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In my thermodynamics notes, the definition of isothermal heat capacity at constant volume is $$C_V := \frac T N \left(\frac{\partial S}{\partial T} \right)_V $$ and at constant pressure is $$C_P := \frac T N \left(\frac{\partial S}{\partial T} \right)_P $$ where $T$ is temperature, $N$ is the number of molecules in the system and $S$ is entropy.

I don't fully understand the derivation symbols. To me, the partial derivative of entropy with respect to temperature would correspond to the "slope in the temperature direction", meaning that all other variables are considered to be "frozen". This would mean that the two partial derivatives above, and consequently $C_V$ and $C_P$, are one and the same object. But it is clearly not so.

Why is there a need to specify that we are keeping volume / pressure constant, when this is already implied in the definition of partial differentiation? How would one make this notation mathematically rigorous, i.e., what is the actual math behind this abbreviated notation? (Possibly taking differential forms into account?)

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  • $\begingroup$ This is because there is an implicit dependency of $S$ on $V$ or $p$ when $T$ changes (such as in $pV=nRT$) $\endgroup$ – Yves Daoust Sep 21 '17 at 9:35
  • $\begingroup$ You might find this useful as well, physics.stackexchange.com/questions/148724/… $\endgroup$ – An aedonist Sep 21 '17 at 9:44
  • $\begingroup$ @YvesDaoust $S$ was first introduced to us (in the entropy formalism) as a function of $U$ (internal energy), $V$ (volume), and the $N_i$'s (amount of substance $i$). $P$ and $T$ were defined by means partial derivatives of $S$ with respect to these variables, i.e. the coefficients in front of the basis forms $dU$, $dV$ etc. How can $P$ and $T$ implicitly define $S$ if they are partial derivatives of $S$ itself? Seems circular. $\endgroup$ – giobrach Sep 21 '17 at 10:00
  • $\begingroup$ (Of course by forms I mean exact linear differential forms on the $(i + 2)$-dimensional state space we're working with, where the scalar field $S$ is defined) $\endgroup$ – giobrach Sep 21 '17 at 10:04
  • $\begingroup$ @giobrach, $P = T (\frac{\partial S}{\partial V})_{U,N}$ (the first law $\mathrm{d}U = \delta Q + \mathrm{d}W$ is my shepard, and $\frac{\delta Q}{T}$), there is a $T$ too. That is where the equation of state sneaks in: I mean, there can be many equations of state, but they must comply to that requirement. $\endgroup$ – An aedonist Sep 21 '17 at 10:19
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Whenever we take a partial derivative, we need to specify what we are keeping constant. We commonly just use the partial derivative notation (say $\frac{\partial }{\partial x_1}$ in $\mathbb{R}^n$) to mean "keeping the other $n-1$ coordinates constant" and if that's what is meant, we usually don't need to say anything more. But that is not the only meaning of the partial notation. You can keep other things constant. For example, in $\mathbb{R}^2$, you can ask, "what is $\frac{\partial f}{\partial x_1}$ keeping $(x_1+x_2)$ constant?".

In the definitions you have given, $C_v$ is the partial derivative w.r.t. T and indeed P is allowed to vary while V is held constant. And in the case of $C_p$, V is allowed to vary as P is held constant.

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