Sound waves is approximately adiabatic; however, for any adiabatic compression, one needs to assume quasi-static process in order to derive $V^rP=\text{Constant}$, and then obtain the speed of sound $v_s=\sqrt{\frac{rP}{\sigma}}$ (from $v_s=\sqrt{\frac{B}{\sigma}}$ , where $B=\frac{\Delta P}{-\Delta V/V}$ , $\sigma$ is density of medium.)
but given sound wave is so fast, how come this become a good approximation?
We know sound wave is a longitudinal wave and it propagates through compression and rarefaction process. Now this compression or rarefaction process takes place in air layer which is very thin. So the time taken by the process is enough for a quasi-static compression or rarefaction. Yes, the velocity of the sound wave is high but the width of the air layers also very small. That's why time taken for this process is enough. It is naturally quite larger than relaxation time.
Actually Laplace first imagine that sound wave propagation occurs in Adiobatic and reversible(or quasi-static) process because the air is not a good conductor of heat and the air layer is very thin as I mentioned above. And then he calculate the velocity of sound wave which approximately match with the velocity of sound experimentally determined. From this he came into conclusion that the propagation process is adiabatic and quasi-static process.
