Tarik makes an excellent point. One should be able to derive from memory all classical thermodynamics. You only need to memorize one equation,  dU = T dS - P dV + Σ μ dN, which combines the first and second law into a single equation. All else is inevitable consequence of straightforward calculus. 

In fact, the conflict arises due to convention difference that geophysicists and statistical physicists use. In the equation you mention, p is the pressure that is applied to the system. If you apply some pressure, the volume difference will be negative, right? (I mean you compress a gas of volume V) so, the minus term  will correct the contribution of that term. Hence, you get a positive add - on energy to the system. 

I agree also with Omer, since PdV is the mechanical work applied to the system.

Dear Omer and Ahmed,

Thanks for your answer.

I partly agree with you for the differential equation, but your answer is unable to explain the integral equation.

Where the pressure p is a state variable of the system, pdV is the state change, for example, pV=nRT.

please see http://en.wikipedia.org/wiki/State_function

Dear Aleksei,

You mentioned grand potential

Ω=U-TS-muN

Ω includes the mechanical potential energy Yx, where Y is the generalized force, dx is the generalized displacement, if consider the mechanical potential energy Yx, the integral equation becomes

U=TS-pV+Yx+ΣμN

Ω=U-TS-ΣμN=-pV+Yx

-pV is a negative definite function. Why the internal energy U or the grand potential Ω contains a negative definite term? For an ideal gas, the internal energy U is equal to the kinetic energy of the particles, then what –pV is?

As a positive definite function, the internal energy U represents the sum of the different types of the energy within the system, so should be expressed as the sum of the functions. For a differential equation, -pdV is understandable from the perspective of energy exchange, but for the integral equation, -pV is illogical.

I agree with your opinion “because terms in the "integral equation" are not independent.” Thus the question becomes

Why the terms in the "integral equation" are not independent” ?

I mean, if the terms in the "integral equation" can be independent, then the negative definite term will be eliminated. In the integral equation, Yx and ΣμN can be confirmed as the two independent terms, and TS is a coupling term.

Dear Tang,

We consider U as a function its natural  (cannonical) variables, i.e. S, V and Ni. For an ideal gas U does not depend on V so if the volume is changed under constant entropy, U will not change unless temperature changes . This may be regarded as a definition of the ideal gas; U is a function of S and Ni but not of V.

For a non-ideal gas, e.g. an elastic solid a change in volume under constant entropy (adiabatic change) U certainly depends on volume, at least if the solid is compressible. Consider compaction of a sample, i.e. dV < 0 and - pdV > 0, i.e. internal energy increases. P depends on the elastic properties of the body. We can then release the pressure and the sample would expand and the internal energy is then lowered.

It should also be realized that U has no natural zero point.

So I really do not see your difficulty with havin the term - pV.

Best regards

John

It is very interesting what wrote Aleksei Bukov. It is important to remember that the total energy in a given condition is the sum of internal energy, kinetic energy, potential energy and other forms of energy.

I like to emphasize that since energy does not have any natural zero point it has no physical meaning to say that the energy is positive or negative. For example, in solid state physics at zero kelvin it is common to define the zero energy as when the atoms are infinitely separated and when they move closer and there is a tendency for binding the energy will be negative.

Essentially John Ågren  answered the question, but let me elaborate a bit. There are a couple misconceptions in the original question that cause confusion:

1. While U may be expressed in terms of physical contributions (kinetic, potential, rotational, vibrational, etc.), the terms TS, -PV, and ΣμN do not represent such physical contributions.

2. There is nothing wrong with negative energy because energy is always measured relative to some reference. Very simple example: the intermolecular potential (say, Lennard-Jones) is actually negative at some distances. This does not mean we cannot add it to the kinetic energy to calculate the total energy of the system.

And if we think about it, PV is not the only negative term in U=TS-pV+ΣμN. The chemical potential can be made to be either positive or negative, depending on the reference state. Which is another way of saying that the signs do not signify physics. 

Dear John,

The actual meaning of the question, as Aleksei mentioned, is that why the “terms in the "integral equation" are not independent” ?

Yes, for an ideal gas U does not depend on V, TdS and the term −pdV can be combined into an independent term. i.e., TdS- pdV= CdT denotes the energy of thermal motion within the system.

