314: Thermodynamics of Materials

Transcription

1 314: hermodynamics of Materials. auhon, K. R. Shull September 30, 2016 Contents Contents : hermodynamics of Materials 4 2 Introduction 4 3 Big Ideas of hermodynamics Classification of Systems hermodynamic State Variables rocess Variables Work Done by an External Force Work Done by an Applied ressure Extensive and Intensive roperties he aws of hermodynamics st aw of hermodynamics nd aw of hermodynamics Reversible and Irreversible rocesses Reversible rocess Irreversible rocess rd aw of hermodynamics Combined statement of 1st and 2nd aws hermodynamic Variables and Relations Enthalpy Helmholtz Free Energy Gibbs Free Energy Other Material roperties hermal expansion coefficient Compressibility Heat Capacity Coefficient Relations General Strategy for Deriving Relations

2 5.6.1 Summary of Relations Derivation he Ideal Gas Specific Heat of Ideal Gas Internal Energy of Ideal Gas U(,) Examples Using Ideal Gas Solids and iquids 27 8 Conditions for Equilibrium Equilibrium in thermodynamic system Finding equilibrium conditions Minima of energies as conditions of equilibrium Maxwell Relations From Microstates to Entropy Microstates and Macrostates ostulate of Equal ikelihood he Boltzmann Hypothesis (to be derived) Conditions for equilibrium Noninteracting (Ideal) Gas Ideal gas aw Diatomic Gases he Einstein Model of a Crystal hermodynamic roperties of a Crystal Unary Heterogeneous Systems hase Diagrams Chemical otential Surfaces Calculation of µ(, ) Surfaces Clasius-Clapeyron Equation Open Multicomponent Systems artial Molal roperties of Open Multicomponent Systems he Mixing rocess: Calculating changes in energy Graphical Determination of artial Molal roperties Chemical otential in Multicomponent Systems roperties of ideal solutions (e.g. ideal gas mixtures) Behavior of Dilute Solutions Case Excess roperties, relative to ideal Beyond the ideal solution: the regular solution Mixtures of Real Gases: Fugacity Multicomponent Heterogeneous Systems Conditions for Equilibrium Gibbs hase Rule

3 16 hase Diagrams Unary hase Diagrams Binary hase Diagrams Interpreting hase Diagrams: ie ines he ever Rule Applications of hase Diagrams: Microstructure Evolution Review ypes of α phase diagrams hermodynamics of hase Diagrams Calculation of G mix from pure component reference states he Common angent Construction A Monotonic wo-hase Field Non-monotonic wo-hase Field roperties of a regular solution Misciblity gap: appearance upon cooling Spinodal decomposition: concave down region in G drives unmixing and uphill diffusion Summary of Chapters 2 and Chapter Chapter Chapter Chapter Chapter Chapter Chapter roblems 89 Contents hases and Components Intensive and Extensive roperties Differential Quantities and State Functions Entropy hermodynamic Data emperature Equilibration Statistical hermodynamics Single Component hermodynamics Mulitcomponent hermodynamics Computational Exercises 95 Nomenclature 97 Index 98 3

4 Catalog Description Classical and statistical thermodynamics; entropy and energy functions in liquid and solid solutions, and their applications to phase equilibria. ectures, problem solving. Materials science and engineering degree candidates may not take this course for credit with or after CHEM (new stuff) Course Outcomes 1 314: hermodynamics of Materials At the conclusion of the course students will be able to: 1. articulate the fundamental laws of thermodynamics and use them in basic problem solving. 2. discriminate between classical and statistical approaches. 3. describe the thermal behavior of solid materials, including phase transitions. 4. use thermodynamics to describe order-disorder transformations in materials. 5. apply solution thermodynamics for describing liquid and solid solution behavior. 2 Introduction he content and notation of these notes is based on the excellent book, hermodynamics in Materials Science, by Robert DeHoff [?]. he notes provided here are a complement to this text, and are provided so that the core curriculum of the undergradaute Materials Science and Engineering program can be found in one place ( hese notes are designed to be self-contained, but are more compact than a full text book. Readers are referred to DeHoff s book for a more complete treatment. 3 Big Ideas of hermodynamics he following statements sum up some of foundational ideas of thermodynamics: Energy is conserved Entropy is produced emperature has an absolute reference at 0 - at which all substances have the same entropy 4

5 3.1 Classification of Systems Systems can be classified by various different properties such as constituent phases and chemical components, as described below. Chemical components Unary system has one chemical component Multicomponent system has more than one chemical component Figure 3.1: Unary system (left) and multicomponent system (right). hases Homogeneous system has one phase Heterogeneous system has more than one phase ice Figure 3.2: Homogeneous system (left) and heterogeneous system (right). Energy and mass transfer: closed and open systems: An isolated system cannot change energy or matter with its surroundings. Closed system can exchange energy but not matter with its surroundings. Open system can exchange both energy and matter with its surroundings. Surroundings Surroundings Surroundings Energy Energy System System System Matter OEN COSED ISOAED Figure 3.3: Open, closed and isolated thermodynamic systems. 5

6 AB AB C AB C C AB+C AB+C B AB A AB C C C BC A A A B B C C AB+C A+BC Figure 3.4: Non-reacting system (left) and reacting system (right). Reactivity Non-reacting system contains components that do not chemically react with each other Reacting system contains components that chemically react with each other Another classification is between simple and complex systems. For instance, complex systems may involve surfaces or fields that lead to energy exchange. 3.2 hermodynamic State Variables hermodynamic state variables are properties of a system that depend only on current condition, not history. emperature,, is an obvious example, since we know intuitively that the temperature in a room has the same meaning to us, now matter what the previous temperature history actually was. Other state variables include the pressure,, and the system volume, V. For example, two states A and A, each with their own characteristic values of, V and : A A, V,, V, (3.1) he values of the pressure difference = = is independent of the path that the system takes to move from state A to A. he same is true for V and. 3.3 rocess Variables Work and heat are examples of process variables. Work (W ) - Done on the system. Note that our sign convention here is that work done ON the system is positive. Heat (Q)- Absorbed or emitted. he sign convention in this case is that heat absorbed by the system is positive. heir values for a process depend on the path. 6

7 3.3.1 Work Done by an External Force he work done by an external force is the force multiplied by distance over which the force acts. In other words, the force only does work if the body it is pushing against actually moves in the direction of the applied force. In differential form we have: δw = F d x (3.2) Note that F and x are vector quantities, but W is a scalar quantity. As a reminder, the dot product appearing in Eq. 3.2 is defined as follows: A B = A B cos θ (3.3) where θ is the angle between the two vectors, illustrated in Figure 3.5. Figure 3.5: Definition of the angle θ between two vectors. he total work is obtained by integrating δw over the path of the applied force: W = }{{} path F d x (3.4) For example, when we lift a from the floor to the table as illustrated in Figure 3.6 we increase the gravitational potential energy, U, of the barbell by an amount mgh, where: m = the mass of the barbell g = the gravitational acceleration (9.8 m/s 2 ) h f = the final resting height of the barbell above the floor Because the potential energy only depends only on the final resting location of the barbell, it is a state variable. he work we expend in putting the barbell there depends on the path we took to get it there, however. As a result this work is NO a state function. Suppose, for example that we lift the barbell to a maximum height h max and then drop it onto the table. he work, W, we expended is equal to mgh max, which is greater than the increase in potential energy, U. he difference between between W and U is the irreversible work and corresponds to energy dissipated as heat, permanent deformation of the table, etc. We ll return to the distinction between reversible and irreversible work later. 7

8 Figure 3.6: Work done on a barbell Work Done by an Applied ressure Another important example of work is the work done by an external force in changing the volume of the system that is under a state of hydrostatic pressure. he easiest and most important example to describe is a gas that is being compressed by a piston as shown in Figure 3.7. Equation 3.2 still applies in this case, but the force in this case is given by the product of the area and pressure: F = A (3.5) It may seem strange to write the area as a vector, but it is just reminding us that the pressure acts perpendicular to the surface, so the direction of A is perpendicular to the surface of the piston. Combining Eq. 3.5 with Eq. 3.2 gives: δw = A d x (3.6) he decrease in volume of the gas is given by A d x, so we can replace A d x with dv to obtain: his result is an important one that we will use quite a bit later on. δw = dv (3.7) Figure 3.7: Work done by force on a piston. 3.4 Extensive and Intensive roperties Extensive properties depend on size or extent of the system. Examples include the volume (V ), mass (m) internal energy (U) and the entropy (S). Intensive properties are defined at a point, with examples including the temperature ( ) and the pressure ( ) and the mass density, ρ. Extensive properties can be defined as integrals of intensive properties. For example the mass is obtained by integration of the mass density over the total volume: 8

9 m = ρ ( r) d r (3.8) V Also, intensive properties can be obtained from combinations of extensive properties. For example C k, the molar concentration of component k, is the ratio of two extensive quantities: n k the total number of moles of component k and the volume: n k C k lim V 0 V (3.9) his approach works for the density as well, with the mass density, ρ, given by the ratio of the sample mass, m, to the sample volume, V : m ρ lim V 0 V (3.10) 4 he aws of hermodynamics Outcomes for this section: 1. State the first, second and third laws of thermodynamics. 2. Write a combined statement of the first and second laws in differential form. 3. Given sufficient information about how a process is carried out, describe whether entropy is produced or transferred, whether or not work is done on or by the system, and whether heat is absorbed or released. 4. Quantitatively relate differentials involving heat transfer/production, entropy transfer/production, and work st aw of hermodynamics he first law of thermodynamics is that energy is conserved. his means that the increase in the internal energy of the system is equal to the sum of the heat flow and the work done on the system and the heat flow into the system. In mathematical terms: U = Q + W + W (4.1) where: U = Internal energy change of the system Q =Heat flow into the system W =Mechanical work done by surroundings on the system ( ext ) 9

10 Surroundings System Internal Energy U Surroundings System du transfer Figure 4.1: Conservation of energy W =All other work done on the system Challenge: How do we evaluate Q, W? }{{} du = δq +δw + δw }{{} state function process variable Integration to find du does not seem advised - process variables are path dependent, destination U is path independent. δ : not an exact differential 4.2 2nd aw of hermodynamics he second law of thermodynamics states that entropy can be transferred or produced, but not destroyed: S uni 0 (4.2) Here S uni is the change in entropy for the universe, which includes the system of interest and its surroundings: Surroundings System Figure 4.2: ransfer of entropy = [ S t + S p] + [ S t + S p ] 10

