# Interpretation of Helmholtz energy and Gibbs energy

I know that this question has many other variations on this site, but I'm trying to see if I understood Helmholtz and Gibbs energy properly or not. The material I'm reading from is Physical chemistry by Thomas Engel and Philip Reid, third edition, and An introduction to Thermal Physics by Daniel Schroeder.

In Schroeder's book, Chapter $$5$$, the author defined Helmholtz energy, $$A$$, as $$A= U-TS$$, where,

$$U$$ is internal energy of the system and $$S$$ is the system's final entropy.

The author says that, It is the total energy needed to create the system minus the heat you can get free from the environment at temperature $$T$$. He further states that it is the available or "free" energy.

Then, for a system in constant pressure $$(P)$$ and temperature $$(T)$$ environment, he defines Gibbs energy, $$G$$, as $$G= U-TS+PV$$, where $$PV$$ is the atmospheric work term that's in enthalpy,

$$H$$ $$=U+PV$$.

Also, from Engel and Reid, chapter $$6$$, we have, for isothermal process,

$$dA$$ $$\le$$ $$đw_{total}$$ $$...(1)$$, including expansion and non-expansion work, where the equality is satisfied for reversible process. Equation (1) allows us a way to calculate maximum work that a system can do on the surroundings.

And in a similar manner $$dG$$ $$\le$$ $$đw_{non-expansion}$$ $$....(2)$$, where the equality is satisfied for reversible process.

It is stated that equation $$(2)$$ allows one to calculate maximum non-expansion work that can be produced.

Now, this is my understanding:

For reversible process, $$A$$ represents total available internal energy, and a part of $$A$$, which is $$G$$, is the available energy to do non-expansion work.For irreversible processes the inequalities (1) and (2) gives the lower bound for expansion and non-expansion work respectively. Is it the correct way of understanding or am I way off? Also, if $$A$$ does represent total available internal energy for reversible process then what does it represent for irreversible process? Thanks in advance

• pV is an atmospheric work term only at isobaric conditions. Commented Jan 23, 2023 at 8:42
• @Poutnik, I did write that the system is in constant temperature and pressure environment. Commented Jan 23, 2023 at 9:06
• OK, but it is misleading formulation there, making impression G is defined for constant pressure (P) and temperature (T), what is not true. It is defined generally, having just special advantages at those conditions, like delta pV being the volume work term. The similar for A and constant T and V. Commented Jan 23, 2023 at 9:43
• I will not probably pay enough attention today, so you may want to wait until US wakes up, there are some TD experts. // Yeat another interpretation of A and G is $\Delta S_\mathrm{tot} \ge -\Delta G_\mathrm{sys}/T$ for T,p const, resp. $\Delta S_\mathrm{tot} \ge -\Delta A_\mathrm{sys}/T$, for T,V const, where $\Delta S_\mathrm{tot}=\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr}$ Commented Jan 23, 2023 at 11:45
• Note that $-\Delta G$ represents the maximal amount of work the system may provide during the respective system state change. For isothermal change of ideal gas state, $\Delta G = W=nRT\ln{\frac{p_2}{p_1}}$ Commented Jan 25, 2023 at 12:42

You made fascinating observations about thermodynamic potentials. Learning thermodynamics is a lifelong task, there are so many philosophical and scientific reflections in these statements that it is difficult for us to exhaust everything we need to know.

In thermodynamics, the four main thermodynamic potentials are defined as functions of various state variables such as temperature, pressure, volume, and entropy. These potentials are used to describe the energy changes in a system and to predict the direction of spontaneous processes. The four potentials are internal energy (U), Helmholtz free energy (A), Gibbs free energy (G), and enthalpy (H).

### Internal Energy (U)

The internal energy (U) is the total energy contained within a system, including the kinetic and potential energies of the molecules. It is a function of entropy (S), volume (V), and the number of particles (N):

$$U = U(S, V, N)$$

The first law of thermodynamics relates changes in internal energy to heat (Q) added to the system and work (W) done by the system:

$$dU = \delta Q - \delta W$$

In a reversible process, where $$\delta Q = TdS$$ and $$\delta W = PdV$$:

$$dU = TdS - PdV$$

### Helmholtz Free Energy (A or F)

The Helmholtz free energy (A or F) is defined as:

$$A = U - TS$$

It is a measure of the useful work obtainable from a closed system at constant temperature and volume. Its differential form is:

$$dA = dU - TdS - SdT$$

Using the expression for $$dU$$:

$$dA = TdS - PdV - TdS - SdT$$ $$dA = - PdV - SdT$$

The Helmholtz free energy is particularly useful in systems held at constant temperature and volume. You could say, as you do, its a lower bound without expansion, because at constant volume ther will be no expansion. You will have to take into account appartly the changens in volume.

