The rms speed of hydrogen molecule at a certain temperature is v. If the temperature is doubled and hydrogen gas dissociates into atomic hydrogen , the rms speed will become

Options

A. v
B. `v/2`
C. `2v`
D. `sqrt2v`

Answer

To solve the problem, we need to analyze the relationship between the root mean square (RMS) speed of gas molecules, temperature, and the number of moles of gas.

### Step-by-Step Solution:

1. **Understand the formula for RMS speed**:
The RMS speed (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the mass of a single molecule of the gas.

2. **Initial conditions**:
Let the initial RMS speed of the hydrogen molecule (H₂) at temperature \( T \) be \( v \). Thus, we can write:
\[ v = \sqrt{\frac{3kT}{m_{H2}}} \]

3. **Doubling the temperature**:
If the temperature is doubled, the new temperature becomes \( 2T \). The new RMS speed (let's denote it as \( v_f \)) can be expressed as:
\[ v_f = \sqrt{\frac{3k(2T)}{m_{H2}}} = \sqrt{2} \cdot \sqrt{\frac{3kT}{m_{H2}}} = \sqrt{2} \cdot v \]

4. **Dissociation of hydrogen gas**:
When hydrogen gas (H₂) dissociates into atomic hydrogen (H), the mass of the individual atoms is half that of the H₂ molecule. Therefore, the mass \( m \) of atomic hydrogen is:
\[ m_H = \frac{m_{H2}}{2} \]

5. **RMS speed after dissociation**:
The new RMS speed for atomic hydrogen at the doubled temperature can be calculated as:
\[ v_{f} = \sqrt{\frac{3k(2T)}{m_H}} = \sqrt{\frac{3k(2T)}{\frac{m_{H2}}{2}}} = \sqrt{\frac{6kT}{m_{H2}}} \]
This can be rewritten using the initial RMS speed \( v \):
\[ v_{f} = \sqrt{2} \cdot \sqrt{2} \cdot v = 2v \]

6. **Final result**:
Thus, the new RMS speed of the atomic hydrogen after doubling the temperature and dissociating is:
\[ v_f = 2v \]

### Conclusion:
The RMS speed will become \( 2v \).