Let two systems *S*_{1}and *S*_{2}be in thermal contact and in diffusive contact. By diffusive contact we mean that particles can flow from *S*_{l}to *S*_{2}and vice versa. We maintain a constant tby placing both systems in thermal contact with a large reservoir, *R*. We saw earlier that for a system in thermal equilibrium with a reservoir, the Helmholtz free energy *F* will be a minimum. Thus for the two systems in contact with *R*, we must have that *F*_{1} + *F*_{2} must be a minimum. The change in *F* = *F*_{1} + *F*_{2} as the number of particles varies is given by

But if we treat *S*_{l}+ *S*_{2}as a closed system, we have that *N*_{1} + *N*_{2} = *N* is a constant.

Thus *dN*_{1} = ‑*dN*_{2} and the requirement that *dF* = 0 becomes

(10.1)

We define the **chemical potential **of a system to be

(10.2)

Then the condition of equilibrium reduces to

m_{1}= m_{2} (10.3)

Strictly speaking, the true definition of mis in terms of the difference

m= *F*(t,*V*,*N*) – *F*(t,*V*,*N*-1)

since particles are a discrete set. However we are usually dealing with such large quantities of particles that we are able to treat mas a continuous function.

Example:

The chemical potential of an ideal gas is found from

*F*= -tln(*Z*_{n})

with *Z*_{n} = *Z*_{1}^{N} / *N*! and *Z*_{1} = *n*_{q}*V*. So

*F*= -t[*N* ln(*n*_{q}*V*) – *N* ln(*N*) + *N*]

and thus

(10.4)

## Connection to Potential Energy

To better understand what a chemical potential is, recall that the Helmhotz free energy is defined to be

*F*= *U* – ts

where *U* is the total energy of the system. If the system is subjected to an external force, this will give rise to either a potential energy or dissipative work. In either case, the Helmholtz free energy can be written as

*F*= *U*_{int} + *NU*_{ext} – ts

where the *N* comes from the fact that the external energy acts on each particle in the system. Now consider two systems *S*_{1}and *S*_{2}which are at the same temperature and are capable of exchanging particles, but which are not yet in diffusive equilibrium. We can assume that initially m_{2}> m_{1}, and we can denote the initial, non‑equilibrium difference in chemical potential as Dm_{i}= m_{2i}‑ m_{1i}. Now let a difference in potential energy be established between the systems, such that the potential energy of each particle in *S*_{1}is raised exactly by Dm_{i}. Such a difference could come from gravitational potential energy, or electrodynamical potential energy, for example.

By establishing a potential energy step, we are adding a potential energy *N*Dm_{i}to the energy of each state in *S*_{l}. Thus the final chemical potential energy of *S*_{1}is

(10.5)

since we didn’t change the energy in *S*_{2}. But now m_{lf}= m_{2f}, so the systems are in

diffusive equilibrium. Thus, we see that **the chemical potential is equivalent to a true potential energy in that the difference in chemical potential between two systems is equal to the potential barrier that will bring the two systems into diffusive equilibrium**. Since the chemical potential has the same properties as potential energy, we see that we cannot define an absolute chemical potential. **Only differences in chemical potential have any physical meanings**. From the argument above, we see that if an external potential step exists, the total chemical potential of a system is the sum

m_{tot}= m_{int}+ m_{ext}

where m_{ext}is the potential energy per particle in the external potential, and m_{int}is the chemical potential that would be present if the external potential were zero. Finally, since we can only physically discuss differences in chemical potential, the physical condition of diffusive equilibrium becomes

Dm_{ext}= Dm_{int} (10.6)

Example:

What is the condition for diffusive equilibrium of a column of ideal gas at temperature t?

