Consider dU=dq-PdV+(gama)dA, where gama is the surface tension and A is the surface area. Then consider different liquids with different surface tension. In general, at a given T and P, the higher the surface tension the lower the molar surface area. This means that the stability of the system at equilibrium requires contraction of the liquid surface molecules together. In other word, tension is like the negative pressure.

Dear Tang, the equation you are writing is an approximation of a more general equation. The minus sign in front of the pressure is there just because, in its origin, thermodynamics has been developped to describe gas properties. The pressure considered in the equation is the ambient one, therefore, the force is in opposition to the surface movement. As I told you before, the equation is just an approximation, because you are considering the pressure as a scalar quantity. It is not! Pressure is a tensorial quantity, represented by a matrix, that in the case of a gas is just diagonal. If you calculate the force due to this tensor, using a rigorous approach, the minus sign comes out naturally. In some cases, however, the pressure is not isotropic, as happens, for example, for a magnetized plasma. In this case the matrix is not diagonal anymore and the pressure of the electrons depends on the direction according to the local magnetic field.

Now to Q2. All the thermodynamic functions are equivalent. Physicists prefer internal energy U or enthalpy becasuse they have a mechanical meaning, while chemists prefer Gibbs free energy, because its direct relation with equilibrium constants and electrochemical potentials. If you use the proper function your problem can be solved more easier.

Enthalpy can be written as dH=TdS+VdP=dQ+VdP, therefore if you work at constant pressure, the enthalpy variation corresponds to the the heat exchanged. The sentence "for a reversible process at constant S. Y, and {N_j}, work can be stored as enthalpy and can be recovered completely", in some sense is valid not only for enthalpy, but also for energy: if you make an isentropic reversible transformation and than go back, you will return to the same final state and therfore anything you excange with the ambient, is given back. This is not true for a cycle in general. I suppose that you extrapolate this sentence from the description of the Carnot cycle, using an ideal working gas for your machine. In this case the two adiabatic branches makes the same work, but with opposit sign. This is valid only for ideal gases, but not in general.

Dear Mohammad,

Yes, if the surface tension (gama) increases, the surface area A will decrease;

But, if the pressure p increases, the volume V will increase.

Why the two are different?

Dear Gianpiero,

Q1. The pressure p and the repulsive force are not in opposition to the surface movement; only the attractive force is in opposition to the surface movement. If the generalized force Y is an attractive force, it will be easy to explain, but the pressure p has different sign from the repulsive force Y, why?

Q2. The explanation “Heat content” for enthalpy H is a prevailing view, in my opinion, the explanation “Heat content” for enthalpy H makes no sense.

Dear Tang Suye: No, as the pressure on a substance is increased its molar volume is decreased and -PdV contribution is to increacse the internal energy. .

Dear Mohammad,

Q1 seems to be a bit perplexing, please carefully read the original.

“the pressure, p, has different sign from other generalized force, if we increase the pressure, the volume increases, whereas if we increase the force, Y , for all other cases, the extensive variable, x, decreases.”

(L.E. Reilchl)

For example, given a PVT system in an equilibrium state, if we increase the pressure p, the volume V will increase. (Where the pressure p is a state variable)

in contrast, given a γAT system in an equilibrium state, if we increase the surface tension γ, the the surface area A will decrease.

Q1 : In a general case, P is the pressure applied to the system by the exterior , not the pressure of the system. It becomes the pressure of the system rigorously for a reversible process.

Q2:1) If you heat a fluid in a close system at constant pressure (and with no other generalized force), the variation of enthalpy is equal to the heat you add to the system. For "classical" fluid system it is the case. To my mind, it is one of the reason the enthalpy was called "heat content". (dU=deltaq-pdV or dH=deltaq)

If you have an other force : du=delatq-pdV+Ydx or dH=deltaq+Ydx

To concluse, mathematically H is an easy way to calculate heat exchange when there is no other force that those of pressure (if you have H value you do not need to calculate the volume change to calculate the heat exchange) but H do not be reduced to heat content for specific applications.

