The speed of sound in a particular gas at a temperature `27""^(@)C` is 340 m/s. Then, the speed of sound in the same gas at a temperature `90""^(@)C` is:

Options

A. 340 m/s
B. 357 m/s
C. 374 m/s
D. 411.4 m/s

Explanation

To find the speed of sound in a gas at a different temperature, 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 adiabatic constant (which remains constant for the same gas)
- \( R \) is the gas constant (which also remains constant)
- \( T \) is the absolute temperature in Kelvin
- \( M \) is the molecular mass of the gas (which remains constant)

### Step-by-Step Solution

1. **Identify the initial conditions**:
- Given temperature \( T_1 = 27^\circ C \)
- Speed of sound at \( T_1 \), \( V_1 = 340 \, \text{m/s} \)

2. **Convert the initial temperature to Kelvin**:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]

3. **Identify the final temperature**:
- Given temperature \( T_2 = 90^\circ C \)

4. **Convert the final temperature to Kelvin**:
\[ T_2 = 90 + 273 = 363 \, \text{K} \]

5. **Use the relationship between the speeds and temperatures**:
Since \( V \) is directly proportional to the square root of \( T \), we can write:
\[ \frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}} \]

6. **Substitute the known values**:
\[ \frac{V_2}{340} = \sqrt{\frac{363}{300}} \]

7. **Calculate \( \sqrt{\frac{363}{300}} \)**:
\[ \sqrt{\frac{363}{300}} = \sqrt{1.21} \]

8. **Calculate \( \sqrt{1.21} \)**:
\[ \sqrt{1.21} \approx 1.1 \]

9. **Substitute back to find \( V_2 \)**:
\[ V_2 = 340 \times 1.1 = 374 \, \text{m/s} \]

### Final Answer
The speed of sound in the gas at \( 90^\circ C \) is approximately \( 374 \, \text{m/s} \).

---