Thermodynamics: Lecture 6

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Nigel Greer
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1 Thermodynamics: Lecture 6 Chris Glosser March 14, OUTLINE I. Chemical Thermodynamics (A) Phase equilibrium (B) Chemical Reactions (C) Mixing and Diffusion (D) Lead-Acid Batteries 2 Chemical Thermodynamics Let s return to the idea of an open system, in which the number of particles for whatever reason is not constant. There are several examples of situations in which the number of particles is not constant. I will discuss three here. 2.1 Phase Equilibrium Let s consider an system isolated by a rigid adiabatic wall. The system is undergoing a phase change. Instead of trying to model this as a single system, it is beneficial to think of this as two systems which are exchanging matter. We have that; n a + n b = n = constant (1) V a + V b = V = constant (2) U a + U b = U = constant (3) If we are in equilibrium, then entropy is maximum; S a + S b = S (4) 1

2 If we imagine that we perturb the system by an infintessimal amount, we find that; ds = ds a + ds b = 0 (maximum). (5) Since; ds a = 1 T a (du a + P a dv a µ a dn a ), (6) and ds a = 1 T b (du b + P b dv b µ b dn b ). (7) Now, since we have the following relations; dn a = dn b, du a = du b dv a = dv b, (8) we must have that; ( 1 1 ) ( Pb du b + P ) ( a µa dv b µ ) b dn b = 0 (9) Tb Ta T b T a T b T a Hence, absolute equilibrium implies that; T a = T b, V a = V b, µ a = mu b. (10) These are the mathematical definitions of thermal equilibrium, mechanical equilibrium, and diffusive equilibrium, respectively. We know that for an isolated system, the energy increases. Therefore; S = S a + S b > 0. (11) let s say that the system has reached thermal and mechancial equilibrium already, and system a has a surplus of mass S. We then have that; System b gains the surplus mass; S a = µ a T ( n a) (12) S b = µ b T (+ n b) (13) Since the sum of these is greater than zero, we must have; µ a > µ b. (14) This shouldn t be too surprising. We derived this for the Gibbs Potential, which is equivalent to the specific Gibbs portential of a uniform system. 2

3 2.2 Chemical Reactions In order to see how we handle chemical reactions in a thermodynamic sense, let s consider the example of the oxidation of hydrogen into water. The reaction is usually written; 2H 2 + O 2 2H 2 O. (15) We may write down a generalization of this relationship; m ν j M j (16) j=1 The M s represent the individual compounds (in this case H 2, O 2, H 2 O.) The coefficients ν are called stoichiometric coefficients (don t ask me how to pronouce this, cause I don t have the slightest). These represent the relative weights of the chemicals in the reaction. For instance, for the water reaction, these are; ν H2 = 2, ν O2 = 1, ν H2 O = +2, (17) Note that the coefficients are negative for the reactants. Now, for any reaction, the change in the number of moles of a particular chemical are proportional to the stoichiometric coefficients by the same factor. Therefore, for chemical equilibrium, we have that; dg = 0 i µ i ν i = 0. (18) Therefore, for our water reaction, we may solve for it s chemical potential; µ H2 O = 1 2 (2µ H 2 + µ O2 ) (19) Relations of these sort are emormously useful, as you are going to find out in this cycle s homework. Now, the stoichiometric coefficients of this reaction are just constants, therefore we must be able to write dn i = ν i dn, (20) for some exact differential dn. Now, We may integrate this equation to obtain; n i = ν i n (21) 3

4 All equilibrium reactions can be classified by specifing this constant of proportionality. In this equation, we may solve for the Final number of moles; (n i ) f = (n i ) 0 + ν i n (22) Note that if n f is to be positive, we must have that the constant n is restricted to a maximum postive value, which we shall denote n max.. Similarly, we can restrict the the low end by examining the reverse reaction, in which case we obtain n min.. These values obviously depend on how much stuff you have. Let s examine the water case again. Let s say we have 2 moles of hydrogen, 3 moles of oxygen, and 3 moles of water initially. then; n max. = 1, (23) n min. = 3 2. (24) These are the bounds on the reaction. As far as calculating what this coefficient is, let s assume that the Water is at a high enough temperature for it to be in a vapor state. Therefore, we may treat all of our chemicals as ideal gasses. Since the chemical potential as a function of pressure varies as; ( ) P µ = kt ln, (25) P0 In the event we have multiple substances, The pressure becomes the partial pressure. We may write, for the water reaction; ( ( ) ( ) ( )) xh2 P xo2 P xh2 OP kt 2 ln + ln 2 ln = 2µ H2 O µ O2 2µ H2. Therefore; Solving, we find; (26) ( ) x 2 RT ln H2 x O2 P = G 0. (27) x H2 O x 2 H 2 x O2 P x H2 O = exp ( ) G0. (28) RT Let s assume atmosphereic pressure, and a temperature which is well within the steam range (400 K) G 0 RT = (228.83/( )) = (29) 4

