H.2 Euler-Maclaur in Sum Formula
9.3 Phase Equilibrium and Miscibility Gap
Armed with a knowledge off, we shall now use several methods to examine the conditions for phase equilibrium, particularly the representation in thev,pplane of the coexistence curve (in theT,pplane) for these phases. Coexistence in thev,pplane is represented by two regions, one in which the equilibrium state is a single phase and the other, known as themiscibility gap, in which the equilibrium state is a composite system consisting of two phases. The spinodal curve derived above lies entirely within the miscibility gap except at the critical point where the two intersect. Outside the miscibility gap, the fluid is stable; between the miscibility gap and the spinodal, the fluid is metastable; and within the spinodal, it is unstable. We shall use several methods to illustrate these points.
9.3.1 Common Tangent Construction
Thecommon tangent constructionis a useful method that provides a graphical solution to phase equilibrium problems. We develop it in general, and then apply it specifically to the van der Waals fluid.
We consider a composite system at uniform temperature T and consisting of two homogeneous phases of the same substance, one having mole numberN1, volumeV1, and molar volumev1 = V1/N1 and the other havingN2,V2 and v2 = V2/N2. The total Helmholtz free energy of the system is
F=N1f(T,v1)+N2f(T,v2). (9.19) If these phases are in equilibrium,Fmust be a minimum with respect to changes of the internal extensivevariablesN1,V1,N2,V2subject to the constraints N1+N2 =constant, andV1+V2=constant.
We first holdN1andN2constant and set the differential dF=0 to obtain N1
∂f(T,v1)
∂v1
T
∂v1
∂V1
N1
dV1+N2
∂f(T,v2)
∂v2
T
∂v2
∂V2
N2
dV2=0. (9.20) Since(∂v1/∂V1)N1 = 1/N1,(∂v2/∂V2)N2 = 1/N2, and dV1 = −dV2 from a constraint, Eq. (9.20) becomes
∂f(T,v1)
∂v1
T
=∂f(T,v2)
∂v2
T
. (9.21)
128 THERMAL PHYSICS
In view of the left-hand equality in Eq. (9.12), Eq. (9.21) is recognized as equality of pressure for two phases at the same temperature, one having molar volumev1and the other having molar volumev2. Thus Eq. (9.21) could be rewritten
p(T,v1)=p(T,v2). (9.22)
This result is not to be unexpected!
Next, we holdV1andV2constant and set the differentialdF =0 to obtain
f(T,v1)+N1
∂f(T,v1)
∂v1
T
∂v1
∂N1
V1
dN1+
f(T,v2)+N2
∂f(T,v2)
∂v2
T
∂v2
∂N2
V2
dN2=0.
(9.23) Since(∂v1/∂N1)V1= −V1/N12,(∂v2/∂N2)V2= −V2/N22, and dN1= −dN2from a constraint, Eq. (9.23) becomes
f(T,v1)−
∂f(T,v1)
∂v1
T
v1=f(T,v2)−
∂f(T,v2)
∂v2
T
v2. (9.24)
We identify the members on the left-hand and right-hand sides of Eq. (9.24) as chemical potentials, that is,
f(T,v)−
∂f(T,v)
∂v
T
v =f(T,v)+p(T,v)v =:μ(T,v). (9.25) This enables Eq. (9.24) to be rewritten
μ(T,v1)=μ(T,v2). (9.26)
From general considerations, Eq. (9.26) is also not to be unexpected!
We seem to have labored to obtain what amounts to Eqs. (9.22) and (9.26), which we might have just written down from general considerations. Nevertheless, the chemical potentialμ, which for a monocomponent system is equal to the Gibbs free energy per mole,g, is ordinarily regarded as a function ofT andp, notTandv. The variablesTand v are the natural variables off, notμ. We therefore return to Eqs. (9.21) and (9.24) and establish the following geometrical interpretation: According to Eq. (9.21), a graph off versusvhas the same slope at two values ofv, namely atv1andv2. There can be many pairs ofv1andv2for which this is true. But either the left or the right member of Eq. (9.24) can be interpreted as theintercept, on thefaxis (v =0), of a tangent to a graph offversusvat v1andv2. So Eq. (9.21) requires parallel tangents and Eq. (9.24) requires equal intercepts. It follows that the simultaneous solution to Eqs. (9.21) and (9.24) requires acommon tangent atv1andv2, as illustrated inFigure 9–5a.
Another important feature ofFigure 9–5a is noteworthy. A composite system consisting of the two phases having molar volumesv1andv2at its common tangent has a total molar free energy that lies along the tangent line joining the points of tangency. To see this, consider first a homogeneous system consisting ofNmoles and having molar volumev∗. For our composite system to have the same volume, we would needN1 +N2 = N and
f μ
v∗ v2
v1
(a)
f
vS1 vS2v2 v1
A B C
D
(b)
FIGURE 9–5 (a) Common tangent construction and (b) chord construction. The curves represent the Helmholtz free energy per mole,f =F/N, versus the molar volumev. For the common tangent construction, the dashed line is tangent to the curve atv1andv2. Its slope is the negative of the common pressure and its intercept isμ, the common value of the chemical potential. The free energy per mole of a composite state having total molar volumev∗lies along the tangent line atv∗and is lower than the free energy of a single phase havingv∗. Hence, the composite system is more stable than a homogeneous system. The chord construction can be used to investigate the local or global stability of a homogeneous phase. If the chord lies above the curve, as for the chordAB, the homogeneous phases along the curveABare stable with respect to a composite, whereas if it lies below the curve, as for the chord CD, the homogeneous phases along the curveCDare unstable.vS1andvS2mark the spinodal pointsS1andS2at which∂2f/∂v2=0.