For the pressure p, our understandings are different: As a state variable, the pressure p is not just a force acting per unit area. An interesting difference shows that “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” [1]. This difference indicates a criterion that can be used to distinguish the source of the pressure p from other generalized force Y, including the case that Y is expressed as a force acting per unit area P, and can be used to distinguish the work done by the pressure p from other generalized force.

If a force acting an elastic solid, it represents that the generalized force Y is expressed as a force acting per unit area P. P in this case is not the state variable of the system, only in reversible process, P is equal to the elastic force of the solid, the latter is also considered as the generalized force Y.

Best regards

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

York: John Wiley & Sons Inc, 1997) p23,

Dear Aleksei,

1). x=2x-x implies that 2x-x can be combined into an independent term x, then -x can be eliminated.

3). T, V, N, are the independent state variables, i think that “independent state variables” mean that can be distinguished.

Best regards

Dear Themis,

-pV is a negative definite function,  here "negative definite" does not depend on some reference.

Best regards

I do not understand your problem. Thermodynamically p

is a potential and, as you know, one rather looks at -p is the potential conjugated to V. The potential -p  then behaves as all other potentials, ie it increases when the conjugated extensive variable V increases if the system is stable.

It seems as my latest comment did not enter. Here it comes again. As pressure can be both positive or negative -PV can have any sign.

Best regards

John

Dear John,

The pressure p and P involves the meaning of two,

1) A force acting per unit area P,

2) Thermodynamic state variable p.

For the first case, the force can be any types of the generalized forces; but it might not be a state variable.

In a differential equation of the first law, the meaning of two is hard to distinguish, in that –pdV or –PdV represents energy exchange, the two both can represent the amount of energy exchange. For an ideal gas, apply a force P to the system, –PdV denotes the amount of energy exchange, –pdV denotes the state change, in a reversible process, the two are equal. But for an integral equation of the first law, pV denotes the state property, so p must be a state variable.

Any types of the generalized forces can be expressed as a force acting per unit area P, but, as my above answer, there is an interesting difference between P and p.

“As pressure can be both positive or negative” only be suitable for P, for example, for an ideal gas, pV=nRT, p cannot be negative.

Best regards

Tang

Hi again!

The external pressure P on a body is a thermodynamic state variable and can be positive or negative dependeing on how loading is applied on the body. There can be internal pressures as well, which are also state variables but internal state variables.  It is true that an ideal gas cannot have negative pressure but all condensed phases certainly can and I guess also non-ideal gas can have negative pressures.

Dear John,

If there is a negative pressure p, it will mean that the statement by L.E. Reichl is incorrect, I have verified this statement, it is correct.

“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."

If there is a negative pressure p, the volume will decreases, here, a negative pressure is consistent with other generalized force.

Best regards

Tang

If I understand you correctly, the answer is that pdV is work done by the system so it decreases its internal energy.

Let F be a thermodynamic potential and X its conjugated extensive properties, then F dF/dX > 0 in a stable system. In a stable system thus -P increases as the volume increases. For an ideal gas  you can see directly, from PV=nRT, that -P increases when volume increases, i.e. d(-P)/dV=nRT/V2 > 0.

Thus -P is the correct thermodynamic potential =dU/dV like T = dU/dS  etc

Ramzi is right PdV is the work done by the system so it decreases its internal energy.

Best regards

John

Tang Suye

In other word, pV is the work required for placing molecules in the energy space pV . So it has a negative contribution to the total energy. 

Dear Ramzi,

Yes, you are right, pdV is the work done by the system, and –pdV denotes the amount of energy exchange.

For a differential equation, -pdV is understandable from the perspective of energy exchange, but for the integral equation, -pV is illogical.

Best regards,

Dear John,

I agree that p and P are the two correct thermodynamic potentials, where p=T(∂p/ ∂T)-(∂U/ ∂V), in general, p≠-(∂U/ ∂V), for an ideal gas, dU/dV=0, p>0, for a photon gas, p=(1/3)αT4, (∂U/∂V)=αT4.

The meaning of the statement by L.E.Reichl is (∂V/∂p)>0, but for the interaction between the particles (∂x/∂Y)

Dear Mohammad,

For a differential equation, -pdV is understandable from the perspective of energy exchange, but for the integral equation, -pV is illogical, as I written in the question, for a photon gas, the integral of –pdV or Helmholtz free energy F=-pV, what is this –pV?

