Speed of sound in gases — Newton's vs Laplace correction

Question

Newton assumed that sound travels through an isothermal process in gases. What value of the speed of sound in air did his formula give, and why was it wrong? How did Laplace correct it? Calculate the speed of sound in air at 0°C using Laplace’s formula.

(NCERT Class 11, Waves)

Solution — Step by Step

Newton proposed that when sound travels through air, the compressions and rarefactions occur isothermally (at constant temperature). For an isothermal process in an ideal gas:

PV = \text{constant} \implies B_T = P

where $B_T$ is the isothermal bulk modulus, equal to the pressure $P$ .

Newton’s formula: $v = \sqrt{B_T/\rho} = \sqrt{P/\rho}$

At STP ( $P = 1.013 \times 10^5$ Pa, $\rho_{\text{air}} = 1.293$ kg/m³):

v = \sqrt{\frac{1.013 \times 10^5}{1.293}} = \sqrt{78328} \approx 280 \text{ m/s}

The experimentally measured value is 332 m/s at 0°C. Newton’s value is about 15% too low.

Laplace realised that sound compressions and rarefactions happen so rapidly that there’s no time for heat exchange with the surroundings. The process is adiabatic, not isothermal.

For an adiabatic process: $PV^\gamma = \text{constant}$

The adiabatic bulk modulus is:

B_{\text{ad}} = \gamma P

where $\gamma = C_p/C_v$ is the ratio of specific heats. For air (diatomic gas), $\gamma = 7/5 = 1.4$ .

v = \sqrt{\frac{B_{\text{ad}}}{\rho}} = \sqrt{\frac{\gamma P}{\rho}}

This is called the Newton-Laplace formula for the speed of sound.

v = \sqrt{\frac{1.4 \times 1.013 \times 10^5}{1.293}} = \sqrt{\frac{1.418 \times 10^5}{1.293}} = \sqrt{109659} \approx \mathbf{331 \text{ m/s}}

This matches the experimental value of 332 m/s at 0°C almost perfectly.

Why This Works

The key physical insight is about how fast the process happens. Sound in air at room temperature has a frequency of, say, 1000 Hz. Each compression-rarefaction cycle takes about 1 ms — far too fast for the compressed air to exchange heat with its surroundings and equalise temperature.

In an isothermal process ( $B = P$ ), the gas is “softer” — it compresses more easily because heat flows out to keep the temperature constant. In an adiabatic process ( $B = \gamma P$ ), the gas is “stiffer” — compression also raises temperature, which increases pressure further, resisting compression more. A stiffer medium transmits sound faster.

The factor $\gamma = 1.4$ for air makes Newton’s value $\sqrt{1.4} \approx 1.18$ times larger — exactly the correction needed.

Alternative Method

Using the ideal gas law $PV = nRT$ , we can write:

v = \sqrt{\frac{\gamma RT}{M}}

where $M$ is the molar mass and $T$ is the absolute temperature. For air ( $M \approx 29 \times 10^{-3}$ kg/mol) at 273 K:

v = \sqrt{\frac{1.4 \times 8.314 \times 273}{0.029}} = \sqrt{109541} \approx 331 \text{ m/s}

The formula $v = \sqrt{\gamma RT/M}$ is more useful for JEE because it shows the temperature dependence directly: $v \propto \sqrt{T}$ . This means if temperature doubles (in Kelvin), speed increases by $\sqrt{2}$ . At 27°C (300 K), $v = 331 \times \sqrt{300/273} \approx 347$ m/s.

Common Mistake

Students sometimes write that Newton’s formula was “completely wrong.” It wasn’t — the formula $v = \sqrt{B/\rho}$ is correct for any mechanical wave. Newton’s error was in using the wrong value of bulk modulus (isothermal instead of adiabatic). The framework was right; only the assumption about the thermodynamic process was wrong. Laplace didn’t change the formula — he corrected the value of $B$ that goes into it.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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