N1v1+N2v2=Nv∗. This results inN1/N =(v2−v∗)/(v2−v1)andN2/N =(v∗−v1)/(v2−v1) which is known as thelever rule. Inserting these values in Eq. (9.19) gives
F/N = [(v2−v∗)/(v2−v1)]f(T,v1)+ [(v∗−v1)/(v2−v1)]f(T,v2) (9.27)
=f(T,v1)+ [(v∗−v1)/(v2−v1)][f(T,v2)−f(T,v1)], composite system.
Comparing this value with that for the homogeneous system having molar volumev∗, we see that the free energy of the composite system is lowerfor all v∗ betweenv1 andv2. Thus, for these values ofv∗, thecomposite system is the equilibrium state.Put another way, given the opportunity, the homogeneous system will decompose to form the equilibrium composite system consisting of two phases that are in equilibrium with each other.
9.3.2 Chord Construction
We can also use the reasoning that led to Eq. (9.27) to establish another valuable construc- tion which we shall call thechord construction. Indeed, Eq. (9.27) is still valid ifv1andv2
correspond to any two points along the curve, provided only thatv1<v2. We can therefore apply it to various points along the curve, such as the pairABor the pairCD, as illustrated inFigure 9–5b. For chordAB, the free energy of a composite lies along the chord, which is above the curveAB, so a homogeneous phase along the curveABis stable with respect to a composite consisting of its end points. But forCD, the free energy of a composite lies on a chord off that is belowf, so the single phase is unstable with respect to that particular
130 THERMAL PHYSICS
composite. Any homogeneous state that lies above the chord corresponding to the com- mon tangent is unstable, but some of those states are locally stable. Local stability or in- stability requires the chord construction to be applied to neighboring points. The spinodal points where−∂p/∂v = ∂2f/∂v2 = 0 separate locally stable states from locally unstable states. States that are locally stable but globally unstable are said to bemetastable.
9.3.3 Summary for f (v) Curves
We can summarize this situation as follows: With respect to a curve off versus v for a monocomponent system, portions of the curve that are convex (as viewed from below) correspond to locally stable states; portions of the curve that are concave (as viewed from below) correspond to locally unstable states. All states that lie above the common tangent betweenv1andv2are ultimately unstable. Thus there are three kinds of states:
unstable locally unstable (locally concave) and also above the common tangent be- tweenv1andv2; therefore, locally unstable and globally unstable;
metastable locally stable (locally convex) but above the common tangent betweenv1and v2; therefore, locally stable but globally unstable;
stable locally stable (locally convex) but outside the common tangent region, i.e., v ≤v1orv ≥v2; therefore, locally stable and globally stable.
The concave and convex regions are separated by points,S1andS2, at which the second partial derivative (∂2f/∂v2)T = 0. But since (∂2f/∂v2)T = −(∂p/∂v)T, these points also correspond to maxima and minima ofp, which means that they correspond to the spinodal curve in thev,pplane. Thus the spinodal curve separates the metastable region from the unstable region. On the other hand, the locus of the points of common tangency in thev,pplane separate the stable region from the metastable region; the region inside this curve is called themiscibility gapbecause within it a compositemixtureof phases located at the ends of the common tangent, rather than a homogeneous phase, is globally stable.
9.3.4 Explicit Equations for van der Waals Miscibility Gap
For the van der Waals fluid, the explicit forms of Eqs. (9.21) and (9.24) are RT
v1−b− a v12 = RT
v2−b− a
v22 (9.28)
and
−2a
v1 −RTln(v1/b−1)+ RTv1
v1−b= −2a
v2 −RTln(v2/b−1)+ RTv2
v2−b. (9.29) For a given value of T, the simultaneous equations (9.28) and (9.29) can be solved for v1 and v2 and thenpcan be evaluated by using Eq. (9.2). In principle, such a solution determines the shape of the miscibility gap, but these equations would have to be solved numerically because they are not tractable analytically. A graphical representation of their
1 2 3 4 -2.2
-1.8 -1.6
f/(RTc) f/(RTc)
v/vc
(a)
2 4 6 8 10 12 14
-2.4 -2.2 -1.8 -1.6 -1.4
v/vc
(b)
FIGURE 9–6 Graphical representation of simultaneous solution of Eqs. (9.28) and (9.29). (a)T/Tc=27/32,v/vc= 0.548 for the liquid and 3.241 for the vapor. The pressure isp/pc=0.183. (b)T/Tc=20/32,v/vc=0.440 for the liquid and 13.585 for the vapor. The pressure isp/pc=0.0411.
solution is presented inFigure 9–6. Since the liquid and vapor have quite different molar volumes, the curve off becomes quite flat at large values ofv and such a solution is not very practical. We therefore turn to other methods to demonstrate the nature of these solutions.