I want to know why we can treat sound waves as an adiabatic process. Precisely, I know that pressure and density vibrations occur so fast that molecules have no time to exchange energy (I might be wrong). But I would like a deeper explanation, not using a mathematical argument, but maybe a physical and numerical one (I haven't found any useful data to help me argue this fact).
thermodynamics – Why sound waves are adiabatic explained
acousticsadiabaticthermodynamicswaves
Related Solutions
To expand on Xcheckr's answer:
The full equation for a single-frequency traveling wave is $$f(x,t) = A \sin(2\pi ft - \frac{2\pi}{\lambda}x).$$ where $f$ is the frequency, $t$ is time, $\lambda$ is the wavelength, $A$ is the amplitude, and $x$ is position. This is often written as $$f(x,t) = A \sin(\omega t - kx)$$ with $\omega = 2\pi f$ and $k = \frac{2\pi}{\lambda}$. If you look at a single point in space (hold $x$ constant), you see that the signal oscillates up and down in time. If you freeze time, (hold $t$ constant), you see the signal oscillates up and down as you move along it in space. If you pick a point on the wave and follow it as time goes forward (hold $f$ constant and let $t$ increase), you have to move in the positive $x$ direction to keep up with the point on the wave.
This only describes a wave of a single frequency. In general, anything of the form $$f(x,t) = w(\omega t - kx),$$ where $w$ is any function, describes a traveling wave.
Sinusoids turn up very often because the vibrating sources of the disturbances that give rise to sound waves are often well-described by $$\frac{\partial^2 s}{\partial t^2} = -a^2 s.$$ In this case, $s$ is the distance from some equilibrium position and $a$ is some constant. This describes the motion of a mass on a spring, which is a good model for guitar strings, speaker cones, drum membranes, saxophone reeds, vocal cords, and on and on. The general solution to that equation is $$s(t) = A\cos(a t) + B\sin(a t).$$ In this equation, one can see that $a$ is the frequency $\omega$ in the traveling wave equations by setting $x$ to a constant value (since the source isn't moving (unless you want to consider Doppler effects)).
For objects more complicated than a mass on a spring, there are multiple $a$ values, so that object can vibrate at multiple frequencies at the same time (think harmonics on a guitar). Figuring out the contributions of each of these frequencies is the purpose of a Fourier transform.
Sound waves are approximately adiabatic, and the sound speed is determined by the adiabatic compressibility. The reason is that long wave length disturbances are approximately solutions to the Euler equation (which conserves entropy), and the Navier-Stokes terms (which generate entropy) are small corrections. Sound is not isothermal, the pressure disturbances induce temperature oscillations.
The amount of heat generated is proportional to the amplitude squared, the frequency squared, and the dissipative coefficients (shear viscosity, bulk viscosity, and thermal conductivity). The sound absorption coefficient (the inverse sound absorption length) is $$ \gamma = (\langle\dot E\rangle/\langle E\rangle)/(2c) $$ with $$ \gamma = \frac{\omega^2}{2\rho c^3} \left[\frac{4}{3}\eta+\zeta + \kappa\left(\frac{1}{c_v}-\frac{1}{c_p}\right) \right] $$ where $\eta$ is shear viscosity, $\zeta$ bulk visosity, and $\kappa$ thermal conductivity.
Under typical conditions the sound absorption length is quite long, and not that much heat is produced. A more efficient mechanism for producing heat is sound absorption by a solid body. This is because the surface of the solid is isothermal, so large temperature gradients can occur at the surface boundary layer.
Best Answer
For starters, to quote Allan Pierce in Acoustics,
That one surprised me when I learned it myself.
In fact, sound is not an adiabatic process for all frequencies. For any medium there is a thermal conduction frequency, $$ f_{\mathrm{TC}} =\frac{\rho c_{p} c^{2}}{2 \pi \kappa}. $$ Frequencies much lower than this value will be well-approximated as adiabatic. However, increasing the frequency through and above this point will transition the process from adiabatic to isothermal. For air, this frequency is $\sim 10^{9} \, \mathrm{Hz}$, well above the range of human hearing, so we almost always treat sound as adiabatic.
The physical reason this occurs is that heat transfer due to conduction is proportional to the temperature gradient. This is just a statement of Fourier's law for heat conduction. Consider what happens as the frequency of a harmonic wave decreases: The wavelength increases, and the slope of the oscillating waveform decreases as it is "stretched out." Assuming equal amplitudes, lower frequency waves will therefore set up smaller temperature gradients, which will conduct heat less effectively. If the heat conduction is negligible, then the entropy is conserved by the process.
So, in summary, the thermal gradients set up by sound waves for typical frequencies of interest are small enough to be neglected, hence sound is a (very nearly) adiabatic process. However, as Thomas pointed out below, in reality frequencies that cross into the potentially-isothermal regime are almost always affected by attenuation first, and the principal effects from conduction and viscosity are actually to damp out the sound wave.
In case you decide you do want to see some math, the energy equation is $$ \rho T \frac{d s}{d t} = \kappa \nabla^{2} T.$$ The previous arguments can be seen mathematically by linearizing about a quiescent base state and assuming harmonic wave solutions for $s$ and $T$. The equation can be rewritten as $$ - i \omega \rho_{0} T_{0} \hat{s} = - \kappa \frac{\omega^{2}}{c^{2}} \hat{T}, $$ $$ \hat{s} = - i \frac{\kappa \omega}{\rho_{0} T_{0} c^{2}} \hat{T}. $$ As the angular frequency $\omega = 2 \pi f \rightarrow 0,$ so must the amplitude of the entropy oscillation, $\hat{s}$. As with the quote at the beginning, much of my answer draws from Acoustics by Allan Pierce.