Where p and V are the two positive definite variables, so -pV cannot be explained as a part of the internal energy.

Best regards,

Dear Aleksei,

As I written to John, the interaction between the particles in the equation

dU=TdS-pdV+Ydx+∑μdN

corresponds to Ydx. indeed, you can express Y as a force acting per unit area P, and in this case, P can be negative if the force is an attraction, but for the interaction between the particles,  the integral equation is positive definite, which denotes the mechanical potential energy within the system.

In the above equation, the interaction between the particles is expressed by Ydx, so for the pressure p and P, our understandings are different. You say P, I say p, the physical content of the two are different, so there is a misunderstanding.

In fact,  U can be considered as "the sum of the free energy and some other type of energy", Yx is the mechanical potential energy within the system, ∑μN is the chemical energy within the system, the two both are the real energy, but this energy classification for the internal energy has not been completed, the trouble is TS and -pV, in my view, these two terms can be combined into an independent term in that TS itself contains a positive pV, similar to y=2x-x, and then the solution of the equation is y=x. Now y=TS-pV is an equation.

Best regards,

I appreciate Alexei's reply! There is no difference in the physical intepretation of p and P. I only would like to add that the combined law (eq. 2.32 in Reichl) directly leads to the expression 2.66 when one takes into account the extensive character of U, S, V and N etc. The only meaningful statement you can make is that if S, N etc are kept constant when you change the volume than the work added to the system is -PdV and if this is positive it leads to an increase in U, if its negative (system performs work) it decreases the energy of the system.

Dear Aleksei,

Please see the link File, I distinguished −pdV from Y dx

Best regards,

I do not understand your comment. -PdV is just an example YdX.

best regards

John

Dear John,

Please see the link File,

Best regards,

Dear John,

I modified my article, this is the new version.

Best regards,

Tang

Your discussion in the attached file is unphysical. Pressure P is a physical quantity that can be meausured with various devices. It is also defined as a thermodynamic potential. However, in general there is no way in a real system to divide it into a part coming from thermal motion and what comes from interactions.  For example if you increase temperature of a solid with positive thermal expansion and keep its size fixed (clamped conditions) it will show an increase in pressure. Is this then your pressure of thermal motion? If you instead take a solid with negative thermal expansion (there are many such solids) you will experience a tensile stress instead for exactly the same temperature change. In fact the thermal expansion comes from the behaviour of the molecular or atomic interactions and if they were symmetric there would be no expansion at all. An increase in thermal energy can thus give a positive or negative pressure dependning on the character of the interactions.

Dear John,

“However, in general there is no way in a real system to divide it into a part coming from thermal motion and what comes from interactions”,

The thermal expansion comes from the behaviour of the molecular or atomic interactions, which also involves the thermal motion of the molecular or atomics, to divide these two parts further can be done by  microscopic theory.

If you compare my equation with classical equation, you will find that the mathematical results of the two equations are same. The difference between them is that my equation makes a finer classification, so which can derive more conclusions. For example, consider a spontaneous reaction or phase transition, we have the criterion diG≤0, where, diG is Gibbs free energy production, for a spontaneous mechanics process, we have the criterion diA≤0, where A=Yx. In addition to these, we also can get another criterion diq≥0, where diq is the heat production, etc.

To distinguish the pressure p of thermal motion from P, we can find that the reversible work –pdV represents heat conversion, this part of work differs from the work by other interactions.

If we like, we can easily back to classical equation in any time, but, there are many conclusions, which need to depend on a finer classification.

Best regards,

Dear Aleksei,

Q is defined as heat in transfer, so δQ depend on the path taken, but the changes of a form of energy involve the two cases, energy transfer and energy conversion, so

heat transfer + heat conversion= the changes of heat energy.

We had already known that the heat transfer δQ cannot represent the state changes, but we don’t know which function can describe the energy property of thermal motion.

Though the heat δQ is thought to be a form of energy in transfer, but this does not mean that the heat energy can only exist in the form of transfer.

Best regards,

Dear Suye\  if one looks at the standard text book of Chemical Thermodynamics |By Guggenheim one finds the following definitions of the characteristic functions for good reasions;

dU=TdS-pdV+ΣμdN.  

