The free energy is "free", because it is the negative change in free energy that can be used in a reversible process to produce work.You can't get more than that. A nice exercise is to calculate the work a Carnot cycle running between a finite heat reservoir T_initial (think of a finite copper block) and a heat reservoir (infinite copper block) of temperature T_R=const

In Thermodynamics the second law applies to processes in isolated systems. Usual systems are not isolated, and therefore, special cases for non-isolated systems are developed. If a given system is in contact with a thermostat (environment) at constant temperature and undergoes a process at constant volume, the second law tells us the system will evolve towards a state in which the potential U-TS reaches a minimum. U-TS is called F, Helmholtz free energy, and its variation is equal to the maximal amount of work that can be obtained through the process. If the process occurs at constant pressure, the second law tells us the system will evolve towards a state in which the potential U-TS+PV reaches a minimum. U-TS+PV=F+PV is called G, Gibbs free energy, and its variation is equal to the maximal amount of work that can be obtained through the process. Threfore, "free" means available or useful for any purpose (e.g. producing work) and not lost through dissipation.

Many times the word "free" is eliminated: Helmholtz energy or Gibbs energy.

The above are the standard explanation in textbook, but the standard explanation cannot explain for “why the Helmholtz free energy is called free”

Helmholtz free energy is a state function, but "the maximal amount of work" is not. The state meaning of Helmholtz free energy F is not “work” or "the maximal amount of work".

There is an age-old question: an adiabatic reversible expansion of an ideal gas can output the maximal amount of work W_m, and W_m is equal to the change of the internal energy the ideal gas, so in this case, the internal energy of the ideal gas is the free energy F?

G is the real free energy of a system, the physic meaning of “the real free energy” means that the energy makes no contribution to entropy. F contains "the real free energy" and −pV , the latter is “the free energy” that can be obtained from heat conversion.

@Tang: In the adiabatic reversible expansion of an ideal gas the system is not in thermal equilibrium with the surroundings, the temperature is not constant, and, therefore, the thermodynamic potential whose variation is equal to the maximal work is the internal energy U(S,V,N), not the Helmholtz energy F(T,V,N) or the Gibbs energy G(T,P,N).

Dear Adrian,

Sorry, there seems to be some misunderstanding. The fundamental definition of Helmholtz free energy is F=Yx-pV+μN(=U-ST).

An adiabatic reversible expansion of an ideal gas can output the maximal amount of work W_m=-pdV, assuming the work W_m drive a piston, we get the work Ydx done by the piston(-pdV=Ydx). In this case, the internal energy of the ideal gas is converted into Helmholtz free energy Ydx by pV work, so there is the question.

The physic meaning of F is Yx-pV+μN, not F(T,V,N) , and not the maximal amount of work . The fundamental definition is F=F(x,V,N), the latters are the equivalent equations in math.

Dear Vahid, Below is my explanation to you question.

The internal energy U

1) U=TS+Yx-pV+μN.

And the two “free energies”

2) F = Yx-pV+μN (=U-TS),

3) G=μN.

By S= (UT-Yx+pV-μN)/T, we know that Yx and μN make no contribution to entropy, the two are the real free energies (I define it by ψ=Yx+μN), “the real free” means that the two have no quality loss. By Eq.(1) we get

4) U=(TS-pV)+(Yx +μN)

(TS-pV) is the heat energy (see below), which has some quality loss and makes contribution to entropy. In Eq.(2) and (4) -pV is not the real free energy, it is a thermodynamic potential, indicates the free energy that can be obtained from heat conversion. The maximal amount of work is equal to the increment in the real free energy+ the work done by the reversible heat conversion (heat energy transform into the real free energy or work).

TS isn’t equal to “the thermal energy”.

For example, consider an ideal gas, we have

1) TdS= cdT+pdV,

TdS is the sum of the two terms. The physical meaning of TdS includes the two (The first term is “the thermal energy”).

2) dU= TdS-pdV.

