A short feedback The first law of thermodynamics (dU=dW+dQ) is simply the restatement of the conservation of energy and it applies to all types of systems, whether in equilibrium or away from equilibrium (Note that however, statistical mechanics tells us that the internal energy slightly fluctuates around its mean; an offset value proportional to 1/sqrt(N) with N being the number of particles)

We can also define characteristic functions such as entropy and internal energy for non-equilibrium systems (NES). However, the basic problem with NES is that these functions are no longer state functions- they depend on the trajectory or 'path' traced by the system in its phase space. In fact, it appears that the trajectory of a non-equilibrium system is highly jagged (like a random walk) and follows no analytic pattern in the usual sense (But we can still handle them with the glory of modern mathematical concepts such as fractals) Therefore there is no known explicit formula which encodes any information about a system away from equilibrium and the term state function in non-equilibrium physics becomes a misnomer. 

Now back to your question Even if our variables assume stochastic values in a hyper-dimensional manifold (or phase space as you wish), they can still been defined point-wise at any allowed regions in the phase space (however their exact values are hidden for us, and we can maximally talk about their expectations)

As I mentioned before, there is no differential in a justified mathematical sense for stochastic variables, but we can still consider infinitesimally small changes of these variables which can be interpreted as differentials and making mathematicians uncomfortable. Practically, this can be achieved by introducing a quasi-static process/change

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

To sum up: There is no exact differential for thermodynamic variables (such as entropy and internal energy) in non-equilibrium state, but we can always talk about very small changes and expected values... 

A short feedback The first law of thermodynamics (dU=dW+dQ) is simply the restatement of the conservation of energy and it applies to all types of systems, whether in equilibrium or away from equilibrium (Note that however, statistical mechanics tells us that the internal energy slightly fluctuates around its mean; an offset value proportional to 1/sqrt(N) with N being the number of particles)

We can also define characteristic functions such as entropy and internal energy for non-equilibrium systems (NES). However, the basic problem with NES is that these functions are no longer state functions- they depend on the trajectory or 'path' traced by the system in its phase space. In fact, it appears that the trajectory of a non-equilibrium system is highly jagged (like a random walk) and follows no analytic pattern in the usual sense (But we can still handle them with the glory of modern mathematical concepts such as fractals) Therefore there is no known explicit formula which encodes any information about a system away from equilibrium and the term state function in non-equilibrium physics becomes a misnomer. 

Now back to your question Even if our variables assume stochastic values in a hyper-dimensional manifold (or phase space as you wish), they can still been defined point-wise at any allowed regions in the phase space (however their exact values are hidden for us, and we can maximally talk about their expectations)

As I mentioned before, there is no differential in a justified mathematical sense for stochastic variables, but we can still consider infinitesimally small changes of these variables which can be interpreted as differentials and making mathematicians uncomfortable. Practically, this can be achieved by introducing a quasi-static process/change

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

To sum up: There is no exact differential for thermodynamic variables (such as entropy and internal energy) in non-equilibrium state, but we can always talk about very small changes and expected values... 

Zhoubin answer is very punctual and appropriate. The only thing I can add is that any system, also in equilibrium, makes jump in the phase space, also in equilibrium, which is the origin of the fluctuations. This is the microscopic view of a macroscopic system.

When you are looking to macroscopic quantities you have that the evolution ii can be expressed as a regular function U(t) plus a random function (neither continuous nor derivable) R(t) with null mean when calculated over  a time interval large enough. 

R(t) oscillates very rapidly, and if we consider the measure operation of the quantities, it is registered in time interval Dtm that is much larger than the mean collision time tc, that is the characteristic time of the fluctuation in R(t). Therefore the measured quantities are the mean value of our quantities and therefore we get only U(t) (by measure). 

Now we have to specify the meaning of dt. From the mathematical point of view the random fluctuation makes the global evolution of the function to be non derivable. But from the physical point of view, thermodynamic derivatives consider dt a small quantities much larger than the fluctuations so that you get only the mean value.

Obviously the amplitude of fluctuations must be small, as it is the case of large systems.

The answers above are statistical and stochstic ones. But of course there is also a phenomenological one: dU in non-equilibrium thermodynamics means the time derivative of the internal energy along the process trajectory of the irreversible process in the state space. There is also a time rate of a non-equilibrium entropy which may be integrable or not, depending of the chosen state space. Details can be found in

Aspects of Non-Equilibrium Thermodynamics, World Scientific Sigapore 1990 Chap.5

Fundamentals of Non-Equilibrium Thermodynamics

in: W. Muschik (Ed.), Non-Equilibrium Thermodynamics with Application to

Solids

CISM Courses and Lectures No. 336

Springer, Wien, 1993, pp. 1 - 63, ISBN 3-211-82453-7

Dear Zhoubin Maneshi and Gianpiero Colonna,

   Thanks for your reply ! Although I do not konw much about non-equilibrium thermodynamics, the information you offered has thrown light on my puzzle. We are getting away with the somewhat uncomfortable mathematical expression. From your reply, 'dU' and 'dS' can still be used in non-equilibrium thermodynamics to denote the infinitesimal change of the corresponding characteristic function although the concept of state function no longer suitable.

In the field of equilibrium thermodynamics, 'dS>dq/T' for an irreversible process is presented in many textbooks. In the intermedia states where the system is not in equilibrium state, how to interpret 'dS>dq/T' from the standpoint of equilibrium thermodynamics? Or 'dS>dq/T' for an irreversible process still holds in non-equilibrium thermodynamics?

Thanks for your time!

The main problem with non-equilibrium system is to find a meaning to quantities commonly used in equilibrium. You can surely define the energy and also the entropy, even if calculating S in non equilibrium is an hard problem, but can you define the temperature?

Obviously it depends on the degree of non-equilibrium if you can still use the formalism of thermodynamics, and therefore a meaning to the temperature can be found (see the book of Prigogine Irreversible thermodynamics). 

In general it is not possible to define the temperature as a state variable and therefore dS>dq/T cannot be written because you do not know what T is. 

Please, read the following attached publications:

@ prof. Muschik

I read the paper but I had some difficulties in following the details, because the formalism is not common in thermodynamics. I have some questions about it. 

I believe that in some cases the theory works, but there is something that I cannot match with my experience.

First of all, the definition of equilibrium state as a steady state is not usual in thermodynamics. What about metastable states? Because the measure of state variables is carried out in a finite time interval, it is no possible to distinguish between a metastable and a stable state in equilibrium. Your theory treat metastable states and stable stats in the same way?

