If the root mean square velocity of hydrogen molecule at a given temperature and pressure is 2km/s" ,the root mean square velocity of oxygen at the same condition in "km/s" is :

Options

A. 1.5
B. 1
C. `2.0`
D. 0.5

Answer

To find the root mean square (RMS) velocity of an oxygen molecule given the RMS velocity of a hydrogen molecule, we can use the formula for RMS velocity:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

where:
- \( v_{rms} \) is the root mean square velocity,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

### Step-by-Step Solution:

1. **Identify the given values:**
- The RMS velocity of hydrogen (\( v_{H_2} \)) is given as \( 2 \, \text{km/s} \).
- The molar mass of hydrogen (\( M_{H_2} \)) is approximately \( 2 \, \text{g/mol} \) or \( 0.002 \, \text{kg/mol} \).
- The molar mass of oxygen (\( M_{O_2} \)) is approximately \( 32 \, \text{g/mol} \) or \( 0.032 \, \text{kg/mol} \).

2. **Set up the relationship between the RMS velocities of hydrogen and oxygen:**
Since both gases are at the same temperature and pressure, we can relate their RMS velocities using their molar masses:
\[ \frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]

3. **Substitute the known values:**
\[ \frac{v_{O_2}}{2 \, \text{km/s}} = \sqrt{\frac{0.002 \, \text{kg/mol}}{0.032 \, \text{kg/mol}}} \]

4. **Calculate the ratio of the molar masses:**
\[ \frac{0.002}{0.032} = \frac{1}{16} \]
Therefore,
\[ \sqrt{\frac{1}{16}} = \frac{1}{4} \]

5. **Substitute this back into the equation:**
\[ \frac{v_{O_2}}{2} = \frac{1}{4} \]

6. **Solve for \( v_{O_2} \):**
\[ v_{O_2} = 2 \times \frac{1}{4} = \frac{2}{4} = 0.5 \, \text{km/s} \]

### Final Answer:
The root mean square velocity of oxygen at the same conditions is \( 0.5 \, \text{km/s} \).