The convention used in the statement of first law p is the external pressure (in agreement with your convention) so that -pdV is the mechanical work exchanged between system and surrounding. In this sense the pdV work of expansion (of mechanical nature with repulsive origin ) could have a corresponding Ydx work of contraction (of thermodynamic nature with attractive origin). This can be used as general statement.

Dear Mohammad and Jean-Paul,

In Q1, the pressure p and the volume V are a pair of conjugate variables (of one system), i.e., the pressure p is the state function of the system, not the pressure external to the system.

Dear Jean- Paul,

Q2: dH is only a partial differential equation, you can calculate the heat (exchange+production) by dH, and you can also calculate the work (the free energy exchange) by dH.

Dear Tang Suye,

in Q1 ; Even if the process is not reversible (or at least quasi-static)? If you have reference of that, please share because it will be interesting to me to clear some ideas I have on thermodynamics

Q2 : If you know that, why did you ask for the physical meaning of H ?

and my last answer on this question ( I did not read the attached paper before but it helps me to understand the reason of the questions): I don't really agree the last sentence of the attached paper. the literal translation of french expression is "remain feet on grounded"

Dear Jean- Paul.

Q1: I agree with L.E.Reilchi’ opinion and had verified on the matter, but I’m not sure why?

Q2: There some different explanations for enthalpy .

In chemical thermodynamics, the main viewpoint is “heat content”.

In physics, "maximum work" and is similar to free energy.

In engineering thermodynamics, I’m not sure.

I need to verify my opinion as you said that "remain feet on grounded"

And please see my article,

https://www.researchgate.net/publication/236983852_arXiv1201.4284v4

I want to make clear some questions that can hardly be explained in current thermodynamics.

Data Internal Heat Energy, Entropy and The Second Law: A New Pers...

Dear Tang Suye

Good, either convention must end up with the same thermodynamic agruements ! agree !

Now, how increasing the internal pressure leads to volume increase ? by performing chemical reaction ? by adding more substabces? by increasing the temperature? or in another way? please let me know if one or all apply to the proposed question. Looking forward for an answer. Thanks

Dear Mohammad,

By δQ, or internal heat conversion such as chemical reaction, except pV work. The original conclusion has no explanation.

The physical meaning of enthalpy is explained in: Mannaerts SH, Energy-balance of the Joule-Thomson experiment: Enthalpy change at decompression. npt-procestechnologie. 2010; 17(4)18-22.

Available on this site. Bas Mannaerts, NL

Dear Bas Mannaerts,

The complete definition of enthalpy is that

H=U-Yx=ST+G.

Where Yx contains-pV (Please see L.E.Reilchl, A Modern Course in Statistical Physics, 2nd. edn,1997, p40), or

H=U-Yx+pV.

Where Yx does not contain-pV, and Yx is the mechanical energy of the system.

Thus, H cannot be explained as “the sum of the internal energy and the pressure energy”.

If you take 100 textbooks on thermodynamics, 99 will define H=U+pV. Apparently Reilchl is not one of the 99. For the sum of enthalpy and mechanical energy some use the 'Bernoulli function', but is not officially established I think.

Dear Bas Mannaerts,

Indeed, most of the textbooks on thermodynamics did not involve the mechanical energy within the system when they define enthalpy, the definition H=U+pV is based on some simple models, and can only apply to these simple models, but is not a complete definition.

Well, that is merely a question of definition. If I were you I would stick to H=U+pV for enthalpy and think of an other name (total energy ?) for the sum of enthalpy and mechanical energy. It will save you a lot of trouble.

I read the attached file and there are some basic problems that need to be clarified.

1) In eq. 2 the work is not an exact differential, but it is the difference between exact differentials. How it is possible? This behavior is reflected to eq. 3. the heat is not an exact differential, but it is expressed as the combination of state variables. It seem something magic this passage.

2) the pressure is defined as the momentum exchange between particles inside the vessel and the boundary. Now this definition is is valid only for a gas. However thermodynamic laws are valid for any kind of system. If you have a solid or a liquid, what is the pressure?

