Is the propagation and generation of sound an isentropic process?

Is it because it is a reversible adiabatic process?

Why would it be adiabatic? Does it happen so fast that there is no time to exchange heat with the environment?


Sound is attenuated in air - this is an irreversible, lossy process that results in the heating of the air. You can conclude that it is not an isentropic process. See my recent answer to another question for some of the math behind this - showing that while the amount of heating is very small, it is not zero.

For practical purposes (for example, for computing the speed of sound in air) the isentropic calculation is quite good - the air heats up during the compression part of the sound wave, and it cools down during the rarification, but this happens so quickly (and for most audible sound, on such a large spatial scale) that there is negligible heat dissipation (which is the loss mechanism that ultimately causes the attenuation of sound).

You can now see why higher frequencies of sound are attenuated more rapidly: the distance between the "hot" and "cold" parts of the sound wave are smaller, and therefore the thermal gradient is bigger…

How big is the temporary heating? For a 120 dB sound wave, the pressure amplitude is not very large. From http://www.sengpielaudio.com/calculator-soundlevel.htm we get the following picture:

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and we conclude that a 120 dB sound has a pressure level of 20 Pa. Given that ambient pressure is about $10^5 Pa$, the fractional pressure change is 0.02%.

Now for an adiabatic gas, the following holds:

$$P^{1-\gamma}T^\gamma = const$$

Which we can manipulate to give

$$\frac{dT}{T} = \frac{\gamma - 1}{\gamma} \frac{dP}{P}$$

For a diatomic gas (air), $\gamma = \frac75$. We conclude that a pressure change of 0.02% corresponds to a temperature change of

$$ \Delta T = \frac27 \cdot 0.0002 \cdot 293 \approx 0.02 K$$

That's a pretty small temperature rise, and helps explain why the isentropic approximation works - there is very little gradient to drive heat dissipation.

What is amazing is that such a small difference is enough to explain a large change in calculated speed of sound: when Newton first tried to derive the speed of sound, he used an isothermal approximation which underestimates the speed of sound by about 15%; Laplace eventually corrected this error. Such a large difference for such a tiny temperature change… Read more about this here (with a nod to this earlier answer where I found this link)

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    $\begingroup$ That it is isentropic is an approximation, which is quite good in most cases. See my original answer. One truly needs to understand kinetics to appreciate this. $\endgroup$ Sep 10 '14 at 0:44

Isentropic processes are ones with constant entropy. Since entropy is defined as dS = dQ/T, then a reversible adiabatic process with dQ = 0 is an isentropic process.

Need to take a step back to understand this. First, the physics of waves in gases come from the fluid equations. These include conservation of mass, momentum and energy. These three equations are simply moment integrals over the Vlasov equation which governs the evolution of a distribution of particles (e.g., air). Since they are moment integrals over the kinetic equation, they do not form a closed set of equations. So your choices are then to solve the exact kinetic equations (hard) or to make simplifying assumptions. One simplifying assumption is to take:

$$ \frac{\partial Q}{\partial x} \ll \frac{\partial p}{\partial t} $$

This assumption simplifies the energy conservation equation into the adiabatic equation of state. So if your wave meets the above condition, namely that the heat flux is much less than the time rate of change of pressure, then you can say that the wave propagation is isentropic.

  • $\begingroup$ This answer agrees with mine in that is states "approximately isentropic". The point I tried to make was that any lossy process is by definition not isentropic - so this answer is looking at "how nearly it is" and I was looking at "the extent to which it isn't". You need to read both to get an appreciation of what is going on. $\endgroup$
    – Floris
    Sep 10 '14 at 0:49

If a system returns to the initial conditions after a process, then that process has been reversed and (assuming one of the initial conditions is the temperature) is isentropic. Once a sound ends, the molecules remain in essentially the same position as before the sound. Drastic increases or decreases in the number of molecules at the source (explosions) could move things around a bit. While we know that sound waves eventually dissipate and turn into heat, they do so slowly and often when interacting with matter bordering the gas. For many purposes, the propagation of sound is isentropic.

It should be clear that some sound generation is isentropic and some isn't (pluck a string, smash some glass).


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