Using 1) in 2), we get

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

For an ideal gas, all of the internal heat energy U is “the thermal energy”, but “the thermal energy” cdT≠TdS.

“the real free” means that the energies have no quality loss and make no contribution to entropy. G=μN is the real free energy, F(x,V,N) is the real free energy+ a thermodynamic potential (indicates the free energy that can be obtained from heat conversion).

The fundamental definition of the two free energies are F=F(x,V,N) and G=μN, not F(T,V,N) and G(T,P,N).

Many textbooks are not designed to study deep enough, to explain your question need a good textbook .

Dear Vahid,

1) The complete thermodynamics involves the theory of heat + mechanics+ chemistry. Yx is the part of the mechanics, it is the potential energies of thermodynamic system. Where Y is the generalized forces, x is the generalized distances, and Ydx is the work done by the generalized force Y (except the pressure p). Such as the surface free-energy, etc.

Ydx=JdL+σdA+E·dP+ H·dM…

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

2) The internal energy U can be classified into the two types based on their contribution to entropy, Yx+μN make no contribution to entropy, and can be transformed freely into other forms of energy (without compensation) , so the two are called as “free energy”. The energies have no quality loss for the terms Yx and μN in that this is a fundamental nature. The heat energy within the system q=(TS-pV) can also be transformed into Yx+μN, but must attach some compensation, i.e., it cannot be transformed freely into other forms of energy, so it is “not free”.

3) The pressure p result from the heat energy within the system, and pdV is the work done by heat, so "pV contributes to energy loss”. We can get the conclusion by S= (UT-Yx -μN)/T+pV/T.

Additional:

Some misunderstanding on F(T,V,N) and G(T,P,N). For instance, we have

dG=d(μN)= Ndμ+μdN

dG=-SdT+Vdp+μdN

By the two equations we get

Ndμ=-SdT+Vdp

Where μdN is the “G flux”, Ndμ is the “G production”, “-SdT+Vdp” denote some heat energy to be transformed into μdN. Where -SdT+Vdp is the negative heat production (+compensation), it is a mathematical relationship, the increment in G production is equal to the negative heat production , but -SdT+Vdp is not the physical content of G. The fundamental definition of G≠G(T,P,N), and similarly, the fundamental definition of F≠F(T,V,N).

These are the explanations based on current thermodynamics. In my paper, these are very simple results. The link please see below. Some parts of the paper seem to be a bit difficult to read because too many details. If you are interested, I suggest to jump over some pages [from page 7 to Eq.(103)]. In Eq.(47), I have redefined “the real free energy”, and separated –pV from Helmholtz free energy.

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

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

The total work performed on/by a system is given by dW=pdV+YdX, where Y and X are generalized forces and displacements different from pressure and volume (that is, any non-mechanical work, dW'). Thus, YdX may correspond to -μdN, -FdL, -σdA, -BdM, -EdP, etc. (-FdL and -σdA are mechanical works, but let's keep the term "mechanical" for the worked associated with pressure).

Let's start from definitions and consider the internal energy U:

U=TS-pV-YX

dU=TdS-pdV-YdX

In a given process: dU≤TdS-pdV-YdX (= if reversible, < if irreversible)

If the process is adiabatic: dU≤-dW

If the process is adiabatic and isochoric: dU≤-dW'

If, in addition, there is no non-mechanical work: dU≤0

Now let's consider the Helmholtz energy:

F=U-TS=-pV-YX

dF=-SdT-pdV-YdX

In a given process: dF≤-SdT-pdV-YdX

If the process is isothermal: dF≤-dW

If the process is adiabatic and isochoric: dF≤-dW'

If, in addition, there is no non-mechanical work: dF≤0

Now let's consider the Gibbs energy:

G=F+pV=U-TS+pV=-YX

dG=-SdT+Vdp-YdX

In a given process: dG≤-SdT+Vdp-YdX

If the process is isothermal and isobaric: dG≤-dW'

