Helmholtz free energy

Helmholtz free energy

In thermodynamics, the Helmholtz free energy is a thermodynamic potential which measures the “useful” work obtainable from a closed thermodynamic system at a constant temperature and volume. For such a system, the negative of the difference in the Helmholtz energy is equal to the maximum amount of work extractable from a thermodynamic process in which temperature and volume are held constant. Under these conditions, it is minimized at equilibrium. The Helmholtz free energy was developed by Hermann von Helmholtz and is usually denoted by the letter A (from the German “Arbeit” or work), or the letter F . The IUPAC recommends the letter "A" as well as the use of name "Helmholtz energy"; [cite book | title = Gold Book | publisher = IUPAC | url = http://www.iupac.org/goldbook/H02772.pdf | accessdate = 2007-11-04] . In physics, the letter F is usually used to denote the Helmholtz energy, which is often referred to as the "Helmholtz function" or simply “free energy”.

While Gibbs free energy is most commonly used as a measure of thermodynamic potential, especially in the field of chemistry, the isobaric restriction on that quantity is sometimes inconvenient for some applications. For example, in explosives research, Helmholtz free energy is often used since explosive reactions by their nature induce pressure changes. It is also frequently used to define so-called fundamental equations of state in accurate correlations of thermodynamic properties of pure substances.

Definition

The Helmholtz energy is defined as:

:A equiv U-TS, [Levine, Ira. N. (1978). "Physical Chemistry" McGraw Hill: University of Brooklyn]

where

* "A" is the Helmholtz free energy (SI: joules, CGS: ergs),
* "U" is the internal energy of the system (SI: joules, CGS: ergs),
* "T" is the absolute temperature (kelvins),
* "S" is the entropy (SI: joules per kelvin, CGS: ergs per kelvin).The Helmholtz energy is the Legendre transform with respect to the entropy, S, of the fundamental relation in the energy representation, U(S,V,N). The natural variables of A are T,V,N.

Mathematical development

From the first law of thermodynamics we have:

: { m d}U = delta Q - delta W,

where U is the internal energy, delta Q is the energy added by heating and delta W is the work done by the system. From the second law of thermodynamics, for a reversible process we may say that delta Q=T{ m d}S. Also, in case of a reversible change, the work done can be expressed as delta W = p dV

Differentiating the expression for A we have:

:{ m d}A = { m d}U - (T{ m d}S + S{ m d}T),:= (T{ m d}S - p,{ m d}V) - T{ m d}S - S{ m d}T,:= - p,{ m d}V - S{ m d}T,

This relation is also valid for a process which is not reversible because A is a thermodynamic function of state.

Minimum free energy and maximum work principles

The laws of thermodynamics are only directly applicable to systems in thermal equilibrium. If we wish to describe phenomena like chemical reactions, then the best we can do is to consider suitably chosen initial and final states in which the system is in (metastable) thermal equilibrium. If the system is kept at fixed volume and is in contact with a heat bath at some constant temperature, then we can reason as follows.

Since the thermodynamical variables of the system are well defined in the initial state and the final state, the internal energy increase, Delta U, the entropy increase Delta S, and the work performed by the system, W, are well defined quantities. Conservation of energy implies:

:Delta U_{ ext{bath + Delta U + W = 0,

The volume of the system is kept constant. This means that the volume of the heat bath does not change either and we can conclude that the heat bath does not perform any work. This implies that the amount of heat that flows into the heat bath is given by:

:Q_{ ext{bath = Delta U_{ ext{bath =-left(Delta U + W ight) ,

The heat bath remains in thermal equilibrium at temperature T no matter what the system does. Therefore the entropy change of the heat bath is:

:Delta S_{ ext{bath = frac{Q_{ ext{bath}{T}=-frac{Delta U + W}{T} ,

The total entropy change is thus given by:

:Delta S_{ ext{bath +Delta S= -frac{Delta U -TDelta S+ W}{T} ,

Since the system is in thermal equilibrium with the heat bath in the initial and the final states, T is also the temperature of the system in these states. The fact that the system's temperature does not change allows us to express the numerator as the free energy change of the system:

:Delta S_{ ext{bath +Delta S=-frac{Delta A+ W}{T} ,

Since the total change in entropy must always be larger or equal to zero, we obtain the inequality:

:Wleq -Delta A,

If no work is extracted from the system then

:Delta Aleq 0,

We see that for a system kept at constant temperature and volume, the total free energy during a spontaneous change can only decrease, that the total amount of work that can be extracted is limited by the free energy decrease, and that increasing the free energy requires work to be done on the system.

