NENG 301 Week 8 Unary Heterogeneous Systems (DeHoff, Chap. 7, Chap )
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1 NENG 301 Week 8 Unary Heterogeneous Systems (DeHoff, Chap. 7, Chap )
2 Learning objectives for Chapter 7 At the end of this chapter you will be able to: Understand the general features of a unary phase diagram Understand the relationship between chemical potential and Gibbs free energy Appreciate how chemical potential surfaces can be used to illustrate the state of equilibrium between phases Understand the differences between stable, unstable and metastable equilibria Derive and understand the significance of the Clausius- Clapyeron (C-C) equation and successfully apply it to systems of physical interest Construct unary phase diagrams by integrating the C-C equation and using database information on thermodynamic quantities
3 Phase, Phase Equilibrium and Phase Diagram A phase is a region of space, throughout which all physical properties of a material are essentially uniform. A simple description is that a phase is a region of material that is chemically uniform, physically distinct, and (often) mechanically separable. In a system consisting of ice and water in a glass jar, the ice cubes are one phase, the water is a second phase, and the humid air over the water is a third phase. The glass of the jar is another separate phase. The term phase is sometimes used as a synonym for state of matter. Also, the term phase is sometimes used to refer to a set of equilibrium states demarcated in terms of state variables such as pressure and temperature by a phase boundary on a phase diagram. However, the state of matter and phase diagram usages are not commensurate with the formal definition given above and the intended meaning must be determined in part from the context in which the term is used. Phase Equilibrium: Left to equilibrate, many compositions will form a uniform single phase mixture, but depending on the temperature and pressure even a single substance may separate into two or more distinct phases. Within each phase, the properties are uniform but between the two phase properties differ. Water in a closed jar with an air space over it forms a two phase system.
4 General features of unary phase diagrams Some general characteristics of unary phase equilibria: Typically phase stability is pictured on a P-T diagram (but remember, we can now convert back and forth between thermodynamic state variables!) The domain of stability for a given phase is represented by an area on a P-T diagram A line between two areas represents the conditions where two phases can coexist (a phase boundary) The coexistence of three phases simultaneously is represented as a point (a triple point) No more than three phases can coexist under the same conditions of temperature and pressure we will prove this later (the Gibbs phase rule)
5 Principles of Maximum Entropy, Minimum Energy/Enthalpy/Helmholtz/Gibbs Free Energies: Criteria for Spontaneous Change In an isolated system the entropy function increases during every spontaneous change In a system constrained to constant entropy and volume the internal energy function decreases during every spontaneous change In a system constrained to constant entropy and pressure the direction of spontaneous change is monitored by a decrease in the enthalpy function In a system constrained to constant temperature and volume the Helmholtz free energy function decreases during every spontaneous change In a system constrained to constant temperature and pressure the Gibbs free energy function decreases during every spontaneous change
6 Criteria & Constraints for Equilibrium Equilibrium Criterion S is a maximum at equilibrium U is a minimum at equilibrium H is a minimum at equilibrium F is a minimum at equilibrium G is a minimum at equilibrium Constant U, V, n S, V, n S, P, n T, V, n T, P, n
7 Condition for equilibrium: An introduction to constraints The condition for maximum entropy means that there are particular values of state variables (x and y, for example) that represent the condition for equilibrium From Chapter 4: we know that there are relations between these variables This brings up the issue of constraints the relations between variables may alter the maximum achievable value of entropy
8 Condition for equilibrium: introduction to constraints The figure shows the impact of constraints The function y = y(x) gives the relation between the two variables Because of this constraint, the maximum value of z(x,y) that can be obtained is z 1 = z(x 1,y 1 ) rather than z max = z(x max,y max ) Obviously: we need to know how to find the minimum or maximum of a function when it is subject to a constraint
9 Dealing with constraints Start with a function (in this case: two variables) z = z(x,y) z z The differential is easy: dz dx dy x y y x The extreme value occurs on the surface z = z(x,y) where the derivatives are simultaneously zero: z z 0 ; 0 x y y x If x and y are not independent, then we will have to apply the constraint: z = z(x,y) where y = y(x) With two variables we substitute so that : z = z(x,y) z(x) Take the derivative, set the coefficient to zero and solve: dz z z dx x x 0
10 A very simple example of constraints Determine the length and width of a rectangle with the maximum area that can be surrounded by a fence 20m long Length (x) and width (y) The area = z(x,y) = xy Introduce the constraint: 2x + 2y = 20 or y(x) = 10 x Express the dependent variable in terms of the independent variable: z(x,y) = xy = x (10 x) = 10x x 2 Form the differential: dz = (10-2x) dx Set the coefficient to zero and solve: 10-2x = 0 or x = 5
11 How to apply this approach to the determination of thermodynamic equilibrium This process can be used to determine the conditions for thermodynamic equilibrium were you use the fact that the entropy will be maximized at equilibrium: 1. Write a differential equation for the change in entropy that the system may experience when taken through an arbitrary process 2. Write differential expressions that describe the constraints that apply to the system (assume it is isolated from its surroundings) 3. Use the isolation constraints to eliminate dependent variables in the description for the change in entropy for the system 4. Set the coefficients of each of the differentials in the expression for ds equal to zero the resultant equations give the conditions for equilibrium for the system
