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.

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\sqrt{2}$
›Reveal solutionSolution
The rms-speed-to-sound-speed ratio depends only on $\gamma$ (the molar mass cancels), so the given ratio $\sqrt2$ pins $\gamma_{mix}=3/2$ . Writing $\gamma_{mix}$ for the He + H $_2$ mixture in terms of $n$ and solving gives $n=2$ .

Concept and Intuition

For any ideal gas, $v_{rms}=\sqrt{3RT/M}$ and the adiabatic speed of sound is $v_s = \sqrt{\gamma RT/M}$ . Taking the ratio, the molar mass $M$ (or the mixture's mean molar mass, since it appears identically in both formulas) cancels entirely, leaving $v_{rms}/v_s = \sqrt{3/\gamma}$ — a purely $\gamma$ -dependent number. This is what lets us pin down $\gamma_{mix}$ directly from the given speed ratio, without ever needing the actual molar masses.

Step-by-Step Solution

Given $v_{rms} = \sqrt2\, v_s$ :

$\sqrt{\frac{3RT}{M}} = \sqrt2\sqrt{\frac{\gamma RT}{M}} \implies 3 = 2\gamma \implies \gamma_{mix} = \frac32$

For the mixture, $\gamma_{mix} = \dfrac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$ . Helium (monatomic): $C_{v1}=\frac32 R$ , $C_{p1}=\frac52 R$ , with $n_1=2$ mol. Hydrogen (diatomic): $C_{v2}=\frac52R$ , $C_{p2}=\frac72R$ , with $n_2=n$ mol. …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$\sqrt{9/5}$
›Reveal solutionSolution
The ratio of sound speed to rms molecular speed in any ideal gas reduces to $\sqrt{\gamma/3}$ ; for a diatomic gas $\gamma=7/5$ , giving $\sqrt{7/15}$ .

Concept and Intuition

Sound in a gas propagates via adiabatic pressure disturbances, so its speed is $v_s=\sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M}$ (Laplace's correction). The rms speed of the molecules themselves comes from kinetic theory: $v_{rms}=\sqrt{3RT/M}$ . Both share the same $RT/M$ factor, so their ratio depends only on $\gamma$ .

Step-by-Step Solution

Write $v_s=\sqrt{\gamma RT/M}$ and $v_{rms}=\sqrt{3RT/M}$ .

Divide: $\dfrac{v_s}{v_{rms}}=\sqrt{\dfrac{\gamma RT/M}{3RT/M}}=\sqrt{\dfrac{\gamma}{3}}$ .

For a diatomic gas, $\gamma=C_p/C_v=7/5=1.4$ . …
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AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
$22.4\times10^{-3}\ \mathrm{kg}$
›Reveal solutionSolution
Laplace's formula for the speed of sound in a gas, evaluated at STP with the given molar mass and $\gamma$ , gives about 402 m/s.

Concept and Intuition

The speed of sound in an ideal gas is given by Laplace's corrected formula $v=\sqrt{\gamma RT/M}$ , where $M$ is the molar mass (mass of one mole) and $\gamma=C_p/C_v$ .

Step-by-Step Solution

$\gamma=1.6$ , $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ , $T=273\ \mathrm{K}$ (STP), $M=22.4\times10^{-3}\ \mathrm{kg\,mol^{-1}}$ .

Numerator: $\gamma RT = 1.6\times8.314\times273 \approx 3631.6$ . …
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AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
$3.4\times10^2\,\mathrm{m\,s^{-1}}$
›Reveal solutionSolution
The familiar value of the speed of sound in air near room temperature ( $\approx20$ – $25^\circ$ C) is about 343 m/s, matching $3.4\times10^2$ m/s.

Concept and Intuition

Sound speed in an ideal gas is $v=\sqrt{\gamma RT/M}$ ; plugging in air's values at room temperature yields the well-known ~340 m/s figure used throughout physics (e.g. in resonance-tube and Doppler problems).

