Find the temperature at which the speed of sound will be double of its value at `0^(@)C`.

Answer

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

### Step 1: Understand the formula for the speed of sound
The speed of sound \( V \) in a gas is given by the formula:
\[ V = \sqrt{\frac{\gamma RT}{M}} \]
where:
- \( \gamma \) = ratio of specific heats (Cp/Cv)
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin
- \( M \) = molar mass of the gas

### Step 2: Determine the speed of sound at 0°C
At 0°C, which is equivalent to 273 K, we can denote the speed of sound as \( V_1 \):
\[ V_1 = \sqrt{\frac{\gamma R \cdot 273}{M}} \]

### Step 3: Set up the equation for the new speed of sound
We want to find the temperature \( T_2 \) at which the speed of sound \( V_2 \) is double that of \( V_1 \):
\[ V_2 = 2V_1 \]
Substituting the expression for \( V_1 \):
\[ V_2 = 2\sqrt{\frac{\gamma R \cdot 273}{M}} \]

### Step 4: Express \( V_2 \) using the new temperature \( T_2 \)
Using the formula for the speed of sound again for \( T_2 \):
\[ V_2 = \sqrt{\frac{\gamma R T_2}{M}} \]

### Step 5: Set the two expressions for \( V_2 \) equal to each other
Now we can set the two expressions for \( V_2 \) equal:
\[ \sqrt{\frac{\gamma R T_2}{M}} = 2\sqrt{\frac{\gamma R \cdot 273}{M}} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{\gamma R T_2}{M} = 4 \cdot \frac{\gamma R \cdot 273}{M} \]

### Step 7: Simplify the equation
Since \( \gamma \), \( R \), and \( M \) are constants, they can be canceled out:
\[ T_2 = 4 \cdot 273 \]

### Step 8: Calculate \( T_2 \)
Now, calculating \( T_2 \):
\[ T_2 = 1092 \text{ K} \]

### Conclusion
The temperature at which the speed of sound will be double its value at 0°C is **1092 K**.

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