11 S uni = S p + S p 0 Entropy change can only be positive in the universe. For infinitesimal changes, ds sys = ds t + ds p where ds sys can be negative. 4.3 Reversible and Irreversible rocesses Irreversible processes produce entropy, whereas reversible processes (an idealization) do not. Entropy cannot be destroyed. Constantly increase in entropy defines the arrow of time. We recognize this intuitively when watching video in reverse. Rate of Entropy roduction Separation from Equilibrium Reversible rocess rocesses are carried out very slowly (very near equilibrium) approach this ideal. he imaginary situation of a perfectly reversible process is extremely useful for calculations. No entropy is produced No permanent changes in the universe o compute differences in state functions, one can chose the simplest (reversible) path. Heat Flow for Reversible rocesses For a very slow process, S p 0. Consider heat absorbed by system δq rev during that very small step. δqrev = 0 for any reversible, cyclic path. herefore, δqrev is a state function. We define δq rev = ds, and Q rev = ds, where S is the entropy. Examine the implications by considering two process paths: 1 reversible and 1 irreversible. Implications for heat transfer heat transferred S rev [A B] = dq rev S rev [A B] = S irr [A B] (these are state functions) However, an irreversible process produces entropy: we are interested in the amount of t :transferred, p : produced S irr,t + S irr,p = S rev,t (4.3) S irr,t = S rev,t S irr,p Entropy transfer is associated with heat transfer S irr,t = δq irr < δq rev. 11

12 For an isothermal process, [Q irr ] < [Q rev ], which is the maximum heat absorbed in any process going from A to B. Clarification: Rev indicates how a process was carried out, i.e. Q rev indicates heat transferred for a reversible process. here is no reversible or irreversible heat. Why does the path matter for us? Reversible -,V,Q,W easier to calculate Irreversible rocess Irreversible -,V, path independent, but Q,W harder to calculate 4.4 3rd aw of hermodynamics hird aw: here is an absolute lower limit to the temperature of matter, and the entropy of all substances is the same at that temperature. For a cyclic process, S cyc = S I + S II + S III + S IV = 0 Experimentally, we find that S I + S II + S III = 0. We conclude that S IV = 0. he entropy of all substances is the same at absolute zero. Utility: S II = ( S I + S III ) rovides information about chemical reaction. Si+C SiC II Si+C I SiC VI IV 0 0 Si+C SiC Figure 4.3: Cyclic process involving formation of silicon carbide, illustrating the third law of thermodynamics he entropy of the compound in Figure is equal to the entropy of pure silicon and carbon at 0 K. 12

13 4.5 Combined statement of 1st and 2nd aws We begin with the following expression for δq rev : he expression for du is: δq rev = ds (4.4) du = δq rev + δw + δw (4.5) he work done on the system by an external pressure, V is included in dw : δw = dv (4.6) hese equations can be combined to give the following combined statement of the first and second laws: du = ds dv + δw (4.7) In our following discussion we will assume that there is no additional work done on the system apart from work done by the external pressure (δw = 0), in which case we have: du = ds dv + δw (4.8) 5 hermodynamic Variables and Relations Outcomes for this section: 1. Derive differentials of state functions (V, S, U, F, H, G) in terms of d and d. 2. Using the Maxwell relations, relate coefficients in the differentials of state functions. 3. Define the coefficient of thermal expansion, compressibility, and heat capacity of a material. 4. Express the differential of any given state function in terms of the differentials of two others. 5. Define reversible, adiabatic, isotropic, isobaric, and isothermal processes. 6. Calculate changes in state functions by defining reversible paths (if necessary) and integrating differentials for (a) an ideal gas, and (b) materials with specified α, β, and C ( ). (Multiple examples are given for ideal gas, solids, and liquids.) 7. Special emphasis (Example 4.13): calculate the change in Gibbs free energy when one mole of a substance is heated from room temperature to an arbitrary temperature at constant pressure. 8. Describe the origin of latent heat, and employ this concept in the calculation of changes in Gibbs free energy G. 13

14 5.1 Enthalpy he enthalpy, H, is defined in the following way: he total differential of the enthalpy is: H U + V (5.1) Now we eliminate du by using Eq. 4.7: dh = du + dv + V d (5.2) dh = ds + V d + δw (5.3) For a process at constant, d = 0 (isobaric), and if no other work is done δw = 0. Under these conditions have: dh p = ds p = δq rev,p (5.4) where the p subscripts indicate that the differential quantities are being evaluated at constant pressure. he enthalpy therefore measures the reversible heat exchange in an isobaric process (δq rev,p = C p d p ). 5.2 Helmholtz Free Energy he Helmholtz free energy is defined in the following way: In differential form this becomes: F U S (5.5) df = du ds Sd (5.6) For a process carried out at constant (isothermal), d = 0 df = Sd dv + δw (5.7) df = dv + δw = δw + δw = δw,tot (5.8) F reports the total (reversible) work done on the system. 14

15 5.3 Gibbs Free Energy he Gibbs free energy, G, is defined in the following way. G U + V S = H S (5.9) dg = du + V d + dv ds Sd (5.10) = ( ds dv + δw ) + V d + dv ds Sd (5.11) dg = Sd + V d + δw (5.12) For processes carried out under isothermal and isobaric conditions (d, d = 0), then dg, = δw, (5.13) G reports the total work done other than mechanical work, e.g. chemical reactions and phase transformations. 5.4 Other Material roperties hermal expansion coefficient Here we define some material properties that relate different thermodynamic state variables to one one another. he first is the thermal expansion coefficient, α, which describes the temperature dependence of the volume: α 1 V ( ) V ( K 1 ) (5.14) Compressibility he compressibility, β, gives the relationship between the pressure and the volume : β 1 V ( ) V ( a 1 ) (5.15) 15

16 5.4.3 Heat Capacity he heat capacity is the amount of energy needed to change temperature by some amount. wo different heat capacities are typically defined. he first of these, C, corresponds to a process performed at a constant pressure:, (J/mol-K) δq rev, C d }{{} (J/mol K) path specification (5.16) δq rev,v C V d V }{{} (J/mol K) path specification (5.17) Empirically, C p is described by four parameters, a, b, c and d. hese four parameters are a convenient way of tabulating the data. C ( ) = a + b + c 2 + d 2 (5.18) Note that C is necessarily larger than C V. At constant, the volume changes, so work is done on the surroundings. herefore, more energy is needed to achieve the same. Below we show how We will show how C V can be calculated from C. 5.5 Coefficient Relations A variety of relationships exist between the different thermodynamic functions. hese relationships are a consequence of some basic mathematical relationships. Suppose that we have a function, Z of two variables X and Y : he differential of Z s given by: Z = Z(X, Y ) (5.19) dz = Z X dx + Z Y Y dy (5.20) X he differential forms of U (Eq. 4.8), F (Eq. ) H (Eq. ) and G (Eq. ) all have the form of Eq he various relationships between the thermodynamic potentials (U, F, G, H) can be summarized by the thermodynamic square shown in Figure 5.1. Our description of how to use it is taken directly from the corresponding Wikipedia page [?] 1. Start in the portion of the square corresponding the thermodynamic potential of interest. In our case we ll use U. 2. he two opposite corners of the potential of interest represent the coefficients of the overall result. In our case these are and. When the coefficient lies on the left hand side of the square, we use the negative sign that is shown in the diagram. In our example we have du = [differential] + [differential]. Now we just have to figure out what the two differentials are. 16

17 -S U V H F - G Figure 5.1: hermodynamic square used to help determine thermodynamic relations. 3. he differentials are obtained by moving to the opposite corner of the corresponding coefficient. In our case we have V opposite and S opposite, so we end up with du = dv + ds, which is what we expect to get from Eq We can then use this same procedure to obtain the expressions given previously for dh, df and dg. Note that two variables on the opposite sides of the thermodynamic square are thermodynamically conjugate variables [?], with units of energy when multiplied by one another. We can also use the thermodynamic square of Fig. 5.1 to obtain a variety of expressions for the partial derivatives of the different thermodynamic potentials. he procedure is as follows: 1. Identify the potential you are interested in, just as before. We ll again take U as our example. 2. We can differentiate with respect to either of the variables adjacent to the potential we chose. For U we can differentiate with respect to S or V, with the other one of these two variables remaining constant during the differentiation. 3. he result of the differentiation is given by the variable that is conjugate to (diagonally across from) the variable we are differentiating by. For example, if we differentiate U by S we have, we have ( U/ S) V =. he following 8 expressions are obtained in this way: = = S = S = ( ) U, = S V ( ) H, V = S ( ) U V S ( ) H S ( ) ( ) F F, = V V ( ) dg, V = ( ) G (5.21) (5.22) (5.23) (5.24) he thermodynamic square can be used to illustrate one more set of relationships, referred to as the Maxwell relationships. M = ( ) Z, N = X Y 17 ( ) Z Y X

18 ( ) M = Y X ( ) N X Y du = ds dv dh = ds + V d df = Sd dv dg = Sd + V d Canchy-Riemann conditions ( ) ( ) = V S S V ( ) ( ) V = S S ( ) S V ( ) S = = ( ) V ( ) V 5.6 General Strategy for Deriving Relations Equation of State Variables, : choose as independent variables V : choose as dependent variable S : dependent Derive: V (, ) and S(, ) Energy Functions U, H, F, G : Derive expressions in terms of, One can then express any function in terms of any two independent variables Summary of Relations Derivation du = ds = dv dz = = Z = Z(X, Y ) ( ) Z dx + X Y ( ) M = Y X ( ) U ds + S V 18 ( ) Z dy Y X ( ) N X Y ( ) U dv V S

19 = ( ) U, = S V ( ) ( ) ( ) U ; = S S V S S V dg = Sd + V d ( ) S = ( ) V = V α (will use later) Another useful example: ds = ( ) S d + ( ) S δq rev, = ds = ( ) S d (from dg) ( ) V = (δq rev, C d ) = V α ( ) S d = C d ds= C d-vαd It will be convenient to derive expressions for the differentials of S, U, V (and G, H, F ) in terms of d,d. Relate S to, V S(, V ) ds = Md + NdV ds = Md + N(V α d V βd ) ds = (M + NV α )d NV βd ds = C d V α d C = M + NV α V α d = N = α ( ) S β = V 19

20 M + N = C, N = α β M = 1 M = C (V α ) 2 β ( ) C (V α ) 2 = β ds = 1 M = ( ) ( ) S S, and N = V V ( ) C (V α ) 2 d + α β β dv }{{} M ( ) S V = C V (4.40) Ideal Gases Example 4.6: C V = C (V α ) 2 C V < C as noted earlier β }{{} positive quantity Find change in as a function of V, S with S held constant. ds = Md + NdV Express in terms of,, then use able 4.5 to solve for M, N We found that ds = C V d + α β dv S, d s = α C V β dv S One can integrate over change in volume to set. 6 he Ideal Gas We begin to apply these principles by considering an ideal gas, which obeys the following constitutive relationship: V = nr (6.1) Here n is the number of moles and R is the gas constant: R = (J/mol K) = (cal/mol K) = (l atm/mol K) (6.2) If n=1, then V = R, where V = V m, where V m is the molar volume. 20