3) Reading the paper I am thinking about an ideal experiment: suppose you have cylinder and you measure the pressure as the force applied from the inside of the cylinder to a piston / the piston surface. Can you distinguish if the force is due to the gas pressure or to a spring or repelling magnets, only considering the first law of thermodynamics?

4) can you make an example of Ydx work you are thinking?

Dear Bas Mannaerts,

Sorry for late reply, a long journey, about 2000 km by car.

On the other hand, U+pV>U, what is “the pressure energy”? Is it a part of internal energy or not? And is it the real energy or not? The sum of the energy within the system cannot be more than the internal energy U.

Dear Gianpiero,

Sorry for late reply, a long journey, about 2000 km by car.

Good questions. Thanks for your reading the attached file.

1) I agree with your opinion, the difference between exact differentials should be an exact differential. eq.2 and eq.3 must be attached with some strict conditions, this is the flaw of current theories, not my approach. The exact differential dU should be expressed as the sum of the exact differentials.

2) The pressure p is defined as the force that root in thermal motion, and is proportional to the average momentum of the microscopic particles (thermal motion), “the momentum exchange between particles inside the vessel and the boundary” is only a measurement external to the pressure p (of equilibrium state).

3) It's easy to distinguish in engineering.

4) Ydx = the dynamics equation of the work (expect pV work).

Ydx=JdL+σdA+E·dP+ H·dM…

Where J (tension or elastic force), σ (surface tension), E (electric field strength), H (magnetic field strength) are the generalized forces …, and dL(length), dA (surface area) …. are the generalized displacements….

An attached question:

I noticed that your research interests include plasma physics, do you find that eq.8 may be an even simpler equation than the equivalent equation of that in current thermodynamics?

"pressure energy" or "energy of displacement" (pV) is not part of U, but a separate form of energy: see for example the Bernoulli equation. It is explained in the paper mentioned (Mannaerts, 2010)

Dear Bas Mannaerts,

Since H is defined as a state function, so pV is the property of the system itself, which cannot be “independent of the content of the system”.

In fact, pV and both –pV (a part of Helmholtz free energy F) can hardly be explained by current thermodynamics. It is a thermodynamic potential related heat conversion.

Dear Tang Suye

Sorry for the late reply but I am very busy in this period.

I think that eq. 8 is correct and equivalent to the traditional form. The problem is that the function q, no matter you call it, cannot be measured.

I do not agree with your definition of pressure. Because of Hesenberg principle, you cannot measure the mean momentum measuring this properties to each particle. The only way to measure it is to put a surface inside the gas. Moreover the mean momentum of particles is not a constant in the vessel, but it is a local property and as a consequence cannot be a state variable. Think about a real gas: the mean momentum close to the surface is different from the momentum in the middle becuse of the difference between interparticle and particle-surface forces.

Moreover, the definition cannot be applied to the solid, where the pressure is due to the repulsive forces between particles in the solid structure, and the same in liquid. In the Pascal Law the pressure is due to the weight of the liquid and not to its temperature. And what about the magnetic pressure?

In conclusion, the pressure should only be considered as the force applied on a surface, independently on what produce it. Because inside a vessel containing gas, it is not possible to distinguish if the pressure exerted on a piston is due to a spring, or to a wheel engine or an idraulic pump.

Dear Gianpiero,

“Measure q” for a generalized thermodynamic system is a problem due to the complexity of the system, maybe it is one of the reasons why we lost the function q, I think, but dq can be calculated by the aid of other variables.

The word “pressure p” has a double meaning: 1) the force applied on a surface, 2) the state variable of thermodynamics. Please see

http://en.wikipedia.org/wiki/Functions_of_state

For a solid, the micro vibration of the atoms and molecules (and other thermal motion) are the superposition effect by thermal motion and interactions, not only the repulsive forces between particles in the solid structure.

