At what temperatures (in $^{\circ}$C) will the speed of sound in air be 3 times its value at $0^{\circ}$C?

At what temperatures (in °C) will the speed of sound in air be 3 times its value at 0°C?

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Concept understanding — Speed of Sound in Gases

Speed of Sound in Gases – From Intuition to Precision

Imagine you're standing at one end of a long, empty hallway. Your friend is at the other end. When you clap your hands, the sound doesn't reach them instantly — it takes a small but noticeable fraction of a second. That delay is the speed of sound in air.

Now think about why sound travels at all. Sound is a mechanical wave — it needs a medium (like air, water, or steel) to travel. When you clap, you push the air molecules near your hands. Those molecules bump into their neighbours, which bump into the next ones, and so on. This chain of collisions carries the disturbance forward. The speed at which this "bump" travels depends on two things:

How stiff the medium is — how quickly it resists being compressed.
How heavy the medium is — how much inertia each molecule has.

In a gas, both of these are linked to temperature and the gas's molecular properties.

The Precise Statement

For an ideal gas, the speed of sound $v$ is given by:

v = \sqrt{\frac{\gamma R T}{M}}

Where:

$\gamma$ (gamma) is the adiabatic index — the ratio of specific heats $C_p / C_v$ . For air (mostly diatomic gases like N₂ and O₂), $\gamma \approx 1.4$ .
$R$ is the universal gas constant ( $8.314\ \text{J/mol·K}$ ).
$T$ is the absolute temperature in Kelvin.
$M$ is the molar mass of the gas (in kg/mol).

v = \sqrt{\frac{\gamma R T}{M}}

This formula tells you three key things:

Speed increases with temperature — hotter gas means faster molecules, so the disturbance propagates quicker.
Speed decreases with heavier molecules — a gas like helium (small $M$ ) has a much higher speed of sound than air. In helium, your voice sounds squeaky because sound travels faster, changing the resonance in your throat.
The factor $\gamma$ matters — it accounts for the fact that compressions and rarefactions in a sound wave happen so fast that heat doesn't have time to flow. The process is adiabatic, not isothermal.

Why Adiabatic? (The "Why" Behind the Formula)

When a sound wave passes through a gas, the pressure and volume change rapidly — hundreds or thousands of times per second. There's no time for heat to flow from the compressed (hotter) regions to the rarefied (cooler) regions. So the gas behaves as if it's thermally isolated. That's why $\gamma$ appears instead of 1 (which would be the isothermal case).

If you used the isothermal assumption, you'd get $v = \sqrt{RT/M}$ , which is about 20% too low for air. The correct adiabatic formula matches experiments beautifully.

A Quick Numerical Check

At room temperature ( $T = 293\ \text{K}$ ), for air ( $M \approx 0.029\ \text{kg/mol}$ , $\gamma = 1.4$ ):

v = \sqrt{\frac{1.4 \times 8.314 \times 293}{0.029}} \approx \sqrt{117,600} \approx 343\ \text{m/s}

That's about 1235 km/h — the familiar value you've probably heard. …

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The speed of sound in a gas is proportional to the square root of its absolute temperature, $v\propto\sqrt T$ . Tripling the speed requires the absolute temperature to increase ninefold; starting from $0^\circ\text{C}=273\ \text{K}$ , this gives $T=2457\ \text{K}$ , i.e. $\boxed{2184^\circ\text{C}}$ .

The governing relationship

For an ideal gas, the speed of sound is $v = \sqrt{\dfrac{\gamma RT}{M}}$ , where $\gamma$ , $R$ , and $M$ are constants for a given gas (air). So for the same gas, $v$ depends on temperature only through $\sqrt T$ : $v \propto \sqrt T$ . Crucially, $T$ here must be the absolute temperature (Kelvin).

Setting up the ratio

Let $v_0$ be the speed of sound at $T_0=0^\circ\text{C}$ , and $v$ the speed at the unknown temperature $T$ , with $v=3v_0$ . The constants cancel in the ratio:

\frac{v}{v_0} = \sqrt{\frac{T}{T_0}} \quad\Rightarrow\quad 3 = \sqrt{\frac{T}{T_0}}

Solving for T

Squaring both sides: $9 = \dfrac{T}{T_0} \Rightarrow T = 9T_0$ .

