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. (a) `u / 2`
B. (b) `2u`
C. (c) `4u`
D. (d) `12u`

Explanation

To solve the problem step by step, we will analyze the changes in the root mean square (rms) speed of nitrogen molecules when the temperature is doubled and the molecules dissociate into nitrogen atoms.

### Step 1: Understand the initial condition
We are given that the rms speed of \( N_2 \) molecules is \( u \). The formula for rms speed (\( v_{rms} \)) is given by:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas.

For \( N_2 \), the molar mass \( M \) is \( 28 \, \text{g/mol} \) (since nitrogen has an atomic mass of \( 14 \, \text{g/mol} \), and \( N_2 \) has two nitrogen atoms).

### Step 2: Write the expression for initial rms speed
The initial rms speed can be expressed as:

\[ u = \sqrt{\frac{3RT}{28}} \]

### Step 3: Analyze the changes when the temperature is doubled
When the temperature is doubled, the new temperature becomes \( 2T \). The new rms speed (\( v'_{rms} \)) can be expressed as:

\[ v'_{rms} = \sqrt{\frac{3R(2T)}{M'}} \]

### Step 4: Determine the new molar mass after dissociation
Since the nitrogen molecules dissociate into nitrogen atoms, the new molar mass \( M' \) will be that of a single nitrogen atom, which is \( 14 \, \text{g/mol} \).

### Step 5: Substitute the new values into the rms speed formula
Now substituting \( M' = 14 \) and \( T = 2T \) into the rms speed formula gives:

\[ v'_{rms} = \sqrt{\frac{3R(2T)}{14}} \]

### Step 6: Simplify the new rms speed expression
We can simplify this expression:

\[ v'_{rms} = \sqrt{\frac{6RT}{14}} = \sqrt{\frac{3RT}{7}} \]

### Step 7: Relate the new rms speed to the initial rms speed
Now we can relate the new rms speed \( v'_{rms} \) to the initial rms speed \( u \):

From the initial condition:

\[ u = \sqrt{\frac{3RT}{28}} \]

Now, we can find the ratio of the new rms speed to the initial rms speed:

\[ \frac{v'_{rms}}{u} = \frac{\sqrt{\frac{3RT}{7}}}{\sqrt{\frac{3RT}{28}}} \]

### Step 8: Simplify the ratio
This simplifies to:

\[ \frac{v'_{rms}}{u} = \sqrt{\frac{28}{7}} = \sqrt{4} = 2 \]

### Conclusion
Thus, the new rms speed \( v'_{rms} \) is:

\[ v'_{rms} = 2u \]

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

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