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=MγRT
Where:
γ (gamma) is the adiabatic index — the ratio of specific heats Cp/Cv. For air (mostly diatomic gases like N₂ and O₂), γ≈1.4.
R is the universal gas constant (8.314J/mol⋅K).
T is the absolute temperature in Kelvin.
M is the molar mass of the gas (in kg/mol).
v=MγRT
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 γ 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 γ appears instead of 1 (which would be the isothermal case).
If you used the isothermal assumption, you'd get v=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=293K), for air (M≈0.029kg/mol, γ=1.4):
v=0.0291.4×8.314×293≈117,600≈343m/s
That's about 1235 km/h — the familiar value you've probably heard. …
The key idea is that the speed of sound in a gas depends on the ratio of pressure to density, which is linked to temperature and molecular mass via the ideal gas law.
Reasoning:
From v=ργP, use the ideal gas law P=MρRT (where M is molar mass). Substituting gives v=MγRT.
(a) Independence of pressure: In the expression v=MγRT, pressure P does not appear. A change in P at constant T causes a proportional change in ρ, so the ratio P/ρ remains constant.
(b) Increase with temperature: From v∝T, raising T directly increases the numerator, so v increases. …
Using the ideal-gas relation ρ=PM/RT turns the formula into v=γRT/M. Pressure cancels (part a); the explicit T gives v∝T (part b); and humid air has a smaller average molar mass M, so v rises (part c).
The key step
v=ργP
For an ideal gas, PV=nRT gives the density ρ=VnM=RTPM, where M is the molar mass. Substituting removes the explicit pressure:
v=PM/RTγP=MγRT
(a) Independent of pressure
At a fixed temperature, increasing the pressure increases the density in exactly the same proportion, so the ratio P/ρ is unchanged. In v=γRT/M the pressure has cancelled entirely, so the speed of sound in air does not depend on pressure.
(b) Increases with temperature
Temperature appears directly under the root, so
v∝T
A rise in temperature raises the molecular speeds, and the disturbance is passed on faster, so the speed of sound increases.
Step 1: Start from v=γP/ρ and eliminate ρ using the ideal gas law P=ρRT/M, so ρ=PM/RT, giving v=γRT/M.
Step 2 (a): Pressure P has cancelled out entirely from this form — at fixed temperature, ρ scales with P so the ratio P/ρ is unchanged, meaning v does not depend on pressure.
Step 3 (b):T appears explicitly under the square root, so v∝T — speed rises as temperature rises. …