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Worked Examples · Example 14.4

Q.Estimate the speed of sound in air at standard temperature and pressure. The mass of 11 mole of air is 29.0×10−3 kg29.0 \times 10^{-3}\ \text{kg}.

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Using Newton's formula v=P/ρv=\sqrt{P/\rho} (the isothermal approximation) with air's density at STP, ρ0=1.29 kg/m3\rho_0=1.29\ \text{kg/m}^3, gives v≈280 m/sv\approx\boxed{280\ \text{m/s}} -- but this is about 15% lower than the measured speed of sound, 331 m/s331\ \text{m/s}. The gap exists because Newton assumed the compressions/rarefactions in a sound wave are slow enough to stay isothermal; Laplace later corrected this, since they're actually adiabatic, and the corrected formula reproduces the measured value almost exactly.

Newton's formula and its assumption

A sound wave in a gas is a longitudinal pressure disturbance. Newton assumed the temperature stays constant as the gas compresses and expands (an isothermal process), which for an ideal gas gives the bulk modulus B=PB=P, and therefore:

v=Pρv = \sqrt{\frac{P}{\rho}}

Step-by-step estimate

1. Find the density of air at STP. 1 mole of any ideal gas occupies 22.422.4 litres (22.4×10−3 m322.4\times10^{-3}\ \text{m}^3) at STP, and 1 mole of air has mass 29.0×10−3 kg29.0\times10^{-3}\ \text{kg}:

ρ0=29.0×10−3 kg22.4×10−3 m3=1.29 kg/m3\rho_0 = \frac{29.0\times10^{-3}\ \text{kg}}{22.4\times10^{-3}\ \text{m}^3} = 1.29\ \text{kg/m}^3

2. Apply Newton's formula, using standard atmospheric pressure P=1.013×105 PaP=1.013\times10^5\ \text{Pa}:

v=Pρ0=1.013×1051.29=7.85×104≈280 m/sv=\sqrt{\frac{P}{\rho_0}}=\sqrt{\frac{1.013\times10^5}{1.29}}=\sqrt{7.85\times10^4}\approx280\ \text{m/s}

v=Pρ≈280 m/s (Newton’s formula, air at STP)v = \sqrt{\frac{P}{\rho}} \approx 280\ \text{m/s (Newton's formula, air at STP)}

Why this is wrong by about 15% -- the Laplace correction

The measured speed of sound in dry air at 0∘C0^\circ\text{C} is 331 m/s331\ \text{m/s} -- Newton's estimate is roughly 15% too low. The error is in the assumption itself: sound's compressions and rarefactions happen far too fast for heat to flow and keep the temperature constant, so the process is actually adiabatic, not isothermal. For an adiabatic process the bulk modulus is Bad=γPB_{\text{ad}}=\gamma P (where γ=Cp/Cv\gamma=C_p/C_v), giving the corrected formula:

v=γPρv = \sqrt{\frac{\gamma P}{\rho}}

For air, γ=7/5=1.4\gamma = 7/5 = 1.4: …

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