### Gibbs Free Energy (G)

The Gibbs free energy (G) is defined as:

$$G = H - TS$$ $$G = U + PV - TS$$

It is a measure of the useful work obtainable from a closed system at constant temperature and pressure. The differential form is:

$$dG = dH - TdS - SdT$$

Using the expression for $$dH$$:

$$dG = d(U + PV) - TdS - SdT$$ $$dG = (TdS - PdV + VdP + PdV) - TdS - SdT$$ $$dG = VdP - SdT$$

Gibbs free energy is used extensively in chemical thermodynamics, especially for predicting the direction of chemical reactions and phase changes under constant pressure and temperature. Here again, as you say, a lower bound for expansion, because at constant pressure volume will expand freely. You will have to take into account appartly the changens in pressure.

### Enthalpy (H)

The enthalpy (H) is defined as:

$$H = U + PV$$

It is a measure of the total energy of a system, including both internal energy and the energy required to make room for it by displacing its environment. The differential form of enthalpy is:

$$dH = dU + PdV + VdP$$

Using the expression for $$dU$$:

$$dH = TdS - PdV + PdV + VdP$$ $$dH = TdS + VdP$$

Enthalpy is useful in processes occurring at constant pressure, such as many chemical reactions and phase transitions. As you say, a lower bound for expansion, because at constant pressure volume will expand freely.

### Chemical potentials in the face of irreversibilities

As you can see, if you want to evaluate transformations by the variation of H or U, you will have to deal with the evolution of entropy throughout the transformation, which is always a difficulty: $$dU = TdS - PdV$$ $$dH = TdS + VdP$$ It is much easier to know if the transformation is constant pressure, temperature, etc. $$dG = VdP - SdT$$ $$dA = - PdV - SdT$$

This is a transformation at constant G (suposing there is just one chemical species). It has constant pressure, because it is at atmospheric pressure and constant temperature, mantained by the water bath, also called Maria bath due to the Egyptian chemist.

If instead of mantaining the flask opened to the atmosphere you close it, mantaining a constant volume, you have a constant A transformation.

### The trouble with irreversibility

If you have to deal with irreversibility you have to use U or H. Instead of talking abbout "maximum" work, I prefer the De Groot and Mazur ("Non-equilibrium Thermodynamics") approach, separating the $$dS$$ in a reversible parcel and an irreversible one:

$$dS = d_e S + d_i S$$ $$d_e S = \frac{dQ}{T}$$ $$d_i S \geq 0$$

The separation between reversible and irreversible entropy has a fundamental reason. The reversible entropy change during a transformation is due to the change in specific entropy between one state and another. Thus, if you have a chemical reaction, combustion, for example, it is impossible to compare the entropy of the products with the reactants, which are completely different. This has no relation to how the transformation is carried out, but only to the fact that there is a transformation from A to B. Irreversible entropy, on the other hand, is due to the way in which the reaction or transformation is carried out, whatever is and is directly related to spontaneity. Irreversible entropy is an excess that promotes transformation, a will, a whim, a bonus that causes transformation to occur. We can imagine that this irreversible entropy can be as small as possible, so that the transformation can be analyzed as reversible, even if it is a theoretical abstraction, i.e. $$d_i S =0$$. In this approximation:

$$dU = TdS - PdV = dQ - PdV$$ $$dH = TdS + VdP = dQ - VdP$$

Of course, things got a little more complicated if you have many chemical species and chemical reactions.

When dealing with many chemical species and reactions, the thermodynamic potentials must account for the chemical potentials and the changes in the number of particles of each species. This leads to modifications in the expressions and differentials of the thermodynamic potentials to include terms related to the chemical composition of the system.

### Entropy as what keeps the system going

So entropy could be seen not as a lose of energy, but what keeps the system going. To be natural the system will change so to create entropy. In this sense, entropy is an "action", in the sense of lagrangian physics, what make the system changing.

$$dU = Td_i S + T d_r S - PdV$$ $$d_{i} S = - \frac{dU}{T} + dS_r - \frac{P}{T} dV$$

Kondepudi and Prigogine present all these potentials as the origem of "forces" that drive the system, creating an action that he represents as $$\sigma$$, the entropy production by unity of volume, such that:

$$\frac{dS}{dt} = \int \sigma dV=$$

This action, as we are used to think in Lagrangian Physics, is to be minimized. Nature always uses the minimum action to drive transformations, which explains inertial movements, heat transfer steady states, and even life, but this is a subject for another occasion.

