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source: http://www.chem.arizona.edu/~salzmanr/480a/480ants/gibbspr/gibbspr.html

找了很久才找到真正詳細易懂的證明,

本文章適合修讀過熱力學,物理化學或材料熱力學等課程的人來閱讀 
 

Let's try to motivate the Gibbs phase rule. Consider a typical "one−component phase diagram".(單化合物相圖)

(P-T diagram; p indicates phase[在該點有幾相], and v indicates variance[變數的數目] )
 

    In the one−phase regions one can vary either the temperature, or the pressure, or both (within limits) without crossing a phase line. We say that in these regions there is a variance of 2. We have indicated in the solid, liquid, and gas regions that there is one phase and the variance is two. 

    Along a phase line we have two phases in equilibrium with each other, so on a phase line the number of phases is 2. However, if we want to stay on a phase line, we can't change the temperature and pressure arbitrarily. If we change the temperature − and keep two phases − then the pressure must change also to keep us on the phase line. Otherwise we go off the line and we no longer have two phases in equilibrium. So on a phase line the number of phases is 2, but the variance is 1.  

    At the triple point there are three phases in equilibrium, but there is only one point on the diagram where we can have three phases in equilibrium with each other. Therefore, at the triple point the variance is zero. 

    Notice that the variance seems to be related to the number of phases, such as, 

v = 3 − p, or                (1)

v = 2 + 1 − p.              (2)

 

Now let's look at another example. Consider the boiling diagram of a "two−component ideal liquid−liquid solution".

    In the one−phase regions (liquid or vapor) we have a variance of three because we can change the temperature, the composition, X, and the pressure (by moving in and out of the plane of the screen).

    In the two phase region the variance is only two because if we change the temperature the compositions of both the liquid and vapor phases must track as indicated by the tie−lines connecting the liquid and vapor composition lines. Here it looks like the variance satisfies, 

v = 4 − p, or                (3)

v = 2 + 2 − p.              (4)

Notice that equations 2 and 4 are the same except for the second term. In Equation 2 there is one component and the second term is 1. In Equation 4 there are two components and the second term in Equation 4 is a 2. It looks like the second term may indicate the number of components in the system. We might guess that the general form for Equations 2 and 4 might be,

v = 2 + c − p.

We shall see below whether or not our guess is correct.

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Before we try to derive a general form for the variance of a system in equilibrium we need to take a look as some mathematical principles.

Take a system where we have three variables, x, y, and z. Let's look at five cases:

Case 1. There is no equation connecting the variables. Then we can pick the values of all three variables to be anything we wish, independently of each other. There are no restrictions on the values of the variables and the variance is 3.

Case 2. There is one equation connecting the variables,

               . (5)

In this case we can select the values any two of the variables, but when we do, the value of the third is fixed by the equation. The number of equations is one and the variance is 2.

Case 3. There are two equations connecting the variables,

                (6a, b)

In this case we can select the value of any one variable at random, but then the values of the other two are fixed by the two equations. (Two equations in two unknowns is a soluble problem.) The number of equations is two and the variance is 1.

Case 4. There are three equations connecting the variables,

                (7a, b, c)

In this case we can not select any of the variables arbitrarily. The values of the variables are fixed because a system of three unknowns and three (linearly independent) equations has a unique solution.

                (8a, b, c)

The number of equations is three and the variance is zero.

Case 5. There are four linearly independent equations connecting the variables.

                (9,a, b, c, d)

This system does not have a solution. There are no values of the three variables which will satisfy all four equations. We conclude that we cannot have a variance less than zero.

The principle we extract here is that

variance = number of variables − number of equations ≥ 0.                (10)

Let us now apply this principle to phase equilibrium. Consider a system which is composed of c components. That is,

c = number of components.                (11)

(We define the number of components as the minimum number of independent substances (compounds and/or elements) required to form the system. If you can make one substance out of the others you can't count it!  For example, a system containing CO₂, CaO, and CaCO₃ has two components because you can form the CaCO₃ by reacting the other two. There is, however, a special case. If our system contains exactly the same number of moles of CaO and CO₂ it is a one component system because it can be made starting with just pure CaCO₃. That is, you can heat the CaCO₃ to generate the other two in equal molar amounts.)

 Number the components 1, 2, 3, . . . c.

Let

p = number of phases.                (12)

(We will have to be careful to distinguish between the number of phases and the pressure since we are using the same symbol.) 
Enumerate the phases by lower case Greek letters, α, β , γ , δ , . . . ζ. Let's now count the number of variables and the number of equations. We will get the variance from

v = variables − equations.                (13)

Our variables are T, p , (pressure) and the mole fractions of the components in each phase.

                (14)

Notice that there are p columns and c rows in the array of mole fraction variables. That is, there are cp mole fraction variables. This gives a total of 2 + cp variables.

Now let's count the equations. First of all the mole fractions in each phase must sum to unity. That is

                (15)

Since there is one such equation for each phase this gives p equations. Next, the chemical potential of each component must be the same in every phase. (In the following we designate ζ as the last phase, that is the phases are denoted α, β , γ , . . . , ζ.)

                (16)

(Note that there is no row of the form μ_1ζ   =   μ_1α, etc. Equations of this form are not linearly independent because they can be deduced from the other p − 1 equations.) So for this array of chemical potential equations there are c columns and p − 1 rows, which yields c(p − 1) equations. The total number of equations is then p + c(p − 1). If, as we have argued above, the variance is given by

v = variables − equations,                (13)

we get,

v = 2 + cp − p − c(p − 1)                (16)

v = 2 + c − p.                (1