to Bruno: That depends on the reference state you are using. The 3rd Law assigns zero entropy to a pure, perfect crystal at zero temperature. But many substances are not really pure, but mixtures of nuclides; e.g., CO2 usually contains a small fraction of the C-13 nuclide. Consequently, there is a mixing entropy. I think most thermodynamic tables assign zero entropy to the naturally occuring nuclidic mixture. Hence, if you work with isotopically pure CO2, you would have a 3rd-Law entropy below zero. Or if you prefer the natural mixture, you might apply a magnetic field at zero temperature and enforce a spin alignment of the C-13 nuclei. This, too would lower the entropy.

Of course, this is not really an argument against the 3rd Law. It only means that one has sometimes to be very cautious when applying Boltzmann's “S = k ln W” to the 3rd Law.

Dear, Faik, as far as i know, negative entropy is impossible. but change in entropy (delta S). if you read chapter three of "Materials Thermodynamics", Hae-Geon Lee, world scientific publisher, 2012. you may understand something.

Yes, it is possible to have a negative entropy. The entropy values on the chart are actually entropy differences calculated relative to a reference state. Depending on the choice of reference state, it may produce values that are both negative and positive.

OK, there are some points, which have to be adressed:

first, physically speaking, you may have some form of negative entropy (as Erwin Schrödinger defined it in the 40ies of the last century), but this is for living systems and does not apply to your problem.

You can have a change in entropy that is negative, as Dinsefa Andoshe said, but you didn't give a very exact description of your problem, but I try to give some thoughts:

your enthalpy and you entropy are connected via:

dH = TdS + Vdp (assuming that only your pressure p changes and you hold the absolute temperature T and the volume V constant).

Therefore: dS = (dH - Vdp)/T

So, in order to obtain the change in entropy from your chart, you will have to look how large your pressure change is in a given intervall an compare it to the change of enthalpy in the same intervall - if it is larger than your change in enthalpy, then you will get a change in entropy which is negative.

I agree with Dinsefa, "negative entropy is impossible. but change in entropy (delta S)"

Negative entropy is simply wrong, S >=0 for all cases.

@Tang Suye: no, there is a negative entropy, also called negentropy: see for example here:

L. Brillouin: (1953) "Negentropy Principle of Information", J. of Applied Physics, v. 24(9), pp. 1152-1163

or

E. Schrödinger, (1944) What is Life - the Physical Aspect of the Living Cell, Cambridge University Press

and

M. Planck, (1945). Treatise on Thermodynamics. Dover, New York

Dear Johannes,

I know these examples, and also know that life consume Gibbs free energy but not "negative entropy".

It is negative entropy - in the first reference it's even in the title. Life does not consume negentropy, it generates it (at least according to Schroedinger).

You can just google it and you will find thousands of papers onto this topic.

Dear Johannes,

But, how "it generates it"? By physical or chemical processes?

Even for S=klnΩ, which also cannot be applied to living system because the principle of equal a priori probabilities cannot be established, not to mention “-(entropy) =klog(l/D)”.

From the third principle of thermodynamics (Nernst principle) isn't it impossible to have a negative entropy (whereas the variation of entropy in an open system can be negative) ?

Dear Tang Suye,

I admit, "generate" was a poor choice of words. Living things keep a storage of negative entropy (this is the more accurate description I think).

The second law of thermodynamics is not violated, because this process happens with an input of energy into the system.

@Bruno: I am not sure why the third principle of TD should forbid that (the Nernst-Theorem just states, that you cannot reach T = 0 K, which makes the entropy also = 0) - maybe you mean the sencond principle?

Although you cannot have T = 0, you may very well get negative, absolute temperatures (this topic was discussed also somewhere here on RG) and I put some explanation for that in that thread (along with some references), but I think this is already quite off-topic here.

In the chart you have the values are for change in entropy form a reference value that it is why you have negative values.

to Bruno: That depends on the reference state you are using. The 3rd Law assigns zero entropy to a pure, perfect crystal at zero temperature. But many substances are not really pure, but mixtures of nuclides; e.g., CO2 usually contains a small fraction of the C-13 nuclide. Consequently, there is a mixing entropy. I think most thermodynamic tables assign zero entropy to the naturally occuring nuclidic mixture. Hence, if you work with isotopically pure CO2, you would have a 3rd-Law entropy below zero. Or if you prefer the natural mixture, you might apply a magnetic field at zero temperature and enforce a spin alignment of the C-13 nuclei. This, too would lower the entropy.

Of course, this is not really an argument against the 3rd Law. It only means that one has sometimes to be very cautious when applying Boltzmann's “S = k ln W” to the 3rd Law.

Entropy: A concept that is not a physical quantity

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

I agree with Prof. Ulrich Deiters

Dear Johannes,

The positive or negative entropy of CO2 depends upon the selection of reference value of entropy. So cowling below that standard will make it negative.

During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

The so-called entropy was such a concept that was derived by mistake in history.

Calculus is not "take for granted".

In fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

It is well known that calculus has a definition.

Any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.

Interestingly, this is incorrect !

The change in entropy of a closed system is always positive. The change in entropy of an open system can be negative with the action of the other system, but then the change in entropy of the other system is positive and the total change in entropy of these systems is positive too.