21 Algorithm 1 Evaluate α, β, C, C V, U, H for an ideal gas. α = 1 ( ) V = ( ) R = V R R R = 1 β = 1 ( ) V = ( ) R = ( ) R V R R 2 = 1 (V α ) 2 β = R 1 2 = R 6.1 Specific Heat of Ideal Gas C V = C (V α ) 2 β = C R (6.3) We will show that C v = 3 2 R for a monatomic ideal gas. herefore, C = 3 2 R + R = 5 2 R. Example 4.8: Calculate the change in when 1 mole of gas is compressed reversible and adiabatically (no heat flow) from V 1 to V 2. (S, V ) In Example 4.6, we found that ds = C V d + α β dv, which simplified to d S = α C V β dv S for an isotropic process. For an ideal gas, α = 1, β = 1, so d S = C V d S = R C V dv S = C V dv S. But = R V dv S V. Evaluate by integrating. 2 1 ln d S ( 2 1 = R C V V2 V 1 dv S V ln ) [ (V2 ) ] R/CV = ln V 1 ( 2 1 ) 2 1 = = R C V ( V1 V 2 ln ( V2 ) R C V V 1 ) C V = 3 2 R ( V1 V 2 ) > 1 for compression, so increases. ( ) R/CV ( V1 V1 2 = 1 = 1 V 2 V 2 ) 2/3 If the gas expands (V 2 > V 1 ), then decreases. One can also derive () and (V). 21

22 6.2 Internal Energy of Ideal Gas U(,) du = (C p V α ) d + V ( β α) d (able 4.5) }{{}}{{} i ii i)c V α = C R = C v ii) β α = = 0 du = C V d, and C V = 3 2 R One can then integrate to find that C V is constant U= 3 2 R( 2-1 ) For an ideal gas, the internal energy depends only on the temperature. 6.3 Examples Using Ideal Gas Example 4.9: Free expansion of an ideal gas Gas Vacuum Gas Vacuum A B A B Figure 6.1: Free expansion of an ideal gas - the valve is opened at t 2 and fills both chambers Walls are rigid: δw = 0 Walls are insulating: δq = 0 System is closed. S t = 0, U = 0 = 0 (ideal gas) he gas expands irreversibly. Entropy is produced, but S is not process dependent because S is a state function. What is S when 1 mole of gas expands freely to twice its volume? (recall = 0). ds = C V d + α β dv, d = 0 22

23 V = R = R V ds = α β dv = dv = R V dv S = V 2 V 1 ( ) ( ) R V dv = R ln V2 V 1 = R ln 2V1 V 1 = R ln 2 S = 5.76J/mol-K his change for a reversible isothermal process is the same as for an irreversible free expansion in an isolated system. S is a state function. Example 4.10a: How much heat is absorbed/released when 1 mole of gas is compressed isothermally from 1 to 25 bar at 300K? o find heat, use S: δq rev, = ds he volume change is unknown, so use V = R ( ) f Q rev, = R ln i ds = C d V α d = V α d ds = R f S = R i 1 d = Rd d ( ) f = R ln i = (8.314J/mol K)298K ln( 25 ) = kj/mol 1 Reversibility requires work/heat to maintain equilibrium. he piston in Figure 6.2 below may be withdrawn very slowly VAC Figure 6.2: iston-cylinder device U = 0 = 0 23

24 Vf S = V i ( ) nr Vf dv = nr ln V V i But no entropy can be produced, so heat flow is needed to maintain (for free expansion, decreases) So actually, U = Q + W = 0 W = Q < 0 he system does negative work! Heat flow is positive (inward). Figure 6.3: Irreversible rocess (top) and reversible process (bottom) Irreversible process Reversible process Q = 0 Q 0 W=0 W 0 U = 0 U = 0 = 0 = 0 able 6.1: Irreversible vs reversible process characteristics U = 0 W = Q W < 0, Q > 0 S t > 0 What is the amount of heat required to raise the temperature to 1000K isobarically? δq rev, = ds p C d C = 5 2 R Q rev, = C = 5 2 R = 5 (8.314J/mol K) ( )K 2 24

25 Note that dh = C d + V (1 α )d. For d = 0, dh = C d Enthalpy measures reversible heat flow in absence of work. his will be useful in the analysis of solids heated under conditions of constant pressure. Some useful relations for heat flow δq rev, = ds = du df δq rev, = ds C d = dh (δw = 0) δq rev,v = ds V C V d V = du V (δw = 0) Example 4.3: Relate S(,V) ds = Md + NdV ds = Md + N(V α d V βd ) = (M NV β )d + NV α d ds = V α d + C d N = C V α = ( ) S V M NV β = V α uijten examples 1. Q for at constant Reversible, δq rev, = ds Use S(, ) like 4.10(a) ds = 2. Find given at constant S M = V α + C V α V β M = C β α Use S(, V ), also idela gas law (~4.8) 3. Free expansion of ideal gas, find S for V S(, V ) (4.9) 4. Adiabatic reversible compression of a solid. (S, ) ds = 0, given (4.14) V α = ( ) S V ( C β α V α ) d + C V α dv 25

26 5. Change in G for heating at constant G(, ), d = 0 (4.13) dg = Sd + V d, need to find S( ) V = R 1 mole of gas expands at an initial temperature constant S from 1l to 2l. What is f? ( i = 300K) ds = Md + Nd = 0 ( ) C (V α ) 2 d + α β dv = 0 α = 1, β = 1 C V d = α β CV dv d = α β CV dv d = CV du d = C V dv Use V = R to get to two variables = R V, C V = 3 2 R d = R C V V dv = d = R dv 3/2R V = 2 dv 3 V 2 1 d = 2 V2 dv 3 V 1 V ( ) ( ) 2/3 ln 2 1 = ln V1 V 2 V 1 V 2 = 1 2, 2 = 1 ( 1 2) 2/3 = 189K Note 1 = 300 R 10 3 m 3 = R 2 = 189R m 3 = R 26

27 ressure drops by threefold V = R 1 mole Fe is compressed reversibly. What is the change in? 1 = 1atm, 2 = 10 5 atm d = V α C d need exponential values Does C vary with? If dc d 0, then ds = C p d V α d ln ( 2 1 ) = V α C α = K 1 Otherwise, C d = V α C 45.3 J/mol K V = 3.1cm3 mol = m 3 /mole 2 = = 300K 7 Solids and iquids et s begin with an example of the calculation of the free energy as a function of temperature for a substance Example: pressure. Solution: Compute G when 1 mole of MgO is heated from 300K to temperature at constant dg = Sd + V d d = 0 dg = Sd = S( )d 27

28 But C depends on temperature so does S. ds = C d V α d = C d ( ) S = C C ( ) S = S o + d C = a + b, S o = 26.8J/mol K Evaluate expression, substitute data at the end 2 ( a ) S = S o + d 1 + b C ( ) 2 ( a ) S = S o + d = S o + d 1 + b S = S o + a ln ( 2 1 ) + b( 2 1 ) 1 = 300K, 2 = ( ) S( ) = S(300K) + a ln + b( 300) 300 a ln + b + [S o a ln b] }{{} constant C ( ) MgO = } {{} }{{} a b G = 300 f dg = S( )d = } {{ } ignored f 300 J ( mol K ) (a ln + b + C)d = [a( ln ) b 2 + c ] 300 f G = a ln b 2 + (C a) = 300a ln b(300)2 (C a)300 28

29 Check this, see also sec 7.3 p. 180 J J a = 48.99, b = mol K molk 2 C = 26.8J/mol K a ln b Consider the following exothermic reaction (referred to as the hermite reaction), where Aluminum reacts with iron oxide to form aluminum oxide and iron. 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe How hot will this system get as the reaction proceeds? We ll do this in two parts. First, we ll calculate the heat generated in the reaction, then we ll figure out how much this heat will increase the temperature of the system. Consider the change in enthalpy for each half-reaction: 2Al O 2 Al 2 O 3 Fe 2 O 3 2Fe O 2 H 1 = kj H 2 = 823 kj mol mol 3 2Al + 2 O Fe 2 O 3 Al 2 O 3 + 2Fe + 2 O 2 H = ( ) kj kj = mol mol he minus sign implies that the reaction is exothermic. Now that we know how much heat is generated, we can figure out what the state of the system will be. Note that Fe melts at 1809 K and boils at 3343 K, and Al 2 O 3 melts at 2325 K. We need to determine to what temperature the system will rise assuming that all heat generated by the reaction is absorbed by the system. Step 1: Heat F e and Al 2 O 3 to the melting point of F e. H Al2 O 3,1 = H F e,1 involves 3 sub-steps: (2 moles) C ( )d = 183, 649 J α γ : C,α d + H α γ γ δ : C,γd + H γ δ δ l : C,δd + H melt C,δ = C, α 29

30 Considering the three integrals of and three latent heats, H F e,1 = 157, 741 J (for 2 moles) So through step 1, we have H F e,1 + H Al2 O 3,1 = J Step 2: Heat F e and Al 2 O 3 to melting point of Al 2 O 3. H Al2 O 3,2 = C,Al d + H Al2 O 3 (S l) = (71, , 000) J H F e,2 = C,F e(l) d = 43, 178 J How much is left? 852, , , 418 = 289, 492 J Step 3: How much energy is needed to get to the boiling point of Fe? { H F e,3 = , 600 J 2325 C,F e(l)d H Al2 O 3,3 = C leaving J,Al 2 O 3 (l)d How much Fe evaporates? Q H F e (l g) = 16,892J 340,159J = 0.05moles or 5% H( ) = a + b 2 2 C 1 + d 3 3 D D = H( o ) [ a o + b 2 2 o C 1 o + d 3 3 o ] 2Al + F e 2 O 3 2F e + Al 2 O 3 Exothermic Compute the final state and temperature assuming adiabataic, reversible reaction at constant pressure. d, Q = 0, d = 0 Need initial S, C Exothermic implies from heat released. ds = Md + N d dh = ds + V d + δw 2Al O 2 Al 2 O 3 : H 1 = 1875kJ/mol 30