This result seems to contradict the equation dA = -S dT - P dV, as keeping T and V constant seems to imply dA = 0 and hence F = ext{ constant}. In reality there is no contradiction. After the spontaneous change the system, as described by thermodynamics, is a different system with a different free energy function than it was before the spontaneous change. We can thus say that Delta A= A_{2} - A_{1}leq 0 where the A_{i} are different thermodynamic functions of state.

One can imagine that the spontaneous change is carried out in a sequence of infinitesimally small steps. To describe such a system thermodynamically, one needs to enlarge the thermodynamical state space of the system. In case of a chemical reaction one would need to specify the number of particles of each type. The differential of the free energy then generalizes to:

:dA = -S dT - p dV + sum_{j}mu_{j}dN_{j},

where the N_{j} are the numbers of particles of type j and the mu_{j} are the corresponding chemical potentials. This equation is then again valid for both reversible and non-reversible changes. In case of a spontaneous change at constant T and V, the last term will thus be negative.

In case there are other external parameters the above equation generalizes to:

:dA = -S dT - sum_{i}X_{i}dx_{i} +sum_{j}mu_{j}dN_{j},

Here the x_{i} are the external variables and the X_{i} the corresponding generalized forces.

Relation to the partition function

A system kept at constant temperature is described by the canonical ensemble. The probability to find the system in some energy eigenstate "r" is given by:

:P_{r}= frac{e^{-eta E_r{Z},

where

:etaequivfrac{1}{k T},

:E_{r}= ext{ energy of eigenstate }r,

:Z = sum_{r} e^{-eta E_{r,

"Z" is called the partition function of the system. The fact that the system does not have a unique energy means that the various thermodynamical quantities must be defined as expectation values. In the thermodynamical limit of infinite system size, the relative fluctuations in these averages will go to zero.

The internal energy of the system is the expectation value of the energy and can be expressed in terms of "Z" as follows:

:U = sum_{r}P_{r}E_{r}= -frac{partial log Z}{partial eta},

If the system is in state "r", then the generalized force corresponding to an external variable "x" is given by

:X_{r} = -frac{partial E_{r{partial x},

The thermal average of this can be written as:

:X = sum_{r}P_{r}X_{r}=frac{1}{eta}frac{partial log Z}{partial x},

Suppose the system has one external variable "x". Then changing the system's temperature parameter by deta and the external variable by "dx" will lead to a change in log Z:

:dleft(log Z ight)= frac{partiallog Z}{partialeta}deta + frac{partiallog Z}{partial x}dx = -U,deta + eta X,dx,

If we write U,deta as:

:U,deta = dleft(eta U ight) - eta, dU,

we get:

:dleft(log Z ight)=-dleft(eta U ight) + eta, dU+ eta X ,dx,

This means that the change in the internal energy is given by:

:dU =frac{1}{eta}dleft(log Z+eta U ight) - X,dx ,

In the thermodynamic limit, the fundamental thermodynamic relation should hold:

:dU = T, dS - X, dx,

This then implies that the entropy of the system is given by:

:S = klog Z + frac{U}{T} + c,

where "c" is some constant. The value of "c" can be determind by considering the limit "T" → 0. In this limit the entropy becomes S = k log Omega_{0} where Omega_{0} is the ground state degeneracy. The partition function in this limit is Omega_{0}e^{-eta U_{0 where U_{0} is the ground state energy. We thus see that c = 0 and that:

:A = -kTlogleft(Z ight),

Bogoliubov inequality

Computing the free energy is an intractable problem for all but the simplest models in statistical physics. A powerful approximation method is mean field theory which is a variational method based on the Bogoliubov inequality. This inequality can be formulated as follows.

