MSE 201A Thermodynamics and Phase Transformations Fall, 2008 Problem Set No. 8. Given Gibbs generals condition of equilibrium as derived in the notes,

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1 MSE 201A hermodynamics and Phase ransformations Fall, 2008 Problem Set No. 8 Problem 1: (a) Let a homogeneous fluid with composition, {x}, be surrounded by an impermeable wall and in contact with a reservoir that fixes its temperature and pressure. Show that the fluid is in equilibrium with respect to the spontaneous formation of a phase with composition, {x'} so long as g' g + k µ k (x' k - x k ) 1.1 where g and g' are the molar Gibbs free energies of the parent and (hypothetical) product phases. Given Gibbs generals condition of equilibrium as derived in the notes, and using the identity, E' - S' + PV' - k µ k N' k g = e - s + Pv 1.3 and the integrated form for the energy, E, we have g' g + (-')s' - (P-P')v' + k µ k (x' k - x k ) 1.4 but, since and P are fixed by the reservoir, = ' and P = P' and 1.3 reduces to 1.1. (b) Show that a homogeneous fluid that is surrounded by an open wall is in equilibrium with respect to the spontaneous formation of a second phase if P' P 1.5 where P and P' are the pressures of the parent and (hypothetical) product phases. Given the condition, 1.2, and substituting the integrated form of the fundamental equation for E', we have ('-)S' + (P - P')V' + k (µ' k - µ k ) N' k 0 1.6

2 In an open system, and the {µ} are fixed by he reservoir. Hence the condition of global equilibrium is which is equivalent to eq (P - P')V' Problem 2: (a) Given a ternary fluid at fixed,p, show that the general condition of equilibrium has the consequence that two or more phases can coexist in equilibrium if and only if their states lie on a plane that is a common tangent to the free energy surfaces of the phases, which does not cut the free energy surface of any other phase. he composition of a ternary solution is specified by the mole fractions of two components (the "solutes"), x 1 and x 2. Since and P are fixed, equilibrium is governed by the Gibbs free energy. Equation 1.4 takes the form: g' g + g x (x' x 1 ) + g x (x' x 2 ) 2.1 which is the equation of a plane tangent to the free energy surface of a phase (å) at the point (,x å 1,x å 2). If this plane cuts the free energy surface of any other phase, the inequality is violated and å is out of equilibrium with respect to the formation of this phase. If this plane is tangent to the free energy surface of another phase ( ), it must touch that surface only at the single point (,x 1,x 2), and the inequality 2.1 is satisfied reciprocally for equilibrium between å of composition x å 1, x å 2 and of composition x 1, x 2. (b) Show that no more than three phases can be in equilibrium at arbitrary and P. Under what conditions can more than three phases coexist? Graphically, a three-phase equilibrium can always be found at, P such that three phases coexist, each of which has the least free energy for some values of x 1 and x 2. he compositions of the three states that are in equilibrium can be found graphically by constructing a common tangent to the free energy surfaces of phases å and, and then rolling the common tangent plane toward the free energy surface of phase until it just touches. It is not, ordinarily, possible for a fourth phase to exist in equilibrium with these three. he tangent plane is fixed by its contact with the three free energy surfaces. A fourth phase is in equilibrium only if its free energy surface is tangent to this fixed plane. his result is in agreement with Gibbs' phase rule, which asserts that a three-phase equilibrium in a three-component system has two degrees of freedom; if and P are fixed, none are left. If, however, the temperature is allowed to vary, there will ordinarily be at least one value of at which four phases are in equilibrium. he simplest visualization is when the free energy surface of a fourth phase passes through a three-phase equilibrium on heating Page 2

3 or cooling. At one temperature, the fourth free energy surface just touches the common tangent to the other three, producing a four-phase equilibrium. he phase rule allows five-phase equilibrium when both and P are varied. o visualize how this can happen, let the system have a four-phase equilibrium, and vary with P so that four-phase equilibrium is preserved. At some P, the fundamental surface of a fifth phase contacts the common tangent plane for four-phase equilibrium, producing a five-phase equilibrium. his happens, of course, at a unique value of P, in accordance with the phase rule. Problem 3: A Van der Waals fluid (in its modern form) has the thermal equation of state P = - F V = N V - Nb - N2 a V and the caloric equation of state F E = 2 = 3 2 N - N2 a V 3.2 (a) Find the fundamental equation in the form F(,V,N). Integrating equation 3.1, F = - Nln(V-Nb) - N2 a V + F'(N,) 3.3 while integrating eq. 3.2 gives F = Nln() a -N2 V + F"(N,V) 3.4 Comparing the two, F = Nln() - Nln(V-Nb) - N2 a V + G(N) 3.5 where G(N) is a function of N only. he function, G(N), can be evaluated from the fact that F is a linear homogeneous function of N: It follows that F(,V,N) = Nf(,v) 3.6 Page 3

