At what temperature will the speed of sound be double of its value at `0@C ` ?

Explanation

To find the temperature at which the speed of sound is double its value at 0°C, we can follow these steps:

### Step 1: Understand the relationship between speed of sound and temperature
The speed of sound in an ideal gas is given by the formula:
\[ v = \sqrt{\frac{\gamma RT}{m}} \]
where:
- \( v \) is the speed of sound,
- \( \gamma \) is the adiabatic index (constant for a given gas),
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the molar mass of the gas.

From this formula, we can see that the speed of sound is proportional to the square root of the absolute temperature \( T \).

### Step 2: Set up the relationship for the given condition
Let:
- \( T_1 \) = 273 K (temperature at 0°C),
- \( v_1 \) = speed of sound at 0°C,
- \( v_2 \) = speed of sound at temperature \( T_2 \).

According to the problem, we want to find \( T_2 \) such that:
\[ v_2 = 2v_1 \]

### Step 3: Use the proportional relationship
Since \( v \) is proportional to \( \sqrt{T} \), we can write:
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]

Substituting \( v_2 = 2v_1 \):
\[ \frac{2v_1}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
This simplifies to:
\[ 2 = \sqrt{\frac{T_2}{T_1}} \]

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

### Step 5: Solve for \( T_2 \)
Rearranging the equation, we find:
\[ T_2 = 4T_1 \]

### Step 6: Substitute the value of \( T_1 \)
Now substitute \( T_1 = 273 \) K:
\[ T_2 = 4 \times 273 = 1092 \text{ K} \]

### Step 7: Convert Kelvin to Celsius
To convert Kelvin to Celsius:
\[ T_2 (\text{°C}) = T_2 (\text{K}) - 273 \]
\[ T_2 (\text{°C}) = 1092 - 273 = 819 \text{°C} \]

### Final Answer
The temperature at which the speed of sound is double its value at 0°C is **819°C**.

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