At constant pressure, calculate the temperatrue at which root mean square speed of a gas becomes double of its value at `0^(@)`

Explanation

To solve the problem, we need to calculate the temperature at which the root mean square (RMS) speed of a gas becomes double its value at 0 degrees Celsius, while keeping the pressure constant.

### Step-by-Step Solution:

1. **Understand the relationship between RMS speed and temperature**:
The root mean square speed (Vrms) 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. **Identify the initial conditions**:
At 0 degrees Celsius, the temperature in Kelvin is:
\[ T_1 = 0 + 273 = 273 \text{ K} \]
Let's denote the RMS speed at this temperature as \(V\):
\[ V_{rms}(T_1) = V \]

3. **Set up the equation for the final condition**:
We need to find the temperature \(T_f\) at which the RMS speed becomes double:
\[ V_{rms}(T_f) = 2V \]

4. **Express the RMS speed at the final temperature**:
From the formula, we can express the RMS speed at the final temperature:
\[ V_{rms}(T_f) = \sqrt{\frac{3RT_f}{m}} \]

5. **Relate the two conditions**:
Since \(V_{rms}(T_1) = \sqrt{\frac{3R \cdot 273}{m}} = V\) and \(V_{rms}(T_f) = \sqrt{\frac{3RT_f}{m}} = 2V\), we can set up the following relationship:
\[ 2V = \sqrt{\frac{3RT_f}{m}} \]

6. **Square both sides**:
Squaring both sides gives:
\[ (2V)^2 = \frac{3RT_f}{m} \]
Therefore:
\[ 4V^2 = \frac{3RT_f}{m} \]

7. **Substituting for \(V^2\)**:
From the initial condition, we have:
\[ V^2 = \frac{3R \cdot 273}{m} \]
Substituting this into our equation gives:
\[ 4 \left(\frac{3R \cdot 273}{m}\right) = \frac{3RT_f}{m} \]

8. **Cancel out common terms**:
The \( \frac{3R}{m} \) cancels out from both sides:
\[ 4 \cdot 273 = T_f \]
Thus:
\[ T_f = 1092 \text{ K} \]

9. **Convert to Celsius**:
Finally, to convert the temperature from Kelvin to Celsius:
\[ T_f (\text{°C}) = T_f (\text{K}) - 273 = 1092 - 273 = 819 \text{ °C} \]

### Final Answer:
The temperature at which the root mean square speed of the gas becomes double its value at 0 degrees Celsius is \(819 \text{ °C}\).

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