Step-by-Step Solution

Recall/derive $v=\sqrt{\gamma RT/M}$ with $\gamma=1.4$ , $T\approx300\,\mathrm K$ , $M=0.029\,\mathrm{kg/mol}$ , $R=8.31\,\mathrm{J\,mol^{-1}K^{-1}}$ . …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$O_2$
›Reveal solutionSolution
Comparing $v=\sqrt{\gamma RT/M}$ for $O_2$ and $He$ at the same temperature reduces to a ratio of $\sqrt{\gamma/M}$ , giving a speed roughly three times higher for the much lighter, monatomic helium.

Concept and Intuition

Speed of sound in an ideal gas depends on both its adiabatic index $\gamma$ (monatomic gases have $\gamma=5/3$ , diatomic $\gamma=7/5$ ) and its molar mass $M$ — lighter, more mono-atomic gases transmit sound faster.

Step-by-Step Solution

$v=\sqrt{\dfrac{\gamma RT}{M}}$ . At the same $T$ , $\dfrac{v_{He}}{v_{O_2}}=\sqrt{\dfrac{\gamma_{He}/M_{He}}{\gamma_{O_2}/M_{O_2}}}=\sqrt{\dfrac{\gamma_{He}}{\gamma_{O_2}}\times\dfrac{M_{O_2}}{M_{He}}}$ .

Using $\gamma_{He}=5/3$ (monatomic), $\gamma_{O_2}=7/5$ (diatomic), $M_{O_2}=32$ , $M_{He}=4$ : …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$0^\circ C$
›Reveal solutionSolution
Since sound speed in a gas scales as $\sqrt{T}$ (absolute temperature), a $10\%$ speed increase from $0^\circ C$ corresponds to a temperature rise to about $57^\circ C$ .

Concept and Intuition

The speed of sound in an ideal gas is $v = \sqrt{\gamma RT/M}$ , so $v \propto \sqrt{T}$ where $T$ is the absolute temperature. A fractional increase in speed corresponds to a squared fractional increase in absolute temperature.

Step-by-Step Solution

Let $v$ be the speed at $T=273\ K$ ( $0^\circ C$ ), and $v' = 1.10v$ the increased speed at temperature $T'$ .

Since $v\propto\sqrt{T}$ :

$\frac{v'}{v} = \sqrt{\frac{T'}{T}} \Rightarrow \left(\frac{v'}{v}\right)^2 = \frac{T'}{T}$

Substitute $v'/v = 1.10$ : $T' = T\times(1.10)^2 = 273\times1.21 = 330.33\ K$ …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$Q. Speed of a sound in air ________ (A) is independent of temperature (B) increases with pressure (C) increases with increase in humidity (D) decreases with increase in humidity$
›Reveal solutionSolution
Humid air is effectively "lighter" (lower average molar mass) than dry air, and since the speed of sound scales as $1/\sqrt{M}$ , sound travels faster through humid air.

Concept and Intuition

The speed of sound in a gas is given by $v=\sqrt{\dfrac{\gamma RT}{M}}$ , where $M$ is the molar mass of the gas. Water vapour molecules ( $H_2O$ , molar mass $\approx18\ g/mol$ ) are lighter than the average nitrogen/oxygen molecules making up dry air ( $\approx29\ g/mol$ ). So adding water vapour (increasing humidity) lowers the overall average molar mass of the air mixture, which increases the speed of sound (since $v \propto 1/\sqrt M$ ).

Step-by-Step Solution

Speed of sound formula: $v = \sqrt{\dfrac{\gamma RT}{M}}$ .

Humid air is a mixture where some heavier $N_2$ / $O_2$ molecules are effectively replaced by lighter $H_2O$ molecules, lowering the mixture's average molar mass $M$ .

Since $v\propto 1/\sqrt M$ , a lower $M$ means a higher speed of sound. …
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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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