Adiabatic Exponent for an adiabatic process

Question

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 $T\left(\frac{V}{N}\right)^{\gamma -1}.$

Answer 1

I'll just write your internal energy $E$ as $U$ and enthalpy $H = U+pV$. To simplify things (less variables) I'll only consider molar quantities and assume extensivity, which gets rid of $N$. The equation of state thus reads: $$ U = V^{1-\gamma}f(S) $$ Let's calculate $\gamma_a = \frac{C_p}{C_V}$. First off: $$ C_V = \frac{\partial U}{\partial T}_V \\ = T\frac{\partial S}{\partial T}_V \\ C_p = \frac{\partial H}{\partial T}_p \\ = T\frac{\partial S}{\partial T}_p $$

so let us convert these partial derivatives into partial derivatives of our basis variables $S,V$: $$ dS = \frac{\partial S}{\partial T}_VdT+\frac{\partial S}{\partial V}_TdV \\ \frac{\partial S}{\partial T}_V = \left(\frac{\partial T}{\partial S}\right)^{-1} \\ \left(1-\frac{\partial S}{\partial p}_T\frac{\partial p}{\partial S}\right)dS= \frac{\partial S}{\partial T}_pdT+\frac{\partial S}{\partial p}_T\frac{\partial p}{\partial V} dV \\ \frac{\partial S}{\partial T}_p = \left(\frac{\partial T}{\partial S}-\frac{\partial T}{\partial V}\frac{\frac{\partial p}{\partial S}}{\frac{\partial p}{\partial V}}\right)^{-1} $$

so:

$$ \gamma_a = \left(1-\frac{\frac{\partial T}{\partial V}}{\frac{\partial T}{\partial S}}\frac{\frac{\partial p}{\partial S}}{\frac{\partial p}{\partial V}}\right)^{-1} \\ = \left(1-\frac{\frac{\partial^2 U}{\partial S\partial V}^2}{\frac{\partial^2 U}{\partial S^2}\frac{\partial^2 U}{\partial V^2}}\right)^{-1} \\ =\left(1-\frac{((1-\gamma)V^{-\gamma}f')^2}{V^{1-\gamma}f''(1-\gamma)(-\gamma)V^{-1-\gamma}f}\right)^{-1} \\ =\left(1+(\gamma^{-1}-1)\frac{f'^2}{f''f}\right)^{-1} \\ $$

so for a general $f$, $\gamma_a\neq \gamma$. Actually, they are equal iff: $$ f''f = (f')^2 \\ f = Ae^{BS} $$ with $A,B$ two constants. Note that this is true in particular for ideal gases which probably motivated your question in the first place.

Hope this helps.

Answer to comment

I’ll detail the calculations of $\frac{\partial S}{\partial T}_p$: \begin{align} dS &= \frac{\partial S}{\partial T}_p dT+ \frac{\partial S}{\partial p}_T dp \\ &= \frac{\partial S}{\partial T}_p dT+ \frac{\partial S}{\partial p}_T \left(\frac{\partial p}{\partial S} dS + \frac{\partial p}{\partial V} dV \right)\\ dT &= \left(\frac{\partial S}{\partial T}_p\right)^{-1} \left(1-\frac{\partial S}{\partial p}_T\frac{\partial p}{\partial S}\right)dS -\left(\frac{\partial S}{\partial T}_p\right)^{-1} \frac{\partial S}{\partial p}_T\frac{\partial p}{\partial V} dV \\ dT &= \frac{\partial T}{\partial S} dS + \frac{\partial T}{\partial V} dV \end{align} By identifying the factors in front of the $dS,dV$ I obtain have a system of equations with two unknowns: $\frac{\partial S}{\partial T}_p, \frac{\partial S}{\partial p}_T $: \begin{align} \frac{\partial T}{\partial S} &= \left(\frac{\partial S}{\partial T}_p\right)^{-1} \left(1-\frac{\partial S}{\partial p}_T\frac{\partial p}{\partial S}\right) \tag{1} \\ \frac{\partial T}{\partial V} &= -\left(\frac{\partial S}{\partial T}_p\right)^{-1} \frac{\partial S}{\partial p}_T\frac{\partial p}{\partial V} \tag{2} \end{align}

which I solve by doing: $$ (1)-\frac{\frac{\partial p}{\partial S}}{\frac{\partial p}{\partial V}}(2) $$

and recover the advertised formula: $$ \frac{\partial S}{\partial T}_p = \left(\frac{\partial T}{\partial S}-\frac{\partial T}{\partial V}\frac{\frac{\partial p}{\partial S}}{\frac{\partial p}{\partial V}}\right)^{-1} $$