H=U+pV . Enthalpy,     F=U-TS  Helmholtz   G=H -TS  Gibbs 

dH=dU+pdV+dpV =    TdS+Vdp+ΣμdN

dF=dU-TdS -dTS =   -SdT- pdV-+ΣμdN.

 dG=dH -TdS -dT S = -SdT+Vdp+ΣμdN  

Euler Theorem for the homogeneous function of first degree in terms of extensive variables one obtains   

U=TS-pV+ΣμN.  after rearranging the terms   U-TS+pV=ΣμN   Left hand side  is nothing but Gibbs function as defined.  G= ΣμN  which is assumed to be valid even for the irreversible processes   accoding to de Groot,

Note  Dr. Suye interpretations or definitions  of characteristic functions are  somewhat out of line. Legendre transformation are done to obtains new functions in terms of desired independent variables or arguments. Above differential forms indicated that

U  is a ftc of  S, V, and  N  open systems

H  is a fct of   S, P and  N        

F  is a fct of  T, V  and  N

G  is a fct of  T, P  and N  

In principle we distinguish between the thermal parts of the energy and the non-thermal parts because  from statistical mechanics temperature is proportional to the average kinetic energy per degree of freedom, i.e we may say that the thermal energy is proportional to kinetic energy of internal motion, i.e. temperature. But already here is a problem because consider a system with fixed volume, i.e. no mechanical work is added, only temperature is changed. Then dU=CVdT. Should you not call this also a thermal energy because it is the part of the energy that is changed by adding heat. Now as the heat capacity in general is not a constant (except for an ideal gas) this part is not proportional to temperature. The reason is that part of the added heat is used also used to increase the potential energy of the atoms.

Energy, I think is a tricky thing. To me the first law or the law of conservation of energy is one of the greatest achievements during the 19:th century. It is really just an accounting system. It introduces the rather abstract concept of energy which is just a kind of currency that allows us to exchange different types of work, mechanical, elctrical and thermal. Although we often talk about electric energy, elastic energy etc such quantities are not defined in a strict sense although their "work" contributions in the combined law

dU=TdS-PdV+...

are well defined. However, once the work has been "added" it simply changes the energy of the system. On your bank account it is well defined how the money was entered, i.e. cash deposit or an electronic payment, but once the money is there it is does not make sense to talk about them as cash money or electronic money.

Problem here we all mixed up with the concept of energy in thermostatic  [Guggenheim hates to call thermostatic as thermodynamics as he mentioned in his book bitterly]-  with the concept of energy in classical as well as quantum mechanics. The kinetic part of the mechanical energy of a dynamical system, which appears as fluctuations in the  phase space of statistical mechanics  of  macroscopic ensemble,  sneaks into the thermostatic  with a new faze name.  and  it is called heat . One meets the same argument if one studies motion of a macro particle in an interactive  media where there is a so-called drag or viscous forces due to the  never ceasing collisions associated with the   Brownian motions of molecular particles,  which are highly erratic and constitute very high frequency random  fluctuations according Landau&Lifshitz (Statistical Physics-I, &121 Dissipation fct.)  and it may  be separable from the macroscopic motion of the  body by redefining the reaction of the media as the friction or drag forces.    Similarly the heat as defined originally in chemical thermodynamics is the transfer of mechanical kinetic energy from the surrounding to the thermodynamic system through the container wall which is impermeable with respect to the mass flow but not the so-called heat that is noting but the  mechanical kinetic energy transfer via  sound waves (phonons) etc.    One should forget thermodynamic doesn't answer the questions such as  how? and what ? but try to answer why? by referring our experience with hot and cold!!!  Actually the that is the our experience, which tells us that the heat flows from the hot object to cold object, and it can   never happen  opposite way,   was  the natural foundation of the second law as rigorously formulated  mathematically by Caratheodory  as an axiom for the integrability of Pfaffian Differential form of the first law. Del Q= SUM  Xi  dxi.   Where Integrability condition is   (mu X) x curl(mu X)=0.  mu is any arbitrary function of state variables. (That turns out to be ınverse absolute temperature, saving an arbitrary constant)

Dear Tarik,

There are some different expressions, they are equivalent.

U=TS-pV+Yx +∑μN

H=U-Yx+pV=TS+∑μN+pV

F=U-TS=Yx-pV +∑μN

U can also expressed as U=U(T,V,x,N), see my paper Eq.(36), this equation is cited from L.E.Reichl A Modern Course in Statistical Physics.