31 i 300 Figure 7.1: hermite reaction 2F e O 2 F e 2 O 3 : H 2 = 823.4kJ/mol 2Al + F e 2 O 3 2F e + Al 2 O 3 2Al O 2 Al 2 O 3 ; H 1 Heat is reversibly absorbed by the system. F e 2 O 3 2F e O 2; H 2 3 2Al + F e 2 O O 3 2 Al 2 O 3 + 2F e + 2 O 2 H 1 H 2 = 852.3kJ/mol δq rev, = ds = dh H F e,1 = dh = C d = 157, 741J for 2 mol Include H due to increasing as well as due to phase transitions. For Al 2 O 3, just C d H Al2 O 3 = 183, 649 J For a total change of 341,390 J Now heat Al 2 O 3 to melting point, and heat F e(l) Step 2 takes: 221,418 J C d + H melt 852,300 J is provided by the reaction: 852, , , 390 = 289, 492 J ds = C V d + α β dv = 3 R 2 d + dv = 3 R 2 d + R V dv Good for constant V: relate S to change. For constant : relate S to volume change. For constant S, ds=0 31

32 Vapor 3343 Fe 16,892 for emp 272, , , K Figure 7.2: Changes in enthalpy in Fe and Al 2 O 3 ( ) 3/2 ( ) ( ) 2/3 2 1 = V1 2 V 2 1 = V1 V = V 2 V ( 2 1 ) 3/2 ( V2 V 1 ) 3/2 = V 1 V 2 ( ) 5/3 2 1 = V1 V 2 ( 2 1 ) 3/2 = ( V1 V 2 ) 5/2 1 V γ 1 1 = 2 V γ 1 2 γ 1 γ 1 1 = γ 1 γ V γ 1 = 2V γ 2 γ = Cp C V = 5/2 3/2 = 5/3 5/3 1 5/3 = 5/3 2/3 = 5/2 32

33 γ 1 = 2/3 du = (C V α )d + V (ρβ α )d du = ( 5 2 R R)d + V (1 1)d = 3 2 R ds = C d V α d = 5 R 2 d V d ds = 5 R 2 d R d ds = C V d + α β dv = 3 R 2 d + dv = 3 R 2 d + R V dv ( ) Constant : S = 5 2 R ln 2 1 Constant : S = R ln ( 2 1 ) S = R ln ( V2 V 1 ) ( ) Constant V: S = 3 2 R ln 2 1 ( ) 5/2 ( ) 5/3 Constant S: 1 2 = 2 1 = V1 V 2 du = dv + ds du = ( ) U V S dv + ( ) U S V ds [ S ( U )] V V = [ V ( U ) ] S V S Note that these are second derivatives of the internal energy, so they should be negative for equilibrium. 8 Conditions for Equilibrium Outcomes for this section: 1. Derive the condition for equilibrium between two phases. 2. Show that at constant temperature and pressure, the equilibrium state corresponds to that which minimizes the Gibbs free energy G. 3. Explain why the evolution to a state of lower Gibbs free energy occurs spontaneously. 33

34 8.1 Equilibrium in thermodynamic system For an isolated system (δq = 0, δw = 0, n fixed, see ) Equilibrium State of Maximum Enropy Maximize S (function of multiple variables) and we will have determined the equilibrium state (regardless of past history). How do we recognize that a system has reached equilibrium? 1. It is in a stationary state (i.e. not changing) 2. If the system is isolated from its surroundings, no further changes occur. Figure 8.1 X Counter example: Cu Rod surroundings system Figure 8.2 After t, system is in steady state. Isolation of rod from heat sources/sinks would result in future evolution. z varies monotonically with x,y - there is no global maximum. Constraints such as y = y(x) may define a maximum in z(x, y) subject to y = y(x) Example 5.1 z = xu + yv For simple cases, eliminate dependent variables. In general, use agrange Multipliers. Challenge: minimize a funciton of several variables when relationships (constraints) exist between those variables. Approach: agrange Multipliers 34

35 z max z(x,y) y(x) intersects z y x Figure 8.3 Minimize: (will be S) Given these constraints (other state functions,variables) f(x 1, x 2,...x n ) Φ 1 (x 1, x 2,...x n ) = 0 function of n variables Φ 2 (x 1, x 2,...) = 0 m constraints degrees of freedom=n-m. Φ m (x 1, x 2...) = 0 Define: df + λ 1 dφ 1 + λ 2 dφ λ m dφ m = 0 λ m : agrange multiplier df = F x 1 dx 1 + F x 2 dx F x n dx n In general, df = F x 1 dx 1 + F x 2 dx F x n dx n. o guarantee that df=0 for all x 1, x 2...x n, the coefficients F x i must all be zero. i.e. F x i + m k=1 λ K Φ K x i = 0 Example 5.1 z = xu + yv Constraints u = x + y + 12 Φ 1 = x + y + 12 u = 0 v = x y 8 Φ 2 = x y 8 v = 0 u, v: dependent x, y: independent dz + λ 1 dφ 2 + λ 2 dφ 2 = 0 Minimize z subject to constraints = xdu + udx + ydv + vdy + λ 1 (dx + dy du) + λ 2 (dx dy dv) = 0 35

36 x λ 1 = 0 y λ 2 = 0 } x + y + 12 u = 0 x y 8 v = 0 = (x λ 1 )du + (y λ 2 )dv + (u + λ 1 + λ 2 )dx + (v + λ 1 λ 2 )dy = 0 λ 1 = x λ 2 = y u + λ 1 + λ 2 = 0 v + λ 1 λ 2 = 0 } u = λ 1 λ 2 v = λ 1 + λ 2 2x + 4 ( x y) ( x + y) = 0 1 y 8 ( ( 1) + y) = 0 4x + 4 = 0 x = y = 0 y = 5 u = x y v = x + y 8.2 Finding equilibrium conditions A unary 2-phase system is in equilibrium when the, and µ for both phases are equal. Consider 2 phases α and β of one component. U = U α + U β, V = V α + V β, S = S α + S β because these are extensive properties. So U α (S α, α, n α ) where n α is the number of moles of α phase. hen dv α = α ds α α dv α + µ α dn α, where µ ( V α n α )V α,s α µ α is the chemical potential of the α phase. o find the equilibrium condition, maximize S. ds α = duα α + α α dv α µα α dn α, ds β = dv β β + β β dv β µ β β dn β ds = ds α + ds β = duα α + α α dv α µα α dn α + dv β β + β β dv β µ β β dn β If the entropy is maximized, then no changes will occur if the system is isolated (rigid walls, insulated). du = du α + dv β = 0 du α = du β ( no heat is exchanged) dv = dv α + dv β = 0 dv α = dv β ( no work is done) dn α = dn β (particle number is conserved) ( ) ( ) ( ) hen ds = 1 α 1 β du α + α α β β dv β µα α µ β β dn α = 0 o guarantee ds=0, these coefficients must be zero independently. Hence, α = β, α = β,µ A = µ B Conditions for Equilibrium 36

37 8.3 Minima of energies as conditions of equilibrium Consider first the internal energy U in an m-component system. du = ds + dv + m i µ i dn i where i : index of components du = δq + δw + δw Consider a reversible process in an isolated system δq rev = ds δw rev = dv δw = i µ idµ i chemical work associated with reversible reactions Claim: For a system of constant S, V, n λ, the equilibrium state is a state of minimum internal energy U. If this is not true, then work can be extracted and reinserted as heat. But then S would increase while nothing else changes. herefore, energy cannot be extracted from a system under these conditions (unchanging/maximized S v,n ). Derive conditions for equilibrium again du = 0 (minimize U) ds = 0 ds α = ds β dv = 0 dv α = dv β dn = 0 dn α = dn β du α = α ds α α dv α + µ α dn α du β = β ds β β dv β + µ β dn α du = ( α β )ds α ( α β )dv α + (µ α µ β )dn α = 0 α = β, α = β, µ α = µ β (Additional condition: second derivative) hermal, mechanical, and chemical equilibrium U will decrease spontaneously if a system of constant S and V is not in equilibrium. Enthalpy H = U + V Hemholtz Free Energy F = U S Gibbs Free Energy G = U + V S he equilibrium conditions for different situations (constant S, ; constant, V ; etc) are outlined below in table xx Variables Held Constant Equilibrium Condition S, dh = 0, V df = 0, dg = 0 S, V du = 0 able 8.1: Equilibrium conditions. 37

38 If and are held constant, a system will seek to minimize the Gibbs free energy, G. Consider two states connected by an isothermal and isobaric path: G 1 = U 1 + V 1 S 1 Consider first a reversible path: W rev, = V and Q rev, = S For any path: G 2 = U 2 + V 2 S 2 G = U + V S, and U = Q + W + W G = [Q + W + W ] W rev Q rev G = (Q Q rev ) + (W W rev ) }{{}}{{} <0 <0 +W <0 In a system at constant and that is not in equilibrium, G will tend to decrease spontaneously to approach chemical equilibrium. Example for two component system with separate environments. 1 V 1 = n 1 R 1 2 V 2 = n 2 R 2 U 1 + U 2 = 0 } et them come to equilibrium in isolation wo isolated systems are defined - each in equilibrium 3 2 n 1R( f i ) n 2R( f i ) = 0 n 1 f n 1 1 = n 2 f + n 2 2 f = n 1 1 +n 2 2 n 1 +n 2 We can also solve for f, V i f = (n 1+n 2 )R f V 1 +V 2 = R(n 1 1 +n 2 2 ) V 1 +V 2 and V 1 = n 1R f f, V 2 = n 2R f f V 1 + V 2 = n 1R f f + n 2R f f f V Final n 1 + n 2 f 8.4 Maxwell Relations For free energies: coefficients in opposite corners derivatives in adjacent corners minus signs apply to coeffiencients For Maxwell Relations: = i.e. ( ) S = ( ) V Note: his does not include the chemical potential. δw = m i 38 µ i dn i

39 Useful relations between S,,,V for an ideal monatomic gas 5 R 2 d V d = 5 R 2 d R d ds = C V d + α β dv = 3 R 2 d + dv = 3 R 2 d + R V dv ds = Cp d V α d = Constant : { S = 5 2 R ln ( 2 2 ) ( S = R ln 2 Constant : ( S = R ln V2 1 ) V 1 ) Constant V: S = 3 2 R ln ( 2 1 ) ( ) 5/2 ( ) 5/3 Constant S: 2 1 = 2 1 = V1 V 2 Reversible (!) path Figure 8.4 ds = Cv d + α β dv = 3 R 2 d + dv = 3 R 2 d + R V dv ( ) ( ) 0 = 3 R 2 d + R V dv 3 2 ln 2 1 = ln V2 V 1 ( ) 3/2 ( ) 5/2 ( ) 5/3 2 1 = V 1 V 2, 2 1 = V1 V 2 = 2 1 One can show this using 1V 1 1 = 2V 2 2 and substituting. 39

