1. Chapter 3 Phase Transitions and Chemical Reactions

2. Phase Transitions and Chemical Reactions 1 Gibbs‘ Phase Rule We now want to return to the important problem of how many state variables are actually necessary to uniquely determine the state of a system. To this end,we start from an isolated system which contains k different particle species(chemical components)and p different phases (solid,liquid,gaseous,...).Each phase can be understood as a partial system of the total system and one can formulate the first law for each phase,

3. ln this formulation of the first law,U(i)of phase i is a function of the extensive state variables ;I.e.,it depends on k+2 variables. Altogether we therefore have P(K+2) extensive state variables.If the total system is in thermodynamic equilibrium,we have in addition the following conditions for the intensive state quantities,cf.Equations (2.45-48)

4. Thermal equilibrium Chemical equilibrium Mechanical equilibrium

5. Thermal equilibrium Chemical equilibrium Each line contains P-1 equations,so that equation is a system of (P-1)(K+2) equations. Thus,we only require extensive variables to determine the equilibrium state of the total system.As we see, this number is independent of the number of phases. Mechanical equilibrium

6. If we now consider that exactly,so,we can substitute for.Here And So that U(i)of phase i can be writen a function of the extensive state variables I.e.,it depends on k+1 variables. Thus,we only require independent intensive variables.Equation(3.4)is named after J.W.Gibbs and is called Gibbs'phase rule. (3.4)

7. Example 3.1: Clausius-Clapeyron equation We want to derive a general equation to determine the vapor pressure of a liquid in equilibrium with its vapor. We have the following equilibrium conditions for two partial systems which can exchange energy,volume,and particles:

8. Because of the Gibbs-Duhem relation these conditions are not independent from each other: if the equation of state is known and if we assume T and p to be given,we can calculate and · The equation The equation (2.74 ） （ 3.12 ） yields a dependence between p and T;i.e.,we can calculate the vapor pressure for a given temperature.

9. If we change the temperature by dT in Equation(3.12), the vapor pressure also has to change by a certain amount dp to account for equilibrium. For the corresponding changes and it must hold that This can be expressed with the Gibbs-Duhem relation in the following way (1) (2)

10. We assume that and analogously for the vapor: (1)=(2) This is the Clausius-Claperon equation.

11. However,in many cases and for not too large temperature differences, this evaporation heat may be considered to be constant. With the evaporation heat per particle we thus obtain (3.13) In many cases and one has (3.14)

12. The intensive quantities and can of course be inserted as measured per mole instead of measured per particle. Figure 3.1 Phase diagram of water

13. Exercise 3.2 vapor pressure of a liquid Determine the vapor pressure of a liquid in equilibrium with its vapor under the assumption that the evaporation heat per particle does not depend on pressure or temperature and that the vapor behaves as an ideal gas. Solution:The best starting point is Equation(3.14): With we obtain

14. We may integrate this after separating variables,for instance from an initial temperature T 0 with vapor pressure P o to a final temperature T with vapor pressure p, Or (3.15)

15. Thus,the vapor pressure increases strongly with temperature( ). Note that Equation (3.15)holds under the same conditions also for the sublimation pressure of a solid: is correct,but the sublimation heat is larger. Thus the pressure curve for sublimation is steeper than the vapor pressure curve.Both curves cross at the triple point.

16. 2 Phase equilibrium and the Maxwell construction When we introduced van der Waals‘ equation of state we already mentioned some inconsistencies of this equation.The isotherms of van der Waals’ equation (Figure 3.2), Figure 3.2 Isotherms of the van der Waals gas.

17. show regions of We now want to show that these contradictions can be resolved by considering the phase transition from gas to liquid. In equilibrium between vapor and liquid, however, a certain vapor pressure is (3.17 ） The vapor pressure is solely a function of temperature and does not depend on the vapor volume V,so that one obtains a horizontal isotherm in the pV diagram.

18. The pressure can be calculated from Equation (3.17), if the temperatures and chemical potentials of the vapor and liquid are known. In the first case we simply have ( =T(S 2 -S 1 ) is the latent heat of the phase transition.

19. and in the case of the van der Waals isotherm we have (3.21) From the condition (3.22)

20. Remark:For a given and T the van der Waals isotherm has also a third (unstable)solution at C,Figure3.3. Figure3.3.Maxwell construction

21. Figure 3.4 Critical point and critical isotherm for CO 2

22. Because of the importance of the critical point we want to calculate the critical state quantities,and from van der Waals ‘ equation.The critical point is characterized by the fact that both derivatives vanish (saddle point): This is the well-known Maxwell construction. k called critical point

23. If one brings the negative terms to the other sides of the respective equations and divides one equation by the other.one obtain,and thus (3.25)

24. If one inserts this into equation (3.24),one gets From and it finally follows with van der Waals ’ equation that

25. The critical state quantities are therefore uniquely determined by the parameters a and b. Hence,for all gases one should have Experimentally one finds for Equation(3.28) numbers between 0.25 and 0.35,which once again confirms the qualitative usefulness of van der waals ‘ equation.

26. 3 Application of the laws of thermodynamics We want to calculate the internal energy U(V,T)of a real gas. The exact differential of U reads {3.49} We have already identified the expression as the heat capacity because of at V=const.

27. In most cases,we want to express in terms of T and p. To this end we denote the exact differential of the entropy S(V,T), for which on the other hand (3.51)

28. By comparing coefficients one finds and Since S has an exact differential it must hold that the result that

29. So that (3.56) Thus we have reached our aim to express by derivatives of the equation of state, since we can readily determine p=p(N,T,V)also for real gases.

30. If we insert Equation (3.56)into Equation (3.49)we have (3.57 ） For example: ideal gases

31. We will see in the next section that such relations are easily derived using the theory of transformations of variables for functions of more than one variable. Since dU is an exact differential one has (3.57 ）

32. However,the righthand side can be as well determined from the equation of state,so that we can calculate the volume dependence of the heat capacity. For an ideal gas one has,for instance, {3.59}

33. and thus {3.60} Therefore,the heat capacity of an ideal gas cannot depend on the volume.As we already how,it is even absolutely constant.

34. Exercise 3.8: internal energy of the van der Waals gas Calculate the internal energy of a van der Waals gas as a function of temperature and volume at constant particle number. Solution The equation of state of the van der Waals gas reads

35. We now evaluate the expression :

36. Hence,as for the ideal gas,the heat capacity of a van der Waals gas cannot depend on the volume because of Thus we have according to Equation (3.57)

37. We can integrate this starting from an initial state T 0 and p o with the internal energy U 0, For temperature differences which are not too large Cv(T)is approximately constant and thus

38. Exercise 3.9: Entropy of the van der Waals gas Calculate the entropy of a van der Waals gas as a function of temperature and volume at constant particle number. Solution According to Equations (3.50),(3.51),and (3.56),we have

39. The quantity for a van der Waals gas was calculated in the preceding Exercise;

40. if we insert Equation (3.61),we obtain Starting from a state T 0 and V 0 with entropy S o we can integrate this equation:

41. For temperature differences that are not too large ( ≈100 k)we have Cv ≈const.and thus The entropy of a van der Waals gas is nearly identical to that of an ideal gas;one only has to reduce the volume by the proper volume Nb of the particles.

42. Exercise : 1.There is a kind of material which the equation of state is, Try to prove that its internal energy cannot depend on the volume. 2.Calculate the entropy of a ideal gas as a function of temperature and volume at constant particle number.