The temperature at which the root mean square velocity of the gas molecules would become twice of its value at `0^(@)C` is

Options

A. `819^(@)C`
B. `1092^(@)C`
C. `1100^(@)C`
D. `1400^(@)C`

Explanation

To find the temperature at which the root mean square (RMS) velocity of gas molecules becomes twice its value at 0°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the RMS Velocity Formula**:
The RMS velocity (C) of gas molecules is given by the formula:
\[ C = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas.

2. **Determine the RMS Velocity at 0°C**:
At 0°C, the temperature in Kelvin is:
\[ T_0 = 0 + 273 = 273 \text{ K} \]
Therefore, the RMS velocity at 0°C is:
\[ C_0 = \sqrt{\frac{3R \cdot 273}{M}} \]

3. **Set Up the Equation for Double the RMS Velocity**:
We want to find the temperature \( T \) at which the RMS velocity is double that at 0°C:
\[ C_T = 2C_0 \]
Substituting the expression for \( C_0 \):
\[ C_T = 2 \sqrt{\frac{3R \cdot 273}{M}} \]

4. **Express the RMS Velocity at Temperature T**:
The RMS velocity at temperature \( T \) (in Kelvin) is:
\[ C_T = \sqrt{\frac{3RT}{M}} \]

5. **Equate the Two Expressions**:
Setting the two expressions for \( C_T \) equal:
\[ \sqrt{\frac{3RT}{M}} = 2 \sqrt{\frac{3R \cdot 273}{M}} \]

6. **Square Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ \frac{3RT}{M} = 4 \cdot \frac{3R \cdot 273}{M} \]

7. **Cancel Common Terms**:
Since \( R \) and \( M \) are constants and can be canceled out:
\[ T = 4 \cdot 273 \]

8. **Calculate T**:
Thus,
\[ T = 1092 \text{ K} \]

9. **Convert Back to Celsius**:
To convert the temperature back to Celsius:
\[ T_{Celsius} = T - 273 = 1092 - 273 = 819 \text{ °C} \]

### Final Answer:
The temperature at which the RMS velocity of the gas molecules becomes twice its value at 0°C is **819°C**.