The rms speed of oxygen molecules at a certain temperature `T` is `v`. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, then the rms speed

Options

A. remains same
B. becomes double
C. increase by a factor of
D. None of these

Explanation

To solve the problem of finding the new root mean square (rms) speed of oxygen molecules after the temperature is doubled and the gas dissociates into atomic oxygen, we will follow these steps:

### Step 1: Understand the rms speed formula
The rms speed of a gas is given by the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{m}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( m \) is the molar mass of the gas.

### Step 2: Identify the initial conditions
Initially, we have:
- The rms speed of oxygen molecules (O₂) at temperature \( T \) is \( v \).
- The molar mass of O₂ is \( 32 \, \text{g/mol} \) or \( 32 \times 10^{-3} \, \text{kg/mol} \).

### Step 3: Calculate the initial rms speed
From the formula, we know:
\[ v = \sqrt{\frac{3RT}{m_{\text{O}_2}}} \]
Substituting \( m_{\text{O}_2} = 32 \times 10^{-3} \, \text{kg/mol} \):
\[ v = \sqrt{\frac{3RT}{32 \times 10^{-3}}} \]

### Step 4: Consider the new conditions
When the temperature is doubled, the new temperature \( T' \) becomes:
\[ T' = 2T \]
After dissociation, the molar mass of atomic oxygen (O) is \( 16 \, \text{g/mol} \) or \( 16 \times 10^{-3} \, \text{kg/mol} \).

### Step 5: Calculate the new rms speed
Using the rms speed formula for atomic oxygen at the new temperature:
\[ v' = \sqrt{\frac{3RT'}{m_{\text{O}}}} = \sqrt{\frac{3R(2T)}{16 \times 10^{-3}}} \]
This simplifies to:
\[ v' = \sqrt{\frac{6RT}{16 \times 10^{-3}}} \]

### Step 6: Relate the new rms speed to the initial rms speed
We can express \( v' \) in terms of \( v \):
\[ v' = \sqrt{\frac{6RT}{16 \times 10^{-3}}} = \sqrt{\frac{6}{16}} \cdot \sqrt{\frac{3RT}{32 \times 10^{-3}}} \]
Since \( v = \sqrt{\frac{3RT}{32 \times 10^{-3}}} \):
\[ v' = \sqrt{\frac{6}{16}} \cdot v = \frac{\sqrt{6}}{4} v \]

### Step 7: Finalize the result
Now we can express the new rms speed:
\[ v' = \frac{\sqrt{6}}{4} v \]

### Conclusion
Thus, the new rms speed of atomic oxygen after doubling the temperature and dissociating is:
\[ v' = \frac{\sqrt{6}}{4} v \]