1. ENZO ZANCHINI Università di Bologna enzo.zanchini@unibo.it +39 051 20 93295 AVAILABILITY FUNCTIONS AND THERMODYNAMIC EFFICIENCY 2. Conditions for mutual stable equilibrium and fundamental relation for an open system Bologna - September 2013 1

2. THEOREM 9: NECESSARY CONDITION FOR MUTUAL STABLE EQUILIBRIUM PROOF IN APPENDIX Consider two closed simple systems A e B, which occupy two fixed regions of space R A and R B with volumes V A and V B. Then, the following conditions are necessary for A and B to be in mutual stable equilibrium: their states A 1 e B 1, with energy values U A 1 and U B 1, are stable equilibrium states; the temperatures of A and B are equal, namely T A 1 = T B 1 ; there exists a positive energy quantity  such that if -  <  <  (2.1) 2 POSTULATE 3 Every closed system, with fixed regions of space occupied by its constituents, is in mutual stable equilibrium with a duplicate of itself which is in the same stable equilibrium state.

3. THEOREM 10 For the set of the stable equilibrium states of a closed simple system with fixed volume, the temperature is a strictly increasing function of the internal energy. Let A se1 be a stable equilibrium state of a closed simple system A, with a region of space R A occupied by its constituents and volume V A, and with internal energy U A 1. Let A d be an identical copy of A and let A d se1 be the stable equilibrium state of A d identical to A se1. By Postulate 3, the state C 1 = (A se1, A d se1 ) is a stable equilibrium state of C. Therefore, by Theorem 9, since A and A d have identical fundamental relations, there exists a positive energy quantity  such that, for every 0 <  <  PROOF Therefore, in the neighborhood of A se1, for a fixed volume the temperature of A is a strictly increasing function of the internal energy of A. Since A se1 has been chosen arbitrarily, the conclusion is proved. (2.4) 3

4. Comment. Theorem 10 shows that the second necessary condition for the mutual stable equilibrium of two closed simple systems with fixed regions of space, given by Eqs. (2.1), is automatically fulfilled. For instance, for  > 0, by employing the equality T A 1 = T B 1 and Theorem 10 one obtains: (2.5) 4 NON EXISTENCE OF THERMAL RESERVOIRS A thermal reservoir should be a system which crosses stable equilibrium states with a fixed region of space and different values of the internal energy and remains in mutual stable equilibrium with a duplicate of itself (therefore, its temperature remains constant). This is in contrast with Theorem 10.

5. THEOREM 11: NECESSARY AND SUFFICIENT CONDITION FOR MUTUAL STABLE EQUILIBRIUM Let C = AB be a composite system, with A and B closed simple systems with fixed regions of space. A necessary and sufficient condition for C to be in a stable equilibrium state is that A and B are in stable equilibrium states with T A = T B. 5 ZEROTH LAW – Traditional statement A system A is in thermal equilibrium with a duplicate of itself (reflexivity); if A is in thermal equilibrium with B, then B is in thermal equilibrium with A (symmetry); if A is in thermal equilibrium with B and B is in thermal equilibrium with C, then A is in thermal equilibrium with C (transitivity). PROOF IN APPENDIX If thermal equilibrium is replaced by mutual stable equilibrium, symmetry is a consequence of the definition.

6. COROLLARY 3 A closed simple system C in a stable equilibrium state with temperature T can be considered as the union of an arbitrary number of closed simple subsystems, in mutual stable equilibrium, with the same temperature T. By definition, C can be transformed into the union of an arbitrary number of closed simple subsystems, without effects on the state. The subsystems are in mutual stable equilibrium; therefore, by Theorem 9, have the same temperature. The latter is T because the partitions do not change the state of C. PROOF ZEROTH LAW If A is in mutual stable equilibrium with B and B is in mutual stable equilibrium with C, then A is in mutual stable equilibrium with C. PROOF The conclusion is a trivial consequence of Theorem 11. 6

7. THEOREM 12 A closed system C in a stable equilibrium state can be considered as the union of an arbitrary number of closed simple subsystems, in mutual stable equilibrium, with the same pressure p. 7 As a consequence of Corollary 3 and Theorem 12, we say that, for a closed simple system in a stable equilibrium state, the properties T and p are intensive, i.e., have the same value for the system and for any subsystem. PROOF IN APPENDIX

