The root-mean-square (rms) speed of oxygen molecules `(O_(2))` at a certain absolute temperature is v.If the temperature is double and the oxygen gas dissociated into atomic oxygen, the rms speed would be

Options

A. v
B. `sqrt2 v`
C. 2 v
D. `2sqrt2 v`

Explanation

To solve the problem, we need to understand the relationship between the root-mean-square (rms) speed of gas molecules, temperature, and mass. The formula for the rms speed \( v_{rms} \) is given by:

\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]

where:
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature,
- \( m \) is the mass of a single molecule of the gas.

### Step 1: Identify the initial conditions
The initial rms speed of oxygen molecules \( O_2 \) at a certain temperature \( T \) is given as \( v \). The molar mass of \( O_2 \) is 32 g/mol, which means the mass of a single \( O_2 \) molecule is:

\[ m_{O_2} = \frac{32 \, \text{g/mol}}{N_A} \quad \text{(where \( N_A \) is Avogadro's number)} \]

### Step 2: Determine the new conditions after doubling the temperature and dissociating the gas
When the temperature is doubled, the new temperature \( T' \) becomes:

\[ T' = 2T \]

When \( O_2 \) dissociates into atomic oxygen, the mass of a single oxygen atom \( O \) becomes:

\[ m_O = \frac{16 \, \text{g/mol}}{N_A} \]

### Step 3: Calculate the new rms speed
Using the rms speed formula for atomic oxygen at the new temperature:

\[ v'_{rms} = \sqrt{\frac{3kT'}{m_O}} = \sqrt{\frac{3k(2T)}{m_O}} \]

Substituting \( m_O \) into the equation:

\[ v'_{rms} = \sqrt{\frac{3k(2T)}{m_O}} = \sqrt{\frac{3k(2T)}{\frac{16 \, \text{g/mol}}{N_A}}} \]

### Step 4: Relate the new rms speed to the initial rms speed
Now we can relate the new rms speed \( v'_{rms} \) to the initial rms speed \( v_{rms} \):

1. The initial rms speed for \( O_2 \) is:

\[ v_{rms} = \sqrt{\frac{3kT}{m_{O_2}}} = \sqrt{\frac{3kT}{\frac{32 \, \text{g/mol}}{N_A}}} \]

2. The new rms speed becomes:

\[ v'_{rms} = \sqrt{\frac{3k(2T)}{\frac{16 \, \text{g/mol}}{N_A}}} \]

3. Simplifying the ratio of the new rms speed to the initial rms speed:

\[ \frac{v'_{rms}}{v_{rms}} = \sqrt{\frac{2T \cdot \frac{32}{N_A}}{T \cdot \frac{16}{N_A}}} = \sqrt{\frac{2 \cdot 32}{16}} = \sqrt{4} = 2 \]

### Conclusion
Thus, the new rms speed \( v'_{rms} \) is:

\[ v'_{rms} = 2v \]

So, the final answer is that the rms speed would be \( 2v \).