I have seen the equation $S(T_2)=S(T_1)+C_p\ln(T_2/T_1)$ where $C_p$ is the molar heat capacity at a constant pressure. I understand that this assumes that the temperature range is sufficiently small that the constant pressure heat capacity does not vary significantly over it.

My question pertains to the derivation of this expression. One of the first steps is substituting $q=C_pdT$ into $dS=\frac{q_{rev}}{T}$. I am not 100% sure as to why these enthalpies are equated. Is it because the heat exchanged by a system at a constant pressure is always reversible heat, or is this not true?

If the above were true, it would explain to me why I never see an expression involving constant volume heat capacity. If the above is not true, could someone also explain why suh a form does not exist?

Finally, is the equation $S(T_2)=S(T_1)+C_p\ln(T_2/T_1)$ only applicable where the pressures are the same at the initial and final temperatures? On the one hand, the derivation assumes a constant pressure in the infintesimal changes from the initial to final temperature (and of course uses a constant pressure heat capacity!); however on the other hand entropy is a state function. So I would think that if one could find how entropy varies wit pressure at a constant temperature, and the change from the final to the initial state involved both a temperature and pressure change, then I would think that the above expression can be used to find the change in entropy due to the temperature change only, and then the additional factor from the pressure could be used?

Thank you in advance.

EDIT: For the second part, from having a quick think about the reversible heat enxchange in and isothermal process, I think $S(T_2,p_2)=S(T_2,p_1)+R\ \ln(p_1/p_2)$. So then one could construct a Hess-like cycle to go from one temperature and pressure/volume, to another temperature and pressure/volume. The use of the cycle and step-wise calulcation works because entropy is a state function I think?

Hopefully I answered the second part of my query (I would appreciate if someone could verify that it is correct), although I am still not sure about why the heats used are the same in the derivation, and whether the expression really does refer to the same pressure at the initial and final temperatures, even though the process in getting from these two temperatures can involve pressure changes (again, thinking about $S$ being a state function, so the process itself when changing states does not matter).


1 Answer 1


It is not true that heat exchanged at constant pressure is always reversible.

But, if you want to determine the change in entropy from thermodynamic equilibrium state 1 at $(T_1,P)$ to state 2 at $(T_2,P)$, you need to forget entirely about the actual irreversible process path that took you from state 1 to state 2. It is of no further use. You instead need to focus exclusively on the two end states. And, in order to apply the entropy equation, you need to devise a reversible path between these exact same two ends states. Any reversible path will do, because they will all give the same value for the entropy change.

In the case of your example, you can choose a constant pressure reversible path for convenience. So, for such a path, from the first law we know that $\mathrm{d}H = \mathrm{d}q$. But, now how do we figure out a reversible path between the two states in which the system temperature is changing reversibly along the path? If we contact the system with at constant temperature reservoir held at the final thermodynamic equilibrium temperature $T_2$, that path would not be reversible because there would be a finite temperature difference between the system and the reservoir over most of the path. But, now, suppose we contact the system, not with a single constant temperature reservoir, but with a sequence of constant temperature reservoirs, each at a slightly different temperature over the range $T_1$ to $T_2$. This would guarantee that, over the entire path, the system would differ only slightly from the reservoir it is currently in contact with. So, in this case, $\mathrm{d}q_\mathrm{rev} = \mathrm{d}H = C_p\,\mathrm{d}T$, where $\mathrm{d}T$ is the change in the system temperature over the present increment of heat exchanged. We use the equation $$\mathrm{d}H = C_p\,\mathrm{d}T$$ because of the precise definition of constant pressure heat capacity in thermodynamics: $$C_p\equiv\left(\frac{\partial H}{\partial T}\right)_P$$ So now, for the change in entropy, we have: $$\Delta S = \int{\frac{\mathrm{d}q_\mathrm{rev}}{T}} = \int_{T_1}^{T_2}\frac{C_p \,\mathrm{d}T}{T}$$

So the key thing to remember here is that, if you want to determine the change in entropy for an irreversible process, you first need to determine the initial and final end states (say, using the first law of thermodynamics) and then devise a reversible path between these same two end states to calculate the integral of $\dfrac{\mathrm{d}q_\mathrm{rev}}{T}$ for that path.

  • $\begingroup$ Hi, I was wondering why you used an exact differential for change in heat instead of an inexact differential? Otherwise it looks great. Thanks for the answer! $\endgroup$
    – cngzz1
    Jan 1, 2021 at 8:13
  • 1
    $\begingroup$ @cngzz1 How would you represent the difference between an exact differential and an inexact differential? $\endgroup$ Jan 1, 2021 at 13:34
  • $\begingroup$ I would use the lower case Greek letter delta q instead. Here's a link on inexact differentials (to see the symbol involved): en.m.wikipedia.org/wiki/Inexact_differential $\endgroup$
    – cngzz1
    Jan 1, 2021 at 14:02
  • $\begingroup$ @cngzz That is just an artifact of how you learned it. I choose not to use that kind of notation. $\endgroup$ Jan 1, 2021 at 14:08
  • $\begingroup$ Ok I understand. Thanks for your reply. $\endgroup$
    – cngzz1
    Jan 1, 2021 at 14:09

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