40 9 From Microstates to Entropy Outcomes for this section: 1. Express the entropy S of the system in terms of equivalent microstates in the equilibrium (most probable) macrostate. 2. Given the energy levels of a system of particles, calculate the partition function. 3. Given the partition function, calculate the Helmholtz free energy, the entropy, the internal energy, and the heat capacity. Destination: S = k B ln Ω where k B =Boltzmann s constant, ev/k, J/K and Ω = number of equivalent microstates in (equilibrium) macrostate. Matter is composed of atoms, and is therefore discrete. Contimuum thermodynamics arises from interactions between discrete components and interactions with the environment. A statistical description is needed to connect the microscopic and macroscopic descriptions. 9.1 Microstates and Macrostates Microstate: Specification of the state (e.g. energy level) of every atom at a particular point in time. Example: 4 identical particles in 2 levels, pour 4 balls in 2 boxes. If balls are placed randomly, which configuration is more likely? # Microstates Distribution 1 (4,0) 6 (2,2) 4 (1,3) able 9.1: Microstates and distribution of balls Macrostate: collection of equivalent microstates. Equivalent means the same numbers of particles within each of the levels. Example: the macrostate with 1, 3R - (1,3) We will use combinatorics to calculate: 1. he distribution function specifies the occupation of atomic states by identical particles, and thus the macrostate. 2. Number of macrostates number of microstates 3. If all microstates are equally likely, then some macrostates are much more likely. (e.g. (2,2)) How do we see this? 40

41 Figure Figure 9.2: Macrostate (1,3) How many microstates correspond to a given macrostate? N 0 : # of balls, r : # of boxes, n i : balls in i th box Ω = N o! n 1!n 2!n 3!...n r! = N! Π i=1 n i! n! n(n 1)(n 2) otal number of possibilities is r No, or 2 4 = 16. here are 5 macrostates: (4,0) (1,3) (2,2) (3,1) (4,0) Example: Four atoms in volume V. How many microstates are there, and how many macrostates? Divide each side into M compartments. Compartments are so small that M 4 and occupation is 0 or 1. 41

42 Number of Microstates Distribution 1 (4,0) Ω 4,0 = 4! 4!0! = 1 6 (2,2) Ω 2,2 = 4! 2!2! = = 6 4 (1,3) Ω 1,3 = 4! 1!3! = = 4 able 9.2: Microstates corresponding to given macrostates V Figure 9.3 R How many ways can we distribute 4 indistinguishable particles over 2M compartments? Case I: Ω I ( 2M 4 ) = (2M)! 2M (2M 1)(2M 2)(2M 3) = (2M)4 4!(2M 4)! 4! 4! Case II: How many configurations have all the particles on one side, i.e. M boxes? Ω II = ( M 4 ) M 4 4! Note that Ω II Ω I = 1 16, Ω II is much less likely. For N atoms, we would find Ω II Ω I = 1 2 N 1 Even a state with 51% on the left would be highly improbable. (1.98M)N (2M) N small.( ry plotting this) for N N A is vanishingly ostulate of Equal ikelihood All allowable microstates (energetically equivalent configurations) are equally likely. j = Ω j rno is the probability of finding the system in macrostate j. he most probable macrostate is the one that represents the most microstates in Ω. 42

43 9.1.2 he Boltzmann Hypothesis (to be derived) most probable macrostates Figure 9.4 j S = k B ln Ω k B = J/K Maximum in Ωcorresponds to a maximum in S. How do we determine the most probable macrostate? What are the conditions of equilibrium? 9.2 Conditions for equilibrium We aim to minimize S, where S = k ln Ω, or What tools do we need to do so? ( ) ln x y = ln x ln y, ln(xy) = ln x + ln y ( ) No! S = k B Π r i=1 n i! }{{} Ω d ln x = dx x Stirling s Approximation: ln x! x ln x x his is very accurate for large x. How large? 1. Express Ω as a sum over logarithms of occupation numbers. Ω = N! Π r i=1 n i! ln Ω = ln N! ln (Πr i=1n i!) Apply Stirling s Approximation, ln(πx i ) = ln x i 43

44 ln Ω = N ln N N r (n i ln n i n i ) Note also that N = r i=1 n i (summing occupation numbers gives the total particle number). So r ln Ω N ln N N n i ln n i + N = r n i ln N i=1 ln Ω r i=1 n i(ln N n i ) Not too bad 2. Differentiate d ln Ω = r i=1 = r i=1 (ln N ln n i)dn i + d ln Ω = r i=1 ln ( n i N ) dni 3. Impose constraints [ dn i (ln N ln n i ) + n i ( dn N dn i n i )] i=1 i=1 r n i ln n i i=1 r ( ni ) r i=1 dn N }{{} dn i i =1 }{{} dn dn For an isolated system, the total particle number and energy are constant as ln Ω is maximized. N = n 1 + n n r = r i=1 n i, dn = 0 U = n 1 Σ 1 + n 2 Σ n r Σ r = r i=1 n iσ i, du = 0 o impose constraints, we use agrange Multipliers. d ln Ω + λ 1 dn + λ 2 du = 0 r i=1 ln ( ) n i N dni + λ r 1 i=1 dn i + λ r 2 i=1 Σ idn i = 0 r [ ( i=1 ln ni ) ] N dni + λ 1 λ 2 Σ i dni = 0 Gives r equations, e.g. ln ( ) n i N λ1 λ 2 Σ i = 0 n i N = eλ 1 e λ 2Σ i, i = 1, 2,..., r 4. Use constraints to eliminate λ 1,λ 2 n i N = eλ 1 e λ 2Σ i r i=1 n i N = 1 r i=1 eλ 1 e λ 2Σ i = 1 e λ 1 = 1 r i=1 eλ 2 i n i N = eλ 2 Σ i r j=1 eλ 2 Σ j 1 Z eλ 2Σ i he fractional occupation number of the i th level. So ln Ω i = r i=1 ( ni ) n i = N 44 r ( e λ 2 Σ i ) n i ln Z i=1

45 What is λ 2? ln Ω = λ 2 = r n i (λ 2 Σ i ln Z) i=1 r n i Σ i + ln Z i=1 r n i = λ 2 U + N ln Z Consider response of system to the reversible absorption of an infinitesimal amount of heat δq. d ln Ω = λ 2 du + d (N ln Z) du = δq rev = ds d ln Ω = λ 2 ds So i=1 S ln Ω S = 1 λ 2 }{{} k B n i N = ln Ω = k B ln Ω λ 2 = 1 k B e Σi/k r i=1 e Σ i/k 1 e Σ i/k }{{} Z }{{} Boltzmann factor artition function We can calculate macroscopic properties from the partition function. Recall ln Ω = λ 2 U + N ln Z, S = k ln Ω (k ln Ω =) S = U + k BN ln Z k ln Z = U S F Hemholtz Free Energy! So F = k N ln Z F = k ln Z N Also, S = ( ) F V = Nk ln Z + Nk ( ) ln Z V hen C V = ( ) U V ( ) ln Z U = F + S = Nk 2 V ( ) ln Z = 2Nk V ( + Nk 2 2 ) ln Z 2 V Note that the pressure dependence of Z is needed to calculate V, H, G, C. 45

46 10 Noninteracting (Ideal) Gas Conditions for ideal gases No interactions (forces/potentials) only kinetic energy. = 1 2 mv2 = 1 2 m(v2 x + v 2 y + v 2 z) he energy distribution of free particles is continuous. For convenience, assume box of dimensions = V Z = } {{ } V dxdydz e 1 2k (v2 x+vy+v 2 z) 2 dv x dv y dv z = V { Gaussians }}{ 1 e 2k mv2 x dvx e 1 2k mv2 ydv y e 1 2k mv2 Z dvz e αx2 dx = π α (for α > 0 ) {}} α { 1 e 2k m v2 x Now we can easily derive many functions., ( 2πk Z = V m ) 3/2 ln Z = ln V + 3 ( ) 2πk 2 ln + 3 m 2 ln ln Z = F = Nk ln Z = Nk ( ) ln Z U = Nk 2 V [ ln V + 3 ( )] 2πk 2 ln m ( ) 3 = Nk 2 1 = Nk U N = 3 2 k Energy per particle 46

47 C V = ( ) ( ) U V = 3 2 NK = 3 N 2 N A kn A = 3 2 nr C Vn = 3 2 R n: moles 10.1 Ideal gas aw df = Sd dv + δw ( ) ln Z = Nk V = = Nk [ V ( ) F V ( ln V ln ( 2πk m ) + 32 ln )] =Nk 1 V = nr V Note that we have ignored interactions between particles/atoms. Attractive interactions lead to, e.g. condensation. We have also ignored repulsive interactions, i.e. atoms occupy finite volumes. Including these leads to the van der Waals equation: 11 Diatomic Gases ( + a V 2 ) (V b) = R Diatomic gases possess rotational and vibrational degrees of freedom. he kinetic energy of a gas molecule with p degrees of freedom is Σ = j=1 b jv 2 j, and Z = V Π j=1 πk b j (e.g. b = m 2 for translations) For rotations v ω, M I Surprisingly, U and C V do not depend on b j : ( ) ln Z Get 1 2R for each degree of freedom C V = 2 Nk = 2 nr Observations: V = ( ) 2 U = Nk 2 = 2 2 Nk Heat capacity measurement reveals the number of degrees of freedom. Monatomic: 3 translational Heteronuclear Diatomic: 3 translational + 2 rotational + 2 vibrational 47

48 1 vib 2 rot Figure 11.1: Diatomic gas 12 he Einstein Model of a Crystal How is the discrete nature of matter manifested in easily observable quantities? Consider atoms in a simple cubic lattice Bounded to six nearest neighbors (vibrating in harmonic potential) F = kx, ω = k m Figure 12.1: Vibration of atoms in harmonic potential 3 bonds/atom, 3N bonds in crystal Figure 12.2: Binding to six nearest neighbors By solving the Schrödinger equation for the harmonic oscillator potential, one can show that i = (i ) ω = (i )hv i = 0, 1, 2... h: lanck s constant, v : frequency, : h 2π Energy is stored in vibration of atoms. o analyze, calculate the partition function for the vibrations: Z = n e Σi/k = i=0 n e (i+ 1 2 )hv/k = e 1 2 i=0 hv k n i=0 e ihv/k Here n is the number of energy levels. If n is large, we can take the limit n and use i=0 e ihv/k = 1 1 e hv/k, Z = e 1/2 hv k 1 e hv/k 48

