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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.