Using $T_0 = 0^\circ\text{C} = 273\ \text{K}$ (the standard exam convention):

$T = 9\times273 = 2457\ \text{K}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$Q. A soap bubble is given a negative charge. The pressure inside the bubble (A) increases (B) becomes equal to atmospheric pressure (C) does not change (D) decreases$
›Reveal solutionSolution
When a soap bubble is given a negative charge, the repulsive forces between the charges create an outward electrostatic pressure. This outward pressure effectively reduces the net inward force that the internal pressure needs to overcome, leading to a decrease in the pressure inside the bubble. The pressure inside the bubble decreases.

Concept and Intuition

A soap bubble is a thin film of liquid enclosing a volume of gas. Due to surface tension, the liquid film tries to contract, which creates an excess pressure inside the bubble compared to the outside atmospheric pressure. This excess pressure is what keeps the bubble from collapsing.

When a charge (negative in this case) is given to a conducting sphere like a soap bubble, these charges distribute uniformly over its outer surface. Since like charges repel each other, there will be an outward electrostatic force acting perpendicularly on every point of the bubble's surface. This outward force can be thought of as an "electrostatic pressure" pushing the bubble's surface outwards.

This electrostatic pressure acts in the same direction as the internal pressure of the bubble. Therefore, it assists the internal pressure in counteracting the inward pull of surface tension and the external atmospheric pressure. To maintain equilibrium, the internal pressure no longer needs to be as high as it was in the uncharged state.

Step-by-step Derivation

Pressure balance in an uncharged soap bubble: For an uncharged soap bubble, the pressure inside ( $P_{in}$ ) is greater than the pressure outside ( $P_{out}$ ) due to surface tension. The excess pressure is given by:

$P_{in} - P_{out} = \frac{4T}{R}$
where $T$ is the surface tension of the soap solution and $R$ is the radius of the bubble. This equation can be rearranged to express the internal pressure:
$P_{in} = P_{out} + \frac{4T}{R} \quad \text{(Equation 1)}$

Effect of charging the soap bubble: When the soap bubble is given a negative charge, these charges distribute uniformly over its surface. The mutual repulsion between these charges creates an outward electrostatic pressure ( $P_e$ ) on the surface of the bubble. This electrostatic pressure is given by:

$P_e = \frac{\sigma^2}{2\epsilon_0}$
where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space. This pressure acts outwards, perpendicular to the surface.

3. New pressure balance in a charged soap bubble:

Now, consider the forces acting on the bubble's surface. The internal pressure ( $P_{in}$ ) and the electrostatic pressure ( $P_e$ ) both act outwards. The external atmospheric pressure ( $P_{out}$ ) and the pressure due to surface tension ( $\frac{4T}{R}$ ) both act inwards. For the bubble to be in equilibrium, the total outward pressure must balance the total inward pressure:
$P_{in} + P_e = P_{out} + \frac{4T}{R}$
Rearranging this equation to find the internal pressure ($P_{in}$) of the charged bubble: …
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$P_1$
›Reveal solutionSolution
The key is to relate the initial and final pressures for each process using the given ratio of specific heats ( $\gamma = 1.5$ ), then equate the final pressures to find the ratio of initial pressures. The result is $\sqrt{2} : 1$ , so option (D) is correct.

Concept and Intuition

We have two different thermodynamic processes—adiabatic and isothermal—each starting from a different initial pressure but ending at the same final pressure after doubling the volume. The ratio of specific heats $\gamma = C_p/C_v = 1.5$ tells us the gas is polyatomic (or diatomic with vibration frozen, but that's not crucial). For an adiabatic process, pressure and volume obey $PV^\gamma = \text{constant}$ . For an isothermal process, $PV = \text{constant}$ . By writing the final pressure in terms of the initial pressure for each process, then setting them equal, we can solve for the ratio of the initial pressures.