49 12.1 hermodynamic roperties of a Crystal Vibrations along each of three bonding axes contribute to energy: F = 3Nk ln Z = 3Nk [ hv 2k + ln(1 e hv/k ) ] = 3 hv 2Nhv + 3Nk ln(1 e k ) Note that the crystal has energy even at zero! U = β ln Z = k 2 ln Z U = 3Nk 2 ln 2 = 2Nk 2 (β = k ) (per particle?) [ 1 hv 2 k ln(1 e hv/k ) ] How does this change with temperature? he heat capacity, C V = U [ ] U = 3 2 Nhv 1 + e hv/k 1 e hv/k C V = 3 2 Nhv hv k 2 e hv/k (1 e hv/k ) + (1 + e hv/k ) (1 e hv/k ) 2 ( ) hv 2 e hv/k C V = 2Nk k (1 e hv/k ) 2 3NK expt Classical limit (3R Figure 12.3: Heat capacity in an atom as a function of temperature J mol K ) he Einstein model captures the qualitative behavior of the heat capacity very well. (But something is missing). Note here regarding information derived from heat capacity. 49

50 13 Unary Heterogeneous Systems Outcomes for this section: 1. State the differential of the Gibbs free energy G as a function of S,, V, and. 2. State the conditions for two phases to coexist in equilibrium. 3. Explain how isobaric section of the chemical potential µ(, ) as a function of temperature and pressure can be calculated from knowledge of the heat capacity and the entropy at room temperature. 4. Given µ(, ) for multiple phases, determine the equilibrium phase at a given temperature. 5. Derive the Clausius-Clapeyron relation: express the general condition for determining the region of two-phase coexistence. 6. Calculate the H given the C ( ), assuming a negligible contribution from thermal expansion. 7. Calculate V α β assuming ideal gas behavior. with the assumption that S and V depend only weakly on the tem- 8. Integrate d d perature. = S V 9. Given the diagram, qualitatively sketch G( ) for a specified region. 10. Given the diagram, specify the sign of V α β. 11. State routon s rule and justify its validity qualitatively. 12. Sketch the diagram for a one-component system. Unary: single chemical component Homogeneous: 1 phase Heterogeneous: 2 phase hases: Gas, iquid, Solid; allotropes are distinct solid phases hase boundary: limit of phase stability multiple phases co-exist in equilibrium at a phase boundary Boundary defined by Clasius - Clapeyron Equation hase transformation: solid liquid, liquid vapor, allotropic process of phase transformation determines microstructure 50

51 solid liquid gas regions 0-2 (K) log -4 S p(atm) hase Boundary -6-8 G Figure 13.1: hase Diagram Cu hase Diagrams hase Diagram of Unary Heterogeneous System hase solid, liquid, gas regions Unary single component Heterogeneous more than one phase (Cu) component he phase boundary defines the limits of phase stability. hases co-exist at a phase boundary. A phase transformation occurs when crossing phase boundaries i.e. s l, l v, s v. triple point Metallic iquid 600 (kbars) allotropic 400 Diamond 200 Graphite transformation (K) Vapor Figure 13.2: hase Diagram of Unary Heterogeneous System (C) 13.2 Chemical otential Surfaces o find µ(, ), integrate dµ = dg = Sd + V d 51

52 10,000 (atm) m.p Ice III iquid b.p 0.1 Ice I Vapor Figure 13.3: Componds can be treated as unary systems (H 2 O) knowns: C, expansion coefficient, compressibility depends on phase hen repeat for each phase (solid α,solid β, liquid, etc. ) M M O O Figure 13.4: Solid α and liquid µ (, ) and µ S (, ) (two surfaces) intersect at a line defined by µ = µ S. M O Figure 13.5: Intersection of µ (, ) and µ S (, ) µ and µ S must be computed from same reference state (e.g. µ S ( o, l o ) Equilibrium conditions are met at intersection α =, α =, µ α = µ AB: solid + liquid COD: solid +gas 52

53 C E A O B D F S G Figure 13.6: Superposition of Solid, iquid, and Gas phase µ EOF: liquid + gas O: triple point S+l+S Relation between chemical potential and the Gibb s free energy (for unary systems). G: intensive molar Gibbs free enrgy G = ng : Extensive du = ds dv + µdn 1, µ = G = U + V S ( ) u n S,V dg = du + dv + V d ds S d Susbstitute du dg = ds dv + µdn + dv + V d ds S d dg = S d + V d + µdn ) µ = µ = ( G n ( G n, ) = (, n (ng)), = G ( ) n, = G 53

54 13.3 Calculation of µ(, ) Surfaces dµ α = dg α = S α d + V α d Approach: compute isobaric sections by integrating S α d. ds α = Cα d Sα ( ) = S α C α (298) + ( ) 298 d µ α ( ) = µ α (298) S α ( )d 298 [ µ α ( ) = µ α (298) S α (298) C α( ] ) d d o compare µ α with µ, we need to calculate µ from the same reference point. Connecitng µ α with µ at the melting point m : µ ( ) µ α (298) = [ µ ( ) µ ( m ) ] + [µ α ( m ) µ α (298)] }{{}}{{} integral of S from just to calculated m this integral [ µ = µ ( m ) S ( m ) C + ( ] ) Mm m d d lot G( ) G α (298) versus for each phase α, i.e. plot of G α begins at 0, do example where =1 bar. * Note the entropy changes with the phase change S ( m ) = S α ( m ) + S α ( m ) S : latent heat = m S ( m ) = S α ( m ) }{{} S298 α + m C α ( )d S α ( m ) }{{} latent heat G ( ) G α (298) = G ( ) G ( m ) + G α ( m ) G α (298) = S α ( m) {}}{ m S α C α (298 + ( ) 298 d + S α ( m ) + }{{} S ( m) m [ +( ) S α (298) C ( ) m d C α( ) ] d d S ( > m) 54

55 + G 0 ref point _ 300 Figure 13.7: Determining the equilibrium state (at fixed ) 13.4 Clasius-Clapeyron Equation Goal: Determine two-phase coexistence lines in order to complete the - phase diagram Areas: single phase ines: 2-phase Intersections: 3 phases (triple point) Consider co-existence of two phases in equilibrium: α = β, α = β, µ α = µ β hase changes change the amounts of two phases: dµ α = S α d α + V α d α dµ β = S β d β + V β d β But we require equilibrium: d α = d β d α = d β dµ α = dµ β S α d + V α d = S β d + V β d (S β S α )d = (V β V α )d S α β d = V β α d S α β : difference in molar entropy V α β : difference in molar volume 55

56 d d = Sα β V α β Classius-Claperyon Equation Slope = But how do we measure entropy change? Entropy Change Volume Change Instead, we measure the heat of fusion of vaporization isobarically under reversible conditions. So, Q α β = H α β In equilibrium, G α = G β H α S α = H β S β ( α = β So S α β = Hα β, and d d = Hα β V α β Can be integrated to get ( ) for 2-phase coexistence. We need H(, ), V (, ) from Chapter 4. Step 1: Find H(, ) d( H α β ) = d(h β H α ) = dh β dh α dh = C d + V (1 α )d d H α β = C d + [V (1 α )d }{{} negligible for pressures up to 10 5 bar o find H, we need the temperature dependence of the heat capacity. C ( ) is often given empirically as C = a + b + c 2 + d 2 (See Appendix B,E) Now we can integrate to get H. C = a + b + c 2 + d 2 56

57 Step 2: Find V α β (, ) V β (, ) V α (, ) and α as liquid or solid. First consider the case of β as a vapor phase, V β V α Assume ideal gas behaviour. V α β = V β V α V β = R hen for sublimation or vaporization to phase β, d d = Hα β V α β = C d R = C R 2 d small, positive slope hase boundaries can be calculated based on knowledge of C. (or vice versa, as we will see) For solid solid transformations, (β is the stable phase at high ) positive ( H, S>0 {}}{ d dt = C α β ( ) d } V {{} determines sign of slope V is usually positive but can be negative (e.g. H 2 O) V ~ constant over few 10 s of atmosphere. An approximate analysis of solid-solid phase boundary: S, H, V are determined by C, α, β, but suppose the variation with temperature is weak. ( ) Since d d = S V = constant, o S V ( o) H o V o he slope d d is generally large since V is small. 1 AM log 0 d/d S V Cu -10 Figure 13.8 d d H V H R 2 d H( ) R 2 d 57

58 If we assume d d H = 0, ( ) ln = H o R ( 1 1 o ) d d S > d d S V Justification for Approximations d H α β ( ) = C p ( )d, and integrate Consider the enthalpy of transformation H( ) = H( o ) + C ( )d }{{} o ypically ~ 100kJ }{{} ypically few kj over few 100K H( o ): enthalpy change at the transformation temperature. Heat to drive phase change is larger than that needed to change the temperature. hus, H( ) H( o ) For vaporization, ( ) ln H( ( o) 1 o R 1 ) o routon s Rule S vap is approximately the same for all materials. S = H high boiling points have larger H vap. materials with ( ) ( ) ( ) H 1 H ln = o R + R o Figure

59 14 Open Multicomponent Systems Outcomes for this section: 1. Given sufficient information about the dependence of a total property B or B versus composition, find the partial molal properties as a function of composition via calculations of derivatives db or d B. 2. Given the partial molal properties as a function of composition, calculate the total properties of the system via integration of the derivatives. 3. Given the differentials of state functions, write partial derivatives that define the partial molal properties. 4. What is B 0 k? 5. What is µ k µ 0 k for (a) a component behaving ideally, and (b) in general? 6. Define the activity of a component. 7. Given µ k (,, X k ), calculate changes in the partial molal properties. 8. State and justify/explain Raoult s law. 9. State and justify/explain Henry s law. 10. Explain how the activity coefficient γ can be used to describe the departure from ideal behavior in terms of excess quantities. 11. Define a regular solution and calculate the partial molal properties and total properties versus composition for binary mixtures. 12. Describe the driving forces for mixing and their origins in statistical mechanics. 13. Interpret changes in state functions with mixing in terms of ideal behavior and departures from ideality artial Molal roperties of Open Multicomponent Systems Volume V is an extrinsic quantity that depends on amount of each component: V = V (,, n 1, n 2,...n c ) C : # components x c : mole fraction n k : # moles of component k hen ( ) ( ) V dv V = d + d +,n m,n k c ( ) V k=1 dn k n k,,n o n k 59

