The temperature at which the average speed of perfect gas molecules is double than at `17^(@)C` is

Explanation

To find the temperature at which the average speed of perfect gas molecules is double that at 17°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between average speed and temperature:**
The average speed \( v \) of gas molecules is given by the equation:
\[ v \propto \sqrt{T} \]
where \( T \) is the absolute temperature in Kelvin.

2. **Set up the ratio of average speeds:**
Let \( v_1 \) be the average speed at the initial temperature \( T_1 \) (17°C) and \( v_2 \) be the average speed at the final temperature \( T_2 \). According to the problem, we have:
\[ \frac{v_2}{v_1} = 2 \]

3. **Express the ratio in terms of temperature:**
Using the proportionality of average speed to the square root of temperature, we can write:
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Substituting the known ratio:
\[ 2 = \sqrt{\frac{T_2}{T_1}} \]

4. **Square both sides to eliminate the square root:**
Squaring both sides gives:
\[ 4 = \frac{T_2}{T_1} \]

5. **Solve for \( T_2 \):**
Rearranging the equation, we find:
\[ T_2 = 4 \times T_1 \]

6. **Convert the initial temperature to Kelvin:**
The initial temperature \( T_1 \) is given as 17°C. To convert to Kelvin:
\[ T_1 = 17 + 273 = 290 \text{ K} \]

7. **Calculate \( T_2 \):**
Now substituting \( T_1 \) into the equation for \( T_2 \):
\[ T_2 = 4 \times 290 = 1160 \text{ K} \]

8. **Convert \( T_2 \) back to Celsius:**
To convert \( T_2 \) back to Celsius:
\[ T_2 = 1160 - 273 = 887 \text{ °C} \]

### Final Answer:
The temperature at which the average speed of perfect gas molecules is double that at 17°C is **887 °C**.

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