Lecture notes on non-equilibrium thermodynamics

Transcription

1 Lecture notes on non-equilibrium thermodynamics by deb shankar ray Indian Association for the Cultivation of Science 2A & 2B Raja S.C. Mullick Road Jadavpur, Kolkata: 32 INDIA

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3 C O N E N S 1 local equilibrium and two laws of thermodynamics Some transport processes Assumption of local equilibrium and two laws change of entropy density in space and time Equation of continuity and conservation laws Equation for entropy density; entropy flow and entropy production reformulation of second law for open system; prigogine equation Reformulation of second law for open system Mathematical details : Gauss theorem linear laws in physics and chemistry Linear laws in physics Linear law in chemistry thermodynamic cross effects Experiments on thermoelectric effects Heat conduction in anisotropic crystals: Cross effects in chemical reactions (Onsager thermodynamic fluctuations 29 7 thermodynamic fluctuations, force and flux Entropy production from fluctuation theory: Calculation of averages < α i X i >: Properties of fluctuation: regression of fluctuation and two laws of irreversible thermodynamics 37 9 application of the laws of irreversible thermodynamics, i: thermoelectric effects application ii: transport of heat and mass principle of minimum entropy production Minimum entropy production Stability of the steady state iii

4 iv contents references 53

5 L I S O F F I G U R E S Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure v

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7 L O C A L E Q U I L I B R I U M A N D W O L AW S O F H E R M O D Y N A M I C S 1 he aim of irreversible thermodynamics is to explain the irreversible processes (such as transport of energy, mass, charge; chemical reactions, etc. under not far-from-equilibrium condition, so that one can apply the principles of equilibrium thermodynamics. However, any direct application of thermodynamic laws is plagued with two difficulties. For example, in the transformation of state from A to B at constant temperature and pressure, nonmechanical work W done on the system is equal to the free energy difference F B F A, i.e., W = F B F A = F if the process is reversible. For an irreversible process W > F. hus we must have to deal with inequalities for the treatment of irreversible processes. Second, transformation of states from A to B takes place in finite time. However, the notion of time is absent in thermodynamics. In applying the laws of thermodynamics to irreversible processes one must address these difficulties. 1.1 Some transport processes We begin by introducing several examples: (i Consider flow of heat through a metallic rod from higher temperature ( 1 to lower temperature ( 2. Heat flux J Q is given by J Q = κ x (Fourier Law (1.1 where x is the coordinate along which heat transport occurs, and x is the temperature gradient along x-direction. κ is the coefficient of thermal conductivity. Unit of J Q is, for example, calorie cm 2 Sec 1. (ii Flow of charge due to potential gradient constitutes an electric current. he flux of charge J e is given by (for example, coulomb cm 2 Sec 1 J e = λ φ x (Ohm s Law (1.2 1

8 2 local equilibrium and two laws of thermodynamics C e ll 1 Q 2 1 > 2 Figure 1.1 (iii Flow of mass due to concentration gradient is responsible for diffusion. he flux of mass J m (unit in gm 1 cm 2 Sec 1 is given by J m = λ C x (Fick s Law (1.3 where C denotes the concentration. In traditional equilibrium thermodynamics time does not appear but irreversible processes involve rate and therefore time has a specific role in irreversible thermodynamics and this poses a problem. Second problem is that since the processes are not in equilibrium one cannot define the state functions, P, etc. for the system in a straightforward way. he question is how to get over these difficulties. 1.2 Assumption of local equilibrium and two laws of thermodynamics Let us consider a very small portion of the rod as shown in Fig 1.1. his is a small cell (which is macroscopically small but microscopically large of volume V compared to the rod, such that the number of molecules or constituents contained in it is of the order (say. As the dimension of the cell is small it can equilibrate very fast so that the cell can be imagined as an isolated locally equilibrated system, on a time scale over which the heat flows from left to right end of the rod. Since the cell equilibrates within a very short time which allows us to register the temperature by means of a thermometer inside it assuming that the temperature remains virtually constant. his assumption may be considered as the postulate of local equilibrium, so that one can assign, P etc for this small cell. hus, P etc. are all constants for this isolated small

