# Traditional definition for heat capacity assume reversible processes?

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.

• 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. – Novice C Aug 14 '17 at 20:02
• 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. – Chet Miller Aug 14 '17 at 20:35
• 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. – Chet Miller Aug 14 '17 at 21:56
• 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. – Novice C Aug 15 '17 at 3:21