Normal Shockwaves and its Thermodynamics Please read this page before going to my question.
When we derive the relation between Mach no. to the left side of the shock wave (M1) and that to the right side of the shock wave (M2),we do that with the help of the pressure wave speed equation i.e. $c=\sqrt{kRT}$. But this relation was obtained by assuming the process to be isentropic and propagation of a normal shock is by no means reversibly adiabatic due to the abrupt discontinuity in properties on either sides. Then how come we are able to use this relation for our derivation please someone explain.
Its said that $c$ is a thermodynamic property and once its derived by any process it can be used for any other process which may or may not be same. But I don't quite understand that.
 A: 
But this relation was obtained by assuming the process to be isentropic and propagation of a normal shock is by no means reversibly adiabatic due to the abrupt discontinuity in properties on either sides.

You are correct.  The fluid on either side of a shock wave is not in a succession of equilibrium states, i.e., energy dissipation is occurring.  Energy dissipation here is just an irreversible transformation of energy.  In this case, the energy being converted is the kinetic energy of the fluid flow across the shock, which is not constant.
You are correct.  We cannot use the normal adiabatic equation of state to describe the pressure and density change across a shock, given by:
$$
\frac{ d }{ dt } \left( P \ \rho^{-\gamma} \right) = 0 \tag{0}
$$
where $P$ is the scalar thermal pressure, $\rho$ is the mass density, and $\gamma$ is the polytropic index or ratio of specific heats.  We must use the Rankine–Hugoniot relations (RHRs), which are just a series of conservation relations.
Note that in the RHRs the immediate fluids on either side of the shock are not in equilibrium however the asymptotic states far from the shock are assumed to be in equilibrium.  The RHRs assume that the shock transition is very abrupt, i.e., that there is no heat flux across the shock ramp.
Now to your question, the derivation of the speed of sound from the adiabatic equation of state is different.  For instance, we can use the adiabatic equation of state to derive the internal energy, $\xi = \tfrac{ P }{ \gamma - 1 }$, and the enthalpy, $\zeta = \tfrac{ \gamma P }{ \gamma - 1 }$, without loss of generality because these functions of the state of the system and don't care about how the system got into that state.  That is, the details for how the system evolved to the state you are trying to describe using $\xi$ and/or $\zeta$ do not matter.
Similarly, the local speed of sound does not care about the impending shock wave, it's just the local speed of sound.  In deriving that, we can assume that the longitudinal oscillations are sufficiently fast and small that we need not worry about heat transfer during one oscillation.  Thus, we can use the adiabatic equation of state to show that:
$$
C_{s} \equiv \frac{ \partial P }{ \partial \rho } = \frac{ \partial }{ \partial \rho } \left( P \ \rho^{-\gamma} \right) = \frac{ \gamma \ P }{ \rho } \tag{1}
$$
A sound wave doesn't care about or even know about the shock or what's on the other side of the shock, so the irreversibility happening within the ramp is not relevant to our formulation of the sound speed here.

Its said that $c$ is a thermodynamic property and once its derived by any process it can be used for any other process which may or may not be same.

I think they are implying that consistency is key.  That is, once you define a form for your speed of sound, as long as you consistently use that form thereafter, the form doesn't necessarily matter (within physically meaningful limits, of course).  It's a roundabout way of saying what I said above, namely that the local speed of sound formula does not care that the shock is irreversible.
