The rms speed of oxygen molecules in a gas in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become

Options

A. (a)`v`
B. (b)`vsqrt(2)`
C. (c)`2v`
D. (d)`4v`

Answer

To solve the problem, we need to determine how the root mean square (rms) speed of oxygen changes when the temperature is doubled and the oxygen molecules dissociate into atoms.

### Step-by-Step Solution:

1. **Understand the formula for rms speed**:
The rms speed (\(v_{rms}\)) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a molecule.

2. **Identify the initial conditions**:
We are given that the initial rms speed of oxygen molecules is \(v\). Therefore, we can write:
\[ v = \sqrt{\frac{3kT}{m}} \]

3. **Determine the new temperature**:
If the temperature is doubled, the new temperature \(T'\) becomes:
\[ T' = 2T \]

4. **Consider the dissociation of molecules**:
Oxygen molecules (O\(_2\)) dissociate into oxygen atoms (O). The molar mass of O\(_2\) is 32 amu, and for O, it is 16 amu. Thus, the mass of one oxygen atom \(m'\) is:
\[ m' = \frac{m}{2} \]

5. **Substitute the new values into the rms speed formula**:
The new rms speed (\(v'_{rms}\)) can be calculated using the new temperature and the new mass:
\[ v'_{rms} = \sqrt{\frac{3kT'}{m'}} = \sqrt{\frac{3k(2T)}{m/2}} \]

6. **Simplify the expression**:
Simplifying the above expression gives:
\[ v'_{rms} = \sqrt{\frac{3k(2T) \cdot 2}{m}} = \sqrt{\frac{6kT}{m}} = \sqrt{2} \cdot \sqrt{\frac{3kT}{m}} = \sqrt{2} \cdot v \]

7. **Final result**:
Therefore, the new rms speed after doubling the temperature and dissociating the molecules is:
\[ v'_{rms} = 2v \]

### Conclusion:
The rms speed will become \(2v\).