8. Consider an open system O having with environment an open system Q: OQ is an isolated system. We say that O is separable from Q at time t if the state (OQ) t can be reproduced as a state of an isolated system AB, in the same external field, such that A and B are closed and separable at time t. If A and B are uncorrelated with each other at time t, namely (AB) t = A t B t, we say that O is uncorrelated with its environment Q at time t. SEPARABLE OPEN SYSTEM UNCORRELATED WITH ITS ENVIRONMENT SET OF ELEMENTARY CHEMICAL SPECIES We call set of elementary chemical species a set of constituents obtained by choosing, among all the chemical species formed by only one kind of atom, those which have the most stable structure in standard conditions of temperature and pressure. Examples: H 2, N 2, O 2, … (set of atomic species) 8

9. Consider O open, with r nonreactive constituents, O 1 state of O with composition n 1 e=e(n 1,en 2, …, n r ) 1. A n1 (auxiliary) closed system, with allowed reactions and compositions compatible with n 1. Let A 1 n1 be a state of A n1 which coincides with O 1. We define as energy and entropy of O, in state O 1, the values of the energy and the entropy of A n1 in the state A 1 n1. ENERGY AND ENTROPY OF A NONREACTIVE OPEN SYSTEM In the following, we will consider only states of an open system in which the system is separable and uncorrelated with its environment. The values of the energy and of the entropy of A n1 in state A 1 n1 are determined by choosing a reference state A 0 n1 and by applying the definitions of energy difference and of entropy difference between two states of a closed system. 9

10. The reference state A 0 n1 and the values of energy and of entropy in that state are chosen as follows. One chooses A n1 as union of s closed subsystems, A 1,eA 2,e…, A s, each of which contains only one elementary chemical species; the subsystem are chosen so that the composition of A n1 is compatible with n 1. Each subsystem A i contains n i particles of the i-th elementary chemical specie and is constrained within a spherical box; each box is very far from the others and is placed in a region of space where the external force field is vanishing. The reference state A 0 n1 is chosen so that each subsystem A i is in a stable equilibrium state A i 0 with temperature T 0 and pressure p 0. The values of energy and entropy in the reference state A 0 n1 are defined as Usually, one chooses: 10 or

11. 11 EXAMPLE 1 mol H 2 0.5 mol O 2 T 0 = 25 °C; p 0 = 1.01325 bar T 0 = 25 °C; p 0 = 1.01325 bar Suppose that the composition of O 1 is one mol of H 2 O, and that H 2 O is liquid at T 0 = 25 °C and p 0 = 1.01325 bar The composition of the auxiliary closed system A n1 must be compatible with one mole of H 2 O; the reference state A 0 n1 is chosen as follows: The internal energy of O 1 is the difference in internal energy between the state A 1 n1 (1 mol of H 2 O at (T 0, p 0 )) and the reference state shown above (internal energy of formation; in practice, the enthalpy of formation is reported in tables). The same holds for the entropy of O 1 (entropy of formation)

12. STABLE EQUILIBRIUM STATE OF AN OPEN SYSTEM A state of an open system O is called a stable equilibrium state if it can be reproduced as a stable equilibrium state of a closed system A in the same external field. FUNDAMENTAL RELATION FOR A SIMPLE OPEN SYSTEM WITH NO REACTIONS Let SE O be the set of the stable equilibrium states of a simple open system O with r nonreactive constituents. Let SE O -n 1 be the subset of the states of SE O which have composition n 1 and let A n1 a closed system with composition n 1 such that its stable equilibrium states coincide with those of SE O -n 1. Every set of equivalent stable equilibrium states of O with composition n 1 coincides with a set of equivalent stable equilibrium states of A n1, which is determined uniquely by the internal energy U and the volume V of A n1. The same argument holds for every composition of O. Therefore, on the whole set SE O is defined a relation of the kind Fundamental relation of O (2.7) 12

13. GIBBS EQUATION FOR AN OPEN SYSTEM Fundamental relation for O(2.7a) By differentiating the fundamental relation (2.7b) one obtains where is called chemical potential of the i-th constituent. Equation (2.8) is called Gibbs equation for the simple open system O. (2.9) (2.8) Fundamental relation for O (2.7b) 13