9 1.2 assumption of local equilibrium and two laws... 3 cell in local sense despite of the fact that they are actually varying quantities over the entire rod. It is in this context the assumption of local equilibrium is essential; otherwise it is very difficult to define temperature, pressure etc. for a non-equilibrium process and one can t use traditional thermodynamics. Now for applying thermodynamics for the cell (, P, V constant we are allowed to write the Gibbs free energy function in terms of internal energy (E, entropy (S, and enthalpy (H. and for which the differential is given by G = H S = E + PV S (1.4 dg = de + VdP + PdV ds Sd (1.5 For the condition of constant temperature, pressure and volume (d = dp = dv = 0 dg,p,v = de ds (1.6 Although temperature, pressure, volume remain constant, yet we have de 0, ds 0, dg 0 due to the change of E, S, and G in space and time. Eq. (1.6, the starting point of our treatment is based on (i the assumption of local equilibrium and (ii the application of first and second laws of thermodynamics for the cell. Now dg can be written as dg,p,v = dg mt + dg ct + dg re (1.7 dg mt, dg ct and dg re are the free energy changes due to the transport of mass, charge and the chemical reaction, respectively. Consider m i the mass of the i-th component of a multi-component system, for which the total mass (m = i m i. he Gibbs free energy change due to mass transport is then given by dg int = i G m i dm i = i µ i dm i (1.8 Here µ i is the chemical potential (per unit mass. he Gibbs free energy change due to charge transport dq is given by dg ct = φdq (1.9

10 4 local equilibrium and two laws of thermodynamics where φ is the potential difference due to an electric field. o calculate the Gibbs energy change due to chemical reaction consider a general reaction scheme, αa + βb = γc + δd (1.10 For the reaction, the free energy change is given by dg re = µ A dn A + µ B dn B + µ C dn C + µ D dn D (1.11 µ A,B,C,D is the chemical potential (per mole. Furthermore we have so that n A = n 0 A αξ, n B = n 0 B βξ, (1.12 n C = n 0 C + γξ, n D = n 0 D + δξ, (1.13 dn A = αdξ, dn B = βdξ, dn C = γdξ, dn D = δdξ, (1.14 If α moles of A combines with β moles of B to produce γ moles of C and δ of D then we may say that the reaction has advanced from equilibrium by one unit. hen for ξ unit of advancement of reaction from equilibrium we write dg re = (γµ C + δµ D αµ A βµ B dξ (1.15 At equilibrium dξ = 0 as a result of which dg = 0 so that equilibrium condition is retrieved. For the system, slightly away from equilibrium, the change of Gibbs energy due to chemical reaction is given by dg re = G ξ dξ = A fdξ (1.16 A f = G ξ is the chemical force or affinity of chemical reaction. [NB. we know that if V(x is potential of a field, then the force due to potential, F = dv(x G ξ dx. hus by analogy we may define the chemical force as A f = ]. Using Eqs ( and Eq. (1.16, we arrive at dg,p,v = i µ i dm i + φdq A f dξ (1.17 Using Eq. (1.17 in Eq. (1.6 we get, ds = de i µ i dm i φdq + A f dξ (1.18

11 1.2 assumption of local equilibrium and two laws... 5 Now dividing Eq. (1.18 by m we have ds = du i µ i dw i φde + A f ρ dξ V (1.19 We now define s = entropy per unit mass, u=energy per unit mass m, w i = m i m is weight fraction of the i-th species, e = charge per unit mass. Multiply Eq. (1.19 by ρ and divide by to obtain d(ρs = d(ρu i µ i d(ρw i φ d(ρe + A f dη [ η = dξ ] V (1.20 where ρs = entropy density, ρu = energy density, ρe = charge density. All the quantities vary in space and time except dη which is a function of time only.

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13 C H A N G E O F E N R O P Y D E N S I Y I N S PA C E A N D I M E 2 How do the density functions ρs, ρu, ρw, ρe change as functions of time and space? Dividing Eq. (1.20 by a time increment we get (ρs t = 1 (ρu t i µ i (ρw i t φ (ρe t + A f V ( ξ t (2.1 o proceed further we must know how a density function ρ(r, t changes with time and space? 2.1 Equation of continuity and conservation laws for charge, mass and energy Consider a box with axes as depicted in the Fig 2.1. he volume of the box dv = dxdydz (2.2 and let some fluid passes from left to right with velocity component v y. hen the amount of mass accumulated in the volume dv in time dt ( ρ dm = dtdxdydz (2.3 t Consider the flow of the mass along y direction normal to xz-plane. Now the amount of mass gathered within the volume (dv in time dt = Z X y y + d y Figure 2.1 Y 7

