Suppose we have some thermodynamic system, not necessarily a classical ideal gas, where $$EV^{\gamma -1} = f(S,N)$$ Then for a process with constant $S, N,$ we have that $$EV^{\gamma -1} = \rm const.$$ Now $$\left(\frac{\partial E}{\partial V}\right)_{S,N} = -P=(1-\gamma)f(S,N)V^{-1}V^{1-\gamma}$$ $$PV=(\gamma -1)E$$ So $$PV+E=\gamma E$$ and for a process with constant $S, N,$ we have that $$PV^{\gamma} = \rm const.$$
Question: Can it be shown that this $\gamma$ is the adiabatic exponent, where the adiabatic exponent is defined by $$\gamma_{a} = \frac{C_P}{C_V}?$$ (This is not a homework question, so I do not know the right answer).
Attempt: I do not think so. $$\gamma_a=\frac{C_P}{C_V} =\frac{\left(\frac{\partial(E+PV)}{\partial T}\right)_{P,N}}{ \left(\frac{\partial E}{\partial T}\right)_{V,N}} = \gamma \cdot \frac{\left(\frac{\partial E}{\partial T}\right)_{P,N}}{\left(\frac{\partial E}{\partial T}\right)_{V,N}}$$ Now $$\left(\frac{\partial E}{\partial T}\right)_{P,N} = \left(\frac{\partial E}{\partial T}\right)_{V,N} + \left(\frac{\partial E}{\partial V}\right)_{T,N}\left(\frac{\partial V}{\partial T}\right)_{P,N}$$ In other words, for $\gamma_{a}=\gamma,$ we need to show that $E=E(N,V,T)$ is only actually only a function of $N, T,$ so that $$\left(\frac{\partial E}{\partial V}\right)_{T,N} =0$$ This doesn't seem doable... $$EV^{\gamma -1} = f(S,N) $$ $$E=f(S,N) V^{1-\gamma}$$ $$\left(\frac{\partial E}{\partial S}\right)_{V,N}=T=\left(\frac{\partial f(S,N)}{\partial S}\right)_{V,N} V^{1-\gamma}$$ $$TV^{\gamma -1}=\left(\frac{\partial f(S,N)}{\partial S}\right)_{V,N}$$ $$S=g(N,TV^{\gamma -1})$$ where $g$ is some function $$\to E = h(N,TV^{\gamma -1})V^{1-\gamma}$$ where $h$ is some function. $E$ needs to be extensive, but this doesn't seem to be enough. For example, $$E=N\tilde{h}\left(T\left(\frac{V}{N}\right)^{\gamma -1}\right) \left(\frac{V}{N}\right)^{1-\gamma}$$ where $\tilde{h}$ is some function of $T\left(\frac{V}{N}\right)^{\gamma -1}.$