14. DIFFERENTIAL OF THE ENTHALPY AND OF THE GIBBS FREE ENERGY FOR A SIMPLE OPEN SYSTEM Since H = U + pV, one has also (2.8) (2.10) Since G = H – TS, one obtains (2.11) 14

15. THEOREM 13 A closed or open system O, in a stable equilibrium state O 1se, can be considered as the union of an arbitrary number of simple open subsystems, in mutual stable equilibrium, which have the same chemical potential as O 1se, for each constituent. 15 As a consequence of Theorem 13, we say that, for an open simple system in a stable equilibrium state, the chemical potentials of the constituents are intensive properties, namely have the same values for the system and for any subsystem. PROOF IN APPENDIX

16. EULER EQUATION Let us consider a portion of an open simple system, in stable equilibrium, with nonreactive constituents. Let us apply the fundamental relation to evaluate the difference in energy between the portion considered and a larger one. Since T, p and  are uniform, one has Let us assume that the second portion is k times the first. Then, the internal energy of the second is k times the first, so that the difference in internal energy  U between the second portion and the first is k-1 times the value of the internal energy U of the first. The same holds for  S,  V,  n i. Therefore, from (2.13) one obtains (2.13) hence, dividing by k-1 Euler equation (2.14) 16

17. Since H = U + pV, from Euler equation one obtains Since G = H - TS, one obtains Euler equation(2.14) (2.15) (2.16) In the case on one constituent, with reference to one mole (2.17) 17

18. CHEMICAL POTENTIAL OF AN IDEAL GAS Consider a set of stable equilibrium states of an ideal gas with the same temperature T, and denote by  0 the chemical potential of the gas at the reference pressure p 0. Then In fact, for two states of an ideal gas, with the same temperature T (2.18) 18

19. CHEMICAL POTENTIAL OF THE i-th CONSTITUENT OF A MIXTURE OF IDEAL GASES In a Gibbs separation process of a mixture of ideal gases, U, H, S, T are conserved, and thus also G = H – TS is conserved. Therefore, one has where G’ is the Gibbs free energy of the mixture and G is that of the constituents, separated, at temperature T and at the partial pressures p i. The additivity of G for the separated constituents and Eq. (2.16) for the mixture yield, respectively (2.19) where  ’ i is the chemical potential of the i-th constituent of the mixture. (2.20b) (2.21) Since, by (2.19), Eqs. (2.20a) and (2.20b) must yield the same value for every choice of the mole numbers n i, one has (2.20a) 19

20. SIGN OF THE STOICHIOMETRIC COEFFICIENTS Consider the reaction, with constituents C A, C B, C M, C N : Usually, one writes Eq. (2.22) in the form (2.22) (2.23) or (2.24) In Eq. (2.24), the stoichiometric coefficients have a sign: they are negative for reactants and positive for reaction products. 20

21. THEOREM 14: CONDITION FOR CHEMICAL EQUILIBRIUM For a closed simple system A in a stable equilibrium state, with an allowed chemical reaction, the chemical potentials of the constituents (evaluated for the corresponding open system without reactions) fulfill the relation PROOF A is closed; the composition of A can change as a consequence of a reaction. Suppose that the reaction is stopped and the system is opened. Then, for every composition, for fixed values of the energy U and the volume V there exists a stable equilibrium state of A; moreover, for any pair of these states the Gibbs equation for A in the absence of reactions holds or: (2.25) 21

22. 22 reactants products elements Instead of the reaction, one can bring back the reactants to elements and build products by elements The changes in energy and entropy can be evaluated through the scheme of a nonreactive open system.