The weight of the liquid and the magnetic pressure = Generalized force Y. Y can also apply on a surface, and customarily called as the pressure, in this case, the word “pressure” has only the meaning 1), not 2).

How an engineer designs an engine if he cannot distinguish the origin of the force? The resultant of two or more forces, for all cases, cannot be distinguished that which part of the resultant force is correspond to a specific sources.

@ Gianpiero

It is true that there is a difference between forces felt by a particle close to the surface and one deep in the bulk. However it is possible to show that the only effect is that the local density at the surface is different from the bulk density. The so called "contact theorem" of the theory of inhomogenous liquids (see Hansen Mc Donals "Theory of simple liquids") ensures that p = rho(wall) kT for a fluid of bulk pressure p confined by a hard wall. rho(wall) is the value of the local (microscopic) density at the wall. This result can be recast in the form "the kinetic pressure at the wall must balance the thermodynamic pressure.

@ Tang Suye: from the previous comment you see that the two definitions you found in wikipedia do not differ.

@ Tang Suye: I read the paper you attached to your original post. I am afraid that you missed some very basic facts about thermodynamics.

Apparently you think you have been able to rewrite the exact differential dU (usually considered as the sum of two nonexact differentials Q and L) as the sum of two differentials.

Provided you make explicit the assumptions which enable your result, you're right. But you have not done anything more than obtaining the well known expression for dU as TdS - pdV + mu dN, just rewriting the differential TdS as dq.

However, in this way you miss entirely the point behind the first principle of thermodynamics. That is the fact that in general neither the work done on a thermodynamic system can be written as -pdV, nor the heat transerred to the system can be written as TdS. Usually, work and heat do depend on the process and this prevents any possibility of considering them as the difference of the values of a function at two different points. It is only in the very special case of a reversible transformation that one has the possibility of writing separately Q and W as differentials.

Dear Giorgio,

Thanks for reading the attached and my original post, and thanks for your question. I’m waiting for this question because I must explain the equation.

The well known equation

dU= TdS-pdV+μdN,

is a correct equation both in Math and Physics, and it is also the sum of the exact differentials, but the equation has a flaw: U has not been expressed as the sum of the independent types of the energy within the system.

In my approach, dq=TdS-pdV, not dq=TdS, so it isn’t “just rewriting the differential TdS as dq”. And TS isn’t equal to “the internal heat energy”.

For example, consider an ideal gas, we have

1) TdS= cdT+pdV,

TdS is the sum of the two terms. The physical meaning of TdS includes the two, which is not one independent term (The first term is dU=dq).

2) dU= TdS-pdV.

Using 1) in 2), we get

3) dU= cdT+pdV-pdV=cdT=dq.

For an ideal gas, all of the internal heat energy U is the internal heat energy q. It isn’t “just rewriting the differential TdS as dq” because dq≠TdS.

In my approach, the well known equation

dU= TdS-pdV+μdN=dq+μdN=dq+dG.

Is it simpler than the original equation?

On the other hand, the heat energy within the system is one fundamental type of the internal energy, we need to define it in a primary concept, S is not a primary concept but derived by q and some other variables. For a fundamental definition, we can define S by q, but cannot define q by S.

Dear Giorgio,

I agree with the view “That is the fact that in general neither the work done on a thermodynamic system can be written as -pdV, nor the heat transerred to the system can be written as TdS.” In my paper, “It is only in the very special case of a reversible transformation”, I established the equation by the aid of “this special case”, and then discuss the state variables.

"Type of internal energy" is not a concept one can assign a formal meaning in thermodynamics. In the general case one has only the internal energy without any possibility of distinguish "forms of internal energy". You should avoid to base your arguments on the very special case of a perfect gas. It may be highly misleading.

Dear Giorgio,

Why don't we just do it? As we've had Gibbs free energy G, which is one "form of internal energy", why can't we distinguish the other forms?

By dU= dq+dG, I can say that “In the general case”, the internal energy include the internal heat energy and Gibbs free energy, U is equal to the sum of q and G. But by dU= TdS-pdV+μdN, you tell me “In the general case one has only the internal energy without any possibility of distinguish ‘forms of internal energy’". Is the new equation simpler than the original equation?

