.

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