4.4 The equation of state and the adiabatic assumption

Continuity gave us $\partial_t \rho' + \rho_0 \nabla \cdot \mathbf{v}' = 0$ . Euler gave us $\rho_0 \partial_t \mathbf{v}' = -\nabla p'$ . Two equations, three unknowns ( $p'$ , $\rho'$ , $\mathbf{v}'$ ). To close the system we need one more relation. That relation is the equation of state: a thermodynamic statement about how pressure depends on density.

Why we need a third equation

A general fluid has more than one independent state variable. For an ideal gas at equilibrium, any two of pressure, density, temperature determine the third (via $p = \rho R T / M$ ). But when a sound wave passes through, all three change. Without an extra constraint, the three perturbations $p'$ , $\rho'$ , $T'$ are independent — and we can’t solve the system.

The extra constraint comes from physics: what process is the gas undergoing as the sound wave compresses and rarefies it?

Isothermal vs. adiabatic

Two limiting cases:

Isothermal: the gas exchanges heat with its surroundings so quickly that its temperature stays at equilibrium. Then $p$ is a function of $\rho$ alone, and the relation is $p / p_0 = \rho / \rho_0$ , i.e. $p' / p_0 = \rho' / \rho_0$ . This is Newton’s 1687 model for the speed of sound. It gives 280 m/s for air — and was known to be wrong by about 20%.
Adiabatic: the gas exchanges no heat with its surroundings over the timescale of the oscillation. Then $p \rho^{-\gamma}$ is conserved (with $\gamma = c_p / c_v$ the ratio of specific heats), giving $p' / p_0 = \gamma\, \rho' / \rho_0$ . This is Laplace’s 1816 correction and gives 343 m/s — the correct answer.

Which one is right for sound? The adiabatic case. The reason: heat conduction in air is slow compared to acoustic oscillation periods. In one period of a 1 kHz tone ( $\sim 1\,$ ms), heat conducts over a distance of about $1\,\mu$ m. The local compression is over a distance of half a wavelength ( $\sim 17\,$ cm). The heat has nowhere near enough time to equalise temperature between compressions and rarefactions. The gas is, to excellent approximation, thermally isolated on the acoustic timescale.

Why the adiabatic curve is stiffer

The two cases differ by where the energy of compression goes, and the kinetic-theory picture of pressure from 1.2 shows it directly. A gas’s pressure is set by both how tightly its molecules are packed and how fast they move:

p \;=\; \tfrac13\, n\, m\, \langle v^2\rangle \;\propto\; \underbrace{n}_{\text{packing}}\,\times\,\underbrace{\langle v^2\rangle}_{\text{speed}},

with $n$ the number density and $\langle v^2\rangle$ the mean-square molecular speed — the quantity temperature measures. There are two ways to raise the pressure: pack the molecules into less volume, or make each one move faster.

Isothermal compression holds the temperature fixed — heat leaks away as fast as the gas is squeezed — so $\langle v^2\rangle$ is unchanged and the pressure rises only through $n$ . Halve the volume, double the density, double the pressure: $p \propto 1/V$ , Boyle’s law.

Adiabatic compression lets no heat escape. The work done squeezing the gas has nowhere to go but into the molecules’ own motion, so the temperature, and with it $\langle v^2\rangle$ , rises as the volume shrinks. Both factors in $p \propto n\langle v^2\rangle$ now climb, and the pressure rises faster than $1/V$ :

p \;\propto\; \frac{1}{V^{\gamma}}, \qquad \gamma = \frac{c_p}{c_v} \approx 1.4 \;\text{for air}.

The extra exponent $\gamma - 1 \approx 0.4$ is the contribution of the heating. The relation $pV^\gamma = \text{const}$ follows from the first law for an adiabatic ideal gas; that derivation, along with the specific heats $c_p$ and $c_v$ , is developed in Physics → Thermodynamics.

V / V₀ = 1.00

isothermal: P/P₀ = 1.00, T/T₀ = 1.00, speed ×1.00 adiabatic: P/P₀ = 1.00, T/T₀ = 1.00, speed ×1.00

Slide the volume down and the two curves separate: the adiabatic curve climbs above the isothermal one because its gas is also heating — the molecules in the adiabatic box redden and speed up while the isothermal box stays cool. This added stiffness is why sound exceeds Newton’s isothermal estimate. A sound wave compresses air far too fast for heat to escape, so the gas heats, stiffens, and carries the wave faster; the factor $\gamma$ that separates the two curves is the same $\gamma$ that turns Newton’s 280 m/s into Laplace’s 343 m/s.

The adiabatic equation of state, linearised

The adiabatic relation $p \propto \rho^\gamma$ can be expanded around equilibrium. Write $p = p_0 + p'$ , $\rho = \rho_0 + \rho'$ , and use $p / p_0 = (\rho / \rho_0)^\gamma$ :

1 + \frac{p'}{p_0} \;=\; \left(1 + \frac{\rho'}{\rho_0}\right)^\gamma \;\approx\; 1 + \gamma \frac{\rho'}{\rho_0} \;+\; O\!\left(\left(\rho'/\rho_0\right)^2\right).

Keeping first order:

\boxed{\;\; p' \;=\; \left(\frac{\gamma p_0}{\rho_0}\right)\, \rho' \;\equiv\; c^2\, \rho'.\;\;}

The constant of proportionality has units of velocity squared:

c^2 \;=\; \frac{\gamma p_0}{\rho_0} \;=\; \frac{\gamma R T_0}{M}.

For air at 20°C: $\gamma = 1.4$ , $T_0 = 293\,$ K, $M = 29\,$ g/mol, $R = 8.314\,$ J/mol/K — giving $c = 343\,$ m/s. The speed of sound has fallen out of the equation of state.

▶ More general fluids:

c^2 = (\partial p / \partial \rho)_s

Derivation

For a general fluid (not just an ideal gas) the adiabatic equation of state is whatever curve $p(\rho)$ describes the locus of constant entropy in the $p$ – $\rho$ plane. Expanding to first order around $(p_0, \rho_0)$ :

p' \;=\; \left(\frac{\partial p}{\partial \rho}\right)_s \cdot \rho',

where the subscript $s$ means at constant entropy. Define

c^2 \;\equiv\; \left(\frac{\partial p}{\partial \rho}\right)_s,

and the linearised equation of state is $p' = c^2 \rho'$ . For an ideal gas this reduces to $c^2 = \gamma p_0 / \rho_0$ as above. For water (much less compressible), $c^2 \approx 2.2 \times 10^6$ in SI units, giving $c \approx 1480$ m/s.

Closing the system

We now have three linearised equations:

\partial_t \rho' + \rho_0 \nabla \cdot \mathbf{v}' \;=\; 0, \qquad \rho_0 \partial_t \mathbf{v}' \;=\; -\nabla p', \qquad p' \;=\; c^2 \rho'.

Three equations, three unknowns ( $p'$ , $\rho'$ , $\mathbf{v}'$ ) — closed system. The next lesson eliminates two of them and arrives at a single second-order PDE for $p'$ .