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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understand the Formula for r.m.s. Velocity**:
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 gas constant, \(T\) is the absolute temperature in Kelvin, and \(M\) is the molar mass of the gas.

2. **Establish the Relationship**:
Since \(V_{rms}\) is directly proportional to the square root of the temperature, we can express this relationship as:
\[ \frac{V_{rms1}}{V_{rms2}} = \sqrt{\frac{T1}{T2}} \]
Given that \(V_{rms1} = 2 \cdot V_{rms2}\), we can substitute this into the equation:
\[ \frac{2 \cdot V_{rms2}}{V_{rms2}} = \sqrt{\frac{T1}{T2}} \]
This simplifies to:
\[ 2 = \sqrt{\frac{T1}{T2}} \]

3. **Square Both Sides**:
Squaring both sides of the equation gives:
\[ 4 = \frac{T1}{T2} \]

4. **Substitute the Known Temperature**:
We know that \(T2\) is the temperature at \(50^\circ C\). To convert this to Kelvin:
\[ T2 = 50 + 273 = 323 \, K \]
Now substitute \(T2\) into the equation:
\[ 4 = \frac{T1}{323} \]

5. **Solve for \(T1\)**:
Rearranging the equation to solve for \(T1\) gives:
\[ T1 = 4 \times 323 = 1292 \, K \]

6. **Convert Back to Celsius**:
To convert \(T1\) back to Celsius:
\[ T1 = 1292 - 273 = 1019 \, ^\circ C \]

7. **Final Answer**:
Therefore, the temperature at which the r.m.s. velocity of the gas becomes double is:
\[ 1019 \, ^\circ C \]