§3 Phase Change of An Unary System

1. §3 Phase Change of An Unary System § 3.1 Criterion of thermodynamic equilibrium 3.1 .1 Entropy criterion Virtual change; Virtual displacement The sufficient and necessary condition of stable equilibrium state for an isolated system reads We also have equilibrium condition stable equilibrium condition maximal maximum corresponding to stable equilibrium smaller maximum corresponding to metastable equilibrium If , this state called neutral equilibrium

2. 3.1.2 Free energy criterion The sufficient and necessary condition of stable equilibrium state for an isothermal and isometric system reads We also have equilibrium condition stable equilibrium condition

3. 3.1.3 Gibbs function criterion The sufficient and necessary condition of stable equilibrium state for an isothermal and isobaric system reads We also have equilibrium condition stable equilibrium condition

4. 3.1.4 The application of criterion of thermodynamic equilibrium Denotes internal energy change with δU and δU0and volume change with δV and δV0for a subsystem and medium respectively, the total system is isolated, so Entropy follows that

5. For an isolated system, it holds that Based on the basic thermodynamics equation with formula (3.1.7) and (3.1.8), we obtain that So

6. Suppose Since , so that Then stable equilibrium condition

7. §3.2 The Thermodynamics Basic Equations of A Open System Unary system(homogeneous); binary system; heterogeneous(monophase) phase system For a open system, we have that where called chemical potential Because where g is mol Gibbs function, then

8. Based on formula (3.2.2), we obtain that According to U=G + TS – pV, the complete differential of internal energy follows that This is the thermodynamics basic equation of a open system. Similarly we have that

9. Define a new thermodynamics function J called grand thermodynamics potential Whose complete differential follows that

10. §3.3 Heterogeneous Equilibrium Conditions of An Unary System Two phases The total energy, total volume and total number of mol are constants. Suppose a virtual change in a isolated system, then

11. Entropy changes of two phases follow that Total entropy change When the system is in equilibrium state , it holds that

12. Namely Stable heterogeneous equilibrium conditions of the unary system also are expresses as follows

13. §3.4 Equilibrium Properties of A Unary Heterogeneous System Equilibrium phase change : If total system is in thermodynamic equilibrium, we have that

14. We want to derive a general equation to determine the vapor pressure of a liquid in equilibrium with its vapor. If we assume T and p to be given, we have that then Due to where s, v is mol entropy and mol volume, so

15. or Due to we thus obtain This is the Clapeyron equation.

16. §3.5 Critical Point and Gas-Liquid Phase Transition 3.5.1 Isotherm and gas- liquid phase transition The van der Waals’ equation of state for one mol reads Figure 3.4.1 Critical point and critical isotherm for CO2

17. 3.5.2 What is stable equilibrium state? With the complete differential of chemical potential The change of chemical potential for two states at isotherm follows that According to the Gibbs function criterion , the Gibbs function of equilibrium state should be least with certain T and P. So every point at line OKBAMR represents a stable equilibrium state .

18. 3.5.3 Maxwell construction State B is a completely vapor state; State A state is a completely liquid state; One reads that It follows that or area(BND)=area(DJA) (3.5.5) This is the well-known Maxwell construction. Remark: (1)Every state at line JDN does not come true because its Gibbs function is maximum with certain T and p. (2) Every state at line BN and AJ satisfies equilibrium stability with smaller Gibbs function and can be observed by experiments response to supersaturation vapor and superheat liquid.

19. 3.5.4 Critical Point Because of the importance of the critical point we want to calculation the critical state quantities Tc, pc ,Vc from van der Waals’ equation. The critical point is characterized by the fact that both derivatives vanish(saddle point): One gets The critical state quantities are therefore uniquely determined by the parameters a and b. Hence, for all gases one should have

20. Introduce new variables called contrastive temperature, contrastive pressure and contrastive volume, the van der Waals’ equation follows that This is van der Waals’ contrastiveequation. Remark! Correspondence state law: the state equations are the same for any matter.

