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Mueiz

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I know that this is found in many textbooks but I see that it is always incomplete.

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- Thread starter Mueiz
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- #1

Mueiz

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I know that this is found in many textbooks but I see that it is always incomplete.

- #2

paweld

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decreasing. Entropy is constant for adiabatically isolated systems only in reversible processes.

In all physically realizable processes (i.e. irreverisble) the entropy of adiabaticaly isolated

system increases. This argument is true however only in case of adiabatic processes -

the system can interact with its surroundings only mechanically. If the system is not

thermally isolated then the entropy of the system might decrease (only the total entropy

of the stsem and surroundings has to increase or at least be constant).

- #3

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I know that this is found in many textbooks but I see that it is always incomplete.

One does not follow from the other- they are equivalent statements. What form of the Second Law are you using?

- #4

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I know that this is found in many textbooks but I see that it is always incomplete.

One can formulate the 2nd law as follows: For a thermodynamical system which is not in a state of thermal equilibrium with the surroundings, the entropy of the system must increase until the state of equilibrium is reached.

This statement can be postulated in thermodynamics (thus has a value of an axiom) and can be proven (thus being merely a theorem) in statistical mechanics (the proof goes under certain assumptions, though).

So the second law you mention can be taken as exactly the statement in question.

- #5

Mueiz

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One does not follow from the other- they are equivalent statements. What form of the Second Law are you using?

or "

- #6

Mueiz

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- #7

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The form of the second law I am using is "No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature"

or "No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. "

That is a very early formulation of the second law- it was written before the first law. In order to proceed, you must first develop a notion of a cycle, define and calculate the flow of heat and work during the cycle, and finally calculate the efficiency of the cycle. I went through the original papers in a thread elsewhere, but the punchline is that Clausius invented the notion of the work equivalence that an amount of heat Q, passing from temperature T1 to T2 is given by Q(1/T2-1/T1).

From that, Clausius showed that for a reversible cycle, [itex]\int dQ/T = 0[/itex], while for an irreversible process, [tex]\int dQ/T \leq 0[/itex].

- #8

Mueiz

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That is a very early formulation of the second law- it was written before the first law. In order to proceed, you must first develop a notion of a cycle, define and calculate the flow of heat and work during the cycle, and finally calculate the efficiency of the cycle. I went through the original papers in a thread elsewhere, but the punchline is that Clausius invented the notion of the work equivalence that an amount of heat Q, passing from temperature T1 to T2 is given by Q(1/T2-1/T1).

From that, Clausius showed that for a reversible cycle, [itex]\int dQ/T = 0[/itex], while for an irreversible process, [tex]\int dQ/T \leq 0[/itex].

I want a proof ..."step by step" as I said in my first post.

This will help us to study thermodynamics as a systematic theory rather than collection of facts and tales.

If we think that two forms are equivalant then we must prove !

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- #9

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I want a proof ..."step by step" as I said in my first post.

This will help us to study thermodynamics as a systematic theory rather than collection of facts and tales.

If we think that two forms are equivalant then we must prove !

Then I suggest you crack a book.

- #10

Mueiz

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Then I suggest you crack a book.

I want a proof in the Forum because I want to discuss some points.

because I think that many of the proofs (at least all those I read) found in textbooks are incomplete.

- #11

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- #12

Andrew Mason

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Start with Clausius' statement of the second law as the premise:I want a proof in the Forum because I want to discuss some points.

because I think that many of the proofs (at least all those I read) found in textbooks are incomplete.

"No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature"

Assume that the change in entropy over the reversible path between two states is less than 0: [itex]\Delta S = \int dS = \int dQ/T < 0[/itex]

Show that this leads to a contradiction of the premise.

Hint: If heat flow occurs isothermally (ie where there is an infinitessimal temperature difference between the surroundings and system), the integral of dQ/T of the system and surroundings is 0. How would you get it to be less than 0?

AM

- #13

netheril96

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- #14

Mueiz

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I read what you wrote in Wikipeadia and found what I expected to find..The proof is clear but you used in it two statements which I think need proof too.

so we need to prove that this relation is independant from the properties of the working fluid and only then it would be logicaly accepted to use any fluid (like ideal gas ) to get the relation.

Of corse I am sure that all forms of the second law are true and equivalant but I think proving this two statements will make the equivalence clear and complete.

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- #15

netheril96

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I read what you wrote in Wikipeadia and found what I expected to find..The proof is clear but you used in it two statements which I think need proof too.

