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\begin{equation}  dG = SdT + \frac{dP}{\rho}  \end{equation}  Now we assume to two  special cases, adiabatic and isothermal flows. First assume an adiabatic flow, $dQ =0$. Then \begin{equation}  \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 equation of state to write $P=(c^{\rm{iso}})^2_s\rho$ and $dP=(c^{\rm{iso}})^2_s d\rho$, then   \begin{equation}  \int{\frac{dP}{\rho}} = G = RT \ln{\frac{P}{P_0}}.  \end{equation}  use the  Gibbs free energy with $dT=0$ to find \begin{equation}  \int{\frac{dP}{\rho}} = G = RT \ln{\frac{P}{P_0}}.  \end{equation}