Question

Speed of sound at \(T_1 = 0^\circ C\) is \(V_0\) and at \(T_2 = \alpha^\circ C\) speed becomes \(2V_0\). Find \(\alpha\):

• A. \(819^\circ C\)
• B. \(918^\circ C\)
• C. \(546^\circ C\)
• D. \(1092^\circ C\)

Hint:

A very common mistake in thermodynamics and gas-related problems is forgetting to convert temperatures from Celsius to Kelvin.
All formulas involving temperature (\(T\)) in gas laws (like ideal gas law, speed of sound) require the absolute temperature in Kelvin.

Answer 1

The Correct Option is A

To solve the problem of finding the temperature \(\alpha\) at which the speed of sound becomes \(2V_0\), we must first understand how the speed of sound in air changes with temperature.

The speed of sound in air is given by the formula: 

\(V = V_0 \sqrt{\frac{T}{T_0}}\)

Where,

• \(V\) is the speed of sound at temperature \(T\).
• \(V_0\) is the speed of sound at 0°C (273 K).
• \(T\) is the temperature in Kelvin at which we want the speed of sound.
• \(T_0 = 273 \text{ K}\) is the reference temperature corresponding to 0°C.\)

Initially, at \(T_1 = 0^\circ C\), the speed of sound is \(V_0\).

At \(T_2 = \alpha^\circ C\), the speed of sound becomes \(2V_0\).

Given: \(V = 2V_0\),

Substitute this into the speed formula:

\(2V_0 = V_0 \sqrt{\frac{T_2}{273}}\)

Cancel \(V_0\) from both sides:

\(2 = \sqrt{\frac{T_2}{273}}\)

Square both sides to eliminate the square root:

\(4 = \frac{T_2}{273}\)

Rearrange to find \(T_2\):

\(T_2 = 4 \times 273\)

Calculating:

\(T_2 = 1092 \text{ K}\)

Convert Kelvin back to Celsius:

\(\alpha = T_2 - 273 = 1092 - 273 = 819^\circ C\)

Therefore, the correct answer is \(819^\circ C\).