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 (Vrms) 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, and \( M \) is the molar mass of the gas.

2. **Set up the initial conditions**:
Let the initial temperature be \( T_1 = 300 \, K \). The corresponding rms velocity is denoted as \( V_{rms1} \).

3. **Express the new condition**:
We need to find the new temperature \( T_2 \) when the rms velocity \( V_{rms2} \) is double that of \( V_{rms1} \):
\[ V_{rms2} = 2 \times V_{rms1} \]

4. **Use the proportionality of rms velocity and temperature**:
From the formula for rms velocity, we can see that:
\[ V_{rms} \propto \sqrt{T} \]
Therefore, we can express the relationship between the two states as:
\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{T_2}{T_1}} \]

5. **Substitute the known values**:
Since \( V_{rms2} = 2 \times V_{rms1} \), we can substitute this into the equation:
\[ 2 = \sqrt{\frac{T_2}{T_1}} \]

6. **Square both sides to eliminate the square root**:
Squaring both sides gives:
\[ 4 = \frac{T_2}{T_1} \]

7. **Solve for \( T_2 \)**:
Rearranging the equation gives:
\[ T_2 = 4 \times T_1 \]
Substituting \( T_1 = 300 \, K \):
\[ T_2 = 4 \times 300 = 1200 \, K \]

8. **Conclusion**:
The temperature when the rms velocity is double that at 300 K is \( T_2 = 1200 \, K \).

### Final Answer:
The temperature will be \( 1200 \, K \).

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