Euler's identity for the enthalpy states that (for a single component and phase)

$ H = TS + \mu N$

What are the implications of this expression in the context of the third law of thermodynamics (consider a "a perfect crystalline substance")? My textbooks describe the behaviour of $\Delta S$, $S$ and $\Delta G - \Delta H$ but not of $\Delta H$ or $H$.

Is it safe to assume that as $C_p$ and $S$ go to zero, so too does $\Delta H$ for any process at 0 K?

  • $\begingroup$ I should add that this problem follows from a "heated" dispute with @Mithoron during my attempt to answer this problem: chemistry.stackexchange.com/questions/103885/… $\endgroup$
    – Buck Thorn
    Commented Nov 6, 2018 at 18:16
  • $\begingroup$ The third law is not my area of expertise and I'm genuinely interested to know the answer. $\endgroup$
    – Buck Thorn
    Commented Nov 6, 2018 at 18:19
  • $\begingroup$ Any decent thermodynamics text has a section on the general thermodynamic relations. The fundamental thermodynamic relation is $dQ = T dS = dE + p dV$ which defines $dE = T dS - p dV$ taking $S$ and $V$ as independent variables. Thus you have $E = E(S,V)$. If instead you take $S$ and $p$ as independent, you can rewrite the fundamental relation as $dH = T dS + V dp$ by defining $H = E + pV$, with $H = H(S,p)$. Notice that this is all defined through differentials, not absolute values. There is no need, nor no reason, to assume that either $S$ or $H$ must go to zero as $T$ approaches zero. $\endgroup$
    – Jon Custer
    Commented Nov 6, 2018 at 18:54
  • $\begingroup$ @JonCuster Well, that is sort of the Planck statement of the 3rd law of thermo, but ok. I've been wrong enough today with my misstatement $H=TS$ that I deserve to be doubted... $\endgroup$
    – Buck Thorn
    Commented Nov 6, 2018 at 19:07
  • 2
    $\begingroup$ The entropy of a system at absolute zero is a constant, but the constant does not have to be zero. Many systems have a degenerate ground state and thus a non-zero entropy at T=0. $\endgroup$
    – Jon Custer
    Commented Nov 6, 2018 at 19:34

1 Answer 1


Short answer: No. There's no reason to expect that this should be the case.

My standard reference for classical thermodynamics is Callen's Thermodynamics and an Introduction to Thermostatistics, 2nd ed. Ch. 11 is relevant here.

Planck's statement of the third law is, according to Callen, "The $T=0$ isotherm coincides with the $S=0$ adiabat." In addition to the theoretical ramifications of having an isotherm and an adiabat coincide, the Planck statement also assigns to this adiabat $S=0$. This is physically grounded in statistical mechanics.

Let $\Delta$ represent the change in a quantity of a reversible isothermal process. Then, at zero temperature:

  • $S = 0$, by the third law.
  • $\Delta S = 0,$ again by the third law.
  • $\Delta G-\Delta H = T\Delta S = 0,$ because $T \to 0$.
  • $$\frac{\text{d}\Delta H}{\text{d}T}-\frac{\text{d}\Delta G}{\text{d}T} = \lim_{T\to0}\frac{\Delta H-\Delta G}{T} = \lim_{T\to0}\Delta S = 0,$$ once again by the third law. (This shows why the Gibbs free energy and the enthalpy are essentially equivalent thermodynamic potentials at low temperatures!)

But there is no reason to expect $\Delta H = \Delta G = 0$. Regardless of temperature, some energy is still required to effect a change in the system.

  • 1
    $\begingroup$ Fair enough, I was about to answer my own question. The "left over" term in H or G is $\mu N$. N obviously can't go to zero. And $\mu$? Presumably not... $\endgroup$
    – Buck Thorn
    Commented Nov 6, 2018 at 19:36

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