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I was reading about the speed of sound in gases. It is clear that the change in pressure and volume of a gas, when sound waves are propagated through it are adiabatic hence $$v=\sqrt{\frac{B}{\rho}} = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma P V}{M}},$$ so why isn't the speed of sound in gas not affected by a change in pressure of gas?

In my textbook, the explanation is that $PV$=constant, and so $\sqrt{\frac{\gamma P V}{M}}$ is not affected by change in pressure. However, as the pressure change in sound wave is adiabatic, it is $PV^\gamma = $ constant, not $PV$.

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The speed of sound, $c=\sqrt{\dfrac{\gamma P}{\rho}}$, depends on the pressure $P$ and the density $\rho$.
However the ideal gas equation, $PV = MRT \Rightarrow P = \left( \dfrac MV\right)RT =\rho RT$, still holds so $P \propto \rho$ and thus $\dfrac P \rho$ is constant.

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  • $\begingroup$ $P\propto \rho$ is only true when $T$ is constant, however we assume the sound propagation to be an adiabatic process, which implies that $T$ does not necessarily need to be a constant. How do you justify the answer taking the above argument into account? $\endgroup$
    – user258881
    Aug 25 '20 at 9:42
  • $\begingroup$ @FakeMod I think I now understand your concern Because the process is adiabatic there is indeed a small variation in temperature about the ambient temperature ans as the variation is very small it is the ambient temperature which affects the speed of sound. Perhaps think of it as the slight increase just balanced by the slight decrease? $\endgroup$
    – Farcher
    Aug 25 '20 at 10:36
  • $\begingroup$ Maybe. It might just work well because the temperature difference is small. However, even in this case, using $PV^{\gamma}=\rm constant$ would still be more accurate. I don't see any reason to not use it, and instead use $PV=\rm constant$. $\endgroup$
    – user258881
    Aug 25 '20 at 10:39
  • $\begingroup$ Should stress that real-world gases are not "ideal," and thus sound speed is pressure-dependent. $\endgroup$ Aug 25 '20 at 11:26

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