Best regards,

Dear Aleksei,

The thermodynamic functions involve a mixed form, including the state variables, such p,V, T, the process variables, such as W, Q, some coupling functions such, as H, F, and implicit functions, such as q, S, it leads to a complex structure of the theories.

The heat energy denotes the energy of thermal motion, which makes contribution to temperature T (and entropy S). It is also the opinion of some physicists, such as J.Fourier, J.C.Maxwell, R.Feynman.

For an ideal gas C=Cv, dU=CdT=CvdT. Am I wrong?

Best regards,

For an ideal gas you are right because there is only kinetic energy of the gas constituents, CV is constant and U=U0+CVT. But in general you are wrong! As I have shown it is in general not possible to define a "heat energy" in any meaningful or useful way.

Dear John,

In your example, dq=deq+diq, where q is the energy of thermal motion, deq denotes the heat added δQ, diq denotes “heat is used to increase the potential energy of the atoms.” diq＜0

In my paper, dU=CdT=CvdT is only used for ideal gas, no “in general.“

The money on my bank account is Swedish kronor, or U.S. dollar?

Best regards,

Your money may be added as Swedish crowns or dollars but once it is on the same account it is in a certain currency. As you understand I am OK with the fact that you may add thermal work, or mechanical work or any other type of work (they are all well defined) but once they are in the system they are just energy. Parts of that energy can then be withdrawn from the system in any form of work (as long as the second law is obeyed).

Best regards

John

Dear John,

My internet connection now is very slow.

But not all the energy makes contribution to temperature T, and not all the energy makes contribution to entropy.

Best regards,

That is right! It dependson  under what conditions you add a certain amount of heat. If you add it under constant volume it goes directly to increase the Internal energy. If you add it under constant pressure a part of it is used to perform pressure-volume work if there is a positive thermal expansion. The increase in energy is therefore smaller than under constant volume.

Dear Aleksei,

I guess, I know what you mean, this is a good question, but it is the problem in current theory, not my approach.

By current theory

dU=TdS-PdV+ΣμdN

For an ideal gas,

dU=TdS-PdV

So by current theory, for ideal gas, the real Gibbs free energy is equal to zero, but we have another equation.

dG=-SdT+VdP+ΣμdN

The equation requires a reasonable explanation.

By definition,

dG=d(ΣμN)= ΣNdμ+ΣμdN

It follows that

-SdT+VdP=ΣNdμ

In my approach

-SdT+VdP=diG=-diq

The equation represents the relation of heat conversion.

-SdT+VdP denotes the amount of heat conversion, it does not mean that dG exist in the form of -SdT+VdP.

Best regards,

Dear Tang,

I cannot see why you say that the Gibbs energy for an ideal gas is zero.

Dear John,

Sorry, I need a long time to open a Web page again.

dU=TdS-PdV+ΣμdN

For an ideal gas,

dU=TdS-PdV

OK! See here:

dU=TdS-PdV

dG=d(U-TS+PV)=-SdT+VdP

At a given T we have dG=VdP

From the gas law V=NRT/P

thus G=G0+NRTln(P/P0)

which in general is non-zero!

dG=d(U-TS+PV)=-SdT+VdP+ΣμdN=ΣNdμ+ΣμdN

G=ΣμN is the primary definition of Gibbs free energy.

U-TS+PV

is an equation, which cannot distinguish energy transfer and energy conversion, and also cannot describe the relation of heat conversion.

At a given T we have dG=VdP, which represents the relation of heat conversion, denotes that if heat to be transformed into G, we can calculate by this relation, VdP is equal to the amount of heat conversion, but VdP is not the real Gibbs free energy.

Dear John,

Can  you explain

dU=TdS-PdV+ΣμdN.

For an ideal gas

dU=TdS-PdV.

If Gibbs free energy of an ideal gas is not equal to zero, why for an ideal gas, dU does not contain ΣμdN ? It is the problem in current theory.

Best regards,

Dear Aleksei,

Same question, can you explain

dU=TdS-PdV+ΣμdN.

For an ideal gas

dU=TdS-PdV.

If Gibbs free energy of an ideal gas is not equal to zero, why for an ideal gas, dU does not contain ΣμdN ? It is the problem in current theory.