Suppose we replace the real Hamiltonian H of the model by a trial Hamiltonian ilde{H} which has different interactions and may depend on extra parameters that are not present in the original model. If we choose this trial Hamiltonian such that

:leftlangle ilde{H} ight angle =leftlangle H ight angle,

where both averages are taken w.r.t. the canonical distribution defined by the trial Hamiltonian ilde{H}, then

:Aleq ilde{A},

where A is the free energy of the original Hamiltonian and ilde{A} is the free energy of the trial Hamiltonian. By including a large number of parameters in the trial Hamiltonian and minimizing the free energy we can expect to get a close approximation to the exact free energy.

The Bogoliubov inequality is often formulated in a sightly different but equivalent way. If we write the Hamiltonian as:

:H = H_{0} + Delta H,

where H_{0} is exactly solvable, then we can apply the above inequality by defining

: ilde{H} = H_{0} + leftlangleDelta H ight angle_{0},

Here we have defined leftlangle X ight angle_{0} to be the average of X over the canonical ensemble defined by H_{0}. Since ilde{H} defined this way differs from H_{0} by a constant, we have in general

:leftlangle X ight angle_{0} =leftlangle X ight angle,

Therefore

:leftlangle ilde{H} ight angle = leftlangle H_{0} + leftlangleDelta H ight angle ight angle =leftlangle H ight angle,

And thus the inequality

:Aleq ilde{A},

holds. The free energy ilde{A} is the free energy of the model defined by H_{0} plus leftlangleDelta H ight angle. This means that

: ilde{A}=leftlangle H_{0} ight angle_{0} - T S_{0} + leftlangleDelta H ight angle_{0}=leftlangle H ight angle_{0} - T S_{0},

and thus:

:Aleq leftlangle H ight angle_{0} - T S_{0} ,

Proof

For a classical model we can prove the Bogoliubov inequality as follows. We denote the canonical probability distributions for the Hamiltonian and the trial Hamiltonian by P_{r} and ilde{P}_{r}, respectively. The inequality:

:sum_{r} ilde{P}_{r}logleft( ilde{P}_{r} ight)geq sum_{r} ilde{P}_{r}logleft(P_{r} ight) ,

then holds. To see this, consider the difference between the l.h.s. and the r.h.s.. We can write this as:

:sum_{r} ilde{P}_{r}logleft(frac{ ilde{P}_{r{P_{r ight) ,

Since

:logleft(x ight)geq 1 - frac{1}{x},

it follows that:

:sum_{r} ilde{P}_{r}logleft(frac{ ilde{P}_{r{P_{r ight)geq sum_{r}left( ilde{P}_{r} - P_{r} ight) = 0 ,

where in the last step we have used that both probability distributions are normalized to 1.

We can write the inequality as:

:leftlanglelogleft( ilde{P}_{r} ight) ight anglegeq leftlanglelogleft(P_{r} ight) ight angle,

where the averages are taken w.r.t. ilde{P}_{r}. If we now substitute in here the expressions for the probability distributions:

:P_{r}=frac{expleft [-eta Hleft(r ight) ight] }{Z},

and

: ilde{P}_{r}=frac{expleft [-eta ilde{H}left(r ight) ight] }{ ilde{Z,

we get:

:leftlangle -eta ilde{H} - logleft( ilde{Z} ight) ight anglegeq leftlangle -eta H - logleft(Z ight) ight angle

Since the averages of H and ilde{H} are, by assumption, identical we have:

:Aleq ilde{A}

Here we have used that the partition functions are constants w.r.t. taking averages and that the free energy is proportional to minus the logarithm of the partition function.

