Simple explanation of relation between speed of sound and r.m.s. speed? In an ideal gas, the speed of sound $v_s$ is related to the r.m.s. molecular speed $v_m$ by
$$\frac{v_s}{v_m}=\sqrt{\frac{\gamma}{3}} \qquad ,$$
where $\gamma=C_p/C_v=7/5$ for a diatomic gas. I understand how to prove this relation from first principles. However, it seems mysterious to me that it pops out like this in the end.
I suppose it's inevitable for dimensional reasons that there will be some relation of this form, since if we want to describe a sample of an ideal gas, a sufficient set of unitful parameters is $m$ and $kT$, and there is a unique way of combining these to give units of velocity. (In a solid or liquid, we have other parameters such as the Young's modulus.) However, this type of dimensional argument doesn't prove that the ratio of the speeds is of order unity.
Is there any straightforward physical plausibility argument for the fact that this ratio of speeds is constant, and for the fact that the ratio is of order unity? I guess it's implausible to have $v_m \gg v_s$, since it seems like then a sound wave would sort of get scrambled and lose its identity because of molecular motion, and this scrambling would take place in much less than a period.
 A: Let us recap how we obtain the quantities in question.
The easiest way to obtain the speed of sound is from continuous mechanics. Assuming isentropic flow and small perturbations of speed and density and linearizing Euler (or Navier-Stokes)  and continuity equations we obtain
$$ v_s^2=\left(\frac{\partial p}{\partial \rho}\right)_s,$$ 
where the derivative of pressure is taken at constant entropy. For an ideal gas that
has adiabatic equation in the form $p/\rho^\gamma = \mathrm{const}$ that means
$$ 
v_s = \sqrt{\frac {\gamma p}{\rho}} = \sqrt{\frac{\gamma\, k \, T}{m}},
$$
where $k$ is Boltzmann constant, $T$ temperature and $m$ is mass of molecule.
The rms speed is obtained from statistical mechanics of gases using Maxwell–Boltzmann distribution:
$$
v_m = \sqrt{\frac{3 \,k \, T}{m}},
$$
here the factor 3 under the square root is consequence of 3-dimensionality of our world.
In order to clarify relationship between those two speeds we can consider the theory that includes both the sound wave propagation and statistical distributions of molecular speeds. That would be the kinetic theory and in particular Boltzmann equation:
$$
\frac{\partial}{\partial t} f+(v\cdot \nabla) f=\mathrm{St},
$$
where $f=f(\vec{r},\vec{v},t)$ is the density function, and  $\mathrm{St}$ in the rhs (from Stosszahlansatz) is the integral collision operator.
If we consider the sound wave propagating along the direction $x$, then its density function would have the form $f(x-v_s\,t,  v_x, v_y, v_z)$ (with the only $t$ dependence in the first argument and without dependence on $y$ and $z$). Substituting this into  Boltzmann equation we obtain
$$ - v_s\frac{\partial }{\partial x} f + v_x  \frac{\partial}{\partial x} f=\mathrm{St}.$$
There, just from the form of equation we could see, that if we assume that rms speed $v_m$ is increased $\alpha$ times (by rescaling the  density function $f' =\alpha^{-3} \cdot f(x-v'_s t,\alpha \vec{v})$), $v_s$ should also be increased accordingly to maintain the solution. So we can conclude that $v_s / v_m = \mathrm{const}$.
This is, of course, valid if we can assume that the collision term in rhs also transforms accordingly or if we could simply neglect the rhs at a first approximation, which should hold for rarefied gases. Another case is rigid-sphere approximation for the collisions that provides appropriate transformation, so $v_s/v_m $ should also be independent on temperature there (but, obviously, if the density is large enough in this case it will be dependent on it).
Here is random example of paper that considers sound-wave propagation in the framework of Boltzmann equation: R.D.M. Garcia, C.E. Siewert, 'The linearized Boltzmann equation: sound-wave propagation in a rarefied gas', DOI:10.1007/s00033-005-0007-8  (just the first from Google scholar results that has online source). 