The scope of non-equilibrium thermodynamics is to predict the evolution of the system toward equilibrium. One of my research activities is to calculate the evolution of internal distributions that strongly departs from the Boltzmann (or maxwell for electrons), macroscopic quantities are not sufficient to the evolution, but only determining the whole distributions we can define the chemical rate coefficients. Therefore, because in your theory you define more state variables to characterize the state of the system, it can help in finding which are these variables? And in the case the number of variables must be equal to the number of internal levels, is the theory equivalent to the kinetic theory?

This question is very important for me because I am trying to find an additional state variables to describe the evolution of "macroscopic quantities" using a reduced number of variables. I have attached one of my papers. 

Operation definition of thermodynamic energy change dU relies on the postulate that 'the work  w  required to bring a thermally insulated system from one completely specified state to a second completely specified state is independent of the source of the work and of its path through which the system passes from the initial to final state'. Namely,  for an infinitesimal change    Δw= dU. Where the only measurable quantity is  the work. This definition doesn't differ whether one has to deal with reversible or irreversible thermodynamics. In fact -following the Max Born – if one introduces the partition of entropy into external and internal parts, and then postulate that the internal entropy change is positive definite quantity for irreversible (natural) changes and zero for equilibrium processes, otherwise it becomes never negative (unnatural). In irreversible thermodynamic, for the infinitesmal process we assumed the validty of the Gibbs’ equation for open as well as the closed systems, and try to formulate the specific entropy variation dS in terms of specific energy, volume and and concentration variaitons. (see: Guggenheim, Prigogine, de Groot and R. Haase). For the systems under thermal equilibrium, one uses Helmholtz free energy dF if they are not exposed to external tractions and body forces to formulate the problem. Where one has dSin=- dF/T, which is positive definite. .

Non-equilibrium thermodynamics is a phenomenological theory. You have to set the state space variables, perhaps by using a microscopic back-ground. The attached paper may be useful ?

In thermodynamics, the equation dU=δQ+δW (+∑μdN) can be applied to all of the energy exchange processes, but dU=δQ+δW cannot describe any process within the system, where Q is the heat in transfer, and W is the free energy in transfer, your question, I think, is that Q and W are not the state variables, and for non-equilibrium state or irreversible process, how do we express dU and dS.

In equilibrium thermodynamics, for a reversible process, we have δQ=TdS and δW=Ydx-pdV, where T, p, and V are the macroscopic quantities of the system. But Y, x are usually not.

In the fundamental equation of thermodynamics

dU=TdS-pdV+Ydx+∑μdN

For the intensive variables T, p, Y, μ, only T and p now are considered, as the macroscopic quantities of a system or a local, Y and μ don’t have to be. Since TdS-pdV denotes an implicit function, the heat energy within the system q (but not the heat in transfer Q or your q), so the first law can be rewrote as

dU=dq+Ydx+∑μdN

The equation does not need to consider the restriction to equilibrium state or reversible process.

In non-equilibrium thermodynamics, the changes in the entropy become

dS=deS+diS

Where deS is the entropy flux, and diS is the entropy production. The second law becomes

dS=deS+diS=δQ/T+diS >=δQ/T

diS>=0

dS>=δQ/T still holds in irreversible process, T in such case is the temperature of δQ. Since δQ is the heat in transfer, so δQ/T is the entropy of δQ, and is the entropy in transfer.

The total differential of diS and the 2nd law please see my paper (v5) eq.(134), and the new statement about the second law becomes: All of the gradients of the thermodynamic forces spontaneously tend to zero.

In general, the definition of an extensive variable does not need to depend on equilibrium state, and an intensive variable comes from the distribution of an extensive variable, so the intensive variable can be instead by the distribution of the extensive variable. For example, you can see in my paper (v5), from eq.(64) to eq.(72), for the equation of the entropy of some systems, T can be instead by the distribution of the extensive variable q, and P/T can be instead by the density of the particle numbers N/V. It implies that the equations of non-equilibrium thermodynamics (the local equilibrium thermodynamics) can be extended outside the restriction to the local equilibrium hypothesis.

The concepts of the state function and the exact differential are still valid for non-equilibrium state; otherwise it will break the rules of exact differential.

Due to my limited knowledge of the non-equilibrium thermodynamics, this topic is a little hard for me. This question originally comes from my reflection on many textbook of equilibrium thermodynamics which use the term 'dS' for both reversible and irreversible process. In the field of equilibrium thermodynamics,  'ΔS' can be suitably used to describe the finite entropy change of the closed system from an initial equilibrium state 1 to a final equilibrium state 2 no matter whether the path is reversible or irreversible. However, dS, I believe, is only suitable for the infinitesimal entropy change in an infinitesimal reversible process in the field of equilibrium thermodynamics because we only talk about the change of the system from one equilibrium state to another equilibrium state in equilibrium thermodynamics.

In thermodynamics, we have ΔU=q+w, always. Well always for closed systems, even if q and w are very small. This is plain physics (energy balance). In thermodynamics we like to imagine a function like U(S,V). If we happen to have this function (we seldom do) we can always (!) differentiate such function and arrive at dU, eg dU=TdS-pdV. This is plain mathematics if (and only if) T=(dU/dS) and p=-(dU/dV). However if we want to identify q with TdS we have to assume reversible heat transfer.  This is were your problem is generated: at the cleavage between physics and math. In doubt follow the physics. 

If one stays in the domain of thermodynamics as proposed by Max Born and later advocated by Guggenheim then it becomes very clear that there is no difference in the definitions of thermodynamic characteristic functions and variables.  Guggenheim uses definition of infinitesimal processes  for those reversible variations  where the system under consideration is in complete thermodynamic equilibrium with its surroundings all along the path of internal equilibrium. In the case of infinitesimal change this constrain drops out.  By definition of thermal energy  denoted by  U  one has   dU= delta w   for  the  adiabatic process\  similarly for the finite change one has ΔU=  w  for adiabatic path where  delta  indicates inexact differential operator. The Fundamental difference between reversible and irreversible thermodynamic is  that the internal entropy part becomes zero for the case of reversible processes  and   is positive definite for the irreversible processes or changes.  There is one important assumption in reversible thermodynamics\ which is the validly of Gibbs Pfafian differential equation as obtained by the thermostatic considerations. 

