Question

Finding conditions for Helmholtz free energy

5.8. The steps in the strategy for finding conditions for equilibrium are: a) Write an expression for the change in entropy of the system when it is taken through an arbitrary process. (b) Write the isolation constraints in differential form. (c) Use the isolation constraints to eliminate dependent variables in the expression for the entropy. (a) Collect terms. (e) Set the coefficients of each differential equal to zero. () Solve these equations for the conditions for equilibrium. Use the example of a unary two phase system presented in Sec. 5.4 to write out each of these steps mathematically.

Answer 1

Because the phase boundary separating two phases is a natural boundary, it is not possible to restrict the flow of matter between the phases. It is necessary to treat each phase as an open system.

The first step requires an expression for the change in entropy experienced by the system when it is taken through an arbitrary process. The entropy for the system is the sum of the entropies of its parts and therefore, the change in entropy of the system is

$dS=d(S^{\alpha }+S^{\beta })=dS^{\alpha }+dS^{\beta }$

where alpha denotes the properties of one phase, designated the alpha phase, and beta designates properties of the other phase called beta phase. It is necessary to compute the changes in entropy experienced separately by the alpha and beta parts of the system during and arbitrary process

Focus first on the behavior of the alpha phase. Its internal energy U is in general a function of its entropy S, volume V, and number of moles n

U = U(S, V, n)

From the combined statement of the first and second laws for the alpha phase

dU = T dS - P dV + u dn, this equation introduces the coefficient

$\mu ^{\alpha }=(\frac{\delta U}{\delta n})$ (Chemical potential)

Then look for an expression for the change in entropy of the alpha phase

$dS^{\alpha }=\frac{1}{T^{\alpha }}dU^{\alpha } + \frac{P^{\alpha }}{T^{\alpha }}dV^{\alpha }+\mu ^{\alpha }dn^{\alpha }$

Identical reasoning for the beta phase produces the analogous expression for the change in entropy of the beta phase

$dS^{\beta}=\frac{1}{T^{\beta }}dU^{\beta } + \frac{P^{\beta }}{T^{\beta }}dV^{\beta }+\mu ^{\beta }dn^{\beta }$

Finally the expression for the change in entropy of the system has six tems

$dS_{sys}=dS^{\alpha }+dS^{\beta }$

The second step. The boundary of an isolated system is rigid, thermally insulting, and impermeable. These characteristics place the following constraints upon the state functions that describe the two phase system

Because there are no exchanges with the surrondings, then by the first law, the internal energy of the system must remain constant.

$dU^{'}_{sys}=d(U^{\alpha }+U^{\beta })=dU^{\alpha }+dU^{\beta }$

Because the boundary is rigid

$dV^{'}_{sys}=d(V^{\alpha }+V^{\beta })=dV^{\alpha }+dV^{\beta }$

Because the boundary is impermeable

$dn^{'}_{sys}=d(n^{\alpha }+n^{\beta })=dn^{\alpha }+dn^{\beta }$

Internal energy, volume, and moles of materials may be exchanged between the phases but are convserved for the isolated system.

$dS^{'}_{sys}=dS^{\alpha }+dS^{\beta }$

$dS^{\alpha }=\frac{1}{T^{\alpha }}dU^{\alpha } + \frac{P^{\alpha }}{T^{\alpha }}dV^{\alpha }-\mu ^{\alpha }dn^{\alpha }+\frac{1}{T^{\beta }}(-dU^{\beta }) + \frac{P^{\beta }}{T^{\beta }}(-dV^{\beta })-\frac{\mu}{T^{\beta }} ^{\beta }(-dn^{\beta})$

Then, Collecting terms

$dS=\frac{1}{T^{\alpha }}-\frac{1}{T^{\beta }}dU^{\alpha }$.... the same expression for P and u

It describes teh change in entropy accompanying an arbitrary change in state of an unary two phase system

There are three conditions for equilibrum.

$\frac{1}{T^{\alpha }}-\frac{1}{T^{\beta }}=0$ Thermal equilibrium

$\frac{P^{\alpha }}{T^{\alpha }}-\frac{P^{\beta }}{T^{\beta }}=0$ Mechanical equilibrium

$\frac{\mu ^{\alpha }}{T^{\alpha }}-\frac{\mu ^{\beta }}{T^{\beta }}=0$ Chemical equilibrium

Its entropy is a maximum and the criterion for equilibrium is satisfied.

Stated in words, a unary two phase system is in equilibrium when the temperatures, pressures and chemical potentials of both phases are equal