Calculate the temperature to which a gas at `0^(@)C` be heated so that the rms speed of its molecules be doubled, keeping other factors constant.

Explanation

To solve the problem of calculating the temperature to which a gas at \(0^\circ C\) must be heated 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**: The rms speed \(v_{rms}\) 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. **Identify Initial Conditions**: The initial temperature \(T_1\) is given as \(0^\circ C\). To convert this to Kelvin:
\[ T_1 = 0 + 273 = 273 \text{ K} \]

3. **Set Up the Equation for Doubling the Speed**: According to the problem, we want to double the rms speed. Let the initial rms speed be \(v_1\) and the final rms speed be \(v_2 = 2v_1\).

4. **Use the Proportional Relationship**: Since \(v_{rms}\) is proportional to the square root of the temperature, we can write:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting \(v_2 = 2v_1\):
\[ \frac{v_1}{2v_1} = \sqrt{\frac{T_1}{T_2}} \]
Simplifying gives:
\[ \frac{1}{2} = \sqrt{\frac{T_1}{T_2}} \]

5. **Square Both Sides**: Squaring both sides of the equation:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \]
This simplifies to:
\[ \frac{1}{4} = \frac{T_1}{T_2} \]

6. **Rearranging the Equation**: Rearranging gives:
\[ T_2 = 4T_1 \]

7. **Substituting the Initial Temperature**: Now substitute \(T_1 = 273 \text{ K}\):
\[ T_2 = 4 \times 273 = 1092 \text{ K} \]

8. **Convert Back to Celsius**: Finally, convert \(T_2\) back to degrees Celsius:
\[ T_2 = 1092 - 273 = 819^\circ C \]

### Final Answer:
The temperature to which the gas must be heated is \(819^\circ C\).

---