If, in addition, there is no non-mechanical work: dG≤0

Now let's consider the enthalpy:

H=U+pV=TS-YX

dH=TdS+Vdp-YdX

In a given process: dH≤TdS+Vdp-YdX

If the process is adiabatic and isobaric: dH≤-dW'

If, in addition, there is no non-mechanical work: dH≤0

Therefore, under certain conditions (constraints) there is a thermodynamic potential that reaches a minimum value in a system evolving spontaneously: U in adiabatic and isochoric processes, H in adiabatic and isobaric processes, F in isothermal and isochoric processes, and G in isothermal and isobaric processes.

In addition, if the variation of that potential is positive, work is done on the system; if negative, work is done by the system. If negative, the variation in that thermodynamic potential is equal to the maximal work done by the system (or obtained from the system) in a reversible process. If the process is irreversible, the variation in the thermodynamic potential is larger than the work done by the system (there are energy losses). And this is "useful" of "free" work that can be obtained from the system, once energy losses are discounted.

Regarding the term "free", nowadays there is a tendency to avoid it. Thus, F is simply called "Helmholtz energy", and G is simply called "Gibbs energy". By the way, G is sometimes called "free enthalpy".

Dear Adrian,

Y and x are the two state functions, U cannot be written as

1) U=TS-pV-Yx

For instance, the surface free-energy σA is one part of the internal energy, by Eq.(1),

U=TS-pV-σA

Then which one is negative? σ? or A?

(there is a logical explanation for -pV)

In the equation dW=…, pdV has different sign from Ydx.

Dear Tang,

I defined the generalized work as dW=ΣYdX, where Y is a generalized force and X is its associated generalized displacement. In this case each of the YdX terms corresponds to one of these types of elemmental work: pdV, -μdN, -FdL, -σdA, -BdM, -EdP, etc. Please, note that all the terms have a negative sign, except the work corresponding to pressure-volume. That is, the generalized forces are p, -μ, -F, -σ, -B, -E, etc. Neither σ nor A are negative, but the corresponding work is -σdA.

From the definition of the internal energy: U=TS-ΣYX, dU=TdS-ΣYdX

we get: U=TS-pV+μN+FL+σA+BM+EP+...

and: dU=TdS-pdV+μdN+FdL+σdA+BdM+EdP+...

and all terms have the appropriate sign so the stability criteria for the system are fulfilled.

Therefore, it is a matter of signs; the final expressions for W and U are right.

Dear Adrian,

Ok, your -Y sometiome = my Y. Sorry, there are some misunderstanding,

But G cannot be weitten as G=-YX, due to G does not Include other terms (compare with U, F and H), it seems to be a bit carelessness.

Dear Tang,

G is the Legendre transform of F regarding pV or the Legendre transform of U regarding TS and pV. Therefore, if F or U contain non-mechanical work terms YX, then, G contains the same terms YX.

If U=TS-pV+μN+FL+σA+BM+EP+...

then G=μN+FL+σA+BM+EP+...

Dear Adrian,

“the Legendre transform” has changed the physical content of G, but not the variables. In “the transform”, G is not the original content μN.

After the Legendre transformation, G keeps the formal definition:

G=μN+FL+σA+BM+EP+...

although the different terms are expressed in terms of T, p, N, L, A, M, P,...

Dear Adrian,

No! I think you are confusing the mathematical relation and physical contents.

In some csae, the differential expression

dG=μdN+FdL+σdA+B·dM+E·dP+.…

is correct. But integral expression should be writen as, and the definition of G must be writen as

G=μN

So, in some cases that differential expression is correct and in some others it is not correct...

G=μN is valid for a simple monocomponent system in the absence of non-mechanical generalized forces. But in general, for a given general system you must include additional -YX terms (-μiNi, -FL, -σA, -BM, -EP, etc.).

You may keep stepping out of Thermodynamics if you want.

*^o^*

Entropy: A concept that is not a physical quantity

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

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