Talk:Clausius theorem

Is the theorem proved in this article?

The article says, "Thus the heat and work in the process a-b and a-c-d-b are the same and any reversible process a-b can be replaced with a combination of isothermal and adiabatic processes, which is the Clausius theorem."

I don't see how the Clausius theorem (which is given formally at the beginning of the article) follows from the conclusion above.

Anyone know?

--Masud 14:51, 12 June 2006 (UTC)

Incidentally, I find the proof given here more understandable (than other proofs based on separating a reversible cycle into many Carnot cycles): http://theory.ph.man.ac.uk/~judith/stat_therm/node30.html.

Anyone up for incorporating that into this article?

--Masud 14:48, 13 June 2006 (UTC)

Hope my edit now makes sense. Aritrop (talk) 06:42, 15 March 2015 (UTC)

Math cleanup

The inequality signs in this article are backwards, the original paper is below:

• Clausius, R. (1865). The Mechanical Theory of Heat – with its Applications to the Steam Engine and to Physical Properties of Bodies. London: John van Voorst, 1 Paternoster Row. MDCCCLXVII.

The whole page will need to be cleaned. --Sadi Carnot 15:35, 30 September 2007 (UTC)

• That so-called "source" is on a vanispamcruftizement and suspect. — Coren ^(talk) 02:30, 20 October 2007 (UTC)

Contradictory information

The beginning of the article uses the ≥ sign, while the second half uses ≤. Which is it? — Preceding unsigned comment added by 166.249.96.45 (talk) 13:28, 18 December 2011 (UTC)

The first one was right. I corrected the other one. The same user also changed the sign in the Second Law of Thermodynamics article. 129.247.247.239 (talk) 14:23, 19 December 2011 (UTC)

As of Sept. 2012, the sign convention switching in this article ruins it for parsing by someone who isn't already familiar. — Preceding unsigned comment added by 173.25.54.191 (talk) 20:05, 11 September 2012 (UTC)

Theorem or Inequality?

The article needs to be tweaked. It could either say about Clausius Theorem or Clausius Inequality. The theorem is a part of and derived from the Inequality. The article misguides and confuses between the Theorem and Inequality. Looking for opinions, before i make the edit. 7Sidz (talk) 09:15, 21 January 2014 (UTC)7Sidz

Dubius step in the proof

The step here

According to the nature of Carnot cycle,

${\frac {\delta Q}{T}}={\frac {\delta Q_{0}}{T_{0}}}$

is questionable. In the Carnot cycle as given, $\delta Q$ and $\delta Q_{0}$ are arbitrary small changes that don't even occur at the same time. How can there be any meaningful equality?
Suspecting a misuse of

${\frac {Q_{H}}{T_{H}}}={\frac {Q_{L}}{T_{L}}}$

Waiting for input before I rewrite the proof. Roamingcuriosity (talk) 16:31, 27 October 2014 (UTC)

the differential equation is used in deriving the theorem because any reversible cycle can be divided into number of small carnot cycle

Hey, sorry, I just made an edit to the proof. I removed the entire reference to the Carnot cycle. Please let me know if this should have to be altered. Aritrop (talk) 01:05, 16 March 2015 (UTC)

Proof not valid?

Isn't $\delta S{=}{\frac {\delta Q}{T}}$ true for reversible processes only? For the total entropy to be $\geq$ 0, the entropy of the reservoir certainly must increase if the entropy of the system were to remain the same, but for irreversible processes, the relationship between $\Delta S$ and $\int {\frac {\delta Q}{T}}$ is not direct.172.9.11.103 (talk) 07:11, 14 July 2017 (UTC)

No. By the definition of entropy, delta S is (always) equal to delta Q divided by T.

The Inequality of Clausius states that the heat transferred (delta Q), divided by T, will always be greater than, or equal to, zero. It will only be zero in a reversible process.

The definition of entropy implies that the change in entropy is equal to delta Q divided by T. Therefore, change in entropy will only be zero in a reversible process. Dolphin (t) 11:54, 14 July 2017 (UTC)

The following sources state otherwise, that is, $dS{=}{\frac {\delta Q}{T}}$ is true for reversible processes only.

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node48.html

http://entropysite.oxy.edu/ — Preceding unsigned comment added by 172.9.11.103 (talk) 06:01, 21 July 2017‎

It can be proved that: For a reversible process $\delta S{=}0$. For an irreversible process $\delta S$ is greater than zero.

Entropy (and change of entropy) can be defined by the following differential equation: $\delta S{=}{\frac {\delta Q}{T}}$

If it is true that this differential equation is only valid for reversible processes, then entropy is only defined for reversible processes. What then is the definition of entropy (or change of entropy) for an irreversible process?