60 We previously defined coefficients of d, d For convenience, define V ( K V n k as partial molal volumes. ),,n j n k We can do the same for any other extensive property B (U, S, F, G, H ) db = Md + Nd + r k=1 B k dn k B K ( B n k ),,n j n k k = 1, 2, >... < Consider a system (or sample) that we create by adding various amounts of the components sequentially (holding others constant). If d = d = 0, dk, = C k=1 V k dn k, which we could integrate. But V k depends on the composition, which changes continuously. However, we can calculate changes in state functions using the simplest possible path: add all components simultaneously and in the proportions of the final mixture. Intensive properties (,, X k ) are constant and C C nk V = V k dn k = V k dn k = k=1 k=1 0 C B = B k n k k=1 C V k n k k=1 Interpretation: We can calculate how much each component contributes to a given extensive quantity. Gibbs - Duhem Equation: Evaluating Differentials [ C ] db = d ( B k n k ) = Apply product rule: k=1 c d( B k n k ) db = C k=1 B k dn k + C k=1 n kd B k, but d, d are held constant, and k=1 C db = Md + Nd + }{{} 0 k=1 B k dn k C n k d B k = 0 k=1 Gibbs Duhem equation relates the different partial molal properties. o be shown: given one, the others can be computed. For two components: n 1 d B 1 + n 2 B2 = 0 60

61 14.2 he Mixing rocess: Calculating changes in energy + + = Reference states Final State Figure 14.1: Mixing of three states into B so ln Reference and Final States have the same and. B o k : value of property B for pure component, per mole. B o = C k=1 Bo k n k: total value of B B mix B so ln B o B mix = C k=1 B k n k C Bk o n k = k=1 k C=1 B k {}}{ ( B k B k ) n k C B mix = B k n k k=1 per mole B = B n = C k=1 How does this vary with changes in composition? B k n k n = C B k X k k=1 C B mix = B k n k k=1 C d( B ) = ( B C k dn k Bndn o k ) = B k dn k k=1 k=1 Alternatively, we could have written C d( B ) = ( Bdn k + n k d B C k ) n k d B k = 0 We can also express these per mole: k=1 k=1 C d B mix = B k dx k k=1 61

62 C X k d B k = 0 k=1 How do we find molal properties? 1. Measurement of total property B or B versus composition. 2. Measurements of M B k versus composition. We will first apply to two component systems. B = B 1 x 1 + B 2 x 2, db = B 1 dx 1 + B 2 x 2 dx 1 = dx 2 x 1 + x 2 = 1, x 2 = 1 x 1 db = ( B 1 B 2 )dx 1 db dx 1 = B 1 B 2 B = B ( ) 1 x 1 + B1 db dx 1 (1 X 1 ) = B 1 db dx 1 + db dx 1 X 1 Solve for B 1 Note also that B 1 = B + (1 X 1 ) db dx 1 and B 2 = B + (1 X 2 ) db dx 2 B 1 = B + (1 X 1 ) d B dx 1 ; B 2 = B + (1 X 2 ) d B dx 2 from Daltoff roblem 8.3: Given behavior of volume change upon mixing V mix = 2.7X 1 X2 2 (??/mol), derive expressions for partial molal volumes of each component as a function of composition. V mix = 2.7X 1 X 2 2 = ax 1X 2 2 a = 2.7 o take derivative, express in terms of X 2 : V mix = a(1 X 2 )X 2 2 d V mix dx 2 = 2aX 2 3aX 2 2 = d V mix dx 1 Now use relations derived in lecture: B 1 = B mix + (1 X 1 ) d B mix dx 1 (B V ) 62

63 C slope B D 0 A 1 Figure Graphical Determination of artial Molal roperties B 2 = B mix + (1 X 2 ) d B mix dx 2 B 2 = AB + B BC B = AB + BC = AC Evaluations of Ms: given B 1 (X) B 2 (X 1 ), B Gibbs - Duhem: X 1 d B 1 + X 2 d B 2 = 0, d B 1 = X 2 X 1 d B 2 B 1 = B 1 (x = 0) + X 2 0 d B 1 = B1 0 + ( X 2 X2 0 X 1 d B ) 2 d B 2 = d B dx 2 dx 2 Given B 2 (X 2 ), we can compute B 1 (X 2 ). hen we can find B = B 1 X 1 + B 2 X 2 Example 8.2: Given H 2, find H 1 and H. H 2 = ax1 2 Compute total derivative: d H 2 dx 1 = 2aX 1 = 2aX 1 dx 2 dx 2 = 1 X 1 + X 2 = 1dX 1 + dx 2 = 1 dx 1 = dx 2 H X2 X 2 1 = X 2 =0 d H 2 X 1 dx 2 X2 X 2 dx 2 = ( 2aX 1 )dx 2 X 2 =0 X 1 H X2 1 = 2a X 2 dx 2 = ax2 2 0 H = X 1 H 1 + X 2 H 2 = X 1 (ax 2 2) + X 2 (ax 2 1) = ax 1 X 2 (X 1 + X 2 ) }{{} =1 H = ax 1 X 2 63

64 Relationships between partial molal properties: relations) (F,G,H,S,U, their derivative, and coefficient Apply differential operator ( n k ),,n j to extensive quantity: Example: Hemholtz free energy F = U S Differentiating, ( ( F n k = U ),,n n k j ),,n j By definition, F k = Ūk S k ( S n K ),,n j See De Hoff for more examples and for coefficient an Maxwell relations Chemical otential in Multicomponent Systems Given µ k (,, X k ), one can find B, B, B(,, X) he thermodynamic state is specified by C+2 variables: du = ds dv + C k=1 µ kdn k, so µ k ( U dn K ) S,V,n j From Chapter 4 then, δw = C k=1 µ kdn k (necessary in open systems) Examining dh, df and dg, we find that Given µ k (, ) = Ḡk, we can express all M s in terms of µ k : S k, V k, H k, Ūk, F k ( ) ( ) Ḡ k µk S k = =,n k,n k ( ) V k = Ḡ k =,n k ( ) µk,n k So ( ) H k = Ḡk + S µk k = µ k,n k ( ) F k = Ūk S µk k = µ k,n k ( ) µk,n k µ k is not measured directly measurement of µ k (, ) enables Ms to be determined. Note that B k can also be determined by substituting µ k µ k = µ k µ 0 k 64

65 ( Activity and Activity Coefficient We found that µ k (, ) = G n k ),,µ Ḡk(, ) but we do j not know how to measure directly. Define µ k µ 0 k = µ k R ln a k, where a k is the activity of component k. Further we define the activity coefficient γ k in a k = γ k X k, where X k is the mole fraction. So Next step: derive this for an ideal solution. µ k = R ln(y k X k ) 14.5 roperties of ideal solutions (e.g. ideal gas mixtures) Imagine removing the partitions in Figure 14.3 below and allowing the gases to mix while keeping, constant. he process is analogous to free expansion. A B C D Mixture A B C D Figure 14.3: Mixture of ideal gases Each component experiences a decrease in pressure from to k. Each gas exerts, through collisions, a partial pressure on the wall of k = X k. he total pressure is = C k = k=1 C X k = k=1 C X k = Consider the change in chemical potential for the component, K which we can find by integrating dµ K : k=1 dµ k = S k d + V k d = V k d since expansion is isothermal. For an ideal gas, V = n R = R = R for which we can find V k = So C k=1 n k, ( V n k ) µ k = µ k µ 0 k = k = R,,n j V k d = k R d = R ln k = R ln X k Since the activity a k = X k, we conclude that the activity coefficient γ k = 1 for an ideal gas. (no interactions) How do the partial molal quantities vary? Ḡk = R ln X k 65

66 V ( ) k = µk = 0,n k S ( ) k = µk = R ln X k,n k H k = µ k + S k = R ln X k + ( R ln X k ) = 0 Ūk = H k V k = 0 F k = Ūk S k = R ln X k he change in total volume, enthalpy, and internal energy is 0, as the changes in partial molal quantities is zero. However, entropy is produced, leading to decreases in the free energies G mix, F mix. 6 All All 0 All Figure 14.4: roperties of an ideal solution including G mix, S mix, H mix, V mix, U mix and F mix. Maximum temperature shown is 1400 K Behavior of Dilute Solutions Case 1 Add a few atoms of component 2 (solute) to a large volume of component 1 (solvent). he solvent atoms still behave as though they are in an ideal solution, regardless of interactions. Raoult s aw: lim X 1 1 a 1 = X 1 he solute atoms interact primarily with solvent atoms, so the average properties will be proportional to X 2 for small X 2. (However, properties depend on solvent-solute combination.) Henry s aw: lim a 2 = γ2 0X 2 : γ2 0 (, ) is Henry s aw constant (in particular range), the solute X 2 0 activity coefficient for a dilute solution. Consider Solute A in Solvent B γ 2 = e aox2 1 /R for X 1 1, γ 0 1 = eα 0/R Example: HW

67 ure A 1 Solute Slope 1 Solvent Slope ure B 0 dilute A dilute B Figure 14.5 One mole of solid Cr 2 O 3 at 2500K is dissolved in a large volume of a liquid Raoultian solution (at 2300K) of Al 2 O 3 and Cr 2 O 3 (20%). Calculate the resulting H and S given m,cr2 O 3 = 2538K, H m,al2 O 3 = 107, 500σ/mol K at m (2324K) S m,al2 O 3 = S m,cr2 O 3 Recall that under isobaric, reversible conditions, S α = Hα β H m,1 m,1 = H m,2 m,2 H m,cr2 O 3 = H m,al2 O 3 = H m,al2 O 3 m,cr 2 O 3 m,al2 O 3 J H m,cr2 O 3 = 107, 500 mol K 2538K 2324K = 117, 400J/mol K Note: H = H S + H mix, but for ideal solution, H mix = 0 Note: if heating of Cr 2 O 3 were required we read C ( ). Next calculate the change in entropy due to 1) melting and 2) mixing. arge liquid volume no change in X Cr2 O 3 H Al2 O 3 = 0 Focus on change in 1 mole of solid Cr 2 O 3. S = S m + S mix S m = H m m = = 46.26J/K S mix = R ln X F ( R ln X i ), where X i Cr 2 O 3 = 1 (assume γ 0 = 1) S mix = R ln(0.2) S = = 59.64J/K 67

68 2500 K Compostition iquid hase Diagram iquid + Solid Solution Solid Solution Compostition Figure 14.6: Cr 2 O 3 Al 2 O 3 hase Diagram 14.7 Excess roperties, relative to ideal µ k = R ln(a k ) = R ln(y k X k ), so we can write Ḡk = R ln γ }{{ k + R ln X }}{{ k } differenceideal behavior If γ k > 1, the excess free energy is positive. If γ k > 1, the excess free energy is negative. = ḠXS k + Ḡid k emperature and pressure derivatives of µ(, ) give the remaining properties in terms of γ k. (See table 8.5) ( ) ( ) H k = Ḡk S k = µ k µk = R 2 ln γk,µ k,n k V ( ) ( ) k = Ḡ n =,n [(R ln γ k + R ln X R )],nk = R ln γk 0 for an ideal gas k,n k 14.8 Beyond the ideal solution: the regular solution 1. he entropy of mixing is the same as for an ideal solution 2. he enthalpy of mixing is not zero, but depends on composition artial Molar Ḡk = µ k = H k S k otal Molar G mix = H mix S mix For ideal solution, H mix = 0, For regular : H mix = H XS mix (X 2,X 3,...X c ) For a regular solution, S reg mix = Sid mix, so SXS mix = 0 G XS mix = HXS mix S XS So ḠXS m mix = H k = R ln γ k We see that γ k can be evaluated from the heat of mixing: γ k = exp ( Hk R et s look at the simplest binary example: ) (check in eq 8.102) et H mix = a o X 1 X 2 = G XS mix, where a o is constant. his corresponds to example 8.1. he change in Gibbs Free energy. 68