23. If we now consider the system closed and the reaction allowed, among all the states with the given set of compatible compositions and the given values of (U, V) only one is a stable equilibrium state: the highest entropy state. For this state, one has (2.25) (2.26) From (2.25) and (2.26) one obtains (2.27) Since A is closed, one has(2.28) By substituting Eq. (2.28) in Eq. (2.27) one obtains the conclusion. 23

24. 24 APPENDIX OMITTED PROOFS

25. 25 PROOF OF THEOREM 9 If either A 1 or B 1 were not a stable equilibrium state, C 1 could be transformed into another state without external effects; therefore, the state C 1 = (A 1, B 1 ) would not be a stable equilibrium state. Let Γ C (E C 1 ) be the set of all the states of C = AB such that: A e B are in stable equilibrium states and are contained in the regions of space R A and R B ; the energy of C has the value E C 1 = U A 1 + U B 1. By the highest entropy principle, necessary condition for C 1 to be a stable equilibrium state is that C 1 is the unique highest entropy state in the set Γ C (E C 1 ). By the additivity of entropy, one has S C = S A + S B. Since in the set Γ C (E C 1 ) the states of A and B are stable equilibrium states, one has S A = S A se (U A, V A ) and S B = S B se (U B, V B ). Moreover, since U A + U B = E C 1 = U A 1 + U B 1 : U A = U A1 +  and U B = U B1 - . Therefore, we can rewrite S C in the form

26. By differentiating with respect to  one obtains (2.2) 26 Necessary conditions for C 1 (which corresponds to  = 0) be the state which maximizes the entropy of C in the set of states Γ C (E C1 ) are (2.3a) (2.3b) Eq. (2.3a) yields T A 1 = T B 1 ; Eq. (2.3b) yields:

27. 27 PROOF OF THEOREM 11 We already proved that the condition is necessary (Theorem 9). Consider a state C 1 of C, with T A = T B. By Theorem 10, T A e T B determine, respectively, U A e U B, and thus U C e= U A + U B. With fixed U C, and fixed regions of space occupied by A and B, there exists a unique stable equilibrium state of C. The state C 1 fulfils the necessary condition T A = T B. No other state of C, with internal energy U C, fulfils the necessary condition. In fact, in order to change U A and U B without changing U C, one must increase U A and decrease U B, or viceversa. But, by Theorem 10, in this way the necessary condition T A = T B is no more fulfilled. Therefore, the state C 1 is the unique stable equilibrium state of C with the fixed regions of space and with the energy U C determined by T A e T B.

28. 28 PROOF OF THEOREM 12 Let C 1 be a stable equilibrium state of C. By introducing an internal movable wall, let us divide C into two closed simple subsystem, A and B, with no effect on the state C 1. Let us consider the following state C 2 of C: A and B are in stable equilibrium; with respect to C 1 : the volumes of A and B have the changes dV A, dV B = – dV A ; the values of the entropy of A and B are unchanged. By the first law, there exists a process  of C, from C 1 to C 2, composed of reversible weight processes for A and B. The work performed by C in process  cannot be positive. In fact, it would be possible: to bring C in a stable equilibrium state without external effects or displacements of walls (Lemma 1), to eliminate the wall, to bring C in a nonequilibrium state by a weight process, thus violating the definition of stable equilibrium state. The work in process  is equal to the sum of the energy decreases of A and B. Thus one has, by employing the Gibbs equation for A and B:

29. 29 Since  W C cannot be positive, either for dV A > 0 or for dV A < 0, it must be p A = p B. Since dV B =  dV A, Eq. (2.6) can be rewritten as (2.6)

30. 30 PROOF OF THEOREM 13 Consider a closed simple system C in the stable equilibrium state C 1se = O 1se, and divide C into two open simple systems, A and B, by a fixed membrane, permeable to the k-th constituent, with no effect on the state. Consider an infinitesimal weight process  for C, in which: the number dn k of particles of the k-th constituent is transferred from B to A; the systems A and B remain in stable equilibrium, with unchanged volume and unchanged entropy. By the first law and the principle of entropy nondecrease, the process exists and is reversible because the entropy of C does not change. The work performed by C in process  cannot be positive. In fact, it would be possible: to bring C in stable equilibrium with no external effect or displacement of the constraints (Lemma 1), to eliminate the membrane, to bring C in nonequilibrium by a weight process, thus violating the definition of stable equilibrium state.

31. 31 Since  W C cannot be positive either for dn k > 0 or for dn k < 0, it must be  k A =  k B. (2.12) The work performed by C in process  is equal to the sum of the energy decreases of A and B. Thus one has, by employing the Gibbs equation for A and B: