At what temperature, the r.m.s. velocity of a gas measured at `50^(@)C` will become double ?

Options

A. `626 K`
B. `1019 K`
C. `200^(@)C`
D. `1019^(@)C`

Explanation

To solve the problem of determining the temperature at which the root mean square (r.m.s.) velocity of a gas, measured at 50°C, will become double, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the r.m.s. velocity formula**:
The r.m.s. velocity \( v_{rms} \) 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 in Kelvin, and \( M \) is the molar mass of the gas.

2. **Convert the initial temperature to Kelvin**:
The initial temperature is given as \( 50^\circ C \). To convert this to Kelvin:
\[ T_1 = 50 + 273 = 323 \, K \]

3. **Set up the equation for the doubled r.m.s. velocity**:
According to the problem, we want the r.m.s. velocity at the new temperature \( T_2 \) to be double that at \( T_1 \):
\[ v_{rms}(T_2) = 2 \cdot v_{rms}(T_1) \]

4. **Substituting the r.m.s. velocity expressions**:
Using the formula for r.m.s. velocity, we can express this as:
\[ \sqrt{\frac{3RT_2}{M}} = 2 \cdot \sqrt{\frac{3RT_1}{M}} \]

5. **Square both sides to eliminate the square root**:
Squaring both sides gives:
\[ \frac{3RT_2}{M} = 4 \cdot \frac{3RT_1}{M} \]

6. **Cancel out common terms**:
The \( 3R \) and \( M \) terms cancel out:
\[ T_2 = 4T_1 \]

7. **Substituting the value of \( T_1 \)**:
Now, substituting \( T_1 = 323 \, K \):
\[ T_2 = 4 \cdot 323 = 1292 \, K \]

8. **Convert back to Celsius**:
To convert \( T_2 \) back to Celsius:
\[ T_2 = 1292 - 273 = 1019 \, °C \]

### Final Answer:
The temperature at which the r.m.s. velocity of the gas will become double is **1019°C**.