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On my professor's lecture notes he claims that $c_V>0$. Intuitively, this would have been my guess as well but he mentions how you can prove this by constructing a setup:

"Construct an isolated system, which is divided into two parts with internal energies $U_1$ and $U_2$ and volumes $V_1$ and $V_2$"

He then argues that maximising entropy at fixed volume will require $c_V>0$.

Here's my attempt at this, but I couldn't get very far. Since the system is isolated then $U=U_1+U_2=\textit{const}$. Also, $V_1$ and $V_2$ must be constant so then I get the total entropy: $$dS =\dfrac{1}{T_1}dU_1+\dfrac{1}{T_2}dU_2 = \left(\dfrac{1}{T_1}-\dfrac{1}{T_2} \right)dU_1 = 0$$ and so $T_1=T_2$. This doesn't seem to help and I'm not really sure how to proceed from here. Any ideas?

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Yes, this is on the right track. By considering two systems like this, you establish that the system is at equilibrium when the two temperatures are equal, where $$ \frac{\partial S}{\partial U}=\frac{1}{T} $$ In other words, transferring a small amount of energy $\delta U$ from one system to the other results in zero change in the total entropy, to first order in $\delta U$. This is what you expect if the total entropy is at a maximum. But it would also apply if the total entropy were at a minimum!

To distinguish between these two cases, you need to consider how the total entropy would change to second order in $\delta U$. You can do this by differentiating the above equation one more time with respect to $U$. This will give an expression on the right hand side involving the temperature and the heat capacity. If the entropy is a maximum (with respect to small transfers of energy between the two systems), and if you are happy that temperature is positive (I'm aware of a few cases where some people argue it could be negative, but normally we take it as positive) then it should follow that the heat capacity is positive.

I won't give the full answer because I believe that this falls into the homework-and-exercises category of this site, but that should be enough for you to derive the answer along the lines suggested by your professor.

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The result can be proved exploiting both the first and the second law of thermodynamics. I found this gem in a Luigi Ettore Picasso's book this july, but I share it with you only now, as a christmas present. Unfirtunately it was written literally in six rows, equations included, and I worked hard to understand it and translate it into this long and clear answer.

Let's put a body, initially at temperature $T_1$, in thermal contact with a resevoir at temperature $T_2 > T_1$, wait equilibrium, and then in thermal contact with a resevoir at temperature $T_1$. In the end the body is again at temperature $T_1$. The cicle is irreversible (thermal contacts between body with finite temperature differences). Let's take the second law in the form of Clausius inequality in its discrete form $\sum_{i=1}^n \frac{Q_i}{T_i} \le 0$ ($T_i$ are temperature of resevoirs: we suppose their tempeture is always approximately constant and all the thermal bustle is in the body). Here we have an irreversible process with two resevoir, so we write \begin{equation*} \frac{Q_2}{T_2} + \frac{Q_1}{T_1} < 0 \end{equation*} Now, what about heat flows? If we don't know details about the processes we cannot reach any conclusions. The body start from state $A$, go to state $B$, and come back again to state $A$, but knowing only initial and final states it is not sufficient to evaluate $Q$ (and $W$, of course), because $Q$ is not a state function. In particular, nobody assures that $|{Q_1}|=|{Q_2}|$. We can go on only giving up some generality and assuming that the volume is constant. If this is the case, the first law assures that $\Delta U=Q$, but $U$ is a state function so $Q$ too can be treated as a state function. So we can be sure (without knowing the details of heating and cooling processes) that heat flows are opposite and we write $Q_2=-Q_1$: \begin{equation*} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) Q_1 <0 \qquad \textrm{(constant volume)} \end{equation*} Because $Q$ in this case can be considered a state function, we can calculate it considering any process connecting the initial and final state. Let's choose a reversibile processes: in any step the temperature of the body is well defined and to find $Q_1$ we can integrate $d Q = C_V dT$ from $T_2$ to $T_1$: \begin{equation*} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \int_{T_2}^{T_1} C_V(T) dT <0 \end{equation*} Thus we arrive at the following important conclusion (please note a change in the extremes of integration) \begin{equation*} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \quad \textrm{ and } \quad \int_{T_1}^{T_2} C_V(T) dT \quad \textrm{ must have same sign} \end{equation*} We are free to choose $T_2$ near $T_1$, so let's suppose $T_2=T_1+\epsilon$ with $\epsilon$ small (positive or negative). We have \begin{equation*} \frac{1}{T_1} - \frac{1}{T_2} \approx \frac{\epsilon}{T_1^2} \end{equation*} and \begin{equation*} \int_{T_1}^{T_2} C_V(T) dT \approx C_V(T_1) \epsilon \end{equation*} In other words, $\epsilon$ and $C_V(T_1) \epsilon$ must have the same sign, so $C_V(T_1)>0$. To be more explicit, we say that

  • If $\epsilon > 0$ we must have $ C_V (T_1) \epsilon >0$, and so $C_V(T_1)>0$.

  • If $\epsilon < 0$ we must have $ C_V (T_1) \epsilon <0$, and so again $C_V(T_1)>0$

But $T_1$ is generic, we didn't make any hypotesis about it, so we conclude that $C_V > 0$ always. With other reasoning (not trivial as it may seem, because exists substance that expand cooling) we can show that $C_p > C_V$, so $C_p$ too is always positive.

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