When the temperature of an ideal gas is increased by 600 K , the velocity of sound in the gas becomes `sqrt(3)` times the initial velocity in it. The initial temperature of the gas is

Options

A. `-73^(@)C`
B. `27^(@)C`
C. `127 ^(@)C`
D. `327 ^(@)C`

Answer

To solve the problem step by step, we will use the relationship between the velocity of sound in an ideal gas and its temperature.

### Step-by-Step Solution:

1. **Understand the relationship between velocity and temperature**:
The velocity of sound \( V \) in an ideal gas is given by the formula:
\[ V = \sqrt{\frac{\gamma RT}{M}} \]
where:
- \( \gamma \) is the adiabatic index (ratio of specific heats),
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

2. **Set up the initial and final conditions**:
Let the initial temperature be \( T_1 \) and the final temperature after an increase of 600 K be \( T_2 = T_1 + 600 \).
The problem states that the final velocity \( V_2 \) is \( \sqrt{3} \) times the initial velocity \( V_1 \):
\[ V_2 = \sqrt{3} V_1 \]

3. **Express the velocities in terms of temperatures**:
Using the relationship of velocities with temperatures:
\[ \frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}} \]
Substituting \( V_2 = \sqrt{3} V_1 \) into the equation gives:
\[ \sqrt{3} = \sqrt{\frac{T_1 + 600}{T_1}} \]

4. **Square both sides to eliminate the square root**:
Squaring both sides results in:
\[ 3 = \frac{T_1 + 600}{T_1} \]

5. **Cross-multiply to solve for \( T_1 \)**:
Rearranging gives:
\[ 3T_1 = T_1 + 600 \]
Simplifying this leads to:
\[ 3T_1 - T_1 = 600 \implies 2T_1 = 600 \]

6. **Solve for \( T_1 \)**:
Dividing both sides by 2:
\[ T_1 = 300 \text{ K} \]

7. **Convert Kelvin to Celsius (if needed)**:
To convert Kelvin to Celsius:
\[ T_1 = 300 \text{ K} - 273 = 27 \text{ °C} \]

### Final Answer:
The initial temperature of the gas is **27 °C**.