The state function changes must be independent of the path taken, it requires that an exact differential does not need to be restricted to any path, including reversible path.

If dS is considered to be only suitable for the infinitesimal entropy change from one equilibrium state to another equilibrium state, and the path only is restricted to reversible process, it means that there has no diS>0, and no the second law, if dS produced via an irreversible path, the function S must be valid for non- equilibrium state.

Irreversible thermodynamics of open composite system with interfaces and surfaces is formulated recently by the author by advocating the  micro -discrete element method. In this approach instead of using Gibbs formula,    Helmholtz Pfafian differential equation is developed, which is very useful for the  treatment  of isothermal irreversible changes may take  place in solids with or without exposed to external surface traction and/or  body forces.  

Dear professor Tarik, I read your papers. In chemical thermodynamics, it is clear that the function S is valid for non-equilibrium state. In my approach, your δS or δ(ΔS) come from  heat production diq, Gibbs free energy is converted into heat, I express this entropy change as diS=(Δμ/T)dN, it is a partial differential of the total entropy production, the driving force of the process is Δμ, there are three other driving forces of irreversible processes, Δ(1/T), ΔY, Δ(p/T).

For Renlong' question, the puzzles are the variables p and T, as I answered before, in the fundamental equation of thermodynamics

dU=TdS-pdV+Ydx+∑μdN.

The variables p and T now are considered as the macroscopic quantities of a system or a local, Y and μ don’t have to be.

Dear Dr. Suye, one should be very careful  by labeling chemical potentials of individual chemical species in multi- component systems as their  molar Gibbs free energies. This is only valid for single component systems. But one   has rather the following connection between Gibbs free energy and the chemical potentials  G=Sum( μi ni)  which is valid also for the irreversible thermodynamics. I should mentioned that Gibbs pfaffian DE as you have in your comment is only practical for those isetropic processes where  dS=0 (See: R. Haase).  On the other hand for isothermal-isobaric and isochoric processes one should employ  Gibbs and Helmholtz representations, respectively. 

The form of the first law that is always correct reads

dE/dt = Qdot -Wdot

where E is the total energy in the closed system, Qdot is the overall heat transfer rate accross the system boundary, and Wdot is the power done by all forces at the system boundary. In an irreversible system, the system is rather inhomogeneous, so it is best to consider E as an integral over the local states, and Qdot and Wdot as integrals over the system boundary.

Internal energy is part of the total energy. Let's ignore potential energy and only consider internal energy U and kinetic energy Ekin, so that

E = U + Ekin

Also U and Ekin are volume integrals over local states. In a system at rest (that is, the center of mass of the total system is resting), the kinetic energy can be non-zero, for instance if the system was stirred. Then

Ekin = Integral (rho/2 v^2 dV)  , where rho is mass density, and v is local velocity.

U = Integral (rho u dV)   ,  where u is local specific internal energy

The above form of the first law is always valid, for instance it can be derived from the Boltzmann equation. In particular it is valid for systems in non-equilibrium, undergoing irreversible changes of state.

Since the system is at rest, the corresponding equilibrium state implies vanishing of the local velocity, so that Ekin_eq = 0. If kinetic energy is non-zero, as in an originally stirred system, an irreversible process will occur, that reduces kinetic energy.

In a reversible process, however, kinetic energy can be ignored, and states can be considered as homogeneous. This implies also that only pressure forces occur, and the first law reduces to the familiar form

dU/dt = Qdot - p dV/dt

For irreversible processes this cannot be used, as should be clear from the above. Irreversible processes depend on the local states (see the integrals above), which must be computed from the partial differential equations of transport theory (Navier-Stokes-Fourer equations for simple fluids, Boltzmann equation for rarefied gases, and so on).

The second law can be considered similarly:

dS/dt = phi_dot + Sgen

where phi_dot is entropy flux and Sgen is entropy generation. S and Sgen are a volume integrals, e.g

S=Integral (rho s dV),

phi_dot is a surface integral. This form is always valid. In simple continua, the entropy flux is surface integral over qn/T, where qn is the local heat flux crossing the surface, and T is the local temperature.

Only if T is homogeneous over the entire non-adiabatic part of the surface do we find phi_dot = Qdot/T.

The entroy flux will deviate from qn/T for more extreme materials, such as rarefied gases.

A large class of materials can be described under the assumption of global non-equilibrium, but local equilibrium, in the sense that the Gibbs equation is valid locally, as

Tds = du - p/rho^2 drho

this is treated well in books on linear irreversible thermodynamics (look at the recent book by Kjelstrup and Bedeaux, and their co-workers).

Even when local equilibrium is violated, the first law and the second law in their above general form are valid, see books on kinetic gas theory. Then, it is problematic, however, to find easy macroscopic constitutive laws (qn/T is not the local entropy flux, the Gibbs equation is not valid, and so on).  

Dear professor Tarik, Thanks for your reminder, I have noticed the difference between the expressions in chemical thermodynamics and thermodynamics (physics). The latter use particle numbers Nj instead of molar numbers ni, it should correspond to μ ->μ’.

I discussed a similar question with some professors, the heat effect of chemical reaction or phase transition mean internal heat production, and this internal heat production is the source of the internal entropy production, but none of thermodynamic functions can describe this internal heat production if there is no heat exchange. In your paper JCP.124.144706.(2006), for eq.(1), d(ΔS_in)/ δt involves the entropy production of the internal heat production, since the heat effect relate to anther gradient Δ(p/T) in general, another part of the internal entropy production should be considered in the equation, otherwise, the equation can only be used for isothermal-isobaric processes.

Dear professor Henning,

δQ is always valid for heat exchange processes, but cannot be applied to internal process, the heat energy has the two sources, the heat flux deq and the heat production diq, if combined these two terms, we will get a state variable, the heat energy q, then flux deq can be considered as heat in transfer both δQ and internal heat flux, the heat flux becomes the flux in a state quantity.

If the system was stirred, it can be considered as the work done by generalized force Y, the work =Ydx. Yx express the mechanical potential energy within the system.

Dear Dr. Suye,  unfortunately you are misinterpreting  in our equation, which is very similar to those equations obtained by Prigogine  ( See:Eq. 3.72)  and de Groot, (See: Eq. 20) using completely different arguments, which do not contain any   divergence of the stress tensor in solids, whether its deviatoric part is missing or not. This is due to the fact for solids under the complete mechanical equilibrium  divergence of these quantities becomes zero. In the case of fluids, the  internal entropy source term involves inner double product of viscosity tensor and the velocity dyadics (See: R. Haase  Eq. 4-37.23). However,   we also observed that our theoretical formulation  inherently  produces rather simple results in the case of  the isothermal -isobaric systems. As you have suggested ad hoc fashion. At the end we found that only the specific Gibbs free energies enter into the computations. We also treated the isothermal-isochoric system, where the system doesn't interact mechanically with the surroundings through the surface tractions or body forces. Where specific Helmholtz free energy densities enter into the picture. See appendix of the second paper.

Dear professor Tarik, I mean, eq.(1) is a very useful equation, but there seems still to be a gradient in (p/T) that should be considered in the equation.

Thanks for your link.

Dear Dr. Suye, There is no any mathematical assumption in the derivation of Eq. (1). If one checks the books related to the internal entropy or the entropy source term by Prigogine, de Groot even the book by Rolf Haase, one can not see any term which involves gradient of pressure p.   Also one can easily show that for elastic solid under mechanical equilibrium the gradient of pressure is identically equal zero in absence of body forces.  More general,   Divergence of stress dyadic becomes zero in the absence of body forces. Also for open systems the Gibbs equation in volumetric density representation doesn't involve pdV/T  term explicitly (See Rolf Haase). 

Dear professor Tarik, the first and the third terms in your Eq.(1) produce heat effect (heat production), the second term involves the internal (heat) energy flux, the three terms may change the temperature, the pressure and the volume of the locals/subdomain and the system. The entropy production related to the change in the pressure, or in other words, in the volume of the locals/subdomain has not been considered. The entropy is the function of the variable V.

Does the first term have already included the third term in math, if the differential was written as “d”? And does it denote the flux in Gibbs free energy (but not the internal entropy flow) + Irreversible chemical processes (Gibbs free energy is converted into heat)?

Eq.(1) includes the internal entropy flux and internal entropy production, it is a bit complicated, since the internal entropy flux between the different locals/subdomains will offset each other, the equation can be simplified in math, then the question may become clearer.

There is no "internal heat change" (see Struchtrup). Heat is an exchange quantity depending on the environment of the system. The temperature change is caused by change of the entropy due to irreversible processes in the interior of the system and due to the external heat exchange which changes the entropy of the system undergoing an irreversible process. Interesting question is: What is the temperature of a discrete systemduring an irreversible process, if yuo do not rely on "local equilibrium" ?

If the entropy is a state variable or not, depends on the choice of the state space. Of course, there is a "dS", that means S^dot along the process, and the differential is not a total one. The entropy becomes a functional depending on the history of the process. See: Handbook of Physics Vol. III/3 C.Truesdell et al. This fact is well known: Rational Thermodynamics, but uncomfortable to handle.

I know that Q denote the heat in transfer, but  please consider following questions.

1) Where had it come from?

2) What is the heat effect?

3) Should the heat of chemical reaction be considered as internal heat change, or an exchange quantity depending on the environment of the system?

4) If we consider a system and “the environment ” as a new system, then what is Q?

5) Which function denotes the energy of the thermal motion within a system? 

I think that If the entropy is a state variable or not, depends on the rule of exact differential.

Dear Dr. Suye; if you wish to get proper answer to your questions please you should go over the book by I. N. Sneddon on 'Elements of partial differential Equations'. Where he treats  Pfaffian DE and the application of Caratheodory's theorem to those  systems, which require  more than two thermodynamic variables for their complete specification. This axiomatic theory is the foundation of Second law thermodynamics as well as the reversibility. I read this book about 54  years ago when I was a first year graduate student at Stanford just for my curiosity on mathematical analysis  but nothing else.  As far as Thermodynamics of Irreversible Processes are concerned I recommend Rolf Haase book. Guggenheim's  'Thermodynamics'  is one of the best text on the chemical thermodynamics, which also follows the axiomatic approach of Max Born. He gives clear cut definition of characteristic function and the Legendre transformation of thermodynamic variables. Best regards.

Dear professor Tarik, Thanks, I have studied most of these theories many times. There are many small points that can hardly be explained by these theories or textbook, for instance, the internal energy U is defined as the sum of the energy within the system, why the equation

U=TS-pV+Yx+∑μn,

contains the difference of functions? Why the pressure p has different sign from other generalized force Y? Everyone knows the first law of thermodynamics but I think no one can answer this question, many had got used to that one does not understand why.

“Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, it doesn't bother you any more”. (Arnold Sommerfeld)

I think Arnold’ “one or two small points” now are still not clear; it is so intriguing: Did we miss out anything in thermodynamics?

(μ_in_I makes no contribution to entropy, and so do its flux, the second term contains an offset term, the first term does not involve the internal entropy flow, the difference Δμdn denotes heat conversion process.)

Best regards.

G=U-TS+PV   is the definition of Gibbs free energy.  If one wishes  T , P  and  nii as independent thermodynamic variables then G should be defined and employ.. By differentiating  this function and and combining with   dU= TdS -PdV + ∑μdn  for the open system  You will get what you wish, namely: dG= -SdT +VdP +∑μdn.  This procedure known as Legendre Transformation in Math. This way you may define Helmholtz free energy F=U-TS  to get  dF= -SdT -PdV +∑μdn.  See: Guggenheim.

One can also show by integrating  Gibbs DE keeping the  T and P constant  G=∑μi ni, , which can be also obtained by Euler's theorem by knowing the fact that  U is homogeneous and of first degree in the extensive variables. 

In the case of generalized forces such as electrostatic or magneto-static forces, one uses the same procedure.

There is noting funny about thermodynamics if you have sound mathematical background and remember that this discipline  relies on the few postulates like algebra or group theory, then the rest is depending upon your imagination.

Dear professor Tarik, in chemically thermodynamics, we have

U=TS-pV+∑μn,

G=U-TS+PV =∑μn.

The real definition of Gibbs free energy should be G=∑μn, but not G=U-TS+PV, the latter is only an equation, and for dG,

dG=-SdT+VdP+∑μdn=∑ndμ+∑μdn,

∑μdn denotes the flux in G, ∑ndμ=-SdT+VdP denotes that the change in G come from the energy conversion, it does not means that G=G(T,p,n) or G=G(S,V,n) but only means that ∑ndμ can be calculated by -SdT+VdP. (in my approach, ∑ndμ=-diq, where diq is the internal heat production)

For Helmholtz free energy F, we have

F=U-TS=-pV+∑μn

The real definition of F should be F=-pV+∑μn, where -pV is a thermodynamic potential, denotes the free energy that can be obtained from heat conversion, it is not the real free energy within the system, and is only a conversion potential. The physical meaning of Helmholtz free energy is that the free energy within a system is equal to the real free energy of the system + the free energy that can be obtained from heat conversion.