You have not broken up my theory. It is the problem in current theory, Just the opposite, my theory gives a reasonable explanation.

Best regards,

Dear Tang,

I will have to down vote your question because it says things that are incorrect. The equation dU=TdS-PdV+ΣμdN applies to all systems, including  the ideal gas. It is wrong to say otherwise. A student would not pass my course with such statement.

I agree completely with Alexei and Themis! As I showed previously it does not matter if you include the last term or not, i.e. if you consider a system with fixed composition or not, the Gibbs energy is not zero for an ideal gas. For an ideal gas with a varying composition the term ΣμdN of course must be included.

I sorry to tell you that you spend lot of efforts to argue about something that stems from a severe misunderstanding from your side.

Best regards

John

Dear professor Themis,

I don't mind you to done vote, but I never down vote because different viewpoint, I understand that peoples are so used to familiar things. If there are no objections, I will feel a bit strange. And I was a straight-A student many years ago, including chemical thermodynamics.

Aleksei' question  I once considered for many years, it only looks like a student' question, the truth is not so simple.

Best regards,

I also would like to add that the primary definition of Gibbs energy is: G=U-TS+PV

The fact that it is also equal to ΣμN  is just a consequence of the extensive property of U that ensures:

U=TS-PV+ΣμN => U-TS+PV=ΣμN, i.e. G=ΣμN

best regards

joHN

Dear Dr. Suye: I will add two comments to your original question, which we can discuss further if need be.

1. Different parts of the internal energy U do not have be positive definite. In classical thermodynamics, S can be negative so the first term in your U can be negative, For negative chemical potentials, the last term is also negative. Indeed, for a bound system in a finite volume, it is easy to see that U can be negative if the potential is unbounded from below.

2. The role of different free energies can be better appreciated by looking into their variations as the system approaches equilibrium. Thus, if the system is in contact with a heat bath at temperature T0, but its volume and the number of particles in it are held fixed, then dF cannot be positive during equilibration. What one usually forgets is that F=U-T0 S which contains the temperature of the heat bath and not of the system is not a property of the system (it contains the heat bath temperature T0). So, it most certainly does not qualify as an intrinsic property of the system. In equilibrium, it achieves its minimum when the system also has its temperature T0 but during its relaxation to equilibrium, its internal energy and entropy have changed. How they individually change depends on relationship of the system with the heat bath. This is why quantities like F, G, etc. are called free energy in equilibrium.

I hope these comments are somewhat useful.

I tried to add a further comment earlier but I probably did not suceed. If U is positive or nehative has no physical meaning because you can define the zero point arbitrarily dependning on your choice of reference. The only thing that has meaning is changes in U.

Dear John:

One must be a bit careful. The arbitrariness in the zero of U comes from the potential energy. But for an ideal monatomic gas, there is no potential energy and one cannot arbitrarily change U without affecting the temperature as it is commonly defined. I am sure that you meant the real case where there is always a potential energy contributing to U but I did not want anyone to misunderstand your statement. as there are some very wrong ideas floating around in this thread.

Puru

Well even for an ideal gas the energy is arbitrary because you can define the velocity any way you want. Suppose you have a gas in a chamber that travels with a constant velocity v relative you, than you can add the kinetic energy to the internal energy of the whole gas without changing the physics.

Dear John:

I do not wish to quibble over a very minor issue. But you should recall that the internal energy is defined in a frame where the center of mass of the body is at rest. In other words, the center of mass kinetic energy must be set equal to zero. So,for U, there is no freedom. The situation is different for the total energy E, which can change from one frame to another. The temperature of the gas is determined by U and not E. This also explains that the entropy S is a function of U and not of E.

Thanks anyway to force me to write the above comment that might help others. I pretty sure that you know all what I just said; you may have just forgotten monetarily.

Puru

Well, I accept what you write but if I want to I could probably write other examples, when there is no absolute value for the internal energy of an ideal gas. One example is when you have an ideal gas as an atmosphere of a planet... I think we understand each other and I will refrain from driving this excercise further.

Best regards

John

Dear Aleksei,

You written

T=dU/dS, p=-dU/dV,…, they are incorrect.

