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\int{\frac{dP}{\rho}} = \frac{\gamma }{\gamma -1} RT  \end{equation}  which we could have also seen from the expression for the enthalpy when $dQ = TdS =0$.  For the isothermal case use the isothermal ideal gas  equation of state $P= RT \rho$ and that $T$ is constant  to write $P=(c^{\rm{iso}})^2_s\rho$ and $dP=(c^{\rm{iso}})^2_s $dP=RT  d\rho$, then \begin{equation}  \int{\frac{dP}{\rho}} = G \int^{\rho}_{\rho_0}{\frac{dP}{\rho}}  = RT \ln{\frac{P}{P_0}}. \ln{\frac{\rho}{\rho_0}}.  \end{equation}  use which we could have also seen from the expression for  the Gibbs free energy with $dT=0$ to find $dT=0$. We can also write out these expressions in terms of the isothermal or adiabatic sound speeds by noting that for the adiabatic case, $dS=0$ implies that  \begin{equation}  \frac{P}{\rho^{\gamma}} = cst  \end{equation}  then the adiabatic sound speed is  \begin{equation}  \int{\frac{dP}{\rho}} \left( c^{\rm{ad}}_{s}\right)^2  = G \frac{dP}{d\rho}  = RT \ln{\frac{P}{P_0}}. \gamma \frac{{P}{\rho}  \end{equation}  $P = \frac{\left( c^{\rm{ad}}_{s}\right)^2}{\gamma} \rho$