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I believe it is widely accepted the general definition for heat capacity is as follows

$$C \equiv \frac{\delta Q}{\textrm{d} T}.$$

One also finds that it is widely taken that

$$C_p = \left(\frac{\partial H}{\partial T}\right)_P,$$

and

$$C_V = \left(\frac{\partial U}{\partial T}\right)_V.$$

It seems to me these latter two results were derived for reversible processes. Is it then that heat capacities are intrinsic to the material and should be path independent, meaning that the path used in the derivation is irrelevant?

From the first law

$$ \textrm{d} U = \delta Q + \delta W,$$

and the Legendre transform variable enthalpy is

$$ \textrm{d} H = \delta Q + \delta W + p\textrm{d} V + V \textrm{d} p.$$

It seems then that you must assume reversible, and ignore composition, so that the work is only $\delta W = -pdV,$ then

$$ \textrm{d} U = \delta Q - p \textrm{d}V,$$

$$ \textrm{d} H = \delta Q + V \textrm{d} p.$$

The two heat capacities clearly follow.

Otherwise it is true that as long as the process is quasi-static

$$ \textrm{d} U = T \textrm{d}S - p \textrm{d}V,$$

$$ \textrm{d} H = T \textrm{d}S + V \textrm{d} p.$$

If one can write the heat capacity as

$$ C = T \frac{\textrm{d} S}{\textrm{d} T},$$

without having to assume a reversible process then again the relations could fall out. However, to do so seems to require a reversible process so that

$$T \textrm{d} S = \delta Q,$$

for it is known that for an irreversible process

$$T \textrm{d} S > \delta Q.$$

I would think that these results aren't path dependent, but I am not sure how to justify this thought.

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2 Answers 2

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The internal energy, enthalpy, and entropy are physical properties of a material, independent of path. So the difference in these quantities between two closely neighboring thermodynamic equilibrium states are independent of the path between these states (which could have been very tortuous and irreversible). Therefore, the heat capacities defined in terms of U, H, and S are path independent.

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  • $\begingroup$ I do not disagree, but my problem is how did we arrive at the heat capacities in terms of U, H and V? We can take the two relations I seek to arrive at as a definition and then there is no issue. But if we start from heat capacity's definition in terms of heat then I am unsure. My issue is for processes I am considering you cannot disobey the fundamental relationships. If you want to say the heat capacities are related to the thermo potentials you seem to enforce δQ=TdS. This just is not the case for irreversible processes, so are heat capacities generally not related to thermal potentials? $\endgroup$
    – Novice C
    Commented Aug 14, 2017 at 20:00
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    $\begingroup$ Imagine going from one state to another through both a reversible process and an irreversible process. Are we in agreement that would result in different amounts of heat, $\delta Q$? Then why wouldn't the heat capacity be path dependent if the states were the same, but $\delta Q$ was not? I'm fine with saying heat capacities are intrinsic to the material, and can be derived through an idealized process; but then it is clear to me that the heat capacities are approximations when stated in that form. $\endgroup$
    – Novice C
    Commented Aug 14, 2017 at 20:02
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    $\begingroup$ This all goes back to freshman physics when they introduced the concept of heat capacity to us in terms of Q. Basically they lied to us, or at least stretched the truth. In thermo we learned that Q is a function of path, while heat capacity should be a physical property of the material, and is thus a function of state. These differences really can not be reconciled, as you have noted. So, in thermo, the definition has been changed slightly by expressing Cv in terms of U rather than Q; then, when volume is constant, $Q=\Delta U$, and the calculated Cv matches the freshman result. $\endgroup$
    – Chet Miller
    Commented Aug 14, 2017 at 20:35
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    $\begingroup$ But, when volume is not constant (and work is done), Q is not equal to $C_v\Delta T$. But, for an ideal gas, irrespective of path, $\Delta U=\int{C_v dT}$. The new definitions of Cv and Cp are mathematically solid. The subscripts v and p refer to how these heat capacities are measured experimentally, not how they are applied in practice. That is, Cv is determined by measuring Q=$\Delta U$ in a test at constant volume, and Cp is determined by measuring Q=$\Delta H$ in a test at constant pressure. But, the application of these quantities in practice is much more general than that. $\endgroup$
    – Chet Miller
    Commented Aug 14, 2017 at 21:56
  • $\begingroup$ Thanks, I am sufficiently satisfied that defining heat capacities relating to the heat is not correct. I've found that Chapter 13 of Landua's statistical physics touches on this issue. Unfortunately he avoids the issue by never saying $C = \delta Q/\textrm{d}T$, but rather starts from the entropy expression I've given. I suppose from now I'll define them as $C = T \frac{\textrm{d}S}{\textrm{d}T}$ until that gets me in trouble again. $\endgroup$
    – Novice C
    Commented Aug 15, 2017 at 3:21
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I think you are right. When we derive those formulas for specific heats, we assume reversibility when we equate dQ with TdS. But I think there is a catch though because of which the specific heat at constant volume, Cv, and contant presuure, Cp are independent of the reversibility or irreversibility of the infinitesimal(quasi-static) process.

Consider any thermodynamic system. It has 3 independent extensive coordinates, say U(internal energy), V(volume) and N(particle number). Assume N is fixed(no matter enters/leaves the system). So, we actually have only 2 extensive coordinates: U and V. We might choose any pair of intensive and/or extensive coordinates apart from (U, V) to describe the state of system e.g. (T, P) or (V, S) or (U, P), etc.

Now, lets say we want to measure Cv. So now, volume is also fixed. In effect, we only need one coordinate to describe the system now. Let that be U(internal energy). Of course, we could choose any coordinate. U is what we need in order to find Cv. So, when we take the material/substance/system from (U, V, N) to (U + dU, V, N), whether reversibly or irreversibly, dT is going to be the same for both reversible and irreversible process(since T depends on U, V, N). And dQ is also going to be the same for both reversible and irreversible process because at constant volume, dQ = dU And dU is same for both reversible and irreversible constant volume process. So we can say that dQ/dT and hence Cv will also be the same.

I think this happens because there is no scope for system to do work. Hence, irreversibility arising out of finite pressure gradients (across movable wall) is ruled out. The only irreversibility that can arise is due to finite temperature differences across a diathermal wall. But that is external irreversibility which only causes the total entropy of the universe(substance + surroundings) to increase. The entropy change of the system/substance remains unchanged regardless of whether the process is reversible or not. Only the entropy change of the surroundings(dQ/T(surroundings)) whose magnitude is small since surroundings and substance/system are at different temperatures during an irrversible process.

Similar argument goes for Cp. Choose (U, P, N) as state variables. Since the process is isobaric and quasi static and pressure of surroundings is a assumed to be constant, there are no finite pressure differences when the work is done. So, the work is always reversible since the process is quasi static. The irreversibility again arises out of finite temperature differences between substance and surroundings which don't affect the dQ or dS or Cp of the system.

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