What will be the temperature when the rms velocity is double of that at 300 K?

Options

A. 300 K
B. 600 K
C. 900 K
D. 1200 K

Explanation

To solve the problem of finding the temperature when the root mean square (RMS) velocity is double that at 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between RMS velocity and temperature**:
The RMS velocity (\(v_{rms}\)) of a gas is directly proportional to the square root of its absolute temperature (T). This can be expressed mathematically as:
\[ v_{rms} \propto \sqrt{T} \]

2. **Set up the proportionality equation**:
If we denote the initial temperature as \(T_1\) (300 K) and the final temperature as \(T_2\), we can write:
\[ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T_1}{T_2}} \]

3. **Define the conditions**:
We know that \(v_{rms2} = 2 \cdot v_{rms1}\) because the RMS velocity is doubled. Therefore, we can substitute this into our equation:
\[ \frac{v_{rms1}}{2 \cdot v_{rms1}} = \sqrt{\frac{T_1}{T_2}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{T_1}{T_2}} \]

4. **Square both sides to eliminate the square root**:
Squaring both sides gives us:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \]
Which simplifies to:
\[ \frac{1}{4} = \frac{T_1}{T_2} \]

5. **Rearrange to find \(T_2\)**:
Rearranging the equation gives:
\[ T_2 = 4 \cdot T_1 \]

6. **Substitute the value of \(T_1\)**:
Now substitute \(T_1 = 300 \, K\):
\[ T_2 = 4 \cdot 300 \, K = 1200 \, K \]

7. **Conclusion**:
The temperature when the RMS velocity is double that at 300 K is:
\[ \boxed{1200 \, K} \]