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My textbook states the following:

If a system were at absolute zero, an additional small amount of heat energy would lead to an infinite increase in entropy. Such a state is impossible. Absolute zero can never be achieved.

It also provides the equation:

$$ \Delta S_\text{surroundings} = \frac{-\Delta H_\text{system}}{T}$$ Where $T$ is given in kelvin.

From the statement, will the entropy of the system or surroundings increase; from what I can deduce, I would say the entropy increase would be within the system as the surroundings is losing energy, but I am not sure.

The second part of my question is, why would this lead to an infinite increase in entropy? Please provide a comprehensible mathematical explanation and also, preferably, an analogy.

Textbook: Pearson Baccalaureate: Higher Level Chemistry, 2nd Edition. By Catrin Brown and Mike Ford Pages: 254-255

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The textbook is referring to the entropy change of the system. While the textbook is correct that absolute zero can never be attained, its statement that the entropy change is infinite is wrong. The authors' rationale for thinking it is infinite likely stems from a misinterpretation of the definition of entropy change:

$dS = \frac{\text{đ}q_{rev}}{T}$

where $\text{đ}q_{rev}$ is the reversible heat flow into the system.

Let's start with a system at $0 K$, and then warm it to a temperature T = T'. Then:

$\Delta S =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$

Let's consider what happens when we flow the first, infinitesimal amount of heat, $\text{đ}q$, into the system. And to make the calculation easy, let's assume the heat flow is reversible (it doesn't have to be; if it weren't, we'd just need to find a reversible process that gets us to the same final state). As soon as we do this, the temperature is no longer zero! And once the temperature rises above zero, the singularity disappears and we no longer have a concern about infinite entropy change.

But, you may ask, what about what about the mathematics right at the very beginning, when the temperature is indeed zero. Here, to properly calculate the entropy change, we need to use a limit, recognizing that both T and $\text{đ}q$ are approaching zero. The easiest way to understand this is to consider the Debye $T^3$ law, which models the constant-volume heat capacity of solids as they approach absolute zero:

$C_v = C_v(T) = k T^3$, where k is a constant, and where I've written $C_v(T)$ as an explicit indicator that $C_v$ is temperature-dependent.

And since, for a constant-volume reversible process, $\text{đ}q_{rev} = C_v(T) dT$, we have:

$\Delta S= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T} = \int_{0}^{T'}\frac{C_v(T)}{T} dT = \int_{0}^{T'}\frac{k T^3}{T} dT=\int_{0}^{T'}k T^2 dT$

I.e., in evaluating what happens to $dS = \frac{\text{đ}q_{rev}}{T}$ as T->$0$, it is necessary to consider what is happening to both the numerator and the denominator. The textbook authors' mistake was in not understanding this. Clearly, once we find a reasonable functional form for $\text{đ}q$ in the limit as T->$0$, the singularity disappears.

Here's another way we can understand that the textbook's statement is wrong: Thermodynamics allows us, in principle, to determine absolute entropies for any substance. The absolute entropy is given by integrating the entropy change from absolute zero to whatever temperature the substance is at:

$\textit{Absolute entropy at T'} \equiv S(T') =\Delta S_{0 \rightarrow T'}= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$

If the entropy change between $0 K$ and any higher temperature were infinite, the absolute entropies of all real substances would (since no substance can be at absolute zero) likewise also be infinite!

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  • $\begingroup$ Your answer seems great, but the mathematics and explanations are way out of my scope of my comprehension. From what I can understand though, your explanation is what I was looking for. However, I will have to revisit it in a few months, once I have learned integral calculus. $\endgroup$ – Liam Jul 29 at 4:29
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    $\begingroup$ It helps to understand integral calculus, but you can get the essence of my argument without it if you can understand limits. Think of it this way: At constant volume, dS=Cv/T dT. As you approach absolute zero, it's not just T that's going to $0$; Cv is going to $0$ as well. And since Cv is going to $0$ faster than T is going to $0$, Cv/T goes to $0$ rather than infinity (for the same reason that x^3/x goes to $0$ rather than infinity as x goes to zero). I.e., the mistake would be in assuming that Cv is a constant; it's not. $\endgroup$ – theorist Jul 29 at 5:08
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    $\begingroup$ I've never seen a differentiation operator denoted with a dyet (đ). Is there a certain meaning attached to this symbol besides using it for derivation? If you would like to use an upright operator, you probably want to use \mathrm d or \mathrm{d} instead. $\endgroup$ – andselisk Jul 29 at 6:29
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    $\begingroup$ @andselisk As I'm sure you know, q and w are not state functions; rather, they are path-dependent. Hence their differential forms are inexact differentials, rather than exact differentials. The dyet is used to indicate that (though typically it is an angled slash rather than a horizontal bar; don't know how to make that with LaTeX). Since they are inexact differentials, any integration with respect to them must be a path integral. While not all physical chemistry/thermodynamics texts use this notation, several well-known ones do, including Callen, Reif, Castellan, and Engel & Reid. $\endgroup$ – theorist Jul 29 at 6:54
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    $\begingroup$ @theorist A-ha, I see, thank you for the explanation. It's just that the symbol looks a bit off due to the MathJax font support. I guess you may leave it as is, or adapt the representation from "d-crossbar" to "d-hat" with \bar{\mathrm d} $(\bar{\mathrm d}).$ Nice answer! $\endgroup$ – andselisk Jul 29 at 7:03

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