The velocity of sound in a gas at temperature `27^@C` is `V` then in the same gas its velocity will be `2V` at temperature.

Options

A. `54^(@)C`
B. `327^(@)C`
C. `927^(@)C`
D. `108^(@)C`

Answer

To solve the problem, we need to determine the temperature at which the velocity of sound in a gas doubles from its initial value at 27°C. The relationship between the velocity of sound and temperature in a gas can be expressed with the formula:

\[ V \propto \sqrt{T} \]

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The initial temperature \( T_1 \) is given as \( 27^\circ C \).
- Convert this temperature to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
- The initial velocity of sound is \( V \).

2. **Establish the Relationship**:
- According to the formula, the velocity of sound \( V \) is proportional to the square root of the absolute temperature \( T \):
\[ V \propto \sqrt{T} \]
- If the velocity doubles, we have:
\[ 2V \propto \sqrt{T_2} \]
- Where \( T_2 \) is the new temperature at which the velocity is \( 2V \).

3. **Set Up the Proportionality**:
- From the proportionality, we can write:
\[ \frac{2V}{V} = \frac{\sqrt{T_2}}{\sqrt{T_1}} \]
- Simplifying gives:
\[ 2 = \sqrt{\frac{T_2}{T_1}} \]

4. **Square Both Sides**:
- Squaring both sides to eliminate the square root:
\[ 4 = \frac{T_2}{T_1} \]

5. **Solve for \( T_2 \)**:
- Rearranging gives:
\[ T_2 = 4T_1 \]
- Substituting \( T_1 = 300 \, K \):
\[ T_2 = 4 \times 300 = 1200 \, K \]

6. **Convert \( T_2 \) Back to Celsius**:
- Convert \( T_2 \) from Kelvin back to Celsius:
\[ T_2 = 1200 - 273 = 927 \, ^\circ C \]

### Final Answer:
The temperature at which the velocity of sound in the gas will be \( 2V \) is \( 927^\circ C \).

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