# Gibbs Free Energy of Thermal Physics in Lecture Notes

The Gibbs free energy is defined to be

G = U ‑ ts+ pV                                              (15.1)

Chemists often call this the free energy, while physicists often call it the thermodynamic potential.  The most important property of the Gibbs free energy is that it is a minimum for a system in equilibrium at constant pressure when in thermal contact with a reservoir.  In order to see this, consider the differential dG,

dG = dU ‑ tds‑ sdt+ pdV + Vdp                                (15.2)

If the system, S, is in thermal contact with a heat reservoir, R1, at temperature tand in mechanical contact with a pressure reservoir, R2, that can maintain the pressure p but cannot exchange heat, then dt= dp = 0.  So dG becomes

dG = dU ‑ tds+ pdV                                          (15.3)

From the thermodynamic identity

tds= dU ‑ mdN+ pdV

we see that

dG= mdN

but dN = 0, so dG = 0, which is the condition of an extremum.  The fact that G is a minimum follows directly from the fact that the entropy has a minus sign associated with it.  Also, from the derivation, we see that G = G(t,p,N).  The general differential of the Gibbs free energy is

Comparing this with (15.2) and using the thermodynamic identity, we can immediately see that

(15.4)

(15.5)

(15.6)

## Intensive and Extensive Quantities

The variables tand p are called intensive quantities;they do not change value when two identical systems are put together.  The variables U, s, V, N and G are called extensive quantities;their values change when two identical systems are put together.  For example, G depends on the number of particles, N.  When two systems are brought together, then the number of particles for the combined system doubles, so the Gibbs free energy also doubles.  Since the Gibbs free energy depends linearly on the number of particles, we can write

G= N j(p,t)

Thus we see that

But we already saw that , so j(p,t) = m, and we get

G = N m(p,t)                                                 (15.7)

Thus, the chemical potential for a system is equal to the Gibbs free energy per particle.  If more than one particle species is present, (15.7) becomes

(15.8)

With this change in definition, the differential dG becomes

(15.9)

## Gibbs Free Energy and Helmholtz Free Energy

What is the difference between the Gibbs free energy and the Helmholtz free energy?  Consider an ideal gas.  We have already seen that the chemical potential for an ideal gas as a function of N, tand V is given by

From the definition of the Helmholtz free energy we know that

Holding tand V constant and integrating, we get

Thus, F is not directly proportional to N if we keep the temperature and volume constant.  From the definition of the Gibbs free energy and the definition of the Helmholtz free energy we can immediately see that

since pV = Ntfor an ideal gas.  Thus

(15.10)

From this we see that the chemical potential as a function of tand p is

(15.11)

## Chemical Reactions

How do we use the Gibbs free energy in a chemical reaction?  We can write a chemical reaction as

(15.12)

where Aj is the jth chemical species, and njis the coefficient of the jth species in the reaction.  For example, in the reaction

H2 + Cl2 = 2 HCl

we have A1 = H2, A2 = C12, A3 = HCl, n1= 1, n2= 1 and n3= ‑2.  We usually discuss chemical equilibrium for reactions that occur under conditions of constant pressure and temperature.  Recalling that the change in the Gibbs free energy is

this reduces to

The change dNj is proportional to the coefficient nj.  This allows us to write

dNj= njdx

where dx indicates how many times the reaction takes place.  Thus dG becomes

But in equilibrium dG = 0, so thus reduces to the condition

(15.13)

## Equilibrium Reactions in an Ideal Gas

If we treat each of the molecules as an ideal gas, we can cast this into a more familiar form.  Recall that the chemical potential of an ideal gas is

mj= t(ln nj ‑ ln cj)

where cj = nq,jZint,j.  Substituting this into the equilibrium condition, we get

This can be rewritten as

or

(15.14)

where  is called the equilibrium constant.  Exponentiating both sides of (15.14), we get the law of mass action,

(15.15)

Example:

Consider the reaction

The law of mass action yields

where FH(int) = ‑I = ‑13.6 eV and I is the ionization potential.  Here [A] is the concentration of the molecule A.  The quantum concentration for H and H + is the same, nq(H) = nq(H+), so the mass action equation reduces to

If charge neutrality is preserved, [e] = [H+], so this finally becomes

(15.16)

This is called the Saha equation.