69 G mix = G XS + G id = a o X 1 X 2 + R X 1 ln X 1 + R X 2 ln X 2 }{{} able 8.3 a o determines the departure from ideal behavior arising from interactions between components. A more realistic model (not symmetric and temperature dependent) of a binary system is H mix = X 1 X 2 [α o ( )X 1 + α 1 ( )X 2 ] = G XS mix et s examine the impact of interactions on H mix, S mix, and G mix, starting with the simpler regular solution. 6 0 All Entropy drives Mixing All All Figure 14.7: roperties of an ideal solution Figure 14.8: roperties of a regular solution For reference, not lecture Gibbs Duhem Equations ( ) S ln γk k = R ln γ k R ln X k R ( ) ln Ūk = R 2 γk R,n k ( ) ln γk ( ) F ln γk k = R ln γ k + R ln X k R,n k,n k,n k X 1 d µ 1 + X 2 d µ 2 = 0 µ k = R ln(a k ) 69

70 X 1 d ln γ 1 + X 2 d ln γ 2 = 0 a k = γ k X k X2 X 2 d µ 2 µ = 0 X 1 X2 ln γ 1 = X 2 X 1 dx 2 dx 2 d ln γ 2 dx X 1 dx 2 2 X2 X 2 d ln a 2 ln a 1 = dx 0 X 1 dx 2 2 ( ) S µk k =,n k ( ) H µk k = µ k +,n k Ḡk = µ k = H k S k G mix = H mix S mix For an ideal solution, H mix = 0, for a regular solution S reg mix = Sid mix G XS mix = Hmix XS Smix XS S XS mix = 0 he entropy of mixing for a regular solution is the same as for an ideal solution (hence S XS = 0) he enthalpy of mixing for a regular solution deciates from that of an ideal solution ( H id = 0, H XS 0). 70

71 Example: HW 8.4 One mole of solid Cr 2 O 3 at 2500 K is dissolved in a large colume of a liquid Raoultian solution (2300 K) of Al 2 O 3 and Cr 2 O 3 (20%). Calculate the resulting H and S given m,cr2 O 3 = 2538K H m,al2 O 3 = 107, m (= 2324 K) S m,al2 O 3 = S m,cr2 O 3 Recall that under isobaric, reversible conditions, S α β = Hα β H m,1 m.1 = H m,2 m,2 H m,cr2 O 3 = H m,al2 O 3 m,cr 2 O 3 m,al2 O 3 J H m,cr2 O 3 = 107, 500 mol K 2538K J = 117, K mol K Note: if heating of Cr 2 O 3 were required, we need C ( ). Next calculate changes in entropy due to 1) melting and 2) mixing. arge liquid volume no change in X Cr2 O 3 H Al2 O 3 = 0 Focus on change in 1 mole of solid Cr 2 O 3 S = S m + S mix S = H m m = = 46.26J/K S mix = R ln X F ( R ln X i ), where X i Cr 2 O 3 = 1 S mix = R ln(0.2) S = = J/K Mixtures of Real Gases: Fugacity Describe real gases in terms fo their deviation from ideal behavior, starting with molar volume V k Define α k = V k R (note that α k = 0 ideal) he change in chemical potential with mixing is then Introduce fugacity f k in µ k = k (α k + R k ( ) )d = α k d k + R ln 71

72 ( ) µ k Rln fk Note: a k = f k aking the exponential, we get [ µ ] f k = k exp α k d As α k 0, f k k Determination of fugacity (or a k ) finding µ(, ) Key goal of Sequence: redict Microstructure Figure 14.9: Calculated and observed microstructures Figure 14.10: Multicomponent heterogeneous (multi-phase) system 72

73 15 Multicomponent Heterogeneous Systems Outcomes for this section: 1. Write a combined statement of the first and second law of thermodynamics for multiple phases. 2. State the conditions for equilibrium for an arbitrary number of phases and components. 3. State the Gibbs phase rule and apply it to the description of unary and binary phase diagrams. 4. Given a single component phase diagram in terms of pressure and temperature, construct alternative representations that enable determination of phase fractions in terms of volume of mole fraction. 5. Given a two-component phase diagram in terms of pressure, temperature, and activity, construct alternative representations that enable determination of phase fractions in terms of volume or mole fraction. 6. Use the lever rule to compute phase fractions from temperature versus composition diagrams. 1. Define conditions for equilibrium 2. Derive Gibbs hase Rule 3. Structure of 1 and 2 component phase diagrams 4. he ever Rule Consider V = R (ideal gas) here are two degrees of freedom: E.g. choose, (independent) V (dependent) is determined Equilibrium between 2 phases implies that the system is in equilibrium. Homogeneous system: du = ds (δq) dv (δw ) + C k=1 µ kdn k (dn) Figure 15.1: Multicomponent heterogeneous and homogeneous systems sum over phases sum over components Heterogeneous System: du sys = α=1 α ds α α dv + C k=1 µα k dnα k 73

74 Figure 15.2: Multicomponent heterogeneous (multi-phase) system 15.1 Conditions for Equilibrium α = β α = β µ α = µ β likewise for all pairs of phases β/γ, γ,..., where =number of phases I = II =... = 1 independent equations I = II =... = 1 independent equations µ I 1 = µii 1 =... = µ 1 µ I 2 = µii 2 =... = µ 2 µ I C = µii C. =... = µ C C = # components Equilibrium conditions generate (C + 2)( 1) = n equations. hese are constraints on the variables 15.2 Gibbs hase Rule F = C + 2 F = m n is the number of degrees of freedom For n equations relating m variables. What is m? I, I, X2 I, XI 3...XI C 2 + C 1 = C + 1 II, II, X2 II, 3...XII C..,, X2, X 3...X C sets of C+1 variables m = (C + 1) m n = (C + 1) (C + 2)( 1) = C + C 2 + C + 2 m-n=c-+2 Gibbs hase Rule 74

75 16 hase Diagrams Outcomes for this section: 1. Apply a common tangent construction to define the regions of phase stability in a G versus X diagram. 0, α β 2. Use a single component G diagram to compute Gk. 3. Given G versus X at different, construct a phase diagram in terms of and X. 4. Describe the influence of interactions (in a simple regular solution model) on the features of a phase diagram, including phase boundaries, their curvature, and regions of two-phase coexistence. 5. Describe the origins and consequences of a miscibility gap Unary hase Diagrams C = 1 F = C + 2 In single phase regions ( = 1), there are two degrees of freedom (e.g.,) E G Figure 16.1: Unary (C=1) phase diagram 1. and required to specify state in single-phase region. 2. For 2-phase region, fixes and (and vice versa) 1 degree of freedom (F = = 1) 3. For triple point, F = 0 Only three phases are possible for a single component diagram. 75

76 Alternative Representations V, X, etc. Not all useful variables (e.g. V,X) are thermodynamic potentials 2-phase coexistence lines areas G riple oint V() Figure 16.2: Alternate representations o describe: 1. Increasing, which increases V 2. Mixture of phases in regions with tie lines, average molar volume. 3. riple points 4. Changes in enthalpy and phase fraction upon heating Binary hase Diagrams C = 2 here are 5 possible phases: α, ɛ, β,, V F = C + 2 For a single phase region, f = 3 lot in,, a 2 space. 2-phase regions are planes, 3-phase regions are lines, and 4-phase regions are points. Q G Figure 16.3: Binary (c=2) phase diagram Alternative Representations: vs. X areas. In Figure 16.4 below, the 2-phase coexistence lines are 76

77 Figure 16.4: vs X for binary phase diagram 16.3 Interpreting hase Diagrams: ie ines For plots with two degrees of freedom, a point on the diagram corresponds to a state of the system. 1. Which phases are present (and #) in equilibrium 2. he state of the phases (,, or other variables). 3. For plots with V or X, the relative amounts. Example: State with average composition X2 o liquid ( Σ ). at temperature. he phase Σ is in equilibrium with the n X o 2 = n Σ X Σ 2 + n X 2 X o 2 = nσ n XΣ 2 + n n X 2 = f Σ X Σ 2 + f X 2 where n= # of moles and f= fraction In terms of X 2, X o 2 = f Σ X Σ 2 + (1 f Σ )X 2 = X 2 + f Σ (X Σ 2 X o 2) f Σ = Xo 2 X 2 X Σ 2 X 2 and f = Xo 2 XΣ 2 X 2 XΣ 2 hese fractions can be derived from the lengths of segments of a tie line he ever Rule f = Xo 2 XΣ 2 X 2 XΣ 2 f Σ = Xo 2 X 2 X Σ 2 X 2 = A AB = B AB 77

78 ie lines A B 0 1 Figure 16.5: ie lines on an X- diagram X Σ 2 X 2 = AB X Σ 2 X o 2 = A X o 2 X 2 = B X- diagrams gives temperature, compositions, and amounts Applications of hase Diagrams: Microstructure Evolution Following an isotherm in the X- diagram of Figure shows the following microstructures: o : homogeneous melt 1 :α + (nucleation) 2 : increase in α phase 3 : solidification - α + β 17 Review G mix = H mix S id mix G mix = G XS mix + R (X 1 ln X 1 + X 2 ln X 2 ) 78

79 Figure 16.6: Microstructure Evolution Figure 17.1: Gibbs free energy of mixing for ideal vs real mixtures G ideal mix = R (X 1 ln X 1 + X 2 ln X 2 ) H mix 0: interactions change Gibbs free energy of mixing : particle kinetic energy increases, influence of interactions decreases, behavior is more ideal ypes of α phase diagrams H mix = a o X 1 X 2 Note the curvature of the phase boundary in Figure 17.2 below - the sign follows that of a 2 o. 18 hermodynamics of hase Diagrams Goal: Generate phase diagrams from G vs. X curves Observe: how interactions (changing potential energy) and temperature (changing kinetic energies) influence characteristics of phase diagrams. Approach: find common tangent lines between G α mix and G mix to determine regions of phase stability. Needed: comparison of solid and liquid reference states 79