We can easily generalize this proof to the case of quantum mechanical models. We denote the eigenstates of ilde{H} by left|r ight angle. We denote the diagonal components of the density matrices for the canonical distributions for H and ilde{H} in this basis as:

:P_{r}=leftlangle rleft|frac{expleft [-eta H ight] }{Z} ight|r ight angle,

and

: ilde{P}_{r}=leftlangle rleft|frac{expleft [-eta ilde{H} ight] }{ ilde{Z ight|r ight angle=frac{expleft(-eta ilde{E}_{r} ight)}{ ilde{Z,

where the ilde{E}_{r} are the eigenvalues of ilde{H}

We assume again that the averages of H and ilde{H} in the canonical ensemble defined by ilde{H} are the same:

:leftlangle ilde{H} ight angle = leftlangle H ight angle ,

where:leftlangle H ight angle = sum_{r} ilde{P}_{r}leftlangle rleft|H ight|r ight angle,

The inequality

:sum_{r} ilde{P}_{r}logleft( ilde{P}_{r} ight)geq sum_{r} ilde{P}_{r}logleft(P_{r} ight) ,

still holds as both the P_{r} and the ilde{P}_{r} sum to 1. On the l.h.s. we can replace:

:logleft( ilde{P}_{r} ight)= -eta ilde{E}_{r} - logleft( ilde{Z} ight),

On the r.h.s. we can use the inequality

:leftlangleexpleft(X ight) ight angle_{r}geqexpleft(leftlangle X ight angle_{r} ight),where we have introduced the notation

:leftlangle Y ight angle_{r}equivleftlangle rleft|Y ight|r ight angle,

for the expectation value of the operator Y in the state r. See here for a proof. Taking the logarithm of this inequality gives:

:logleft [leftlangleexpleft(X ight) ight angle_{r} ight] geqleftlangle X ight angle_{r},

This allows us to write:

:logleft(P_{r} ight)=logleft [leftlangleexpleft(-eta H - logleft(Z ight) ight) ight angle_{r} ight] geqleftlangle -eta H - logleft(Z ight) ight angle_{r},

The fact that the averages of H and ilde{H} are the same then leads to the same conclusion as in the classical case:

:Aleq ilde{A}

Generalized Helmholtz energy

In the more general case, the mechanical term (p{ m d}V) must be replaced by the product of the volume times the stress times an infinitesimal strain: [cite book
last = Landau
first = L. D.
authorlink = Lev Landau
coauthors = Lifshitz, E. M.
languange = English
others = (Translated from Russian by J.B. Sykes and W.H. Reid)
year = 1986
title = Theory of Elasticity (Course of Theoretical Physics Volume 7)
edition = Third ed.
publisher = Butterworth Heinemann
location = Boston, MA
id = ISBN 0-7506-2633-X
]

:{ m d}A = Vsum_{ij}sigma_{ij},{ m d}varepsilon_{ij} - S{ m d}T + sum_i mu_i ,{ m d}N_i,

where sigma_{ij} is the stress tensor, and varepsilon_{ij} is the strain tensor. In the case of linear elastic materials which obey Hooke's Law, the stress is related to the strain by:

:sigma_{ij}=C_{ijkl}epsilon_{kl}

where we are now using Einstein notation for the tensors, in which repeated indices in a product are summed. We may integrate the expression for { m d}A to obtain the Helmholtz energy:

:A = frac{1}{2}VC_{ijkl}epsilon_{kl}^2 - ST + sum_i mu_i N_i,

: = frac{1}{2}Vsigma_{ij}epsilon_{ij} - ST + sum_i mu_i N_i,

Application to fundamental equations of state

The Helmholtz free energy function for a pure substance (together with its partial derivatives) can be used to determine all other thermodynamic properties for the substance. See, for example, the equations of state for water, as given by the IAPWS in their [http://www.iapws.org/relguide/IAPWS95.pdf IAPWS-95] release.

See also

* Gibbs free energy for thermodynamics history overview and discussion of "free energy"
* Grand potential
* Work content - for applications to chemistry
* Statistical mechanics
* This page details the Helmholtz energy from the point of view of thermal and statistical physics.

References

Further reading

*Atkins' "Physical Chemistry", 7th edition, by Peter Atkins and Julio de Paula, Oxford University Press


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