At what temperature, the average speed of gas molecules be double of that at temperature, `27^(@)C`?

Options

A. `120^(@)C`
B. `108^(@)C`
C. `927^(@)C`
D. `300^(@)C`

Explanation

To solve the problem of finding the temperature at which the average speed of gas molecules is double that at 27°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between average speed and temperature**:
The average speed of gas molecules (V_avg) is given by the formula:
\[ V_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \]
where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molecular weight of the gas.

2. **Set up the equation for the two temperatures**:
Let \( T_1 = 27°C = 300 K \) (since we need to convert Celsius to Kelvin by adding 273).
We want to find \( T_2 \) such that:
\[ V_{\text{avg}}(T_2) = 2 \times V_{\text{avg}}(T_1) \]

3. **Substituting the average speed formula**:
Using the formula for average speed, we can write:
\[ \sqrt{\frac{8RT_2}{\pi M}} = 2 \times \sqrt{\frac{8RT_1}{\pi M}} \]

4. **Squaring both sides to eliminate the square root**:
\[ \frac{8RT_2}{\pi M} = 4 \times \frac{8RT_1}{\pi M} \]

5. **Cancel out common terms**:
Since \( \frac{8R}{\pi M} \) is common on both sides, we can cancel it out:
\[ T_2 = 4T_1 \]

6. **Substituting the value of \( T_1 \)**:
\[ T_2 = 4 \times 300 K = 1200 K \]

7. **Convert the temperature back to Celsius**:
To convert Kelvin back to Celsius, we subtract 273:
\[ T_2 = 1200 K - 273 = 927°C \]

### Final Answer:
The temperature at which the average speed of gas molecules is double that at 27°C is **927°C**.