We've already had Gibbs free energy G for “the general case”.

With regard to the first question, I would suggest to read the classical book:

H.B. Callen, Thermodynamics and an Introduction to Thermostatics, Wiley, New York, 1985.

The internal energy takes care of all the energy that is transferred to the system either in the form of work or of heat. If we take a solid, we can change its internal energy by either stressing it or heating it up. Remember that heat and work are forms of energy transfer and not forms of energy

As far as the second question is concerned, it helps quite a lot in thinking of enthalpy as the quantity that replaces internal energy (which is introduced for closed systems) in open systems.

In fact, the first principle of thermodynamics (which is nothing else than the principle of energy conservation) is written in its simplest form for closed systems:

M du = M T ds - p dV

where M and V are the total mass and volume of the system, T is temperature and u and s are respectively the specific internal energy and entropy.

For steady state conditions in an open system, the energy equation can be written as:

m DeltaH = Q - L

where m is the mass flow rate (mass per unit time) that goes through the system, H is the total specific enthalpy (h + Vsquare/2) and Q and L are respectively the heating and the working (shaft) powers (energy per unit time).

A comparison between the two formulas may help in understanding enthalpy and why in turbines, compressors etc. (open systems) one uses total enthalpy and not internal energy.

For low Mach nuber flows, the total enthalpy H coincides with the (static) enthalphy h = cp T (for calorically perfect substances).

So much confusion over such simple things:

In

dU = q+w = TdS -pdV (1)

the work is defined as the work done on the system. If I decrease the volume by exerting pressure, I do work on the system, and the internal energy increases. Since in that case dV is negative, -pdV is indeed positive. For the rest -pdV is just the expression for mechanical work: force x displacement, or for a piston: (force/surface area) x (displacement x surface area)

In contrast: by stretching a rubber band you do work by lengthening it. The sign is positive in that case. I don't know if it is true for all other cases, but can't offhand think of another one with negative sign.

The difference between enthalpy H and internal energy U is because you can do two things: keep the volume constant during the addition of heat, or keep the pressure constant. If you keep the volume constant, all the heat entering the system goes to raising the temperature. C_V tells you how much.

However, if you keep the pressure constant , the volume will increase, and some of the heat will go to the work done by the system. The other part goes to raising the temperature. That is why for gases C_p is always larger than C_V: you need to add more heat to get the same increase in temperature when the pressure is constant.

Keeping the pressure constant is much easier than keeping the volume constant. Therefore chemists usually work with H. The difference is relevant in systems where gases play a role, For most other systems the difference is small and often negligible.

In mathematical terms: written as the fundamental equation (1) U is a function of V and S. H is a function of p and S.

dU = TdS -pdV

H = U+pV

therefore

dH = TdS -pdV +pdV+ Vdp = TdS + Vdp

This transformation between functions of different variables is sometimes called a Legendre transform.

Note in particular that when dp = 0, dH = TdS = dq

As for the difference between TdS and dq: S is a state function, and therefore independent of the path between states. q is not a state function and depends on the road taken. Along a reversible path dS = dq/T, for all others dS > dq/T.

There is confusion also on symbols.

Probably Mr. Van der Zwan in his equation

dU = q+w = TdS -pdV (1)

means:

dU = dq + dw = TdS -pdV

A differential quantity cannot be equal to a finite one.

As far for the other comments they are clearly written in my books:

CARLOMAGNO G.M.: "Fluidodinamica", 1-371, Liguori, Napoli ISBN 88 207 3279 3 (2001)

CARLOMAGNO G.M.: ”Elementi di gasdinamica”, 1-475, Liguori, Napoli ISBN-13 978-88-207-4663-6 (2009)

unfortunately both in italian.

Also H is a state function

Sorry for forgetting the d's in the equation.

Also, referring to the original question: you cannot speak of "heat content". Heat is not a state function, enthalpy indeed is.

