The temperature at which the average speed of the gas molecules is double to that at a temperature of `27^(@)C` is

Options

A. `54^(@)C`
B. `198^(@)C`
C. `327^(@)C`
D. `927^(@)C`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Relationship Between Speed and Temperature**:
The average speed (V) of gas molecules is given by the equation:
\[ V \propto \sqrt{T} \]
This means that the average speed is directly proportional to the square root of the temperature.

2. **Set Up the Proportionality**:
Let \( V_1 \) be the average speed at temperature \( T_1 \) (27°C), and \( V_2 \) be the average speed at temperature \( T_2 \). Since \( V_2 \) is double \( V_1 \):
\[ V_2 = 2V_1 \]

3. **Convert the Given Temperature to Kelvin**:
The temperature \( T_1 \) in Celsius is 27°C. To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

4. **Use the Proportionality to Set Up the Equation**:
From the proportionality, we can write:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting \( V_2 = 2V_1 \):
\[ \frac{V_1}{2V_1} = \sqrt{\frac{T_1}{T_2}} \implies \frac{1}{2} = \sqrt{\frac{T_1}{T_2}} \]

5. **Square Both Sides**:
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \implies \frac{1}{4} = \frac{T_1}{T_2} \]

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

7. **Substituting the Value of \( T_1 \)**:
Now substitute \( T_1 = 300 \, K \):
\[ T_2 = 4 \times 300 = 1200 \, K \]

8. **Convert \( T_2 \) Back to Celsius**:
To convert \( T_2 \) back to Celsius:
\[ T_2 = 1200 - 273 = 927 \, °C \]

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