21. §3.6 Liquid –Drop Formation 3.6 .1 The equilibrium condition of Liquid –drop Formation Assume liquid –drop at α phase; vapor at β phase; surface at γ phase, the basic thermodynamics equations of three phases read Thermal equilibrium condition for three phases reads Suppose one virtual change at the system, it holds that

22. The changes of free energy of three phases follow that Therefore we have Suppose the shape of liquid-drop is ball,

23. It follows that According to free energy criterion, it holds that The formula (3.6.6) is mechanics equilibrium condition. The formula (3.6.7) is phase change equilibrium condition.

24. 3.6.2 The question about Liquid –drop formation Assume p expresses the equilibrium pressure of the vapor- liquid when liquid surface is even. This formula can determine the relation between saturation vapor pressure and temperature. Suppose p’ expresses the pressure of vapor when vapor and liquid phases reach equilibrium and liquid surface is curving. The condition of phase transition equilibrium follows that

25. It holds that Suppose vapor is the ideal gas. According to the formula (2.4.15) chemical potential follows that then Insert formulas (3.6.10) and (3.6.12) into the formula (3.6.9) , with formulas (3.6.8) one gets

26. In general speaking, ,thus it approximatively gets Under certain pressure, the radii of liquid –drop at equilibrium state reaches This is called relevance radii. Remark! • When r>rc, then , liquid –drop will be increscent; r<rc, then , liquid –drop will be disappear (2) The condensation core is needed for the liquid condensation; otherwise supersaturated vapor would be formed.

27. For a bubble in the liquid, it gets Base on the formulas (3.6.16) and (3.6.17) one can understand the super-heating phenomenon.

28. §3.7 The classification of phase transitions One order phase transition be characterized by (1)Potential heat of phase transition (2) Specific volume break (3) Possibly appear metastable. (4)

29. two order phase transition be characterized by (1) Chemical potential continuous (2) One order partial derivative continuous (3) No potential heat of phase transition and specific volume break (4) Two order partial derivative with break

30. Alan-Faith equation

31. §3.8 Critical Phenomenon and Critical Exponent Let 3.8.1 Liquid-gas fluid system at critical point • the density of critical point β , with value 0.34 given by experiment,iscalled a critical exponent. (2) The isothermal compressibility but the proportion coefficients of the two formulas are not the same.

32. 3.8.2 Ferromagnet at critical point (1) magnetization β , with value given by experiment,iscalled a critical exponent. m=0 (T>Tc) (2) susceptibility ( with zeroth magnetic field) but the proportion coefficients of the two formulas are not the same.

33. (3) When t=0, the relation between magnetization m and in addition weak magnetic field follows that , with value , given by experiment,iscalled a critical exponent. (4) When , specific heat(H=0) given by experiment . but the proportion coefficients of the two formulas are not same.

34. §3.9 Landau Continuous Phase Transition Theory T> Tc: perfect symmetry and low order with ordered parameter zero T< Tc: low symmetry and high order with ordered parameter non-zero

35. 3.9.1 monaxial ferromagnetic Duo to ordered parameter m is a small quantity, the free enegy of the sysytem near Tc follows that For stable equilibrium state, we have The formula (3.9.2) have three solutions:

36. m=0, expresses non-ordered state, corresponds to the case with T> Tc and a>0. , expresses ordered state, corresponds to the case with T< Tc , a<0 and b>0. thus it gets a=0 with T= Tc . suppose and Thus we have Comparing the formula (3.8.1’), we have .

37. Based on the formulas (3.9.1) and (3.9.4), one gets with specific heat formula , it follows that At critical point there is a break of specific heat with critical exponent α=0.When , the Gibbs function G holds that

38. Assume H is very weak, and neglect a and b change with H , then Derive the partial derivative for H, susceptibility follows that Comparing the formula (3.8.2’), we have γ= γ’ =1.

39. When T=Tc, a=0, the formula holds that Comparing the formula (3.8.3’), we have δ =3. Remark! The critical exponents given by Landau theory are not good agreement with experiments. Fluctuating need to be considered.