By the equation

S=U/T-∑μn/T+pV/T.

Since ∑μn is one part of the internal energy U, it means that ∑μn makes no contribution to energy; ∑μn is the real free energy of the system, and the entropy of this part of the internal energy is equal to zero.

In most cases, Legendre Transformation means that in energy conversion processes, the increment in one energy function can be calculated by the decrement in another energy function due the internal energy U is a conserved quantity, we should consider the physical meaning behind these relational expressions of Legendre Transformation. In current thermodynamics, there is a state function that was lost, and replaced by an implicit function; so it is difficult to understand clearly.

Dear Dr. Suye;  There is one very useful equation which is extensively used in metallurgy and chemistry it is called;' Gibbs-Duhem equation'  That is;

SdT-VdP+∑ndμ=0

As you obtained and misinterpreted in your letter.  It is particularly useful in its application to changes at constant temperature and constant pressure (in general stress tensor) when it may be written as   ∑xi Dμi =0 . Where  D operator denotes variation in composition at constant  T and P.   xi  is the mole fraction. Legendre transformation is a mathematical device to go from one set of state variables to another  by reproducing new functions, which are called characteristic functions.  You are using a lots of physical and engineering terminologies and concepts,which have no use and   place in chemical or mathematical thermodynamics.  All these make more confusion then any thing else!

There have been many replies to the original question so I would only add comments that have not been really addressed. As several have noted, the relation dU=dQ-dW is always valid, even for an open system. One has to carefully define dW. While dU and dW are mechanical concepts, dQ is a thermodynamic concept. In the notation now becoming common, dQ is really d_eQ, the heat exchanged with the surrounding medium and dW is really d_eW, the work exchanged with the medium. Again, for any process, d_eQ=T_0d_eS in terms of the temperature T_0 of the medium and the exchange entropy d_eS with the medium. One cannot replace d_eS with dS as the latter also includes the irreversible entropy generated in the system of interest.

One can make S a state function by properly including what are called internal variables to enhance the state space. This happens for a special class of nonequilibrium states, which I have called internal equilibrium states. In this case, dS does not depend on the path.  In general, S for a general nonequilibrium state will also be an explicit function of time and so it will not be a state function so dS will depend on the path just as d_eQ and d_eW do.

If you want to learn more about this approach, contact me and I will direct you to some of my papers..

Dr. Professor Gujrati, your first paragraph is the standard presentation  of the first law of thermodynamics. Better to say restatement of conservation of energy in general. d should be replace by Greek delta symbol for  Q  and  W, which means they are inexactly differential operator. The definition of work term is very important. Landau calls this reversible work. Similarly Callen rather prefers to call as 'quasi-static work', that means the application of surface traction and body forces on the body should be vanishingly  slow rate  that in every instant  of the process there should be complete mechanical equilibrium or quenching equilibrium.

del Q  is not the heat exchange but the heat received by the system through its boundaries (external or internal). The exchange is  not the proper terminology here. actually one should rewrite  the first law as:  delQ= dU-delW.   del W  also should include body forces (See: Sokolnikoff's book on Math Theory of Elasticity). Clapeyron's theorem tells us that the strain energy of deformation done by external forces (taking the system from one equilibrium state to another) is one halve the total worked done. Therefore other halve goes to dissipation of energy. I used this fact in the developed  of the irreversible thermodynamics of deformed  bodies with inner and outer surface.

Prigogine uses the terminology such as 'non-equilibrium stationary states' while dealing with chemical reaction. That means at least one of the  observable property of the system after the initial transient state reaches to a constant asymptotic value.

I  am little bid puzzled about your definition of  particular states where the entropy dS doesn't depend on the path as internally equilibrium state. That means  dSi=0,   dSex is path independent.  In reality  dSi /dt should be positive definite for any conceivable process. Actually, in the domain  validity of the classical  irreversible thermodynamics,  which means all the changes taking place in the vicinity of the equilibrium path all the characteristic functions besides entropy assumed to be path independent.  Real dynamical system only the expectation values of the dynamical variables and their changes have mathematical meanings. 

In our computer simulation studies we can tract down the global Helmholtz free in internally deformed system which is mechanically isolated.  

Dear Professor Ogurtani: Thanks for your response. I do not think that we are saying anything different. What you call the heat received from the medium is what I call the heat exchanged with the medium. What I am trying to say is that it does not include heat that is generated within the system. Similarly, I said that dW requires careful definition for an open system. If you want to include the traction forces, you can do it; all you have to do it to define the system's boundary carefully.

I did not say that d_iS=0 for a system in internal equilibrium. The presence of internal variables generate irreversible entropy d_iS, which is path dependent just as d_eS is. But their sum dS for a system in internal equilibrium is path independent as the entropy is a state function in the larger state variable space. If the system is not in internal equilibrium, then S is not a state function as it will explicitly depend on time t.

How do you define a system is in internal equilibrium?  Up to now no body made any serious attempt to handle those cases where entropy function is  an explicit function of time.  But we assume; the  entropy is a function of state variables, which are time dependent.  Actually, we make on more simplification step by assuming that the  entropy is a positive define quadratic form in terms of state variables.  This means all the kinematics paths wonders around  the  stable state configurations(absolute stable or meta-stable).  Gibbs (Gibbs ;page. 56) gives two  different version of  internal equilibrium and stability of isolated system in terms of entropy and energy. My claim relies on the first version.

Dear Professors,

Thanks for your replies which have really thrown light on the question and aroused active discussion over the issue. Maybe it's more interesting to trace back and go through the original papers for a clear development of thermodynamics from its starting point.

Many Regards!