The d's are important.

Then, I never spoke of "heat content" but I rather clearly said:

Remember that heat and work are forms of energy transfer and not forms of energy.

So heat, as work, are energies in transit.

As far as your sentence:

In contrast: by stretching a rubber band you do work by lengthening it. The sign is positive in that case. I don't know if it is true for all other cases, but can't offhand think of another one with negative sign.

is concerned, you can add internal energy to a solid system by either stretching it or compressing it. the result is that in both cases you are adding internal energy because you must provide some work. The minus sign before p is because pressure is always considered as a compression stress while tensile stresses have a positive sign.

Dear Giovanni and Gert,

We all know the explanations that you mentioned, the questions are not that simple.

To be honest: yes they are.

By the way, compressing a car spring (google "car springs" images) gives a negative Ydx term.

Giovanni: The d's are important, I said I was sorry. "Heat content" was mentioned in the original question, I did not refer to your answer.

Dear Gert,

I'm afraid you've just entirely misunderstood the question, where we “increase the pressure” or “increase the force, Y” not by doing work, not by “compressing a car spring”, for instance, we change the temperature of “a car spring”. Then if the force Y increase, x decrease”.

Dear Gert,

the sign is always positive because you change both the sign of force and the one of displacement. You have to do work both to stretch as to compress the spring and this work goes into internal energy.

OK, point taken, but I think this is quibbling over rather irrelevant things, because we know what increases or decreases the internal energy, and how that the determines the sign of the work function. Let's go back to the original problem.

Tang Suye claims that traditional thermodynamics is incomplete and a new state function, and a new way of thinking, is useful or even necessary. Apart from the fact, which was already pointed out, that pressure is only relevant in a minority of systems, and that the available state functions have allowed us to study, and derive useful relations for almost any conceivable system, it is of course not impossible that for more than 150 years people working in the field missed something essential, or that a new point of view can shed light on previously unappreciated problems.

So, in stead of just deriving a relation for this new function, why not apply it? We know all the existing functions (U, S, A, G, H, ...) exactly, or to a good approximation, for a large variety of systems. So here is my suggestion: take your new function, calculate it for a number of systems in which pressure does play a role (ideal gas, van der Waals gas, photon gas, compression of springs or rubber blocks, deformation of crystals, chemical reactions in the gas phase, curvature of a soap bubble, gas-liquid phase transitions, vapour pressure of solutions, to name just a few), and show us how this new function, internal heat energy, gives additional insights that we did not yet have. Once we have these expressions we will be able to discuss the merits of this new state function, and see if the information is not already contained in the others. If there is a new state function, there will be new Maxwell relations between new thermodynamic derivatives, what do those tell us? I can think of an almost limitless number of questions and problems that it can be applied to, so why wait?

By the way, the author of the book you refer to is called Reichl, not Reilchl. And could you point out where in her book I can find the background to your problem?

What Tang Suye is trying has to do with Thermodynamics Potential.

If you want to know more and improve your italian language try to read the attached

https://dl.dropboxusercontent.com/u/3580191/Potenziali%20Termodinamici.pdf

In the attachment, right after the formula, you should read in parenthesis (G1,.....Gm-1, Ym,.....,Yn)

Yes, I understand that idea, and that for other forms of work there are other choices for potentials than the standard G and A (or F, depending on your field) you have for an ideal gas (I did a lot of work on polarizable and charged systems in the past) but that does not relieve him from the requirement to show the usefulness of a newly proposed potential. It greatly helps understanding if you take a non-trivial system and calculate quantities. Preferably not the ideal gas, since many problems I encountered with students came from the fact that almost all Physical Chemistry textbooks use that as the primary example, and at the end the students apply the simple formula's you get for the ideal gas to almost anything else.

So I stick with my suggestion: take the new potential or state function, whatever you want to call it, and calculate it for a non-trivial system. He claims that there is a new state function q(T,V) so I simply ask, what is it for the ideal gas, the van der Waals gas, and the photon gas (three gases for which the internal energy, the entropy, and the pressure as a function of V,T are known)? If it works in those cases, and is not a simple extension of the things we already know, and leads to new insights, we can have further discussions. If it exists it can be calculated.

