If the oxygen `(O_2)` has root mean square velocity of C `ms^(-1)` then root mean square velocity of the hydrogen `(H_2)` will be

Options

A. C `ms^(-1)`
B. `1/Cms^(-1)`
C. `4Cms^(-1)`
D. `C/4ms^(-1)`

Explanation

To find the root mean square (RMS) velocity of hydrogen (H₂) given that the RMS velocity of oxygen (O₂) is C m/s, we can use the relationship between RMS velocity and molecular mass. The formula for RMS velocity (v_rms) is given by:

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

Where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step-by-step Solution:

1. **Identify the RMS velocity for O₂**:
Given that the RMS velocity of oxygen (O₂) is \( v_{rms(O_2)} = C \) m/s.

2. **Determine the molar mass of O₂**:
The molar mass of oxygen (O₂) is approximately 32 g/mol.

3. **Determine the molar mass of H₂**:
The molar mass of hydrogen (H₂) is approximately 2 g/mol.

4. **Set up the ratio of RMS velocities**:
Since the temperature (T) and the gas constant (R) remain constant for both gases, we can set up the ratio of the RMS velocities:
\[ \frac{v_{rms(H_2)}}{v_{rms(O_2)}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} \]

5. **Substituting the values**:
Substitute the known values of the molar masses:
\[ \frac{v_{rms(H_2)}}{C} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \]

6. **Solve for \( v_{rms(H_2)} \)**:
Rearranging the equation gives:
\[ v_{rms(H_2)} = 4C \]

### Final Answer:
The root mean square velocity of hydrogen (H₂) will be \( 4C \) m/s.

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