### Molar Quantities

When dealing with thermodynamic potentials per mole, volume per mole, and entropy per mole, we use specific quantities, often denoted with a bar or tilde to indicate they are molar quantities. Here’s how we can rewrite the equations with these molar quantities:

• Molar Internal Energy ( $$\bar{U}$$): $$\bar{U} = \frac{U}{n}$$
• Molar Helmholtz Free Energy ( $$\bar{A}$$): $$\bar{A} = \frac{A}{n}$$
• Molar Gibbs Free Energy ( $$\bar{G}$$): $$\bar{G} = \frac{G}{n}$$
• Molar Enthalpy ( $$\bar{H}$$): $$\bar{H} = \frac{H}{n}$$
• Molar Volume ( $$\bar{V}$$): $$\bar{V} = \frac{V}{n}$$
• Molar Entropy ( $$\bar{S}$$): $$\bar{S} = \frac{S}{n}$$
• Molar Number of Particles ( $$\bar{N_i}$$): $$\bar{N_i} = \frac{N_i}{n}$$
• Chemical Potential ( $$\mu_i$$): remains $$\mu_i$$, as it is already an intensive quantity.

where $$n$$ is the number of moles.

### Priviledged perspective of G

From the four potentials, you will notice that G in the unique that has all intensive properties (S and V) outside the differential:

$$dU = TdS - PdV$$ $$dH = TdS + VdP$$ $$dG = VdP - SdT$$ $$dA = - PdV - SdT$$

This permits:

$$dG = d\bar{G}N = N\bar{V}dP -N\bar{S}dT + \bar{G} dN$$ $$dG = d\bar{G}N = VdP -SdT + \bar{G} dN$$

Look what happens when you try to do this with, for instance, U (we get back to $$\bar{G}$$):

$$dU = dN\bar{U} = TdN\bar{S} - PdN\bar{V}$$

$$Nd\bar{U} = T\bar{S}dN + TNd\bar{S} - NPd\bar{V} - P\bar{V}dN - \bar{U}dN$$

$$dN\bar{U} = TNd\bar{S} - NPd\bar{V} + (-\bar{U}+T\bar{S}- P\bar{V})dN$$

$$dN\bar{U} = TNd\bar{S} - NPd\bar{V} -\bar{G}dN$$

So, the chemical potential is always $$\bar{G}$$ and we, strangely, give it another name: $$\bar{G} \equiv \mu$$

### Gibbs Free Energy (G) multiple chemical species

The Gibbs free energy for a multi-species system is defined as:

$$G = H - TS = U + PV - TS$$

The differential form is:

$$dG = d(U + PV) - TdS - SdT$$

$$dG = (TdS - PdV + \sum_i \mu_i dN_i + PdV + VdP) - TdS - SdT$$ $$dG = VdP - SdT + \sum_i \mu_i dN_i$$

This form shows how the Gibbs free energy changes with pressure, temperature, and the number of particles of each species.

### Internal Energy (U) multiple chemical species

For a system with multiple chemical species, the internal energy depends on the entropy (S), volume (V), and the number of particles ( $$N_i$$) of each species $$i$$:

$$U = U(S, V, \{N_i\})$$

The differential form includes terms for the change in the number of particles: $$dU = TdS - PdV + \sum_i \mu_i dN_i$$

where $$\mu_i$$ is the chemical potential of species $$i$$, representing the change in internal energy with respect to the change in the number of particles of species $$i$$.

### Helmholtz Free Energy (A) multiple chemical species

The Helmholtz free energy in a system with multiple species is defined as:

$$A = U - TS$$

Its differential form includes the chemical potential terms:

$$dA = dU - TdS - SdT$$ $$dA = TdS - PdV + \sum_i \mu_i dN_i - TdS - SdT$$ $$dA = -PdV - SdT + \sum_i \mu_i dN_i$$

### Enthalpy (H) multiple chemical species

The enthalpy for a system with multiple chemical species is:

$$H = U + PV$$

The differential form is:

$$dH = dU + PdV + VdP$$ $$dH = TdS - PdV + \sum_i \mu_i dN_i + PdV + VdP$$ $$dH = TdS + VdP + \sum_i \mu_i dN_i$$

