The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociates into two atoms. The new root mean square speed of the atom is

Options

A. `sqrt(2)v`
B. y
C. 2v
D. 4v

Answer

To solve the problem, we need to analyze the changes in the root mean square (RMS) speed of the gas molecules when the temperature is doubled and the diatomic molecules dissociate into two atoms.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The root mean square speed of the diatomic gas is given as \( v \).
- The formula for the root mean square speed \( v \) of a gas is given by:
\[ v = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

2. **Identify the Molar Mass**:
- Let the molar mass of the diatomic gas be \( M \).
- Therefore, we can write:
\[ v = \sqrt{\frac{3RT}{M}} \]

3. **Change in Conditions**:
- The problem states that the temperature is doubled. Thus, the new temperature \( T' \) becomes:
\[ T' = 2T \]
- The diatomic molecules dissociate into two monatomic atoms. This means the new molar mass \( M' \) becomes:
\[ M' = \frac{M}{2} \]

4. **Calculate the New RMS Speed**:
- The new root mean square speed \( v' \) of the atoms can be calculated using the same formula:
\[ v' = \sqrt{\frac{3RT'}{M'}} \]
- Substitute \( T' \) and \( M' \) into the equation:
\[ v' = \sqrt{\frac{3R(2T)}{\frac{M}{2}}} \]
- Simplifying this gives:
\[ v' = \sqrt{\frac{3R \cdot 2T \cdot 2}{M}} = \sqrt{\frac{6RT}{M}} = \sqrt{4} \cdot \sqrt{\frac{3RT}{M}} = 2 \cdot v \]

5. **Final Result**:
- Therefore, the new root mean square speed of the atom is:
\[ v' = 2v \]

### Conclusion:
The new root mean square speed of the atom after the temperature is doubled and the diatomic gas dissociates into two atoms is \( 2v \).