Dear Dr. Renlong Ye

I have been always trying to go through the early works **mostly before 1960* and to see how those grandfather were treating the subjects. Not only for the natural sciences and mathematics but also for the social sciences and humanities, I have spent years in reading and reflecting on the every aspect of theoretical physics excluding the high energy nuclear physics and cosmology. All these efforts were just for curiosity but nothing else, For example I have presented two distinct mathematical formulations of irreversible thermodynamics of surfaces and interfaces with triple junction singularities by firstly advocating discrete micro element technique by utilizing the few thermodynamic concepts and postulates augmented by generalized conservation laws, The second approach relies on the functional formulation of irreversibly using the string model, This model is highly sophisticated in mathematical context and results completely consistent answers compared to the first method, However the physical interpretation of the system parameters requires definitely feed backs from the first approach, Best Regards,

Dear Professor Ogurtani:Think of an isolated system with fixed E,V and N to be denoted by X. Its out of equilibrium entropy cannot be a state function S(X); the latter is a constant. It must be written as S(X,t) to account for the law of increase of entropy. The explicit time dependence may come from some additional time-dependent variables that cannot be controlled by an observer. It may refer to a variable describing the inhomogeneity of the system. These are normally called internal variables. If we have the right number of internal variables, to be denoted by y(t) in that the entropy can be expressed as a state function of X and y(t) only with no additional explicit time dependence, then the state is called an internal equilibrium state. The number is not right, we will still have an explicit time dependence. The system is not in internal equilibrium. Even in the latter case, progress can be made. For example, one can show that the nonequilibrium Gibbs free energy decreases in time. Therefore, it is not correct to state that one cannot study the latter states in thermodynamics.

Hope this helps.

Dear Professor Tarik, 

Thermodynamics of materials and physical chemistry are my Undergraduate Courses, so I had known Gibbs-Duhem equation  when I was a student.

Thermodynamics now is Grey box theory. I think it is too complex, for example, “there are as many different forms of free energy in a thermodynamic system as there are combinations of constraints,”[1] Such as H, G, F, Ω… but only a few of these thermodynamic potentials are independent. No one can make sure that  how many independent thermodynamic potential there are in thermodynamics, no one! It does not seem like a perfect foundation in the usual sense.

δQ is the heat flux δQ=deq, and there is another source of heat diq, comes from heat conversion, I called as heat production, then we can get the total differential

dq= deq+diq.

For chemical thermodynamics, the first law becomes

dU=dq+dG= deq+diq+deG+diG

diG=-diq

If you want back to the form of current theory, instead of dq by

dq=TdS-pdV.

I understand people are so used to a familiar way. Best Regards!

[1] L.E.Reichl A Modern Course in Statistical Physics, 2nd ed.

The opinion, “dQ is really d_eQ”, means that Q is a physical quantity in transfer, “dQ” implies the total differential of Q. dQ=d_eQ denotes that Q can only be a flux.

The opinion, "dW is really d_eW", needs to be considered carefully, in addition to above, δW involves inter-conversion between heat and work (pV work), it not only a flux.

GUGGENHEIM :  In a finite process which is not adiabatic the system is said to absorb a quantity of heat  Q or alternatively to give off a quantity of heat  Q  where    Q= Del U - w.    This definition doesn't tell us  Q is  originated externally  or internally. Otherwise one would exclude the heat generated by chemical reaction and phase transformations.  Also the electrostatic and magnetic interactions would be excluded. This law is the simple restatement of the energy conservation principle.

Similarly w  work  may be done externally through the system boundaries and/or internally through the body forces originated by  graviational or electrostatic and magneto static long range force fields. 

Dear professor Gujrati,

I now can compare and contrast my approach with that of your approach [PRE. 81, 51130 (2010)] for the definition dS and the variables q/Q, One wrote that you “introduced an identity dS = dQ/T and T represents the temperature of the system and not the heat bath”. and “dQ is not merely the heat exchange d_eQ with the bath”.

In my approach, δQ/T is the entropy of δQ, and T is the temperature of δQ, neither that of the system nor that of the heat bath. And you can see in my approach, dS=dq/T+pdV/T, differs form your dS = dQ/T, and by your eq.(15), your Q is still a process variable, my q is a state function, dq=deq+diq, deq can also be applied to describe the internal heat flux.

I introduced the function q in my  book published in 2008, it was written as dU_T=dU-∑Ydx at that time (where Ydx contains μdn but does not contains –pdV), denotes the energy within the system that make contribution to temperature T. Eventually it becomes dU_T->dq.

W has already been defined as “work”, and pV work involves inter-conversion between heat and the free energy, it is not a flux. Except for pV work, W is the free energy flux, and it should be considered as the flux in the state quantity. In my approach (http://arxiv.org/pdf/1201.4284v5.pdf , section 1), using A denotes Yx that A=Yx, it follows that

dU=dq+dA+dG= deq+deA+deG+diq+diA+diG

The increment in U is equal to the sum of the energy fluxes + the productions, for internal energy productions

diq+diA+diG=0.

Best Regards.

Dear Dr. Suye, I sorry to say that you guys are creating your own thermodynamics, which takes many years to find acceptance by the international community. First of all you should define what types of thermodynamics systems (homogeneous, heterogeneous (discontinuous) or continuous) you have in your mind. Of course work term w may be considered composed of three terms; w= wa+wi+w*. wa signifies the acceleration work done on the entire system and work against external forces fields, wi the reversible deformation work as well as reversible electrification and magnetization work done on the entire system, and w* the rest work done(friction work, electrical work with external current sources, etc.). This last term can also contain reversible portions (i.e. reversible electrical work in galvanic cells).

Sorry! The flux is vector quantity; on the other hand u energy of any type is a scalar quantity. One can not add vector and scalar quantities together to form an equation. Therefore it is nonsense to talk about the work as a flux. You are trying to develop a theory which is internally inconsistent, and mathematically improper.

Here we may use the following fundamental hypothesis: The energy neither can be created nor it can be destroyed , it can only change its form.

Dear professor Tarik, I understand people are so used to a familiar way.

1) About flux and production, there are some recognized expressions such as

dS=deS+diS

where deS is the entropy flux, and diS is the entropy production, my flux and production are similar to.

2) I agree with that pV work is not a flux.

3) I agree with that “The energy neither can be created nor it can be destroyed, it can only change its form”. But it only means that the total energy of the system U cannot be created or be destroyed, due to energy is a conserved quantity.

It does not mean that such as mechanics potential energy, Gibbs free energy cannot be created or be destroyed. Both the two are non-conserved quantities, so we have diq, diA, diG.

The words “for internal energy productions” follows that diq+diA+diG=0, it means that no diU.

Best Regards.

Dimension of the entropy flux should be    energy per degree per area and per time/  Is this the dimension of ds/

Dear professor Tarik,

dS=deS+diS is a scalar equation, i.e, here we only consider quantitative relations.

