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 determine the new root-mean-square (rms) speed of oxygen molecules when the temperature is doubled and the gas dissociates into atomic oxygen.

### Step-by-Step Solution:

1. **Understand the formula for rms speed**:
The root-mean-square speed (Vrms) is given by the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Identify the initial conditions**:
For diatomic oxygen (\( O_2 \)):
- The molar mass \( M \) is 32 g/mol.
- The initial temperature is \( T \).
- The initial rms speed is \( V \).

3. **Determine the new conditions**:
- The temperature is doubled: \( T' = 2T \).
- The oxygen gas dissociates into atomic oxygen, which has a molar mass of \( M' = 16 \) g/mol (since each \( O_2 \) molecule splits into two \( O \) atoms).

4. **Substitute the new values into the rms speed formula**:
The new rms speed \( V'_{\text{rms}} \) can be expressed as:
\[ V'_{\text{rms}} = \sqrt{\frac{3R(2T)}{M'}} \]
Substituting \( M' = \frac{M}{2} \) (since the molar mass of atomic oxygen is half that of molecular oxygen):
\[ V'_{\text{rms}} = \sqrt{\frac{3R(2T)}{16}} = \sqrt{\frac{6RT}{16}} = \sqrt{\frac{3RT}{8}} \]

5. **Relate the new rms speed to the initial rms speed**:
We can express the new rms speed in terms of the initial rms speed:
\[ V'_{\text{rms}} = \sqrt{\frac{3RT}{32}} \cdot \sqrt{2} = V \cdot \sqrt{2} \]
Thus, we have:
\[ V'_{\text{rms}} = 2V \]

6. **Conclusion**:
The new rms speed of the dissociated oxygen atoms at the doubled temperature is:
\[ V'_{\text{rms}} = 2V \]
Therefore, the correct answer is \( 2V \).