# How can mean free path affect the propagation of sound in a gas?

I was thinking, since mean free path is a measure of "how long" a particle has to travel to collide with another, that this surely must influence sound propagation somehow. I would say this since sound propagation, as far as I know, literally happens because vibrating particles bounce against their neighbour. Does this mean that there is a relationship between speed of sound and mean free path? Does this also mean that a denser material would see sound propagating faster because the mean free path would be shorter? And since densities get lower at higher temperatures, does this mean that if the temperature is higher then the speed of sound is slightly slower? Or would it be faster since particles have more energy and thus more possibilities of "bumping" into each other?

• To the best of my knowledge, I can only say yes. Regarding temperature, greater temperature means higher sound travel speed. Apr 12, 2021 at 17:48

Without going into a full derivation (which is actually quite fun, in my opinion), the linear sound speed may be written as $$c_0 = \sqrt{\left( \dfrac{\partial P}{\partial\rho} \right)_s},$$ where $$c_0$$ is the sound speed, $$P$$ is the thermodynamic pressure, $$\rho$$ is the mass density, and the subscript $$s$$ means that the partial derivative is performed assuming constant entropy. Assuming an ideal gas, we may then write $$P = \rho R T,$$ where $$R$$ is the universal gas constant and $$T$$ is the absolute temperature. Furthermore, for an adiabatic process we may write $$P\rho^{-\gamma} = \text{constant},$$ where $$\gamma$$ is the ratio of the specific heats. Implicit differentiation of the adiabatic process expression yields $$\dfrac{\partial P}{\partial\rho} = \gamma \dfrac{P}{\rho} = \gamma R T.$$ Thus, the sound speed in an ideal gas is given by $$c_0 = \sqrt{\gamma\dfrac{P}{\rho}} = \sqrt{\gamma RT}.$$ The sound speed then decreases with increasing density, assuming constant pressure, and is proportional to the square root of the temperature.
According to Wikipedia the mean free path of an ideal monatomic gas may be given by $$l = \dfrac{k_BT}{\sqrt{2}\pi d^2 P},$$ where $$l$$ is the mean free path, $$k_B$$ is the Boltzmann constant, and $$d$$ is the particle diameter. Combining this expression with the expression for the speed of sound then yields $$c_0 \propto \sqrt{Pl}.$$ Thus, given a constant ambient pressure (not sound amplitude), the speed of sound is proportional to the square root of the mean free path.