The temperature at which the root mean square velocity of a molecules will be double of its value at `100^(@)C` is

Options

A. `1492^(@)C`
B. `1219^(@)C`
C. `546^(@)C`
D. `273^(@)C`

Explanation

To solve the problem of finding the temperature at which the root mean square (RMS) velocity of gas molecules will be double that at 100°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the RMS Velocity**:
The root mean square velocity (Vrms) of gas molecules is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

2. **Initial Conditions**:
At \( T_1 = 100°C \), we first convert this temperature to Kelvin:
\[ T_1 = 100 + 273 = 373 \text{ K} \]

3. **Setting Up the Equation**:
We know that we want the RMS velocity to be double at some temperature \( T_2 \):
\[ V_{rms,2} = 2V_{rms,1} \]
This can be expressed in terms of temperature:
\[ \sqrt{\frac{3RT_2}{M}} = 2\sqrt{\frac{3RT_1}{M}} \]

4. **Simplifying the Equation**:
Since \( R \) and \( M \) are constants for the same gas, we can simplify:
\[ \sqrt{T_2} = 2\sqrt{T_1} \]
Squaring both sides gives:
\[ T_2 = 4T_1 \]

5. **Calculating \( T_2 \)**:
Substituting \( T_1 = 373 \text{ K} \):
\[ T_2 = 4 \times 373 = 1492 \text{ K} \]

6. **Converting to Celsius**:
Finally, we convert \( T_2 \) back to Celsius:
\[ T_2 = 1492 - 273 = 1219°C \]

### Final Answer:
The temperature at which the root mean square velocity of the molecules will be double that at 100°C is **1219°C**.

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