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

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.

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: $$\frac{\partial S}{\partial T}_p = \left(\frac{\partial T}{\partial S}-\frac{\partial T}{\partial V}\frac{\frac{\partial p}{\partial S}}{\frac{\partial p}{\partial V}}\right)^{-1}$$
• I checked the math again, and it looks correct, no missing factor. I'm just expressing $dS$ with respect to $dp,dT$, then expanding $dp$ with respect to $dS,dV$ and finally regrouping. Could you elaborate please? How would it affect the final for result for $\frac{\partial S}{\partial T}_p$?
• Thank you, I realized how you did that step, and I believe it is correct. The step after it though, seems incorrect. $dS= \left(1-\frac{\partial S}{\partial p}_T\frac{\partial p}{\partial S}\right)^{-1}\frac{\partial S}{\partial T}_pdT+...$ Implying $\left(1-\frac{\partial S}{\partial p}_T\frac{\partial p}{\partial S}\right)^{-1}\frac{\partial S}{\partial T}_p = \frac{\partial S}{\partial T}_V$ So $\frac{\partial S}{\partial T}_p = \left(1-\frac{\partial S}{\partial p}_T\frac{\partial p}{\partial S}\right)\frac{\partial S}{\partial T}_V$. At this point, I am not sure how to proceed. Dec 1, 2022 at 4:10