# Where is the error in this change of entropy derivation?

The thermodynamic definition of entropy is $\mathrm{d}S \equiv \frac{\delta Q_{\mathrm{rev}}}{T}$ where $\delta Q_{\mathrm{rev}}$ is a strictly reversible "change" in heat. The second law of thermodynamics tells us that, for some irreversible change in heat $\delta Q_{\mathrm{irrev}}$, we will have $\mathrm{d}S \gt \frac{\delta Q_{\mathrm{irrev}}}{T}$. Additionally, the definition of heat capacity with some state variable $X$ held constant is $C_X \equiv \left(\frac{\partial Q}{\partial T}\right)_X$.

Now, I've been lead to believe that the above definition for heat capacity holds in all circumstances (with reversible heat flow and irreversible heat flow). With that in mind, let's look at the change in entropy $\Delta S_\textrm{rev}$ resulting from some reversible, constant $X$ process that starts in some state $A$ and ends in some other state $B$: $$\mathrm{d}S = \frac{\delta Q_{\mathrm{rev}}}{T} \implies \Delta S_\textrm{rev} = \int_A^B \frac{\delta Q_{\mathrm{rev}}}{T} = \int_A^B \frac{C_X \mathrm{d}T}{T}.$$ Consider also the change in entropy $\Delta S_\textrm{irrev}$ resulting from some irreversible, constant $X$ process that also starts in state $A$ and ends in state $B$: $$\mathrm{d}S > \frac{\delta Q_{\mathrm{irrev}}}{T} \implies \Delta S_\textrm{irrev} > \int_A^B \frac{\delta Q_{\mathrm{irrev}}}{T} = \int_A^B \frac{C_X \mathrm{d}T}{T}.$$ Entropy is a state variable and both the reversible process and irreversible process above begin in state $A$ and end in state $B$. Therefore, $\Delta S = \Delta S_\textrm{rev} = \Delta S_\textrm{irrev}$. However, my logic is clearly faulty somewhere as this means I've just shown $$\Delta S = \int_A^B \frac{C_X \mathrm{d}T}{T}$$ and $$\Delta S > \int_A^B \frac{C_X \mathrm{d}T}{T}.$$ Obviously these cannot both be true at the same time so where have I gone wrong? My statistical mechanics professor suggested that the definition for heat capacity at the top of my post is invalid for irreversible processes so my substitution $\delta Q_\textrm{irrev} = C_X \mathrm{d}T$ is incorrect. Is her explanation right? If it is, why is the usual definition for heat capacity invalid for irreversible processes? Is there any definition of heat capacity that is true for both reversible and irreversibly processes?

## 1 Answer

The problem is that heat Q is path dependent while heat capacity C is supposed to be a function of state (i.e., a unique physical property of the material), independent of path. So, even though they taught us in freshman physics that dQ=CdT, this relationship is no longer generally correct in the more complicated situations encountered in thermodynamics. The inconsistencies are overcome however when we modify the definition of heat capacity a little for use in thermodynamics, by defining heat capacity in terms of internal energy U and enthalpy H, both of which are functions of state: $$mC_v=\left(\frac{\partial U}{\partial T}\right)_V$$ $$mC_p=\left(\frac{\partial H}{\partial T}\right)_P$$ Another issue with your analysis is that, in applying the Clausius inequality to an irreversible process, the temperature T in the denominator is supposed to be the temperature at the boundary of the system through which the heat transfer $\delta Q_{irrev}$ is occurring (The temperature of the system is typically non-uniform spatially, and varies from location to location). So, more precisely, the Clausius inequality should read: $$\Delta S\geq\int{\frac{dQ}{T_B}}$$where $T_B$ is the temperature at the boundary through which the heat is passing.

For an irreversible process, you can't simply determine the change in entropy by integrating dQ/T, with dQ representing the actual irreversible heat and with T being the temperature at the boundary. You must devise an alternate path between the same initial and final states of the system that is reversible, and carry out the integral for that path (even though it is not for the actual irreversible heat and boundary temperature).

• But if you perform that integral between the same states on the reversible path then you still end up with the problem the OP is having. Namely that the change in entropy is equal to some value in one case but larger than that value in another case. – Aaron Stevens Sep 6 '18 at 2:51
• @Aaron Stevens There is only one unique value for the integral for a reversible path between the initial and final states. This is true even though there may be more than one reversible path. All reversible paths will give exactly the same value for the integral. But, as I said, you can't use the heat for the irreversible path in evaluating the integral if you are trying to get the entropy change. – Chet Miller Sep 6 '18 at 2:52
• I completely agree with you. But that doesn't resolve the final issue the OP is having. The OP seems to understand this point. It doesn't resolve the contradiction of equality in one case and inequality in the other when they should both be the same (assuming $\Delta S$ should be change in entropy of just the system). – Aaron Stevens Sep 6 '18 at 9:17
• @AaronStevens That's because, for the irreversible case, he is not using the boundary temperature in the denominator of the integral, as required by the Clausius inequality (assuming that T in the numerator is the spatially averaged temperature of the body in the irreversible case). If he had (properly) used the boundary temperature in the denominator, there would be no inconsistency. For the irreversible case, the boundary temperature is not equal to the spatially averaged temperature within the body. – Chet Miller Sep 6 '18 at 11:34
• Oh ok I see that makes more sense. Thank you for the clarification. Does this mean that my answer is incorrect? My area of research is not primarily focused on thermodynamics, so my answer is based on what I remember from classes I have taken. – Aaron Stevens Sep 6 '18 at 11:43