# Thermodynamic potentials

# Basics

From Callan, we have two equivalent principles of thermodynamics:.

**Entropy maximum principle**. The equilibrium value of any unconstrained internal parameter is such as to maximize the entropy for the given value of the total internal energy.**Energy minimum principle**. The equilibrium value of any unconstrained internal parameter is such as to minimize the energy for the given value of the total entropy.1

The basic formula for the internal energy as a function of S and V (and the chemical potentials, Ni) is

$latex U = TS – pV – \mu N$

where N stands for the sum over different chemical potentials

$latex \sum_{i} \mu_{i}N_{i}$

representing the quasi-static chemical work and

$latex dU = TdS – pdV – \mu dN$

Depending on the problem at hand, It may be simpler to solve for the equilibrium state if the formula had different independent variables. That is what Legendre transformations do.

# Legendre transformations

Given a formula

$latex Y = Y(X)$

then

$latex P=\frac{\partial Y}{\partial X}$

is the slope of the curve at a given point. We would like an equivalent representation of the system but one where the independent variable is P rather than X. We will use the intercept of the Y axis of a tangent of constant slope at X.

Then the slope is given by

$latex P = \frac{Y-\psi}{X-0}$

which can be arranged to give the **Legendre transformatio**n

$latex \Psi = Y – PX$

Then

$latex d\Psi = dY – PdX -XdP = -XdP$

or

$latex -X = \frac{\partial \Psi}{\partial P}$

so the inverse transformation is just

$latex Y = XP + \Psi$

In general, for

$latex Y = Y(X_0,X_2…X_i)$

the partial slope of this hypersurface is given by

$latex P_k = \frac{\partial Y}{\partial X_k}$

and the Legendre transformation is

$latex \Psi = Y -\sum_{k} P_kX_k$

Example from Callan, problem 5.2-1:

$latex y = \frac{x^2}{10}$ gives $latex P = \frac{\partial Y}{\partial X} = \frac{x}{5}$

so

$latex \Psi (P) = y – Px = -\frac{5}{2}P^2$

Inversely,

$latex -X = \frac{\partial \Psi}{\partial P} = -5P$

so

$latex Y(X) = \Psi + XP = \frac{X^2}{10}$

# Principle thermodynamic potentials

The most used thermodynamic potentials are the following and are all Legendre transforms of the internal energy U as shown in the table.

Name | Symbol | Formula | Natural variables | Constant | Interpretation |

Internal energy | U | TS – pV + μN_{i} |
S, V, N_{i} |
– | Total internal energy |

Helmholtz free energy | F (or A) | U – TS | T, V, N_{i} |
T | Max energy at constant T, V after system pays “entropy tax” |

Enthalpy | H | U + pV | S, p, N_{i} |
p | Total heat (caloric) energy availabe in system at const. P after V change |

Gibbs free energy | G | U + pV -TS | T, p, N_{i} |
T and p | Max energy available at const. T, P after V change and “S tax” |

We can identify each term with a type of energy:

- TS with heat (or caloric) energy,
- pV with volume-change energy (as when a piston is pushed) and
- the sum $latex \sum_{i} \mu_{i}N_{i}$ as the energy of chemical bonding.

We will ignore such things as electrical energy.

The above-cited energy-minimum principle applies to all these potentials under specific circumstances.

**Helmholtz Potential Minimum Principle**. The equilibrium value of any unconstrained internal parameter in a system in diathermal contact with a heat reservoir minimizes the Helmholtz potential over the manifold of states for which T = Tr [at constant T].[ref]Callan, 155[/ref]

**Enthalpy Minimum Principle**. The equilibrium value of any unconstrained internal parameter in a system in contact with a pressure reservoir minimizes the enthalpy over the manifold of states of constant pressure (equal to that of the pressure reservoir).[ref]Callan, 156[/ref]

**Gibbs Potential Minimum Principle**. The equilibrium value of any unconstrained internal parameter in a system in contact with a thermal and a pressure reservoir minimizes the Gibbs potential at constant temperature and pressure (equal to those of the respective reservoirs).[ref]Callan, 157[/ref]

We can attribute physical meanings to these potentials.

In the case of enthalpy, a system at constant P can expand or contract, therefore losing energy to move around molecules. This volume-change energy is exactly what is subtracted from U in order to derive H. So enthalpy is the energy released as heat by the system at constant P. Or, in the other direction,

…heat added to a system at constant pressure and at constant values of all the remaining extensive parameters (other than S and V) appears as an increase in the enthalpy.[ref]Callan, 161[/ref]

As for the Helmholtz free energy,

… the work delivered in a reversible process, by a system in contact with a thermal reservoir [so at constant T], is equal to the decrease in the Helmholtz potential of the system.[ref]Callan, 159[/ref]

The Helmholtz free energy is thus the available work at constant temperature and volume. The term TS represents the “entropy tax” which the system must pay in order for the total entropy of the universe to remain zero.

The Gibbs free energy is useful when T and P are constant, which is the usual case of chemical reactions open to the atmosphere, which acts as a reservoir of T and P. Gibbs free energy is thus much beloved of chemists. Callan gives a more interesting example, pointing out that it may also be true “… in a small subsystem of a larger system that acts as both a thermal and a pressure reservoir (as in the fermentation of a grape in a large wine vat).”[ref]Callan, 167[/ref]

# Spontaneous reactions

Suppose there is only one thermodynamic system in the universe and that its T and V are constant. Then the sum of the entropy changes of the system and of its surroundings is the entropy change of the universe.

$latex \Delta S_{univ} =\Delta S_{surr} +\Delta S_{sys}$

If the energy of the system changes by an amount U, then the surroundings will change by -U, and

$latex \Delta S_{surr} = – \frac{\Delta U}{T}$

so

$latex \Delta U_{univ} = \Delta S_{sys} – \frac{-\Delta U}{T}$

which means

$latex \Delta U_{univ} = T\Delta SU_{univ} = -(\Delta U -T\Delta S) = -\Delta F$

So the entropy maximum principle says that for constant T and V, the change in the Helmholtz free energy must be negative, so that $latex \Delta S_{univ} = \frac{\Delta U_{univ}}{T} > 0$.

Similarly, if only a spontaneous chemical reaction occurs at constant T and P, then a change in heat energy (enthalpy) of the system means the surroundings change by the negative of that., i.e., $latex -\Delta H_{sys}$. So equation (1) gives

$latex \Delta S_{univ} = -\frac{\Delta H_{sys}}{T} + \Delta S_{sys}$

The only change in the universe is due to the dispersal of energy in a quantity we may call -ΔG. Therefore

$latex \Delta S_{univ} = -\frac{\Delta G}{T} = -\frac{-\Delta H_{sys}}{T} + \Delta S_{sys}$

which may be rewritten as

$latex \Delta G = \Delta H_{sys} – T\Delta S_{sys}$

which is the Gibbs free energy.

# Bibliography

Atkins, Peter, *The laws of thermodynamics: A very short introduction*. Oxford: Oxford University Press, 2010. Print.

Callan, Herbert B. *Thermodynamics and an introduction to thermostatistics*. New York: John Wiley and Sons. 1985, 2005. Print.

Shankar, R. *Fundamentals of Physics. Mechanics, Relativity, and Thermodynamics*. New Haven: Yale UP, 2014. Print.