where dp/dρ is the derivative of pressure p with respect to density ρ.

The presentation below is a derivation of this relationship. This derivation is more pedestrian than the usual derivation and does not require so many feats of mental agility.

Consider a tube of gas with constant cross-sectional area A. Let u be the fluid velocity, p the pressure and ρ the density. The momentum equation for this flow is:

ρdu/dt = -∂p/∂x
 

which says that the force on a parcel is equal to the negative of the pressure gradient. The rate of change of the velocity du/dt is the instantaneous rate of change ∂u/∂t plus the advection term u∂u/∂x. When the left is divided by density and the terms are replaced the result is:

∂u/∂t + u∂u/∂x + (1/ρ)∂p/∂x = 0
 

Since mass is conserved the equation of continuity holds, which for this case is:

∂ρ/∂t + u∂ρ/∂x + ρ∂u/∂x = 0
 

These may be moved toward linearization by assuming u is small and ignoring products of small terms. The form of the equations is then

∂u/∂t + (1/ρ)∂p/∂x = 0
 
∂ρ/∂t + ρ∂u/∂x = 0
 

It is unnecessary to have both p and ρ in the equations in as much as they are connected through the equation of state. Then

∂p/∂x = (dp/dρ)(∂ρ/∂x)
 

Note that the derivative of p with respect to ρ is a total derivative.

With this change the equations become

∂u/∂t + (1/ρ)(dp/dρ)∂ρ/∂x = 0
 
∂ρ/∂t + ρ∂u/∂x = 0
 

To make the equations completely linear the coefficients of the partial derivatives must frozen; i.e. take on constant values equal to their average values. Thus the fully linear equations would be expressed as:

∂u/∂t + (1/ρ₀)(dp/dρ)₀∂ρ/∂x = 0
 
∂ρ/∂t + ρ₀∂u/∂x = 0
 

Now we can look for wave-like solutions of the form u=Ae^k(x-ct) and ρ=BAe^k(x-ct). When these forms are substituted into the equation and the results divided by e^k(x-ct) we obtain the two homogeneous equations in two unknowns

-ickA + ik((1/ρ₀)(dp/dρ)₀)B = 0
 
-ikcB + ikρ₀ A = 0
 

This set of equations will have a nontrivial solution only if the determinant of the coefficient matrix is equal to zero.

The determinant equation is

-(ck)² - (-k²)(dp///dddρ)₀- 0
 

The solution to this equation is simply

c² = (dp/dρ)₀
 

If the ideal gas law holds and temperature is constant then

p = R_mTρ
and thus
(dp/dρ) = R_mT
so the speed of sound is
c = (R_mT)^1/2
 

When adiabatic conditions hold then the Poisson equation can be used for evaluating dp/dρ; i.e.,

p = Kρ^γ
 

where K is a constant and γ = c_p/c_V.

The derivative is then

dp/dρ = γKρ^γ-1 = γp/ρ
but p/ρ = R_mT so
dp/dρ = γR_mT
and thus
c = (γR_mT)^1/2
 

One implication of the above formula is that if the absolute temperature increases 1 percent then the speed of sound will increase 1/2 of 1 percent. The formula is

lim_ΔT→0[(Δc/c)/(ΔT/T)] = (T/c)(∂c/∂T) = (∂(ln(c))/∂(ln(T)) = 1/2