The r.m.s. speed of the molecules of a gas at `100^(@)C` is v. The temperature at which the r.m.s. speed will be `sqrt(3)v` is

Options

A. `546^(@)C`
B. `646^(@)C`
C. `746^(@)C`
D. `846^(@)C`

Explanation

To solve the problem, we need to find the temperature at which the root mean square (r.m.s.) speed of gas molecules is \( \sqrt{3}v \), given that the r.m.s. speed at \( 100^\circ C \) is \( v \).

### Step-by-Step Solution:

1. **Convert the temperature from Celsius to Kelvin**:
\[ T_1 = 100^\circ C = 100 + 273 = 373 \, K \]

2. **Write the formula for r.m.s. speed**:
The r.m.s. speed \( v_{\text{rms}} \) of a gas is given by:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas.

3. **Set up the relationship between the two states**:
Since the gas is the same in both conditions, we can write:
\[ \frac{v_{\text{rms,1}}}{v_{\text{rms,2}}} = \sqrt{\frac{T_1}{T_2}} \]
where \( v_{\text{rms,1}} = v \) and \( v_{\text{rms,2}} = \sqrt{3}v \).

4. **Substituting the known values**:
\[ \frac{v}{\sqrt{3}v} = \sqrt{\frac{373}{T_2}} \]
This simplifies to:
\[ \frac{1}{\sqrt{3}} = \sqrt{\frac{373}{T_2}} \]

5. **Square both sides to eliminate the square root**:
\[ \frac{1}{3} = \frac{373}{T_2} \]

6. **Rearranging to find \( T_2 \)**:
\[ T_2 = 3 \times 373 = 1119 \, K \]

7. **Convert the temperature back to Celsius**:
\[ T_2 = 1119 - 273 = 846^\circ C \]

### Final Answer:
The temperature at which the r.m.s. speed will be \( \sqrt{3}v \) is \( 846^\circ C \).

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