## Reaction Speed

The law of mass action expresses the condition satisfied by the concentrations once the reaction has gone to equilibrium, but it tells nothing about how fast the reaction proceeds.  Consider the reaction

with concentrations nA, nB and nC = nAB.  What is dnAB/dt?  We can write

where C(t) and D(t) are constant with respect to rate.  In equilibrium, dnAB/dt = 0, so

In general, a graph of the potential energy required for a reaction looks like

Here DEis called the activation energy.  This is the potential energy required for the reaction to take place.  DHis called the heat of reaction and is the energy that evolves out of the reaction.

## Isotherms

One of the uses of the Gibbs free energy is to plot the curve of pressure verses volume for a quantity of matter at constant temperature.  This curve is called an isotherm.

Different regions of the isotherm correspond to different forms of the matter, e.g. solid, liquid, or gaseous.  The phase is a portion of a system that is uniform in composition.  Two phases may coexist, with a definite boundary between them.  Liquid and vapor (a gas that is in equilibrium with its liquid or solid form) may coexist on a section of an isotherm only if the temperature of the isotherm lies below a critical temperature, tc.   Above that critical temperature only a single phase ‑ the fluid phase ‑ exists, no matter how great the pressure.  Consider a system that is originally only a liquid at a constant temperature.

Raise the piston.  As the piston is raised, more vapor is formed until there is only vapor in the chamber.  Plotting the pressure verses volume for this transformation we get

The thermodynamic conditions for the coexistence of two phases are the conditions for the equilibrium of two systems that are in thermal, diffusive and mechanical contact.  These conditions are that

(15.17)

At a general point in the p‑tplane the two phases do not coexist: if m1< m2, the first phase alone is stable, and if ml> m2the second phase alone is stable.

We see that ml(p,t) = mg(p,t), so there must exist some form of coexistence curve.   This is the curve that divides the phases on a ptdiagram.

Consider a small segment of the curve.  Then the condition for coexistence is that

(15.18)

and

Since the changes are small, we can expand the second condition to get

(15.19)

Subtracting (15.18) from (15.19) and rearranging the terms,

(15.20)

Now recall that the Gibbs free energy could be written as

If we define the volume and entropy per molecule as v = V/N and s = s/N respectively, then

and

so (15.20) becomes

(15.21)

Note that this derivative refers to the very special interdependent change of p and tin which the gas and liquid continue to coexist.  The number of molecules in each phase will vary as the volume is varied, subject only to the condition that Nl + Ng = N.

## Enthalpy

Recall that the quantity of heat added to a system was related to the entropy by

Thus the quantity of heat added by the transfer of one molecule is

(15.22)

where L is called the latent heat of vaporization.  If we write vgvl = Dv, then (15.21) can be rewritten as

(15.23)

This is known as the Clausius‑Clapeyron equation, or the vapor pressure equation

Finally, the latent heat of a phase transition is, as we have seen, equal to t times the entropy difference of the two phases at constant pressure.  It is also equal to the difference in the enthalpy,H, of the two phases, where H = U + pV.  To see this, consider the differential

by the thermodynamic identity.  But at constant pressure this becomes

dH= tds+ mdN

Considering the change across the coexistence curve, we see that the last term in dH becomes (mg‑ ml)DN, while  and .  But mg= mlon the curve, so

(15.24)

Using the definition of the heat capacity at constant pressure, this can be written in a more useful form.  The heat capacity at constant pressure is given by

Integrating this yields

(15.25)

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