MS212 Thermodynamics of Materials ( 소재열역학의이해 ) Lecture Note: Chapter 7

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1 2017 Spring Semester MS212 Thermodynamics of Materials ( 소재열역학의이해 ) Lecture Note: Chapter 7 Byungha Shin ( 신병하 ) Dept. of MSE, KAIST Largely based on lecture notes of Prof. Hyuck-Mo Lee and Prof. WooChul Jung 1

2 Course Information Syllabus Chap 1. Introduction and Definition of Terms (1 lecture) Chap 2. The First Law of Thermodynamics (2 lectures) Chap 3. The Second Law of Thermodynamics (4 lectures) Chap 4. The Statistical Interpretation of Entropy (2 lectures) Chap 5. Auxiliary Functions (2 lectures) Chap 6. Heat Capacity, Enthalpy, Entropy and the Third Law of Thermodynamics (3 lectures) Chap 7. Phase Equilibrium in a One-Component System (2 lectures) Chap 8. The Behavior of Gases (2 lectures) Chap 9. The Behavior of Solutions (3 lectures) Chap 10. Gibbs Free Energy Composition and Phase Diagram (3 lectures)

3 Equilibrium 1. The condition of a system must be time independent; the system must have achieved a stationary state. 2. The internal state of a system must not change if it is isolated from its surroundings.. If the first condition is obeyed, but the second one isn t, then the system is in a steady state rather than in thermodynamic equilibrium. Since, for an isolated system, no entropy can be transferred across the boundary, any entropy change must result from production inside the system. The second law guarantees that this change is positive, i.e., the total entropy of the system can only increase while the system approaches equilibrium. à criterion for equilibrium For an isolated system, the equilibrium state corresponds to the state with maximum entropy

4 Equilibrium δq=0 δw=0 δn i =0 T S U, P, V, µ i, n i T S U, P, V, µ i, n i du (*) + du (,) = 0 dv (*) = 0 = dv (,) dn 6 (*) = 0 = dn6 (,) (total system = isolated) (rigid) (impermeable) 1. Thermal contact via diathermal wall ds) $,&,' = ds (*) + ds (,) = 1 T (*) du(*) + 1 T (,) du(,) = 1 T (*) 1 T (,) du (*) 0 T (*) > T (,) du (*) < 0 T (*) < T (,) du (*) > 0 T (*) = T (,) ds = 0 (max. entropy) (Thermal equilibrium) (T: escaping tendency of U)

5 Equilibrium δq=0 δw=0 δn i =0 T S U, P, V, µ i, n i T S U, P, V, µ i, n i du (*) + du (,) = 0 T (*) = T (,) dv (*) + dv (,) = 0 dn 6 (*) = 0 = dn6 (,) 2. Mechanical contact via non-rigid wall ds) $,&,' = ds (*) + ds (,) = P(*) T dv(*) + P(,) T dv(,) = 1 T P(*) P (,) dv (*) 0 P (*) > P (,) dv (*) > 0 P (*) < P (,) dv (*) < 0 P (*) = P (,) ds = 0 (max. entropy) (Mechanical equilibrium) (P: escaping tendency of -V)

6 Equilibrium δq=0 δw=0 δn i =0 T S U, P, V, µ i, n i T S U, P, V, µ i, n i du (*) + du (,) = 0 T (*) = T (,) dv (*) + dv (,) = 0 P (*) = P (,) dn 6 (*) + dn6 (,) = 0 3. Contact via impermeable wall ds) $,&,' = ds (*) + ds (,) = μ * 6 T dn * 6 μ, 6 T dn, 6 = 1 T μ6*, μ 6 μ 6 * > μ 6, dn 6 * < 0 μ 6 * < μ 6, dn 6 * > 0 μ 6 * = μ 6, ds = 0 (max. entropy) (Diffusive equilibrium) dn 6 * (µ i : escaping tendency of n) 0

7 Equilibrium Criteria V (*) n 6 (*) A (*) G (*) T R V (,) n 6 (,) A (,) G (,) T (*) = T (,) = T < T R, P R At constant T, da SdT PdV μ 6 dn 6 ; da) C = da (*) + da (,) = P * dv * μ 6 (*) dn6 (*) = P * P, dv * μ 6 * μ 6, da) C PdV μ 6 dn 6 + P, dv, μ 6 (,) dn6 (,) dn 6 (*) 0 At equilibrium, da) C = 0 P 1 = P 2 ; μ i 1 = μ i 2 At constant T and P, V (*) n 6 (*) A (*) G (*) V (,) dg SdT + VdP μ 6 dn 6 (,) n 6 A (,) dg) C,J = dg (*) + dg (,) μ * 6, μ 6 (*) dn 6 0 G (,) At equilibrium, dg) C,J = 0 μ 1 i 2 = μ i T (*) = T (,) = T < P (*) = P (,) = P <

8 Component and Phase Component: substance with fixed proportion of elements e.g., Water(H 2 O): 2 Elements (H & O) 1 Component (H 2 O: composition fixed) Phase: any portion that is physically homogeneous within itself, bounded by a surface so that mechanically separable P=1atm K K ice얼음 water 물 수증기 steam (single phase) ( 단일상 ) (single ( 단일상 phase) ) (single ( 단일상 phase) ) ice 얼음 + water + 물 (two phases) water 물 + 수증기 + steam (two phases) (2 상 ) (2 상 )

9 Phase Diagram of H 2 O (water): Graphical representation of the equilibrium state of a system (P, T as variables in a one-comp system) Heat Capacity, C P : C P, solid = 38 J/mol K at 273K C P, liquid = 76 J/mol K at 273K C P, gas = 36.5 J/mol K at 373K Heat of Formation, DH o H2O: DH o H2O, solid DH o H2O, liquid DH o H2O, gas = kj/mol = kj/mol = kj/mol Heat of Transition, DH trans : DH melt DH vap = 6.01 kj/mol = kj/mol Entropy of Substance, S o H2O: S o H2O, solid S o H2O, liquid S o H2O, gas = 41 J/mol K = J/mol K = J/mol K Entropy Change of Transition, DS trans : DS melt DS vap = 22.0 J/mol K = J/mol K

10 Equilibrium between Ice and Water Ice and water are in equilibrium with one another at 0 o C and 1 atm. à G of ice + water has its minimum value! Heat in: some ice water at 0 o C and 1 atm, no change in G Heat out: some water ice at 0 o C and 1 atm, no change in G If, by the addition of heat, 1 mole of ice is melted, H 2 O (solid) à H 2 O (liquid) at 0 o C and 1 atm DG = G H2O(l) G H2O(s) = 0 G H2O(l) = G H2O(s) For the system of ice + water containing n moles of H 2 O, ( n = n H2O(s) + n H2O(l) ) G = n H2O(s) G H2O(s) + n H2O(l) G H2O(l) = [ n at 0 o H2O(s) + n H2O(l) ] G H2O(s,l) C and 1 atm the ratio of sol/liq changes but no change in G

11 Equilibrium between Ice and Water μ 6 = G n 6 O C,J G L n 6 μ 6 ; or dg ) C,J = μ 6 dn 6 integrate drop-by-drop of species i at constant T, P (for a single component system, G = nμ ) The chemical potential of a species in a particular state (such as fixed T, P) equals the molar Gibbs free energy of the species in the particular state. G = n H2O(s) µ H2O(s) + n H2O(l) µ H2O(l) ; dg ) T,P = µ H2O(s) dn H2O(s) + µ H2O(l) dn H2O(l) At 1 atm and T > 0 o C, ice spontaneously melts: dg ) T>273K,P=1atm = µ H2O(s) dn H2O(s) + µ H2O(l) dn H2O(l) = [ µ H2O(s) + µ H2O(l) ] dn H2O(l) < 0 µ H2O(s) > µ H2O(l) ( or G H2O(s) > G H2O(l) ) At 1 atm and T < 0 o C, water spontaneously freezes: dg ) T<273K,P=1atm = µ H2O(s) dn H2O(s) + µ H2O(l) dn H2O(l) = [µ H2O(s) µ H2O(l) ] dn H2O(s) < 0 µ H2O(s) < µ H2O(l) ( or G H2O(s) < G H2O(l) )

12 Shape of a Function from 1 st & 2 nd Derivatives If the first derivative f' is positive (+), then the function f is increasing ( ) If the first derivative f' is negative (-), then the function f is decreasing ( ) If the second derivative f'' is positive (+), then the function f is concave up (U) If the second derivative f'' is negative (-), then the function f is concave down ( ) f < 0; f > 0 f > 0; f > 0 f < 0; f < 0 f > 0; f < 0

13 Variation of G with T at constant P (µ) G T O J, G T,R J = S < 0 = S T O J = c T T < 0 S H2O(l) > S H2O(s) The liquid phase is more disordered than is the solid phase. à steeper slope in G vs. T

14 Variation of G with T at constant P G H TS G (W) = H (W) TS W G (X) = H (X) TS X G (X W) = H (X W) T S X W

15 Variation of G with P at constant T G P O C = V > 0, G P,R C = V P O C = Vκ T < 0 G (X) P O C = V (X) and G (W) P O C = V (W) G (X W) O P C = V (X W) For most solids, V (X W) > 0 Very unusual, H 2 O, Sb, Bi, I 2, V (X W) < 0

16 G as a function of T and P For equilibrium to be maintained: For any infinitesimal change in T and P dg (W) = dg (X) dg (W) = S W dt + V W dp dg (X) = S X dt + V X dp G (W) = G (X) at equilibrium, H = T \] S dp dt \] = S (W) S (X) V (W) V (X) = S (X W) V (X W) = Clapeyron Equation H (X W) T \] V (X W) For H 2 O, V (X W) < 0 & H X W > 0 dp dt \] < 0 Ice-skating is possible. Why?

17 Equilibrium b/w Vapor & Condensed Phase V (_`ab cde) = V cde V _`ab V cde ; dp dt O \] = H (_`ab cde) TV cde if the vapor behaves ideally, e.g., PV = RT dp dt O \] = P H RT, dp P = d ln P = H RT, dt Clausius-Clapeyron if H is independent of T, ln P = H RT + constant Here, P is equilibrium pressure for solid (liquid) vap. P is saturated vapor pressure exerted by the condensed phase at T. P relates to the tendency of particles to escape from the solid (or a liquid). A substance with a high vapor pressure at normal temperatures is often referred to as volatile. The vapor pressure of a single component in a mixture in the system is called partial pressure.

18 Equilibrium b/w Vapor & Condensed Phase if DC P f(t), H = H,op + c J T 298 = H,op 298 t c J d ln P = H RT, dt ln P = H,op 298 c J R 1 T + c J R + T c J ln T + const ln P = A + B ln T + C T

19 Phase Diagram of H 2 O dp dt O \] curve for 2-phase equilibrium dp dt = H T V AOA : s l BOB : l v COC : s v At a point O (triple point): 3-phase coexistence, s + l + v Gibbs phase rule: F = C P + 2 (C: # of component, P: # of phase, F: # of degrees of freedom) e.g. at the triple point, F = = 0

20 Course Information Gibbs-Duhem equation du = TdS PdV μ 6 dn 6 (*) intensive extensive Integrate by bringing together many small identical systems with the same intensive variables: U U 0 = TS PV μ 6 n 6 ; U 0 = 0 Take total differential: du = TdS + SdT PdV VdP dμ 6 n 6 μ 6 dn 6 Comparison with (*): w 0 = SdT VdP n 6 dμ 6 Gibbs-Duhem Equation intensive extensive 6x*

21 Course Information Gibbs-Duhem equation 0 = SdT VdP n 6 dμ 6 w 6x* Gibb-Duhem equation says intensive variables are not all independently variable within a phase. If you change T and P and all µ i except one, the last one is set automatically. This equation can be applied to each phase separately, or to the whole system consisting of multiple phases. From this comes the simplest derivation of the Gibbs phase rule One Gibbs-Duhem equation for each of P phases Variables are T, P, and one chemical potential for each of C species P equations & (C + 2) unknowns # of variations (F) in intensive parameters which may be made arbitrarily while remaining in equilibrium is C + 2 P F = C P + 2 Gibbs Phase Rule

22 Phase Diagram (H 2 O) G G T T

23 Phase Diagram (H 2 O) d G T

24 Phase Diagram (H 2 O) G G P P

25 Phase Diagram (H 2 O) G P

The Second Law of Thermodynamics (Chapter 4)

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