At what temperature, rms velocity of `O_(2)` molecules will be `1//3` of `H_(2)` molecules at `-3^(@)C`

Options

A. `90 K`
B. `1167^(@)C`
C. `-3^(@)C`
D. `207^(@)C`

Explanation

To solve the problem, we need to find the temperature at which the root mean square (rms) velocity of \( O_2 \) molecules is \( \frac{1}{3} \) of the rms velocity of \( H_2 \) molecules at a temperature of \(-3^\circ C\).

### Step-by-Step Solution:

1. **Understand the formula for rms velocity**:
The rms velocity \( v \) of a gas is given by the formula:
\[ v = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas.

2. **Identify the known values**:
- For \( H_2 \):
- Molar mass \( M_{H_2} = 2 \, \text{g/mol} \)
- Temperature \( T_{H_2} = -3^\circ C = 273 - 3 = 270 \, \text{K} \)
- For \( O_2 \):
- Molar mass \( M_{O_2} = 32 \, \text{g/mol} \)
- We need to find the temperature \( T_{O_2} \) such that:
\[ v_{O_2} = \frac{1}{3} v_{H_2} \]

3. **Set up the equation**:
Using the rms velocity formula, we can write:
\[ v_{H_2} = \sqrt{\frac{3R(270)}{2}} \]
\[ v_{O_2} = \sqrt{\frac{3RT_{O_2}}{32}} \]
According to the problem:
\[ \sqrt{\frac{3RT_{O_2}}{32}} = \frac{1}{3} \sqrt{\frac{3R(270)}{2}} \]

4. **Square both sides**:
Squaring both sides to eliminate the square root gives:
\[ \frac{3RT_{O_2}}{32} = \frac{1}{9} \cdot \frac{3R(270)}{2} \]

5. **Simplify the equation**:
Cancel \( 3R \) from both sides:
\[ \frac{T_{O_2}}{32} = \frac{270}{18} \]
\[ \frac{T_{O_2}}{32} = 15 \]

6. **Solve for \( T_{O_2} \)**:
Multiply both sides by 32:
\[ T_{O_2} = 15 \times 32 = 480 \, \text{K} \]

7. **Convert to Celsius**:
To convert Kelvin to Celsius:
\[ T_{O_2} = 480 - 273 = 207^\circ C \]

### Final Answer:
The temperature at which the rms velocity of \( O_2 \) molecules will be \( \frac{1}{3} \) of the rms velocity of \( H_2 \) molecules at \(-3^\circ C\) is \( 207^\circ C \).

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