Pressure, entropy and enthalpy For an ideal compressible flow, this relation holds:
$$P = P(s, h)$$
where $s$ is the specific entropy and $h$ is the specific enthalpy.
I don't know why: I know that $s = s(e, v)$, but even using Maxwell relations, the density or the temperature is still inside the definition of $P$, so $P=P(s, h, T)$. Could anyone help?
 A: In Thermodynamics, it is quite common to have some perplexity on the possibility of expressing state functions as a function of other state functions. It is a safe attitude since it is not true that everything can be considered a function of everything else. A counterexample is the case of $P=P(T,v)$ (here, I am using the usual Thermodynamic convention of expressing with the same symbol the value of the pressure and its functional dependence on its independent variables). In general, it is impossible to obtain $v=v(P,T)$ for temperatures below the critical temperature due to flat isotherms in the coexistence region.
The case of $P(s,h)$ is different from the previous example. To show that it is possible to express pressure as a function of entropy and enthalpy, it is enough to start from enthalpy as a function of its natural variables (entropy and pressure) and notice that the relation
$$
h=h(s,P)
$$
can always be inverted to provide $P=P(s,h)$, based on the implicit function theorem, since
$$
\left.\frac{\partial{h}}{\partial{P}}\right|_s=v>0.
$$
