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.
Tip
A handy rule: For dry air at 0°C, the speed of sound is about 331 m/s. For every 1°C rise, it increases by roughly 0.6 m/s. So at 20°C, it's about 331 + 12 = 343 m/s.
Common Mistake to Avoid
Watch out
Do not use the formula v=P/ρ (which comes from the bulk modulus) without remembering that for a gas, the bulk modulus is γP, not P. The correct form is v=γP/ρ, which is equivalent to the formula above because P=ρRT/M for an ideal gas.
Why This Matters for Exams
You'll often be asked to compare speeds in different gases (e.g., He vs. air, or H₂ vs. O₂) at the same temperature. Use v∝1/M.
Temperature dependence is a favourite: v∝T.
The adiabatic nature is a conceptual question — be ready to explain why γ appears.
Sound in gases is a beautiful blend of thermodynamics and wave physics. The formula isn't just a plug-and-chug tool — it tells a story about how molecules jostle and pass along energy.
This is exactly the kind of concept that turns up under searches like "Speed of Sound in Gases class 11 physics syllabus" or "Speed of Sound in Gases solved examples" — and it belongs squarely in the Class 11 Physics NCERT/CBSE curriculum. Beyond board exams, it's a dependable scoring topic in JEE Main, NEET and state engineering/medical entrance exams once the core logic clicks.
For the He+O2 mixture, γmix=19/13, Mmix=136/6g/mol; v=γmixRT/Mmix≈401 m/s.
✓Final answer
v≈400.9m/s
Step 1. For a gas mixture, the effective specific heats are mole-weighted averages: for helium (monatomic, CV1=23R) and oxygen (diatomic, CV2=25R), with n1=2mol He and n2=4mol O2: CV,mix=n1+n2n1CV1+n2CV2=62(1.5R)+4(2.5R)=63R+10R=613R.
Step 2. Correspondingly CP,mix=CV,mix+R=613R+R=619R, so γmix=CP,mix/CV,mix=19/13≈1.4615.
Step 3. The mean molar mass is Mmix=n1+n2n1M1+n2M2=62(4)+4(32)=68+128=6136≈22.67g/mol=0.02267kg/mol.
Step 4. Speed of sound: v=MmixγmixRT=0.022671.4615×8.314×300=0.022673645.6=160834≈401.0m/s.
✓Final answer
v≈400.9m/s
Compute mole-weighted γmix and Mmix for the mixture, then apply v=γRT/M.
Using a simple average of the two individual gas speeds instead of computing the mixture's own effective γ and M.