The R.M.S. speed of oxygen molecules at temperature T (in kelvin) is `"v m s"^(-1)`. As the temperature becomes 4T and the oxygen gas dissociates into atomic oxygen, then the speed of atomic oxygen

Options

A. remains the same
B. becomes 2v
C. becomes `sqrt2v`
D. becomes `2sqrt2v`

Answer

To solve the problem, we need to find the new root mean square (R.M.S.) speed of atomic oxygen after the temperature changes and the molecular form of oxygen dissociates into atomic oxygen.

### Step-by-step Solution:

1. **Understand the R.M.S. Speed Formula**:
The R.M.S. speed (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

2. **Initial Conditions**:
For molecular oxygen (O2) at temperature \( T \):
- The molar mass \( M_{O2} = 32 \, \text{g/mol} \).
- The R.M.S. speed is given as \( v \).

3. **Change in Temperature**:
The temperature is increased to \( 4T \).

4. **Dissociation of Oxygen**:
When oxygen gas (O2) dissociates into atomic oxygen (O), the new molar mass becomes:
- \( M_{O} = 16 \, \text{g/mol} \).

5. **Calculate New R.M.S. Speed**:
We need to find the new R.M.S. speed \( v' \) for atomic oxygen at the new temperature \( 4T \):
\[ v' = \sqrt{\frac{3R(4T)}{M_{O}}} \]

6. **Substituting Values**:
Substitute \( M_{O} = 16 \, \text{g/mol} \) into the equation:
\[ v' = \sqrt{\frac{3R(4T)}{16}} \]

7. **Relate to Original Speed**:
We can express \( v' \) in terms of the original speed \( v \):
\[ v = \sqrt{\frac{3RT}{32}} \]
Therefore, we can write:
\[ v' = \sqrt{\frac{4 \cdot 3RT}{16}} = \sqrt{\frac{3RT}{4}} = \sqrt{4} \cdot \sqrt{\frac{3RT}{16}} = 2 \cdot \sqrt{\frac{3RT}{16}} = 2 \cdot v \]

8. **Final Result**:
Since we have a factor of \( 2 \) from the temperature change and the factor of \( \sqrt{2} \) from the change in molar mass, we can combine these:
\[ v' = 2 \sqrt{2} v \]

### Conclusion:
The new R.M.S. speed of atomic oxygen after the dissociation and temperature change is:
\[ v' = 2\sqrt{2} v \]