Step-by-step solution

Adiabatic process For an adiabatic change, $P V^\gamma = \text{constant}$ . Let the initial pressure be $P_{1i}$ and initial volume $V$ . After doubling the volume to $2V$ , the pressure becomes $P_1$ . So:

$P_{1i} V^\gamma = P_1 (2V)^\gamma$ $P_1 = P_{1i} \left(\frac{V}{2V}\right)^\gamma = P_{1i} \left(\frac{1}{2}\right)^\gamma$
With $\gamma = 1.5 = \frac{3}{2}$ :
$P_1 = P_{1i} \left(\frac{1}{2}\right)^{3/2} = \frac{P_{1i}}{2^{3/2}} = \frac{P_{1i}}{2\sqrt{2}}$

Isothermal process For an isothermal change, $P V = \text{constant}$ . Let the initial pressure be $P_{2i}$ and initial volume $V$ . After doubling the volume to $2V$ , the pressure becomes $P_2$ . So:

$P_{2i} V = P_2 (2V)$ $P_2 = \frac{P_{2i}}{2}$

Given that $P_1 = P_2$ Equate the expressions: …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$v$
›Reveal solutionSolution
The pressure of an ideal gas depends on the product of molecular mass and the square of the rms speed. Using $P \propto m v^2$ , the ratio of initial to final pressure is $\frac{9}{2}$ , so the correct option is (B).

Concept & Intuition

For an ideal gas, pressure arises from molecules colliding with the walls. The kinetic theory shows that pressure is proportional to the average kinetic energy per molecule times the number density. Since the number density is constant here (the gas amount and volume are unchanged), pressure depends only on the average translational kinetic energy: $P \propto \frac{1}{2} m v_{\text{rms}}^2$ . So when mass and rms speed change, the pressure changes as $P \propto m v^2$ . This direct proportionality lets us compare pressures without needing the gas constant or temperature.

Step-by-step reasoning

Write the proportionality for pressure From kinetic theory, for a fixed number of molecules and fixed volume:

$P \propto \frac{1}{2} m v^2 \quad \Rightarrow \quad P \propto m v^2$
The constant factor $\frac{1}{2}$ cancels in a ratio.

Define initial and final conditions

Initial: mass $m$ , rms speed $v$ → $P_i \propto m v^2$

Final: mass $2m$ , rms speed $v/3$ → $P_f \propto (2m) \left(\frac{v}{3}\right)^2$

Compute the final proportionality
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TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$\dfrac{C_p}{C_v} = \dfrac{5}{3}$
›Reveal solutionSolution
For an adiabatic process, $TV^{\gamma-1}$ is constant. Using $\gamma = \frac{5}{3}$ and initial temperature $T_1 = 300\ \text{K}$ , the final temperature after expanding to $3V$ is $T_2 \approx 144.2\ \text{K}$ , which matches option (A).

The key concept here is the adiabatic process — a thermodynamic change where no heat is exchanged with the surroundings ( $Q = 0$ ). For an ideal gas undergoing a reversible adiabatic expansion, the relation between temperature and volume is given by $TV^{\gamma-1} = \text{constant}$ , where $\gamma = \frac{C_p}{C_v}$ is the adiabatic index. This relation comes directly from combining the ideal gas law with the first law of thermodynamics and the condition $dQ = 0$ .

Why does temperature drop when volume increases adiabatically? Because the gas does work on its surroundings while expanding, and with no heat coming in, that work is done at the expense of its internal energy. For a monoatomic gas, internal energy depends only on temperature, so a decrease in internal energy means a drop in temperature. The formula $TV^{\gamma-1} = \text{constant}$ captures exactly this trade-off.

Now let’s apply it step by step.

Identify the given data

Initial volume: $V_1 = V$

Final volume: $V_2 = 3V$

Initial temperature: $T_1 = 27^\circ\text{C} = 27 + 273 = 300\ \text{K}$

Adiabatic index: $\gamma = \frac{5}{3}$

We need $T_2$ in Kelvin.

Write the adiabatic relation

For a reversible adiabatic process:

$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$

Substitute the values $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$ So:

$300 \cdot V^{\frac{2}{3}} = T_2 \cdot (3V)^{\frac{2}{3}}$

Simplify the volume ratio $T_2 = 300 \cdot \left( \frac{V}{3V} \right)^{\frac{2}{3}} = 300 \cdot \left( \frac{1}{3} \right)^{\frac{2}{3}}$ …
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Q.At what temperatures (in ∘^{\circ}∘C) will the speed of sound in air be 3 times its value at 0∘0^{\circ}0∘C?

The governing relationship

Setting up the ratio

Solving for T

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Q.At what temperatures (in $^{\circ}$ C) will the speed of sound in air be 3 times its value at $0^{\circ}$ C?

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