1)when you say :"Suppose a system absorbs heat δQ at temperature T. Since the value of does not depend on the details of how the heat is transferred, we can assume it is from a Carnot engine, which in turn absorbs heat δQ0 from a heat reservoir with constant temperature." and use the result in arbitrary cycles that means you assume that any cycle can be devided into number of carnot cycles..but this statement need a proof.

2)The relation of carnot cycle dQ/dT = dQo/dTo is calculated using the the properties of ideal gas while the desired result should hold for any working fluid.

so we need to prove that this relation is independant from the properties of the working fluid and only then it would be logicaly accepted to use any fluid (like ideal gas ) to get the relation.

Of corse I am sure that all forms of the second law are true and equivalant but I think proving this two statements will make the equivalence clear and complete.

The Carnot engine is a hypothetical attachment to the system. It is only a supplement to my proof. The system's cycle is not broken down into Carnot cycles; it is coupled with Carnot cycles.

- #16

Mueiz

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The Carnot engine is a hypothetical attachment to the system. It is only a supplement to my proof. The system's cycle is not broken down into Carnot cycles; it is coupled with Carnot cycles.

Carnot cycle is a special case , you can use it in a proof of something related general cycles only in two cases:

1)The result we want to arrive at does not depend on the details of the cycle.

2)The system's cycle can be broken down into Carnot cycles.

If you agree with this, which one you choose and how can you proof it before using it ?

If not ..please explain to me what do you mean by the words: "

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- #17

netheril96

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Carnot cycle is a special case , you can use it in a proof of something related general cycles only in two cases:

1)The result we want to arrive at does not depend on the details of the cycle.

2)The system's cycle can be broken down into Carnot cycles.

If you agree with this, which one you choose and how can you proof it before using it ?

If not ..please explain to me what do you mean by the words: "hypothetical attachment", "supplement to my proof", "coupled with Carnot cycles".

I agree with neither of the options. I have stated clearly in that article that each infinitesimal process of the system can be coupled with a Carnot cycle specially tailored to the forgoing process, which provides the system its heat.

- #18

Mueiz

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I agree with neither of the options. I have stated clearly in that article that each infinitesimal process of the system can be coupled with a Carnot cycle specially tailored to the forgoing process, which provides the system its heat.

"

This is a copy-paste of what you say in wikipeadia ! it agree with the first option.

Now, I want you to prove this statement.

- #19

netheril96

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"Since the value of dQ/dT does not depend on the details of how the heat is transferred, we can assume it is from a Carnot engine"

This is a copy-paste of what you say in wikipeadia ! it agree with the first option.

Now, I want you to prove this statement.

What difference does it make to the system where it drains heat from? Will the attributes of the heat reservoir affect the value of dQ? So why can't the heat come from a Carnot engine?

- #20

A. Neumaier

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For a systematic theory of thermodynamics that develops everything from a few basic assumptions (any theory must start some place), see Chapter 4 of my book at http://lanl.arxiv.org/abs/0810.1019 . I'd welcome pointers to things that are incomplete.I want a proof ..."step by step" as I said in my first post.

This will help us to study thermodynamics as a systematic theory rather than collection of facts and tales.

- #21

Mueiz

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For a systematic theory of thermodynamics that develops everything from a few basic assumptions (any theory must start some place), see Chapter 4 of my book at http://lanl.arxiv.org/abs/0810.1019 . I'd welcome pointers to things that are incomplete.

Thanks...this is interesting and benefitial ,but here you mix somehow macroscopic and microscopic methods to prove and that is ok, but I want a pure macroscopic proof ..you know that if one want to use microscopic method then all the laws of thermodynamics are simply not more than results of statistical calculation.

- #22

Pythagorean

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I like your styleThanks...this is interesting and benefitial ,but here you mix somehow macroscopic and microscopic methods to prove and that is ok, but I want a pure macroscopic proof ..you know that if one want to use microscopic method then all the laws of thermodynamics are simply not more than results of statistical calculation.

This is essentially the same question I asked in an earlier thread. I just ended the semester so I was going to actually look into Clausius' reasoning from "On a Mechanical Theory of Heat" as Andy Resnick has suggested in both threads.

But from what I saw in my thermo course, I'm also skeptical about macroscopic models.

- #23

A. Neumaier

Science Advisor

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Thanks...this is interesting and benefitial ,but here you mix somehow macroscopic and microscopic methods to prove and that is ok, but I want a pure macroscopic proof ..you know that if one want to use microscopic method then all the laws of thermodynamics are simply not more than results of statistical calculation.

But statistical mechanics predictions are highly accurate when the number of particles is in the macroscopic regime. The _only_ way to understand the macroscopic world from first principles is via statistical mechnaics, since it is well-known that on small scales, most observations are strongly fluctuating.

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