# Proving that $c_V$ is positive

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?

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.

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.