I'm tempted to answer in the affirmative, depending upon BOUNDARY CONDITIONS. Without getting into it too far
Suppose the ideal gas is contained within (for simplicity) a cubic box of length L. One has that the wavenumbers $$k_{n}=\frac{n\pi}{2L}$$ are then quantized, such that there are phonon-like modes of propagation.
One will then obtain something like:
$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$
with momenta:
$$p_{nml}=\hslash k_{nml}$$
and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.
In a sense then, normal modes of sound are phonons.
EDIT: a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles
It's also worth noting that, for phonons in a crystal, it is in actuality, the boundary of the whole crystal that cause quantization of phonon wavevectors http://users.physik.fu-berlin.de/~pascual/Vorlesung/SS09/slides/EPIV-09SS-SolSt_K3-Lattice%20vibrations.pdf (page 2) For example, in an infinite crystal, the allowed wavevectors would be continuous, and momentum wouldn't be quantized at all! Any good solid state text explains this, I like Kittel's intro to solid state physics