If there is something like q(T,V), then the thermodynamic derivatives exist, and relations between them. Are these new physical results? Since all existing state functions are related by Legendre transforms, where does this one relate to the existing ones? If you need a system with more than one form of work than just volume work, make one up (ideal gas of polarizable molecules inside a capacitor or something like that), and do the calculations for it. Don't propose an exciting new function trying to overthrow standard thermodynamics without at least telling us how it can be calculated, used, or measured.

What you call "a new state function q(T,V)" is the already existing Helmholtz's function. Please, see in the attachment formulas (2.42) and (2.46) with their differential (2.52). You do not need to know italian language to look for them.

It is difficult to find something new in thermodynamics. People like Fermi dealt with it.

The above formulas are per unit mass.

The perfect gas model has nothing to do with them. If you assume this model, by means of its equations of state (in this case two) you may derive their explicit formulas.

It is not what I call a new state function, it is what he calls it in the attached pdf. I quote

"The internal heat energy q =q(T, V ) is the function of the two independent variables T and V , and can be proved to be a state function in mathematics[1]. This conclusion will rewrite the theoretical form of the current thermodynamics."

But looking at it more carefully now, I guess you are right, it is just the Helmholtz free energy. I don't expect this to rewrite the theoretical form of current thermodynamics.

Well. at least it wasn't a perpetual motion machine.

Dear Gert,

you are right in your last sentence, but it got very close to it.

Seasonal greetings to all participants.

This is the expression for the free energy for the perfect gas

https://dl.dropboxusercontent.com/u/3580191/doc1.pdf

Dear Gert,

Thanks for your suggestion, and thanks your “By the way…”

I have verified most cases you suggested (my research interests include theoretical physics and physical chemistry), and compared my approach with the fundamental equation of traditional thermodynamics, the conclusion is, the two are equivalent in mathematical results.

Below are some parts of the new theoretical form of thermodynamics.

1) The first law

dU=dq+dψ=dq+(Ydx+dG)

Where q is the internal heat energy, ψ is the free energy, correspond to the potential energy of a thermodynamic system, Y dx does not contain −pdV . q makes contribution to entropy, ψ makes no contribution to entropy.

2) Entropy

dS=dq/T+pdV/T

The additional restriction to reversible processes is not an essential requirement for the definition of the state function. dQ/T is only the entropy of the heat flux δQ, and basing the definition of the concept of entropy on imaginary engine cycles is not a perfect foundation in the usual sense.

3) The second law

The equation please see my paper Eq.(104) and Eq.(105).

The statement about the second law: All of the gradients of the thermodynamic forces spontaneously tend to zero.

4) The basic sources of irreversibility

There are three basic sources of the total entropy production diS: i). Heat flux only occurs from higher temperature to lower temperature; ii). Free energy is converted into heat energy without compensation; iii). The independent process caused by the pressure.

5) The building mechanisms of the second law.

The total entropy production always increases monotonically, but the partial differential of which can be less than zero, the second law itself contains the building mechanisms.

……

Others,

1) For Helmholtz free energy, -pV is not the real free energy, F contains the free energy ψ and −pV , the latter is “the free energy” that can be obtained from heat conversion.

q=q(T,V) is not F, which can be converted into F.

2) In Reichl’s book you cannot find the background to my question 1.

Dear Giovanni,

Thanks, I don't know Italian language; I can only read it by google translator.

Dear Tang Suye,

if what you call:

"thermal motion of the particles"

is not

"internal energy"

what is for you the internal energy ?

Dear Giovanni,

It's not my point, or it is a slip of the pen, where?

The energy of the "thermal motion of the particles within the system is one part of the internal energy.

Dear all, I am quitting this conversation. It is obvious that no progress is made. I read in the above answer by Tang Suye that q(T,V) is not the Helmholtz energy but also that he has verified all the examples I mentioned (whatever that means and without showing) and shown that they are "equivalent in mathematical results". One brief remark and a long one follow.

