If the speed of sound at `0^(@)C` is 330 m/s, then the speed of sound at `20^(@)C` will be

Options

A. 330 m/s
B. 340 m/s
C. 342 m/s
D. 324 m/s

Explanation

To find the speed of sound at 20°C given that the speed of sound at 0°C is 330 m/s, we can use the relationship between the speed of sound and temperature. The speed of sound in a gas is given by the formula:

\[ v = \sqrt{\frac{\gamma RT}{M}} \]

Where:
- \( v \) is the speed of sound,
- \( \gamma \) is the ratio of specific heats,
- \( R \) is the gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molecular mass of the gas.

### Step 1: Convert temperatures from Celsius to Kelvin
- The temperature at 0°C in Kelvin is:
\[ T_1 = 0 + 273 = 273 \, K \]
- The temperature at 20°C in Kelvin is:
\[ T_2 = 20 + 273 = 293 \, K \]

### Step 2: Write the equations for the speed of sound at both temperatures
- Speed of sound at 0°C:
\[ v_1 = \sqrt{\frac{\gamma R T_1}{M}} = \sqrt{\frac{\gamma R \cdot 273}{M}} \]
- Speed of sound at 20°C:
\[ v_2 = \sqrt{\frac{\gamma R T_2}{M}} = \sqrt{\frac{\gamma R \cdot 293}{M}} \]

### Step 3: Find the ratio of the speeds
To find the speed of sound at 20°C in terms of the speed at 0°C, we take the ratio:
\[ \frac{v_2}{v_1} = \frac{\sqrt{\frac{\gamma R \cdot 293}{M}}}{\sqrt{\frac{\gamma R \cdot 273}{M}}} \]

### Step 4: Simplify the ratio
Since \(\gamma\), \(R\), and \(M\) are constants and cancel out, we have:
\[ \frac{v_2}{v_1} = \sqrt{\frac{293}{273}} \]

### Step 5: Solve for \(v_2\)
Now, we can express \(v_2\) in terms of \(v_1\):
\[ v_2 = v_1 \cdot \sqrt{\frac{293}{273}} \]
Given that \(v_1 = 330 \, m/s\):
\[ v_2 = 330 \cdot \sqrt{\frac{293}{273}} \]

### Step 6: Calculate \(\sqrt{\frac{293}{273}}\)
Calculating the square root:
\[ \sqrt{\frac{293}{273}} \approx 1.036 \]

### Step 7: Final calculation
Now, substituting back:
\[ v_2 \approx 330 \cdot 1.036 \approx 342 \, m/s \]

### Conclusion
Thus, the speed of sound at 20°C is approximately **342 m/s**.

---