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. `14 u`

Explanation

To solve the problem step by step, we will use the formula for the root mean square (RMS) speed of gas molecules and analyze the changes that occur when the temperature is doubled and nitrogen molecules dissociate into atoms.

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

### Step 2: Calculate the RMS Speed of \( N_2 \)
Given that the RMS speed of \( N_2 \) molecules is \( u \), we can express this as:
\[ u = \sqrt{\frac{3RT}{M_{N_2}}} \]
The molar mass \( M_{N_2} \) of nitrogen gas is 28 g/mol. Thus, we can write:
\[ u = \sqrt{\frac{3RT}{28}} \]

### Step 3: Determine the New Conditions
When the temperature is doubled, the new temperature \( T' \) becomes:
\[ T' = 2T \]
When nitrogen gas dissociates into nitrogen atoms, the new molar mass \( M_{N} \) of nitrogen atoms is 14 g/mol.

### Step 4: Calculate the New RMS Speed for Nitrogen Atoms
Now, we need to find the new RMS speed \( u' \) for nitrogen atoms:
\[ u' = \sqrt{\frac{3RT'}{M_{N}}} \]
Substituting \( T' \) and \( M_{N} \):
\[ u' = \sqrt{\frac{3R(2T)}{14}} = \sqrt{\frac{6RT}{14}} = \sqrt{\frac{3RT}{7}} \]

### Step 5: Relate the New RMS Speed to the Original RMS Speed
Now, we can relate \( u' \) to \( u \):
\[ u' = \sqrt{\frac{3RT}{7}} = \sqrt{\frac{3RT}{28}} \cdot \sqrt{4} = u \cdot \sqrt{4} = 2u \]

### Conclusion
Thus, the new RMS speed of nitrogen atoms when the temperature is doubled is:
\[ u' = 2u \]

### Final Answer
The RMS speed becomes \( 2u \).

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