The short one: Energy of thermal motion is what we call kinetic energy,

The longer one: Over the years I, and I am sure Giovanni as well, received many letters from people claiming to overthrow thermodynamics, relativity theory, quantum mechanics or other things that are mathematically complex and sometimes counter intuitive, which apparently makes them unhappy; or to claim that they have the concept of a working perpetual motion machine. I usually don't spend time anymore to find the error. In the latter case, a pmm, I tell them to build one and see if it works. Or to find vulnerable people who are willing to donate money to build a prototype. Then they can at least make some money.

In the other cases I mostly ignore it, or, if the error is glaring tell them where the error is. That almost never helps. In the present case the first and last line of the manuscript already made clear that it probably was not useful to spend time on this, but what the heck, I just love thermodynamics. So even though Tang Suye does not tell us when the long lost function was lost (first I thought maybe he meant caloric, useful to Carnot, but a wrong idea and since then lost), and what it was, and who lost it. In the last line he tells us about a new function (presumably the one that was lost, but again he does not show us that) that will overthrow the way we think about thermodynamics. I did not go over the stuff in the middle, and although the T,V dependence should have tipped me off, it was Giovanni who pointed out after more careful reading that it was just the Helmholtz energy. Tang Suye agrees with that and he doesn't.

Thermodynamics is one of the older (classical dynamics is the oldest) fields in physics. It was developed over a period of more than 100 years. At the end of the 19th century it was well established, and great scientists like Boltzmann, Maxwell, and Planck studied it in great detail, wrote books about it, and used it to develop their own ideas further. Einstein considered it the greatest theory of all that would never be overthrown, It was at the basis of new developments like quantum mechanics. Almost every great scientist wrote about it, Fermi, as Giovanni pointed out, but most of the other people we now consider founders of quantum mechanics and other fields of physics in the early 20th century wrote books about it. There were axiomatic developments (Caratheodory), and the theory was applied to almost every conceivably system. Black holes now have entropy (Bekenstein) and temperature (Hawking) to keep the laws of thermodynamics valid. In addition the theory is, since Boltzmann and Gibbs, validated by statistical mechanics. You have Reichl's book, among many others, to study that as well.

Why I am telling you this: well, if you're planning to overthrow something, you should at least show how your own theory is better, or at least as good, at explaining all those things, and that in addition your theory can shed light on previously unexplained phenomena. I don't say it is impossible. Theories are sometimes overthrown, but rarely. Most of the time limits are set on them: classical mechanics is still valid in a more restricted area. It is possible that thermodynamics and the second law have their limits. In particular when we start looking at very small systems. (Google "Maxwell's demon", or "Feynman's Brownian motor", if you want to learn more about that). Stating that you found a new function that overthrows the way we think about thermodynamics just does not cut it. It was already pointed out in the beginning of this thread (by Giorgio if I remember correctly) that maybe you should devote somewhat more time to study the topic. You are smart enough, you already were able to show without the help of a book that there is another Legendre transform; U-TS. If you don't (devote more time to study) you may end up like so many of the people referred to above: writing to scientists trying to explain your theory, complaining that "established science" does not want to listen to you, arguing with editors of famous journals, but never listening to those who do spend time explaining things and never getting anywhere. For a nice example google "Carel van der Togt" and find his website.

With these remarks I stop contributing to this thread, and start looking for another one that rouses my interest. I also extend my seasonal greetings to all of you.

Dear Gert,

Thanks for your writing that long, my English is a bit poor, so I need to seek the physical content in you answer, it seems to be very few, and I found that you seems to like to change the topic.

I don't mind your rejection of my ideas, but we had known well the story that you written.

I got tired of the discussion (I am quite old) so I am quitting as well. Good luck with your findings. Hope to eventually see you in Stockolm.

Dear Giovanni,

I can understand, thanks for your contribution to the discussion, with best regards.