### Chemical Potentials and Reactions

In systems with chemical reactions, the change in the number of particles of species $$i$$ is related to the stoichiometry of the reaction. For a reaction involving species $$i$$ with stoichiometric coefficients $$\nu_i$$, the extent of reaction $$\xi$$ describes the progress of the reaction. The change in the number of particles of each species is:

$$dN_i = \nu_i d\xi$$

The Gibbs free energy change for a reaction is:

$$dG = \sum_i \mu_i dN_i = \sum_i \mu_i \nu_i d\xi$$

At equilibrium, the Gibbs free energy change for the reaction is zero:

$$\sum_i \mu_i \nu_i = 0$$

### Summary multiple chemical species

In multi-species and multi-reaction systems, the thermodynamic potentials include additional terms to account for the chemical potentials and changes in the number of particles of each species:

• Internal Energy (U): $$dU = TdS - PdV + \sum_i \mu_i dN_i$$
• Helmholtz Free Energy (A): $$dA = -PdV - SdT + \sum_i \mu_i dN_i$$
• Gibbs Free Energy (G): $$dG = VdP - SdT + \sum_i \mu_i dN_i$$
• Enthalpy (H): $$dH = TdS + VdP + \sum_i \mu_i dN_i$$

### Thermodynamics beyond expansion

You should think of work in thermodynamics much beyond expansion. it could be, for instance, work against a spring: $$dU = TdS - k dx$$

It could be eletromagnetic work, with $$D$$ being the magnetic induction and $$H$$ $$H$$ magnetic field strenght, $$E$$ the electric field strenght and $$D$$ the electric displacement.

$$dU = TdS - PdV + Ed(DV) + Hd(BV) + \sum_{i} \mu dN_i$$

So you shouldn't think at the work as solely an effect on volume. Actually, heat is residual, it all that can't be thought as work. It is our limitation to deal and control the system, as the Heisemberg principle.

### References:

• De Groot and Mazur - "Non Equilibrium Thermodynamics"
• Tester and Modell - "Thermodynamics and its Applications"
• Koundepudi and Prigogine - "Modern Thermodynamics"
• Thank you so much. It will take time to understand everything that you wrote and I'll message you once I have done understanding it. But thank you so much! Commented Jul 3 at 13:13
• Nice answer, i would suggest to add a reference to Prigogine while talking about irreversible thermodynamics, he, along with De Donder, De Groot and Mazur, was a pioneer in the field! Commented Jul 3 at 18:29
• Good idea, Giuseppe. Commented Jul 3 at 18:37

The first Law sums the internal energy for a defined amount of matter at a defined state. It establishes that U, internal energy and several other properties specifically S, PV, T, and by extension the composites, H, enthalpy = U+PV; A = U-TS; and G = H-TS are also state functions. This means that a defined state has the same value for each of these functions regardless of how the state was established. The parameters needed to define a state: T, P, V, N etc. and methods of measuring these are many and are described in P Chem texts and engineering manuals and in a previous post. Suffice that they are done with care, diligence, and accuracy.

It is difficult to sum the exact internal energy of a system; the First Law say that it is a constant: the solution is to define the exact Energy by defining the energies of Standard states and working from there; again textbook and handbook material. The only actual calculations are entropies that have been calculated for many substances from 0Kelvin using the 3rd Law of Thermodynamics. Careful measurement of physical parameters such as P, T, V, N, controlling the physical processes and relating the measurements to a state function allows the determination of the state function at another state because the zero always subtracts out.

Things were great! The equivalence of heat and work were established: work could be turned into heat but heat could not be turned back into work at least not efficiently. There was another pesky problem. The first Law works on complete reactions: A = B. Most reactions behaved differently; they went to equilibrium. This problem as discussed by many chemists etc. Carnot, Descartes, Clausius, Helmholtz and Gibbs come to mind. Entropy entered the picture.

An equilibrium reaction proceeds in both directions; at equilibrium the rates are equal. In one direction the reaction is driven by loss of energy to the environs with an increase of entropy of the environs, in the other direction the reaction is drivenby gain of energy aand entropy by the system. Entropy change is best described by the increased distribution of energy and particles in the system.

Gibss and Helmholtz describe this with their functions. At equilibrium the free energy changes are zero DeltaA = U - TS = Zero, DeltaG = H -TS = Zero. The Helmholtz function is at constant volume, The Gibbs function is at constant pressure for easiest results. These functions evaluate the useful energy available from a given state to equilibrium.