A sample of gas is at `0^(@)C`. To what temperature it must be raised in order to double the rms speed of the molecule.

Options

A. `100^(@)C`
B. `273^(@)C`
C. `819^(@)C`
D. `919^(@)C`

Answer

To solve the problem of determining the temperature to which a gas must be raised in order to double the root mean square (rms) speed of its molecules, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between RMS Speed and Temperature**:
The rms speed (v_rms) of gas molecules 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,
- \( M \) is the molar mass of the gas.

2. **Establish the Initial Conditions**:
The initial temperature \( T_1 \) is given as \( 0^\circ C \), which is equivalent to:
\[ T_1 = 273 \, K \]

3. **Set Up the Equation for Doubling the RMS Speed**:
If the rms speed is to be doubled, we can express this as:
\[ v_2 = 2v_1 \]
Squaring both sides gives:
\[ v_2^2 = 4v_1^2 \]

4. **Relate the RMS Speeds to Their Temperatures**:
Using the formula for rms speed, we can write:
\[ v_1^2 = \frac{3RT_1}{M} \quad \text{and} \quad v_2^2 = \frac{3RT_2}{M} \]
Therefore, substituting these into the equation gives:
\[ \frac{3RT_2}{M} = 4 \left(\frac{3RT_1}{M}\right) \]

5. **Cancel Common Terms**:
Since \( R \) and \( M \) are constants and appear on both sides, we can simplify:
\[ T_2 = 4T_1 \]

6. **Calculate the Final Temperature**:
Substituting \( T_1 = 273 \, K \):
\[ T_2 = 4 \times 273 = 1092 \, K \]

7. **Convert the Final Temperature to Celsius**:
To convert from Kelvin to Celsius, we subtract 273:
\[ T_2 = 1092 - 273 = 819^\circ C \]

### Final Answer:
The temperature to which the gas must be raised in order to double the rms speed of the molecules is:
\[ \boxed{819^\circ C} \]