Dear Professor Suye, I am not going to say anything because we are playing from completely different tunes when it comes to thermodynamics. What the way you introduce thermodynamics ad hoc fashion one may not go too far in treating complicated problems such as the deformable solids with surfaces and interfaces exposed to external elastostatic and electrostatic force fields. These are most important problems scientifically as well as technologically. Best Regards

Dear Dr. Suye: This is in response to your comment addressed to me two days ago. In my approach, dQ=deQ+diQ, where deQ is the exchange heat with the medium and diQ is the heat generated within the system. It then follows that dS=deS+diS=dQ/T. I still have dU=dQ-dW with dW=deW+diW; here deW=P0dV in terms of the medium 's pressure P0 and diW is the work done within the system. It can be shown that diW=diQ always and that dW=PdV in terms of the pressure of the system.

I hope this will clarify my point of view.

Dear professor Gujrati,

I have not found the expression diQ and the equation dQ=deQ+diQ in your paper (maybe you know why I write so, I’m not sure), and, 1) the equation dS=deS+diS=dQ/T is incorrect (I had made a similar mistake many years ago), you may check it by the process to convert heat into work. 2) dW=deW+diW is inexact, deW=P0dV corresponds to my diq, it is not a flux, 3) δW=pdV only can be applied to pV work in a reversible process, ie., dW=deW+diW=pdV is incorrect, you cannot distinguish the flux and production for your W (or my free energy) in that you have not considered the work done by other generalized force, only Yx can be distinguished into these two parts, pV cannot, and it should be considered as the flux and production in the quantity of state but not the quantity of process.

The original idea I introduce the function, the internal heat energy q is to eliminate “the difference of functions" in Euler equation, then it will become easier to prove that the entropy is a state function in math (it was 9 years ago). I noticed “it denotes the energy within the system that makes contribution to temperature T”, it follows that the question, the relation with δQ must be considered, and I noticed “the pressure p has different sign from other generalized force”, it prompted me to consider that the physical meaning of pV work differs from the work done by other generalized force, it is the conversion between heat and work, pV work means diq, and, to express the total differential equation for diS need to eliminate the flux δQ/T from dS, therefore that I get diq=dq-δQ=dq-deq, similar distinction was later extended to A and G.

This is my train of thought.

Dear Dr. Suye, for an infinitesimal change dS=δQ/T is the definition of entropy variation in any given text book in thermodynamics in the Western World. Similarly, δ Q= dU-δW where δQ is the heat received by the system (internal and/or external), and δW is the reversible work done on the system. dU is the increase in the thermal energy of the system (See the definition of U in Guggenheim or Sneddon). 1/T is the integration factor of the Pfaffian differential equation denoted by δQ= dU-δW which means dS=δQ/T is an exact differential. Where S is called entropy of the thermodynamic system.(See: Ian Sneddon's famous book on Partial Differential Equations, McGraw-Hill, New York 1957, page 40).

Dear Professor Suye: The issues are elaborated in the following two papers.

arXiv:1105.5549

arXiv:1206.0702

It appears that we are using various terms differently, which may be the cause of the disagreement. I believe that dQ=deQ+diQ and dW=deW+diW are well defined concepts. In the paper you had cited, diQ=0 as we were only considering heat transfer in Eq. (15); Eq. (12) is the general result. In Eq. (18), the temperature of the system is that of the medium so that the second term in the equation just above it represents diQ=diW. I did not discuss these points in that paper. But if you, after looking at the two papers, still dispute them, we can discuss it further. Please contact me at [email protected] directly.

With best regards,

Puru

Dear professor Gujrati,

The first time I read your paper due to a Referee wrote some words in his Report (for my manuscript arXiv:1201.4284), I received the Report in 2.Sep.2011

“Gujrati in a paper in PRE last year has also introduced an identity dS = dQ=T [sic] but his T represents the temperature of the system and not the heat bath and his dQ is not merely the heat exchange d_eQ with the bath. The author must compare and contrast his approach with that of Gujrati.”

…

The paper that Referee suggested (only one) is PRE. 81,051130.2010. I found that it is difficult to compare and contrast these two, so I have not continued to pay attention to your research.

I have read the paper your mentioned above (arXiv:1105.5549), I agree with that “we are using various terms differently”, but more important is in different ways, and with different conclusions, I would like to point out the differences between the two approaches.

1) (my) dq=deq+diq≠(your) dQ=deQ+diQ, (my) dq=TdS-pdV, (your) dQ=TdS.

2) (my) diq=-δW_v≠(your) diQ=diW, including for all of my diq.

3) (my) dA=deA+diA≠(your) dW=deW+diW=PdV, in general cases.

4) My approach does not change any mathematical results of current thermodynamics, your approach need to be verified, it is easy to find some examples, ie., your approach may change the mathematical results of current thermodynamics.

I don’t mind to explain these differences in detail, private or public.

I’m not a professor.

Best Regards,

Dear professor Gujrati,

I have carefully read your article arXiv:1105.5549, I’m sure that our approaches are different, a simple example please see bellow.

Consider an idea gas, the initial state is that the two locals was in local equilibrium, the final state is equilibrium state. Your diQ denotes “the irreversible heat”, diQ=ΔPdV≥0, where ΔP is the pressure difference of the two locals. My q denotes the energy of thermal motion within the system, so dq=dU=0, and diq=0. In this case, diQ=ΔPdV≥0 makes no sense, due to “diQ” does not represent the increment in the energy of thermal motion within the system, and there is no “irreversible heat” production. Thus, your diQ cannot be merged with deQ to define dQ, the physical meaning of “the irreversible heat” diQ is unclear. My diq≠diQ.

Heat can be converted into work if attached with some compensation, it implies that the pressure difference Δp can be increased in coupling processes, and in this case, your diQ=ΔPdV can be less than zero, ie., d(ΔP)≥0, dV≤0 and diQ=ΔPdV≤0, your diS= diQ/T is not the total differential of the entropy production, so dQ=deQ+diQ=TdS is inconsistent with the second law. This issue cannot be explained by the extended equation “diQ = diW≡dW0 represents the last three terms ….”

Best Regards.

