We now will state fundamental laws regarding the entropy when the temperature gets close to its absolute zero, i.e. \(T \rightarrow 0\) K (later on we discuss the consequences). First we have the Nernst theorem:
| \begin{equation*} \lim_{T \rightarrow 0} \Delta S_{trans} = 0 \label{eq:Nernst_equ} \end{equation*} | (3.14) |
Some observations regarding the Nernst theorem:
For \(T \rightarrow 0\) K: \(\Delta G \approx \Delta H\)
The DIFFERENCES of the heat capacities of educts and products of any reaction / transition goes to zero for \(T \rightarrow 0\) K (e.g. rhombic and monoclinic sulfur)
Diagrams \(\Delta G / T\) and \(\Delta H / T\): slope of both state functions goes zero for \(T \rightarrow 0\) K
According to the Nernst theorem \(S(T = 0)\) need not be zero!
BUT: According to the Planck theorem (= Third law),
| \begin{equation*} \lim_{T \rightarrow 0} S = 0 \label{eq:Planck_theorem} \end{equation*} | (3.16) |
”The entropy of any homogenous substance, which is in complete internal equilibrium, may be taken as zero at 0 K”, e.g. if the heat capacity change goes to zero, the structure essentially remains the same; thus, entropy change = 0.
This implies absolute values of the ”third-law entropies”.
BUT: Residual entropies (\(S \gt 0\)) based on configurational contribution (disorder) may exist. A) molecules (CO), B) presence of isotopes, C) spin configurations: \(S =k\; \ln(2^n)\).
The most important consequence of the third law is that \(T = 0\) K can never
be reached; according to Planck: ”If the heat capacity goes to zero, each minimum action serves for an enhancement
of \(T\) inside the sample; thus, it appears practically impossible to approach 0 K”.
\(S = 0\) is never fulfilled in real systems due to the intrinsic disorder of the crystals. As we will see
later from the statistical approach of Boltzmann \(S = 0\) means that only one configuration is possible, however
real crystals contain defects. They cannot be removed at \(T = 0\) because of missing thermal activation.
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© J. Carstensen (TD Kin I)