The correct expressions are

T=(∂U/∂S)v,N

p= T(∂p/ ∂T)V-(∂U/ ∂V)T

…

For an ideal gas

dS=CvdT/T+pdV/T=dU/T+pdV/T

It implies

TdS=dU+pdV

Such that

dU=TdS-pdV

No ΣμdN.

Dear Gujrati,

I cannot agree with the concept of negative energy.

Best regards

Dear Aleksei,

dS=CvdT/T+pdV/T=dU/T+pdV/T

is the total differential of the entropy for an ideal gas.  ΣμdN make no contribution to the entropy.

Best regards

Dear John and Themis,

For an ideal gas

dS=CvdT/T+pdV/T=dU/T+pdV/T

It implies

TdS=dU+pdV

Such that

dU=TdS-pdV

No ΣμdN.

Best regards

ATTENTION!!!

dU=TdS-pdV  FOR CLOSED SYSTEMS

dU=TdS-pdV +ΣμdN   FOR OPEN SYSTEMS

ONLY  GIBBS FREE ENERGY HAS THE PROPERTY OF;

G=ΣμN  and    dG= Σμ dN  + ΣN dμ

THAT GIVES FAMOUS   GIBBS- DUHEM RELATION

SdT - VdP +  ΣNi dμi =0     That is the desired connection between T, P and  μi

BEST REGARDS

Dear Aleksei

For

S=N(ln(V/N)+1+i/2*(lnT+ln(C)+1))

Where is μ? where is the gas constant R or Boltzmann constant k?

pV=NkT.

Agree with Tarik!/john

Dear Tang  and Aleksei:  

dU=TdS-pdV +ΣμdN  FOR OPEN SYSTEMS  give  a little kick to terms you get,

dS =dU/T+pdV/T -1/TΣμdN, EQD.

Similarly  giving another kick:  using  G=ΣμN  and the definition of  G=U-TS+PV  one can get;

S= U/T +PV/T - 1/T ΣμN,    EQD.

In the case of deformed elastic and hyper elastic solids  pdV   should be replaced by the double inner products of   [ -Stress: d Strain] ,     tensors.

Best Regards

Dear Professor Tarik,

As a general equation,  I agree with you, for some special systems, the third term is equal to zero, so in such cases, we have

dS=dU/T+pdV/T

Such as an ideal gas, or a photon gas.

Best Regards,

Dear Tang,  But you have to specify those cases concisely!   dS=dU/T + pdV/T  would be true if the system is bounded by a wall which is impermeable to  mass transfer, and have no chemical and nuclear reactions are going on in the system,  plus no internal entropy source and sinks are present in the system.  Don't forget in the case of reactions eve though system would be closed in the sense mass transfer through the boundaries,  a term such as enters into above equation, which  involves  Affinities  Ai  and the extent of the reactions  Ei ,  namely [Σ (Ai d Ei)].  

dS=dU/T + pdV/T+1/T Σ(Ai d Ei)   (Prigogine)      Where  dSinterior=1/TΣ(Ai d Ei)>0

dSexterior=dQ/T=dU/T + pdV/T       (Prigogine)     page. 23

Above system are closes of mass exchange through the external boundaries

Regards.

Dear all,

Please see

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

--> Entropy

dS=CvdT/T+(∂p/∂T)vdV

For an ideal gas, CvdT=dU, and  (∂p/∂T)v=p/T.

Best regards,

Dear Tang, dS=CvdT/T+(∂p/∂T)vdV,  where the last term is right if only if one has the ideal gas represented by  PV=nRT relationship,  otherwise it is wrong.

Note: Wikipedia's treatment is for the ideal gas, and it  made extremely simple case horribly complicated.

Best Regards

For an ideal gas I would start with its Gibbs energy:

G=G0 + NRTln(P/P0)

where the first term is a function of temperature only. From (dG/dP)T=V , which is a general relation, we  find

(dG/dP)T=V=NRT/P, i.e. the gas law is recovered. We may further use the general reelation

(dG/dT)P= -S  and we find that

S=(dG0/dT)P+NRln(P/P0)

The first term is

CPln(T/T0)+S0

because CP=TdS/dT=-Td2G/dT2

Thus

S=CPln(T/T0)+S0+NRln(P/P0)

I  am sorry some minus signs are missing

Dear all,

Please look for a thermodynamic textbook (physics or engineering thermodynamics）as you choose.

dU=TdS-PdV+ΣμdN is an general equation, it is correct, but for an ideal gas or a photon gas the third term is equal to zero, so dU=TdS-PdV.

dU=TdS-PdV implies that μ=0.