One consequence of the second law of thermodynamics is that entropy increases during any irreversible process. This statement is reliant on the ability to determine the change of entropy in an irreversible process. Clearly the change of entropy during an irreversible process can be determined. Therefore it cannot be valid to say entropy (or change of entropy) is only defined for a reversible process. Dolphin (t) 12:54, 29 November 2017 (UTC)

My professor basically said that for an irreversible process, the change in entropy is determined from experiments where the process from the initial state to the final state is almost completely reversible, and $\Delta S=\int {\frac {\delta Q}{T}}$ is measured. However, this change in entropy may not be equal in value to the $\int {\frac {\delta Q}{T}}$ of an irreversible process with the same initial and final states. I'm not sure whether a sufficiently similar reversible process exists for measuring the change in entropy in the case where different starting temperatures ultimately reach the same equilibrium temperature. D4nn0v (talk) 05:09, 30 November 2017 (UTC)

@D4nn0v: Consider four experiments in which four identical systems moves from the same initial state to the same final state in four different ways. In one experiment, the change of state is reversible and entropy doesn't change. In another, the entropy increases a small amount and we say the change of state is slightly irreversible. In another, the entropy increases more than in the previous case and, again, we say the change of state is irreversible. In the fourth case, the entropy increases significantly and we say the change of state is highly irreversible. Perhaps this is what your professor had in mind - irreversible processes don't incur the same increase in entropy.

However, the above explanation doesn't lend any support to the idea that the definition of entropy change for irreversible processes is different to the definition of entropy change for a reversible process. Dolphin (t) 04:14, 1 December 2017 (UTC)

I have just noticed that at Introduction to entropy#Explanation Wikipedia says:

${{\rm {\delta }}S}\geq {\frac {{\rm {\delta }}q}{T}}.$

This inequality was not previously known to me so obviously there is more to this than meets the eye. I shall have to study the topic more closely! Dolphin (t) 22:42, 1 December 2017 (UTC)

This is what I have in mind at the moment. There can be a change in a system's entropy during either a reversible process or an irreversible process. For example, when two bodies simply reach equilibrium tempreature, the process is irreversible. Yet, for the initially hotter body, it has lost entropy. Conversely, if a heat engine were to run at 100% efficiency, enough heat would be converted to work so that there would be no total increase in entropy. During this reversible process, however, the cooler reservoir has gained entropy, albeit by the same amount that the hotter reservoir has lost. We may recover the initial condition by using work, perhaps stored in a flywheel, to re-cool the cooler reservoir and re-heat the hotter reservoir. The question is whether $\delta S$ is equal to $\delta Q/T$.

Going back to the experiments you mentioned, we can run them either on an isolated system or on a system interacting with a surrounding. Let's consider scenario (A) for isolated systems first. There is no 'outside', so the system is everything that we have. $\delta Q=0$ always; therefore, $\delta Q/T=0$. If the initial and final states have the same entropy, then the process must be reversible, thus $\delta S=\delta Q/T=0$. Otherwise, if the initial and final states were different in entropy, then there can be no reversible process connecting the two. In fact, as far as we know, the total entropy can only increase. Therefore, it seems that $\delta S>\delta Q/T=0$. However, for the entropy of an isolated system to increase, the temperature may be undefined before equilibrium is reached, so what $\delta Q/T$ exactly means is debatable. For now, what we can say is that in an isolated system, defining $\delta S$ as $\delta Q/T$ may be invalid for irreversible processes, but such a definition may be possible for reversible processes.

Now, consider scenario (B) for non-isolated systems. First, (1) let's assume that the system does not generate any entropy by itself; that is, any change in its entropy must result from exchanging heat with the surrounding. Say we know what $\delta S$ is, but do we know what $\delta Q/T$ is? I'm not sure, not even for the reversible process. Of course, we can always define $\delta S$ to be $\delta Q/T$ for the reversible process. But what about the irreversible processes? What if work were done to or by the system during the process (without affecting the initial and final states), varying its temperature in addition to whatever effect the flow of heat is having? What if the process were cyclic, $\Delta S=0$, for the system at least? In any case, there doesn't seem to be any constraints on how $\delta S$ must be defined macroscopically.

(2) If the system itself were to generate additional entropy, then does that mean all processes would be considered irreversible? If the process were cyclic, then certainly entropy must be extracted from the system/gained by the surrounding. But again, would $\delta S$ be equal to $\delta Q/T$? It seems that without additional arguments, for a non-isolated system, $\delta S$ can be defined as $\delta Q/T$ either for reversible processes exclusively or regardless of a process's reversibility. D4nn0v (talk) 09:05, 2 December 2017 (UTC)

Thanks for this comprehensive argument. It is persuasive. I will think about it for a while but I’m sure I will agree with you. Dolphin (t) 20:19, 2 December 2017 (UTC)

What is the integral  :$\oint$ with respect to?

In :$\oint {\frac {\delta Q}{T}}\leq 0,$ what does the :$\oint$ refer to? please explain what we are integrating over sorry if this question would be better on stack exchange. — Preceding unsigned comment added by 70.64.25.205 (talk) 18:34, 22 December 2017 (UTC)

It's integrated over the inexact differential heat term $\delta Q$. Is this what you meant to ask about? D4nn0v (talk) 05:04, 27 December 2017 (UTC)

I think a more accurate answer would be: we are integrating an inexact form over some closed path in the thermodynamic states space Kelvynwelsch (talk) 19:25, 11 May 2021 (UTC)

Tone needs improvement

This article reads like a textbook, not like an encyclopedia, especially because of the use of "we" and "our". Please see Wikipedia:What_Wikipedia_is_not#Wikipedia_is_not_a_manual,_guidebook,_textbook,_or_scientific_journal.

Ira

Ira Leviton (talk) 01:39, 16 August 2020 (UTC)