Dear Dr Suye; Let us assume we have two sub-system separated by a mobile impermeable membrane. The whole system is enclosed by a rigid insulating walls that means dU=du1+dU2=0. Let them they have in mutual thermal equilibrium initially (T1=T2=T). Let the in-conductive membrane move dV1=- dV2. Write down the changes in thermal energy for this infinitesimal process:

dU1=T1dS1- P1dV1, and dU2=T2dS2- P2dV2 and sum them up:

dU=T(dS1+dS2) -( P1-P2)dV1=0 rearranging

T(dS1+dS2) =( P1-P2)dV1 where dS= (dS1+dS2) Hence one has:

TdS=( P1-P2)dV1 where dS is nothing but internal entropy of the whole system since the process is adiabatic, dSi, which is positive for the natural process and zero for the equilibrium process, never becomes negative, according to Max Born.

TdS=TdSi=( P1-P2)dV1 ≥0

What this second law tells us that membrane moves from the high pressure site to the low pressure site. QED

Dear Colleagues,

I think that the discussion is on textbook level. All these topics are well known also

in non-equilibrium thermodynamics. Perhaps, these two papers may help:

J. Non-Equilib. Thermodyn. 29 (2004) 237-255

J. Non-Equilib. Thermodyn. 34 (2009) 75-92

Dear Dr. Muschick you are hundred per cent right. I am sometimes embarrassed to write it. But I think it is useful for those who are not familiar with the basic concepts of chemical thermodynamics even though they seem to us trivial or self evident. For me thermodynamics like algebra or quantum mechanics, which relies on the few (plausible) hypothesis and the rest is the mathematics (logic) manipulations. I would appreciate it very much if you could send me your above mentioned two articles. I would like to see your approach to the subject. Best Regards

I think your are definitely aware of  Caratheodory' axion.  Which is nothing but  the rephrasing  of the  mathematical theorem that puts the necessary and sufficient conditions for the integrability of Pfaffian d,referential  forms invlolving more than two variables.

 By the way Caratheodory born in İstanbul, and like me he studied in Naval Academy in Belgium.  He comes from  a well known Byzantine family, who took extremely high bureaucratic positions in the foreign affairs  during Ottoman time.

Dear professor Tarik, I agree with your opinion, I mean, (P1-P2)dV cannot be explained as “the irreversible heat” .

Dear professor Muschik.

It only looks like “on textbook level”, in fact, the question involves the structure of thermodynamics. For example, “there are as many different forms of free energies in a thermodynamic system as there are combinations of constraints”. But no one can make sure that how many forms of the free energies are independent, I bet you cannot find answer in any textbook. This is a question only on the first law of thermodynamics, so we cannot say that “All these topics are well known”.

Dear Tang, what you are writing related to thermodynamics, they are all perfectly alright , and I can easily transform them  into the formats, which were  how introduced into the chemical thermodynamics by our grand fathers in Western World  during the last hundred fifty years or more. Such  as:  Joule and Thomson (Lord Kelvin) Gibbs, Rankine, Maxwell, Planck, Max Born etc.   You are trying to visualize them by creating their mental pictures with the help of our   five senses .   But I feel that I could only menage what is hot and what is cold; what is heavy and what is light, what is hard ans what is soft   etc. Which are all I can perceive not more not less with my healthy five senses.  Here all those characteristic functions are defined for the mathematical conveniences to deal with the  thermodynamics systems under the specific constrains and symmetries if any.

But they should be  defined operationally  as pointed out by Percy W.  Bridgmann, who believed  in  by  taking  a coherentist approach to science in which a theory is validated if it makes sense of observations as part of a coherent whole. Best Regards 

Dear professor Tarik,

I have no doubt for the quantitative analysis of current thermodynamics, yes, it is correct, but I don’t think they are all perfectly alright, there are still something that have not been clarified. I noticed, for example, for “what is the heat energy”, J.Fourier, J.C.Maxwell, and R.Feynman have disagreements. I think they are right.

Best thing we all included above mentioned  ingenious comments,   should read Emanuel Kant. That wouldn't  be refreshing?  Best Regards.

Dear Renlong Ye, you are mixing up two concepts - irreversibility and non equilibrium.

Thermodynamics deals with systems in equilibrium states only - in other words, the initial and final states of the system must be equilibrium states with well defined values for any of the properties of the system.

As you very well did realize, no definite value can be assigned to any property of a system in a sate of non equilibrium. As such, no definite value can be assigned to the change in the value of a property when either the initial or final or both initial and final states are nonequilibrium states. Therefore, you don't have any values for dU, dS etc when either or both states of a system before and after a process are nonequilibrium states. It is important to note that in thermodynamics, no process whether reversible or irreversible, connects two states both of which are not equilibrium states. Stated differently, thermodynamics concerns only with processes that connect two states that are equilibrium states. So much for nonequilibrium states.

Coming to irreversibility, an irreversible process connects two EQUILIBRIUM states of a system. If we have two equilibrium states A, B of a system, A and B may be connected by a reversible or irrversible path (or process). The change in the value of the properties of the system such as ΔU or ΔS have unique values whether the path connecting them is traversed reversibly or irreversibly. Note, we don't ever use word nonequilibrium here.

Coming now to your concern about the interpretation of dU and dS, dU cannot by itself indicate whether a given process is reversible or irreversible (spontaneous or not). On the other hand, value of ΔS (or dS) for a process indicates whether a process is reversible or not. A positive value for ΔS asserts the process to be spontaneous, a negative value asserts a process to be non spontaneous and a zero value asserts the system is at equilibrium and has no tendency to change. The ΔS involved in these statements corresponds to the change of entropy of the universe.

Dear Zhoubin Maneshi, If you are interested in the rigorous mathematical  treatment of the subject of irreversibility, I refer again section 7-8  entitled 'as ' Caratheory's Theorem (1909)  and Application to Thermodynamics''  in  E. Partial Differential Equations by Ian N. Sneddon, where Author puts envises on the Caratheory's Theorem  for the Integrability of Pfaffian Differential Equations and the existence of two functions in terms of state variables in multi dimensional space, which are called ENTROPY s and Inverse TEMPERATURE.  The basis relies on ''Whether or not the inaccessibility of points in the neighborhood of a given point along any path for which Pfaffian differential form is satisfied''

For me the weakness of this theorem in practice is related to its  basic assumption that the system should follow adiabatic path!!  How about  for the non-adiabatic processes?   Most of the irreversible processes are dissipative processes with and without the work term. 

To sum up: There is no exact differential for thermodynamic variables (such as entropy and internal energy) in non-equilibrium state, but we can always talk about very small changes and expected values... 

Entropy: A concept that is not a physical quantity

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