12 Extensive Properties (Z ) versus Molar Properties (Z) We will start with a unary (single component) two-phase nonreacting system (i.e., ice/water) In what will follow below, we/dehoff will use a prime ( ) to denote an extensive quantity, an unprimed variable corresponds to the intensive (molar) quantity Thus for a unary two phase system In general, for a multicomponent, multiphase (n-phase) system: Z Z Z Z sys Z Z Z sys n
13 The chemical potential The chemical potential plays a central role in the thermodynamics of materials systems In words, the chemical potential gives the change in energy that is associated with the transfer of matter into or out of a phase For the time being we will define the chemical potential of a phase using: U remember: du = T ds P dv n S, V You can view this as the change of energy that accompanies the transfer of matter into a phase when everything else is held constant: an infinitesimal amount of a component into one mole of phase
14 Application of the Entropy Criterion for Equilibrium The expression for entropy change of the system: ds sys = ds + ds Using combined 1 st & 2 nd laws: Rearrange: Sum: du = T ds - P dv + dn du = T ds - P dv + dn ds = du /T + P dv /T - dn /T ds = du /T + P dv /T - dn /T ds sys = ds + ds = du /T + P dv /T - dn /T + du /T + P dv /T dn /T
15 Application of the Entropy Criterion for Equilibrium Now consider the constraints for an isolated system: du = -du ; dv -dv ; dn -dn Eliminate variables: ds sys = [1/T -1/ T ] du + [P /T -P /T ] dv - [ /T - /T ] dn Solve for conditions for equilibrium: Thermal: Mechanical: Chemical: T = T P = P = The first two are no surprise, but the last one will continue to have importance to this: thermodynamic equilibrium implies the equality of chemical potential across all phases
16 Application of the Internal Energy Criterion for Equilibrium Definition: du = dq + dw = T ds - P dv Expression for internal energy change of the system: Using combined 1 st & 2 nd laws: du sys = du + du du = T ds - P dv + dn du = T ds - P dv + dn Sum: du sys = du + du = T ds - P dv + dn + T ds - P dv + dn
17 Application of the Internal Energy Criterion for Equilibrium Constraints on variables (S, V, n) ds = -ds ; dv -dv ; dn -dn Eliminate variables: du sys = [T - T ] ds + [V - V ] dv [ - ] dn Solve for conditions for equilibrium: Thermal: T = T Mechanical: P = P Chemical: =
18 Enthalpy Criterion for Equilibrium Definition: H = U + PV dh = T ds + V dp Expression for enthalpy change of the system: dh sys = dh + dh Using combined 1 st & 2 nd laws: dh = T ds + V dp + dn dh = T ds + V dp + dn Sum: dh sys = dh + dh = T ds + V dp + dn + T ds / + V dp + dn
19 Enthalpy Criterion for Equilibrium Constraints on variables: S, P, n ds = -ds dp dp 0 dn -dn Eliminate variables: dh sys = [T - T ] ds + [V V ] dp [ - ] dn 0 Solve for conditions for equilibrium: Thermal: T = T Mechanical: P = P (assumed) Chemical: =
20 Helmholtz Free Energy Criterion for Equilibrium Definition: F = U TS df = - S dt - P dv Expression for Helmholtz F.E. change of the system: df sys = df + df Using combined 1 st & 2 nd laws: df = S dt + P dv + dn df = S dt + P dv + dn Sum: df sys = df + df = S dt + P dv + dn + S dt + P dv + dn
21 Helmholtz Free Energy Criterion for Equilibrium Constraints on variables T, V, n dt = dt 0 dv -dv dn -dn Eliminate variables: df sys = [S S ] dt + [P - P ] dv [ - ] dn 0 Solve for conditions for equilibrium: Thermal: T = T (assumed); Mechanical: P = P Chemical: =
22 Chemical potential and the Gibbs free energy We have already defined the chemical potential as the change in energy associated with the transfer of matter into or out of a phase: Now use the definition of Gibbs f.e.: G = U + PV - TS Differentiate this expression: dg = du + PdV + V dp - TdS - S dt Compare with: So we get: du = TdS - PdV + dn du = TdS - PdV + dn dg = - S dt + V dp + dn or primed: extensive quantity unprimed: intensive (molar) quantity G n PT,
23 Gibbs Free Energy Criterion for Equilibrium Definition: G U + PV - TS = H - TS dg = - S dt + V dp Expression for free energy change of the system: dg sys = dg + dg Using combined 1 st & 2 nd laws: dg = -S dt + V dp + dn dg = - S dt + V dp + dn Sum: dg sys = dg + dg = -S dt + V dp + dn - S dt + V dp + dn
24 Gibbs Free Energy Criterion for Equilibrium Constraints on variables T, P, n dt = dt 0 dp dp 0 dn -dn Eliminate variables: dg sys =- [S S ] dt + [V V ] dp [ - ] dn 0 0 Solve for conditions for equilibrium: Thermal: T = T (assumed); Mechanical: P P (assumed) Chemical: =
25 Overview of the problem at hand Now that we know how to describe the conditions for equilibrium, let s start applying that knowledge to real systems Start with unary systems: a single chemical component The most convenient way to look at equilibrium between phases is in a P-T diagram Consider the cases of copper (left) and carbon (right)
26 Metastable to stable equilibrium: Carbon Diamond Graphite
27
28 Stable, metastable, and unstable equilibrium Metastable: Can remain in its current configuration until sufficient force applied, at which time it can transform to stable Unstable: Perfectly balanced, but an infinitesimal force will transform it to metastable or stable Stable: the lowest energy configuration with no change 28
29 Metastable to stable equilibrium Rather than moving blocks, we are more interested in moving atoms Let A and B denote two different configuration of atoms, and assume that the Gibbs free energy as a function of position is shown by the red line For configurations at both A and B, dg = 0 (the derivative of the Gibbs F.E. is zero) Configuration B represents a metastable equilibrium; Configuration A represents the stable minimum Note that the transition from B to A will require the system overcome an energy barrier Gibbs free energy G dg = 0 B dg = 0 A 29
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