4 G(N) = Nln(N) + Nζ 3.7 where ζ is a constant. Hence the free energy per mole is, f(,v) = ln() - ln(v-b) - a v + Ω 3.8 (b) Show that there is a critical temperature, c, such that the fluid is stable everywhere on any isotherm in the P-v plane for which > c, but has instabilities for < c. Show that the critical point on the isotherm, c, falls at a point (P c, v c ) in the (P,v) plane. Find expressions for c, P c and v c. he critical state of a Van der Waals fluid is determined by the equations P v = 2 P v 2 = and has the coordinates c = 8a 27b 3.10 P c = a 27b v c = 3b 3.12 he condition of stability for the fluid on an isotherm in the P-v plane is that its compressibility not be negative, or - P v 1 = v = (v-b) 2-2a v he instability points occur at zeros of the left-hand side of he homogeneous equation is a cubic equation in v that can be solved, but has no simple solution. We can best approach the solution by expanding about the critical point, as was done in Section 11.4 of the notes. Equation shows that the instability points near c, v c are the solutions to the equation Îv 0 = (v 0 - v c ) = ± - P 2 Î 3P where the coefficients P 2 and P 3 are evaluated at the critical point: Page 4

5 P 2 = P v = 1 (v c - b) 2 = 1 4b P 3 = 1 2 P vvv = - a 162b Hence v 0 = 3b ± - 81b 3 Î 2a 3.17 his equation only has solutions for Δ < 0, that is, the fluid has no instabilities near v c for > c, but has two symmetric instabilities for < c and Δ small. (c) Given eq. 3.1, describe how you would find the boundaries of the two-phase region and the region of instability in the p-v plane. (You do not have to do the calculations, just set them up.) An example of an isotherm of the Van der Waals gas is shown in Fig his particular isotherm is for the case *=/ c = 0.8, that is, the temperature is below the critical temperature. he plot is in terms of the reduced pressure and volume, as described in section of the notes. Note that the curve has two extrema. hese are the points at which - P v 1 = v = (v-b) 2-2a v 3 = and are, hence, the points at which the fluid becomes unstable. he instabilities (spinodal points) must lie within the two-phase region. o find the boundaries of the two-phase region at this temperature we note that when the liquid and vapor are in equilibrium, their pressures and their chemical potentials must be equal: P L = P V 3.19 µ L = µ V 3.20 Hence the two equilibrium states must lie on a line of constant pressure, suh as tne blue line in the figure. he condition of constant chemical potential can be applied as follows. he change in chemical potential with the pressure, P, is given by Page 5

6 µ P = v 3.21 where v is the specific volume. Fig. 8.1: P-v isotherm for the Van der Waals fluid at a temperature of *=0.8 (relative to c ). he reduced pressure and volum are plotted. Note that a constant pressure line passes through the isotherm at three points: v L, v 0 and v v, where v 0, the intermediate intersection, is within the unstable segment of the curve. We find µ(v) by integration: µ(v) µ(v L ) = P(v) vdp 3.22 P(v L ) By inspection, this integral is the net area under the curve v(p) (it may help to visualize this if you turn the graph on its side). It follows that [µ L (v 0 ) - µ L (v L )] is just the area bounded by the isotherm and the blue line in the figure between v L and v 0, and is positive since P is increasing along the segment of the isotherm at larger v. Similarly, [µ L (v v ) - µ L (v 0 )] is the area bounded by the isotherm and the blue line in the figure between v 0 and Page 6

7 v v, and is negative since P is decreasing along the segment at higher v. When the two areas are equal, so [µ L (v v ) - µ L (v 0 )] = - [µ L (v 0 ) - µ L (v L )] 3.23 µ L (v v ) = µ L (v L )] 3.24 It follows that we can find the equilibrium volumes at graphically by moving a horizontal, constant pressure line vertically until the two areas are the same. Since this method is correct for any, we can generate the equilibrium lines that bound the twophase region. (d) How would you compute the tensile strength of the fluid? he tensile strength of the fluid is the pressure at which it breaks, or spontaneously cavitates. his minimum possible pressure is the instability limit on the fluid side, a reduced pressure of about in the example shown in Fig. 8.1 Problem 4: Assume that the phase diagram of a binary solution contains a miscibility gap in which a high-temperature solution (α) separates into two solutions (α +α ) with the same structure but different compositions. Find the instability points and equilibrium compositions of the two phases just below the critical point at the top of the miscibility gap. From Sec of the Notes) he critical point lies at the top of the miscibility gap in the -x phase diagram, at the temperature, c, and composition, x c. Above c a solution with the critical concentration, x c, is a stable, homogeneous solution. Below c the homogeneous solution decomposes into two solutions with different compositions. In order for a miscibility gap to appear in the phase diagram of a binary solution the two phases must differ only in composition; they must, therefore, have the same structure. he general thermodynamic criteria for a critical state in a binary solution follow from the general theory of crtitical points given in Chapter 16. Since the solution is stable in the immediate vicinity of the critical point, the second derivative of g(x) vanishes there (creating the instability), but to ensure stability in the limit that drops to c the third derivative must also vanish and the fourth derivative must be positive. In terms of the relative chemical potential, these conditions are: µ x = Page 7

