The RMS velocity of nitrogen gas molecules is 'V cm/sec at a certain temperature. When the temperature is doubled, the molecules dissociated into individual atoms. Now the RMS velocity of nitrogen atoms is x V cm/sec. What is the value of 'x'?

Answer

To solve the problem, we need to find the value of 'x' when the RMS velocity of nitrogen atoms is expressed in terms of the RMS velocity of nitrogen molecules. Here's a step-by-step solution:

### Step 1: Understand the RMS Velocity Formula
The root mean square (RMS) velocity (U_RMS) of gas molecules is given by the formula:
\[ U_{RMS} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

### Step 2: Define Initial Conditions
Let:
- The initial temperature be \( T_1 = T \),
- The initial molar mass of nitrogen gas (N₂) be \( M_1 = M \),
- The initial RMS velocity of nitrogen gas molecules be \( U_{RMS1} = V \).

Using the formula, we can express this as:
\[ V = \sqrt{\frac{3RT}{M}} \]

### Step 3: Analyze the New Conditions
When the temperature is doubled, we have:
- New temperature \( T_2 = 2T \),
- The molar mass of nitrogen atoms (N) is half that of nitrogen gas, so \( M_2 = \frac{M}{2} \).

### Step 4: Calculate the New RMS Velocity
The new RMS velocity of nitrogen atoms (N) can be expressed as:
\[ U_{RMS2} = \sqrt{\frac{3RT_2}{M_2}} = \sqrt{\frac{3R(2T)}{\frac{M}{2}}} \]
This simplifies to:
\[ U_{RMS2} = \sqrt{\frac{3R(2T) \cdot 2}{M}} = \sqrt{\frac{6RT}{M}} = \sqrt{4} \cdot \sqrt{\frac{3RT}{M}} = 2 \cdot U_{RMS1} \]

### Step 5: Relate the New RMS Velocity to the Original
Since \( U_{RMS1} = V \), we can substitute:
\[ U_{RMS2} = 2V \]
Thus, we can express this as:
\[ U_{RMS2} = xV \]
From the above, we find that:
\[ x = 2 \]

### Final Answer
The value of \( x \) is \( 2 \).

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