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 root mean square (R.M.S.) speed of atomic oxygen after the dissociation of molecular oxygen (O2) at a temperature of 4T.

### Step-by-Step Solution:

1. **Understand the R.M.S. Speed Formula**:
The R.M.S. speed (v_rms) of gas molecules 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, and \( M \) is the molar mass of the gas in kg/mol.

2. **Initial Conditions for O2**:
For molecular oxygen (O2), the molar mass \( M \) is 32 g/mol, which is equivalent to \( 32 \times 10^{-3} \) kg/mol. The R.M.S. speed at temperature \( T \) is given as \( v \):
\[ v = \sqrt{\frac{3RT}{32 \times 10^{-3}}} \]

3. **New Conditions at Temperature 4T**:
When the temperature is increased to \( 4T \) and the gas dissociates into atomic oxygen (O), the molar mass of atomic oxygen is 16 g/mol, or \( 16 \times 10^{-3} \) kg/mol. We need to find the new R.M.S. speed \( v' \):
\[ v' = \sqrt{\frac{3R(4T)}{16 \times 10^{-3}}} \]

4. **Simplifying the Expression for v'**:
Substitute \( 4T \) into the formula:
\[ v' = \sqrt{\frac{3R \cdot 4T}{16 \times 10^{-3}}} = \sqrt{\frac{12RT}{16 \times 10^{-3}}} = \sqrt{\frac{3RT}{4 \times 10^{-3}}} \]

5. **Finding the Ratio of v to v'**:
Now, we can find the ratio \( \frac{v}{v'} \):
\[ \frac{v}{v'} = \frac{\sqrt{\frac{3RT}{32 \times 10^{-3}}}}{\sqrt{\frac{12RT}{16 \times 10^{-3}}}} = \frac{\sqrt{3RT} \cdot \sqrt{16 \times 10^{-3}}}{\sqrt{12RT} \cdot \sqrt{32 \times 10^{-3}}} \]
Simplifying this gives:
\[ \frac{v}{v'} = \frac{\sqrt{16}}{\sqrt{12}} \cdot \frac{\sqrt{32}}{\sqrt{10^{-3}}} \]
\[ = \frac{4}{\sqrt{12}} \cdot \frac{4}{\sqrt{10^{-3}}} = \frac{4 \cdot 4}{\sqrt{12}} = \frac{16}{\sqrt{12}} = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3} \]

6. **Final Calculation of v'**:
Rearranging gives:
\[ v' = \frac{v \cdot 4}{\sqrt{3}} = \frac{4v}{\sqrt{3}} \]

### Final Result:
The speed of atomic oxygen after dissociation at temperature \( 4T \) is:
\[ v' = \frac{4v}{\sqrt{3}} \]