5 Hence, x 2 H 2 x O2 x H2 O We note that we may eliminate the oxygen fraction; = (30) x O2 = 1 x H2 x H2 O. (31) We can then plot the fraction of water as a function of the mole fraction of hydrogen. It is actually maximized for this process at about.1. I made a graph with Mathematica for this process. As you may suspect, the peak is when the mole fraction of hydrogen is slightly less than 2/3 of the mixture. In fact, we can maximize the fraction of H 2 O with the choice x H 2 =.59. The fraction of mole fraction of O 2 turns out to be roughly.29. Divideing these, this is exactly two, as you would expect. 2.3 Mixing and Diffusion Most mixing phenomenea happen at constant temperature and pressure, and therefore, the Gibb s energy is a central component. Let s define the partial pressure of a system component; P i = n i n tot. P. (32) As an example, if a system contains n i moles of a certain type of ideal gas, It s partial pressure is P i = n i R T. (33) V Any Gibbs function can be written in terms of the specific Gibbs energies of its components; G = n i g i (34) What exactly are these g i? Well, we can calculate them. First, for an ideal gas, we note; dt ds = c P T RdP (35) P we may integrate this to get the entropy function for an ideal gas; ( ) ( ) T P s s 0 = c P ln R ln (36) T0 P0 Now, if we recall that; dh = c p dt, (37) 5

6 we have; Now, this is related to the Gibbs energy by ( h = g + T s = g + T s 0 + c P ln h h 0 = c p T (38) ( T T0 ) R ln ( P P0 )), (39) Therefore, the specific gibbs energy is; ( ( ) ( )) T P g = c P T T s 0 + c P ln R ln, (40) T0 P0 This horrible thing can be written concisely as; P g = RT ln + φ(t ). (41) P0 Where φ(t ) is a function of T only. To see what all of this means, let s take the case of two ideal gases separated by a partition. The total Gibb s energy is just the sum of the components; G initial = n 1 g 1 + n 2 g 2. (42) The specific gibbs energy per molecule for the two systems are; P P g 1 = RT ln + φ 1 (T ), g 2 = RT ln + φ 2 (T ). (43) P0 P0 Therefore; G initial = n 1 RT P P ln + φ 1 (T ) + n 2 RT ln + φ 2 (T ). (44) P0 P0 OK, if we remove the partion, then the gasses mix, and the gasses now only contribute to their particular partial pressure. That is; P1 P2 G final = n 1 RT ln + φ 1 (T ) + n 2 RT ln + φ 2 (T ). (45) We have assumed that the process happens isothermally. Now, recalling that we may write; G final = n 1 µ 1 + n 2 µ 2, (46) we then have an expression for the chemical of the combined system; Pi µ i = RT ln + φ i (T ) where x i is the mole fraction of the gas, n i /n. (47) = g i + RT ln x i, (48) 6

7 2.4 Lead-Acid Batteries A typical Lead Acid battery undergoes the following reaction to generate charge; Pb + PbO 2 + 4H + + 2SO 2 4 2PbSO 4 + 2H 2 O. (49) we shall assume that the reaction produces charge at a much greater rate than the rate that charge is taken off the battery, so that there is basically a constant supply of charge. The reaction will procede until the Gibbs free energy balences out. By this I mean that it will cease to produce charge when the battery potential equals the intial Gibbs free energy. A quick calculation establishes that at room temp and pressure, the change in the gibbs energy is -390 kilojoules/mole. Taking a closer look at the reaction itself, we have the following three subprocesses; Pb + HSO 4 PbSO 4 + H + + 2e (50) occurs at the - terminal. This is the source of the electrons and the relavent reaction for calclulating the potential. At the + electrode, the reaction is; PbO 2 + HSO 4 + 3H+ + 2e + PbSO 4 + 2H 2 O (51) in solutions, th following reaction occurs; 2SO H + 2HSO 4. (52) Therefore, each mole of reactant pushes two electrons around the circuit. The electrical work per electron provided turns out to be; W = 390 = 2.02 ev. (53) Therefore the effective voltage of teh battery is 2.02 Volts. 7

Physics 360 Review 3

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