Let *n*(0) be the number of particles in the bottom chamber, and *n*(*h*) be the number of particles in the top chamber. We want m(0) = m(*h*), where we must have m = m_{ext}+ m_{int}. Then from (10.4), we have that

(10.7)

Once we have the number of particles at a height *h*, we can determine the pressure by recalling that the pressure of an ideal gas is proportional to the concentration of particles,

*p*= *n*t

Thus the pressure of an ideal gas at height *h* is

(10.8)

Example:

What is the chemical potential of an ideal gas in a magnetic field *B*?

In a magnetic field we can view the gas as being made up of two types, or species, of particles; those with spin up, and those with spin down. If the spin is up, the potential energy is ‑*mB*, while the potential energy of the spin down particles is *mB*. For the container out of the magnetic field the chemical potentials are

while those in the magnetic field have

we also have that

and

Using the condition of diffusive equilibrium, we get

and

so

*n*(*B*) = *n*(0) cosh(*mB*/t)

Example:

For a lead acid battery, the relationship between chemical potentials and potential steps is even more apparent. In a normal lead acid battery, the negative electrode consists of metallic lead, Pb, and the positive electrode is a layer of lead oxide, Pb0_{2}, on a Pb substrate. The electrodes are immersed in diluted sulfuric acid, H_{2}SO_{4}, which is partially ionized into H^{+} ions and S0_{4}^{‑‑} ions.

In the discharge process, both the metallic Pb of the negative electrode and the Pb0_{2} of the positive electrode are converted to lead sulfate, PbS0_{4}, via the two reactions

Negative electrode:

Positive electrode:

Thus, the negative electrode acts as a sink for the S0_{4}^{—} ions, and the positive electrode acts as a sink for the H^{+} ions. If the battery terminals are not connected, electrons are depleted from the positive electrode and accumulate in the negative electrode, thereby charging them both. As a result, electrochemical potential steps develop at the electrode‑electrolyte interfaces, steps of exactly the correct magnitude to equalize the chemical potential steps and stop the diffusion of ions, which stops the chemical reactions from proceeding further. If an external current is permitted to flow, the reactions resume.

If we denote by DV_{‑} and DV_{+} the differences in electrostatic potential of the negative and positive electrodes relative to the common electrolyte, diffusion of the sulfate ions will stop when (remembering that the sulfate carries two negative charges)

-2eDV_{–} = Dm(SO_{4}^{—})

Similarly, diffusion of the H^{+} ions will stop when

eDV_{+} = Dm(H^{+})

The electrostatic potential as a function of position is seen to be

The two potentials, D*V*_{‑}and D*V*_{+}, are called **half‑cell potentials, **and have magnitudes of D*V*_{‑}= ‑0.4 V and D*V*_{+}= 1.6 V. Thus, the total electrostatic potential that develops across the cell in order to stop the diffusion reaction is

D*V*= D*V*_{+}‑ D*V*_{‑}= 2 V

This is the **open‑circuit voltage, **or EMF, of the battery.

## Chemical Potential and Entropy

Recall that we defined the chemical potential as

Substituting in the definition of *F*, we get

We can replace the right hand side of this by considering the following argument.

Let the entropy be a function of the independent variables *U*, *V* and *N*, s= s(*U*,*V*,*N*).

Then the differential of sis given by

Let *dV* = 0 for the process under consideration, and require the changes in *d*s, *dU* and *dN* be inter‑related in such a way that *d*t= 0. Thus we get that

or, after dividing by (*dN*)_{t}and using the definition of 1/t,

(10.9)

Returning to m, we can solve (10.9) for to yield

(10.10)

What is the difference between this expression and the one involving *F*? Recall that *F* is a function of the variables t, *V* and *N*, so m= m(t,*V*,*N*) in this form. In (10.10), m = m(*U*,*V*,*N*), i.e. it has a dependence on a different set of variables. A third formulation that we will not prove in class is

(10.11)

We can now derive a more general form of the thermodynamic identity we encountered earlier. Once again consider an infinitesimal change in entropy, where now the entropy depends on *U*, *V* and *N*

but, as before, we recall that , and . So the change in entropy becomes

or

(10.12)