8 Let 2 µ x 2 = µ x 3 > x = x - x c 4.4 = - c 4.5 hen, expanding the compositional derivative of the relative chemical potential about ( c, x c ) and keeping only the lead terms in and x we have µ x = 1 ( ) + 2 ( x) where 1 and 2 are constant coefficients. It follows from equations that both 1 and 2 are positive. Integrating equation 4.6 gives the relative chemical potential near the critical point µ(,x) = µ( c,x c ) + 1 ( x) ( x) he locus of instabilities near the critical point can be found from equation 4.1. Since the equation is quadratic in x, there are two solutions to the homogeneous equation µ x = When > 0 both solutions are imaginary; the binary solution is stable for all concentrations near x c. When < 0 instabilities occur at x i = ± which shows that the system must separate into two distinct phases. he compositions of the two phases that are in equilibrium at < 0 (call them å' and å") are fixed by the condition µ(,x å ) = µ(,x å ) 4.9 Page 8

9 Since µ(,x) is symmetric about x c, x å' = - x å" and x å' = - x å" = It follows that the miscibility gap has a parabolic shape near the critical point, and that a parabolic instability or spinodal curve lies within the miscibility gap and converges at the critical point. Problem 5: where Let a binary solution have the fundamental equation g(x) = g'(x) + [ x ln(x) + (1 x)ln(1 x) ] 5.1 g' ( x) = µ 0 1 x + µ 0 2 (1 x) + h 1 x(1 x) 2 + h 2 x 2 (1 x) 5.2 (a) Find the chemical potentials of the two species and show that they have the correct forms in the dilute solution limits. o find the chemical potentials we use the relation g(x) = µ 1 x + µ 2 = g x x + µ Ignoring the logarithmic terms (we can add then in later), we have from which it follows that when x ~ 0, µ = (µ 1 0 µ 2 0 ) + h 1 (1 x)(1 3x) + h 2 x(2 3x) 5.4 µ 2 = g µ x = µ x 2 [ 2h 1 (1 x) + h 2 (2x 1) ] + ln(1 x) 5.5 µ 1 = µ + µ 2 = µ (1 x) 2 [ h 1 (1 2x) + h 2 (2x)] + ln x 5.6 µ 1 = µ h 1 + ln x µ 2 = µ ln(1 x) 5.7 when x ~ 1, Page 9

10 µ 1 = µ ln x µ 2 = µ h 2 + ln(1 x) 5.8 showing that the potentials have the correct forms in both limits. (b) Find the condition of stability with respect to changing composition. he condition of stability is µ x = From eq. 5.4, x µ = (µ 0 1 µ 0 2 ) + h 1 (1 x)(1 3x) + h 2 x(2 3x) + ln 1 x 5.8 and hence µ x = 2(h 2 2h 1 ) + 6x(h 1 h 2 ) + x(1 x) 5.9 his is a cubic equation, whose homogeneous form can have three solutions for given. he only solutions that matter are those for positive with x in the range (0,1). It can be easily shown (though you don t have to) that, for given, h 1 and h 2 there are either two or zero real solutions in the range (0,1). he former case shows the existence of a miscibility gap. (c) Assume h1=h2=h. Show that the solution has a miscibility gap at low temperature if and only if h > 0. Derive expressions for the critical temperature at the top of the miscibility gap, and the instability and equilibrium phase boundaries at lower. In this case, from eq µ x = 2h + x(1 x) 5.10 which defines instabilities at 2hx(1 x) = 5.11 If h 0 this has no solutions with > 0. When h >0 there are solutions for low. Since the maximum of x(1-x) falls at x=0.5, the solutions are confined to the range Page 10

11 0 < c = h For < c there are two solutions, symmetric about x=0.5. In the limit ~0 the solutions converge on the values 0,1. As ~ c the two values converge toward x=0.5. It follows that this case yields a -x phase diagram that has a symmetric spinodal gap with a peak at c, and branches that go to x=0 and 1 as ~0, in agreement with the hird Law. (d) Why do you only expect immiscibility if h > 0, and then only at lower? When h>0 the enthalpy of mixing is positive. Essentially, the two species in the solution would prefer to bond to themselves rather than to one another. his leads to chemical segregation at low temperature where the enthalpy is dominant, with solution at high temperature where the entropy is dominant. When h<0 the species prefer to bond to one another, which leads to the formation of ordered compounds at low. Page 11

Chapter 17: One- Component Fluids

Chapter 17: One- Component Fluids Chapter 17: One-Component Fluids...374 17.1 Introduction...374 17.2 he Fundamental Surface...375 17.2.1 he fundamental equation...375 17.2.2 Behavior at a stability limit...377