The `rms` speed of `N_(2)` molecules in a gas in `u`. If the temperature is doubled and the nitrogen molecules dissociate into nitrogen atom, the `rms` speed becomes

Options

A. u/2
B. 2u
C. 4u
D. 14u

Explanation

To solve the problem, we need to understand how the root mean square (rms) speed of gas molecules changes with temperature and molecular mass.

### Step-by-Step Solution:

1. **Understanding the rms Speed Formula**:
The rms speed (u_rms) of gas molecules is given by the formula:
\[ u_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

2. **Initial Conditions**:
Let the initial rms speed of \( N_2 \) be \( u \). Therefore, we can express it as:
\[ u = \sqrt{\frac{3RT_1}{M_{N_2}}} \]
where \( T_1 \) is the initial temperature and \( M_{N_2} \) is the molar mass of nitrogen gas.

3. **Doubling the Temperature**:
If the temperature is doubled, the new temperature \( T_2 \) becomes:
\[ T_2 = 2T_1 \]

4. **Dissociation of Nitrogen Molecules**:
When \( N_2 \) dissociates into two nitrogen atoms, the effective molar mass changes. The molar mass of a nitrogen atom is \( M_N \), and since \( N_2 \) dissociates into two \( N \) atoms, the new molar mass \( M_{N} \) becomes:
\[ M_{N} = \frac{M_{N_2}}{2} \]

5. **Calculating the New rms Speed**:
Now, substituting the new temperature and molar mass into the rms speed formula, we get:
\[ u'_{\text{rms}} = \sqrt{\frac{3R(2T_1)}{\frac{M_{N_2}}{2}}} \]
Simplifying this:
\[ u'_{\text{rms}} = \sqrt{\frac{3R \cdot 2T_1 \cdot 2}{M_{N_2}}} = \sqrt{\frac{12RT_1}{M_{N_2}}} \]

6. **Relating New rms Speed to Initial rms Speed**:
We can relate this back to the initial rms speed \( u \):
\[ u'_{\text{rms}} = \sqrt{4} \cdot \sqrt{\frac{3RT_1}{M_{N_2}}} = 2u \]

### Final Answer:
Thus, the new rms speed after doubling the temperature and dissociating the nitrogen molecules into nitrogen atoms is:
\[ u'_{\text{rms}} = 2u \]