Vanadium 50 said:
Velocity is the derivative of position, i.e. number density, i.e. pressure.,
That doesn't make sense. The flow field, ##\vec{v}(t,\vec{x})## is the velocity of a fluid element, being momentarily at position ##\vec{x}## at time, ##t## (Eulerian description).
To answer the question in the OP: You start with Euler's equation of a free particle
$$\rho (\partial_t \vec{v} +(\vec{v} \cdot \vec{\nabla} \vec{v})=-\vec{\nabla} P$$
and the continuity equation for mass
$$\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{v})=0.$$
In addition you need an equation of state, which we take as a polytrope
$$P=P_0 \left (\frac{\rho}{\rho_0} \right)^n.$$
Here ##(\rho_0,P_0)## is the mass density and pressure of the fluid at rest, and we assume that the deviations from these values and the velocity field itself are small, so that we can linearize the equations:
$$\rho_0 \partial_t \vec{v} = -\vec{\nabla} P, \qquad (1)$$
$$\partial_t \rho + \rho_0 \vec{\nabla} \cdot \vec{v}=0. \qquad (2)$$
In this approximation
$$\vec{\nabla} P=v_{\text{s}}^2 \vec{\nabla} \rho \quad \text{with} \quad v_{\text{s}}^2=\left .\frac{\mathrm{d} P}{\mathrm{d} \rho} \right|_{\rho=\rho_0}= n \frac{P_0}{\rho_0}.$$
For an adiabatic equation of state (which is consistent with the assumption of a perfect fluid made above), you have ##n=C_p/C_v##. For an ideal gas that's ##n=(f+2)/f##. The air, consisting mostly of molecules with two atoms, ##f=5##, and thus ##n=1.4##. In any case we get from (1)
$$\rho_0 \partial_t \vec{v}=-v_{\text{s}}^2 \vec{\nabla} \rho. \qquad (3)$$
Taking the divergence gives
$$\rho_0 \partial_t \vec{\nabla} \cdot \vec{v}=-v_{\text{s}}^2 \Delta \rho.$$
Taking the time-derivative of (2) finally leads to the wave equation for the density,
$$\frac{1}{v_{\text{s}}^2} \partial_t^2 \rho + \Delta \rho=0.$$
The same equation holds for ##P## since at the same linear order of the deviations from the equilibrium
$$P=P_0 + v_s^2 (\rho-\rho_0).$$
Concerning ##\vec{v}## we have to assume that it is irrotational, i.e., there's a potential for it
$$\vec{v}=-\vec{\nabla} \Phi. \qquad (4)$$
Then from (2)
$$\partial_t \rho - \rho_0 \Delta \Phi=0. \qquad (5)$$
Plugging (4) in (3) we find
$$\rho_0 \partial_t \Phi=v_{\text{s}}^2 \rho \; \Rightarrow \; \partial_t \rho=\frac{\rho_0}{v_{\text{s}}^2} \partial_t^2 \Phi$$, and finally plugging this again into (5), also the velocity potential obeys the same wave equation as ##\rho## and ##P##:
$$\frac{1}{v_{\text{s}}^2} \partial_t^2 \Phi-\Delta \Phi=0.$$
So all these quantities obey the wave equation with the phase velocity ##v_{\text{s}}##, which thus is the sound velocity.