Best Regards

Dear John,

As an equation, dG=NRTdln(P/P0) is also correct, so there is a conflict, it need an explanation: What is NRTdln(P/P0)?

Best Regards

Dear Tang,

The statement that μ=0 for ideal gas is simply wrong. No book makes this claim. We should not be making things up. 

Dear Themis,

Yes, No book makes this claim that μ=0 for ideal gas, in that there is a conflict that has not yet clarified.  

But in many books, you can find that for an ideal gas dU=TdS-PdV, and then according to dU=TdS-PdV, it implies that for an ideal gas, μ=0.

As I written to John, As an equation, dG= NRTdln(P/P0) is also correct, so there is a conflict, it need an explanation: What is NRTdln(P/P0)?

I look for the physical picture for each concept and each expression.

Best Regards,

Dear professor Tarik,

Otherwise, the last term is also correct, in current theories

(∂S/ ∂V)T= (∂p/ ∂T)V

is a general relation for all cases.

Best Regards,

Dear Tang,  that is true.  This relationship is known as the Maxwell Relationship, which can be obtained from the second derivative of Helmholtz Free energy with respect to temperature and volume or volume and temperature.

d2F/ dVdT= -dP/dT      d2 F/ dTdV=-dS/dV    QED.

Best Regards.

Dear Aleksei,

Yes, the entropy depend on the particle numbers, but for an ideal gas, N dose not mean ΣμdN

dS=CvdT/T+pdV/T=[(i/2)Nk]dT/T+(Nk/V)dV.

The two terms both involve N.

Best Regards,

Dear John,

Whether we start form dU=TdS-pdV or dG= NRTdlnP, it does not change the conflict, the two equations both are correct, but the conflict can hardly be explained in current theories.

dG has the two sources: energy transfer and energy conversion, these two processes have not yet been distinguished in current theories, when we distinguish these two sources of dG, there will be no such conflict. dG= NRTdlnP represents the relation of the energy conversion between the different energy types, but NRTlnP/P0 is not the real G within the system.

Best Regards,

As I see it a major achievment of thermodynamics is the introduction of the energy concept which relates the change in energy to the work performed on the system (postive or negative), thermal mechanical, electric etc. From a thermodynamic point of view it is not meaningfull to distinguish between different forms of energy once it is stored in the system as energy. Of course from a more detailed physical picture it may still be meaningful to divide the energy in kinetic energy of the atoms and molecules and perhaps divide that in other ways translational rotational etc, and potential energy comming from Coloumb interactions, stresses etc.

I think the reason you run into all these problems is that you do not understand the first law of thermodynamics properly As has been pointed out the combined law for a system with fixed composition is dU=TdS-pdV  and the size enters in each term in this equation. If the size is zero it say just 0=0.  You should observe also that dU=TdS-pdV  only holds for reversible changes. For spontaneous (irreversible changes) you have

dU=TdS-pdV - TdiS

where the last term is the internal entropy producion and TdiS>0 from the second law and vanishes for reversible changes.

Best regards

John

Dear Tang,

We obtain dU=TdS-PdV by setting dN=0: this refers to a system of fixed number of molecules. This equation is again general (i.e it is not specific to ideal gas). And of course it does not mean μ = 0. 

I understand that you want to gain physical insight. When it comes to thermodynamics, my advice is this: if you want physical insight, concentrate on the first and second law, that's where all the physical meaning lies. After that, it's all math. If you try to interpret every step of thermodynamic calculus in "physical" terms, you will very quickly become confused. 

Agree with Themis!/john

Dear John,

dS=deS+diS

δQ/T=deS

dU=TdeS-pdV+TdiS=TdS-pdV

Best regards

Dear Themis,

OK, now we consider an isolated system, 

dS=CvdT/T+pdV/T=CvdT/T+nRdV/V

dU=TdS-pdV

The number of molecules is fixed, we get the integral equations

S=Cvln(T/T0)+nRln(V/V0)

U=CvT

Where is μ?

“This equation is again general (i.e it is not specific to ideal gas).”

No! dU=TdS-pdV is just and only specific to ideal gas and a few simple systems.