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

Options

A. `34^(@)C`
B. `68^(@)C`
C. `162^(@)C`
D. `887^(@)C`

Explanation

To solve the problem of finding the temperature at which the average speed of perfect gas molecules is double that at \(17^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Convert the Given Temperature to Kelvin**:
The first step is to convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
For \(17^\circ C\):
\[ T_1 = 17 + 273 = 290 \, K \]

2. **Understand the Relationship Between Speed and Temperature**:
The average speed of gas molecules is related to the temperature by the equation:
\[ v \propto \sqrt{T} \]
This means that the ratio of the speeds at two different temperatures can be expressed as:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]

3. **Set Up the Equation for the Given Condition**:
Since we need the average speed to be double, we can express this as:
\[ v_2 = 2v_1 \]
Therefore, the ratio becomes:
\[ \frac{v_1}{2v_1} = \frac{1}{2} \]
Substituting this into our earlier equation gives:
\[ \frac{1}{2} = \sqrt{\frac{T_1}{T_2}} \]

4. **Square Both Sides**:
To eliminate the square root, we square both sides of the equation:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \]
This simplifies to:
\[ \frac{1}{4} = \frac{T_1}{T_2} \]

5. **Rearrange to Solve for \(T_2\)**:
Rearranging the equation gives:
\[ T_2 = 4T_1 \]
Substituting \(T_1 = 290 \, K\):
\[ T_2 = 4 \times 290 = 1160 \, K \]

6. **Convert \(T_2\) Back to Celsius**:
Finally, we convert \(T_2\) back to Celsius:
\[ T(°C) = T(K) - 273 \]
Thus:
\[ T_2(°C) = 1160 - 273 = 887^\circ C \]

### Final Answer:
The temperature at which the average speed of perfect gas molecules is double that at \(17^\circ C\) is \(887^\circ C\).

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