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. (a)340 m/s
B. (b)357 m/s
C. (c)374 m/s
D. (d)411.4 m/s

Explanation

To solve the problem of finding the speed of sound in a gas at a temperature of 90°C, given that the speed of sound at 27°C is 340 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin**:
- For \( T_1 = 27°C \):
\[ T_1 = 27 + 273 = 300 \, K \]
- For \( T_2 = 90°C \):
\[ T_2 = 90 + 273 = 363 \, K \]

2. **Use the relationship between the speed of sound and temperature**:
- The speed of sound in a gas is proportional to the square root of the absolute temperature:
\[ v \propto \sqrt{T} \]
- Therefore, we can write:
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]

3. **Substitute the known values**:
- Given \( v_1 = 340 \, m/s \), \( T_1 = 300 \, K \), and \( T_2 = 363 \, K \):
\[ \frac{v_2}{340} = \sqrt{\frac{363}{300}} \]

4. **Calculate the right side**:
- First, calculate the fraction:
\[ \frac{363}{300} = 1.21 \]
- Now take the square root:
\[ \sqrt{1.21} = 1.1 \]

5. **Calculate \( v_2 \)**:
- Now substitute back to find \( v_2 \):
\[ v_2 = 340 \times 1.1 = 374 \, m/s \]

6. **Final Answer**:
- The speed of sound in the gas at 90°C is \( 374 \, m/s \).