14 8 change of entropy density in space and time (amount of mass arriving at y normal to dxdz plane - (amount of mass leaving at y + dy normal to dxdz plane. he amount of mass arriving at y ρ(yv y dxdzdt. he amount leaving y + dy ρ(y + dyv y (y + dydxdzdt. hen the amount of mass accumulated in the volume dv in time dt is given by dm = ρ(yv y dxdzdt ρ(y + dyv y (y + dydxdzdt (2.4 Now using aylor series expansion and keeping the first order term only in the second term of the right hand side we have ρ(y + dyv y (y + dy = ρ(yv y (y + (ρv y dy (2.5 y Using Eqs ( and Eq (2.5 one obtains ρ t = (ρv y y (2.6 his is due to change of density along y direction normal to xz plane. Considering the density changes along other two directions we may write ρ t [ (ρvx = + (ρv y x y + (ρv ] z z =.(ρ v ( divergence of (ρ v =.( ( j since flux ( j = density(ρ velocity( v (2.7 Eq. (2.7 is a continuity equation, which shows how the mass of fluid remains conserved. In our case the mass, energy and charge are conserved. herefore we have (ρu t (ρw i t (ρe t =. j u ( j u = energy flux (2.8 =. j i ( j i = mass flux of i-th component (2.9 =. j e ( j e = charge flux (2.10 where (ρu, (ρw i and (ρe are the energy density, mass density of the i-th component and charge density, respectively.

15 2.2 equation for entropy density; entropy flow and entropy production Equation for entropy density; entropy flow and entropy production Such continuity equations are not applicable for entropy density since entropy is not a conserved quantity. Using the continuity Eqs ( in Eq (2.1 we obtain (ρs t = 1. ju + i µ i. ji + φ. je + A f V ( ξ t (2.11 he last term on the right hand side of Eq (2.11 is due to chemical reaction and does not satisfy conservation law of the form ( Each of the first three terms on right hand side is a product of a scalar and divergence of some vector quantity. Now considering A as a scalar and B as a vector we have (.A B = î x + ĵ y + ˆk. ( A îb x + A îb y + A îb z z = x (AB x + y (AB y + z (AB z ( = A x (B x + y (B y + z (AB z ( A + B x x + B A y y + B z A z = A. B + B. A (2.12 herefore we have A. B =.A B B. A (2.13 Using the relation Eq (2.13 in Eq (2.11 we obtain (ρs =. 1 ( t 1 j u + j u. +. µ i j i i i ( ( φ φ +. j e j e. + A ( f ξ V t [ ] 1 j e or =. j u i [ ( 1 + j u. i + A ( f ξ V t (ρs t µ i j i ( φ ( µi j i. j e. ( ] φ ( µi j i. (2.14 =. J s + σ (2.15

16 10 change of entropy density in space and time It looks like a continuity equation but with an additional source term σ. he presence of this additional source term indicates that entropy is not conserved because something is produced inside the system (σ. σ is the entropy production. J s on the other hand, is due to the flow of entropy across the volume and may be termed as entropy flow. he time scale over which the time derivative (ρs t is meaningful is relatively long since the time scale of equilibration of the isolated cell is very short.

17 R E F O R M U L A I O N O F S E C O N D L AW F O R O P E N S Y S E M ; P R I G O G I N E E Q U A I O N Reformulation of second law for open system he entropy flux J s and the entropy production (σ in Eq (2.15 are given by J s = 1 j u i µ i j i φ j e (3.1 and σ = j u. ( 1 i ( µi j i. j e. ( φ + A f V ( ξ t (3.2 Eq (2.15 expresses the entropy density (ρs as a function of time and space and thus it is a local equation. Herein lies an important difference between traditional thermodynamics and present one. he former deals with average properties of a system over the entire available space and time. Earlier we considered an infinitesimal parallelepiped of volume dv to obtain the local equation (2.15. Now in order to have its global form (over the finite V we have to integrate it over the total volume of the cell V. t S +. J s dv = σ dv (3.3 V where V dv(ρ s = S, the total entropy on integration. V. J s dv is an integral over volume. It can be reduced to an integral over surface A that encloses the volume V using Gauss theorem (details are given at the end of this section V. J s dv = A V d A. J s (3.4 A is the area of the surface that encloses the volume V. [Note that area is a vector that can be oriented in arbitrary directions]. 11

18 12 reformulation of second law for open system... n d A Figure 3.1 S u rfa c e Figure 3.2 An infinitesimal area is given by d A or nda, where n is the unit vector normal to the plane of da. Using Eq (3.4, Eq (3.3 yields the global form t S + J s.d A = σ dv (3.5 Consider now the following cases: A (a Closed system: Here there is no mass and charge transport ( j i = 0 and j e = 0 and also we assume that no chemical reaction occurs so that ξ = 0. From Eq (3.1, we therefore write V J s = 1 j u (3.6 Hence the internal energy flux is the heat flux since we have considered P and V constant for the cell. As a change of notation it is convenient to write j u = j Q and Eq (3.5 reduces to the following form t S + 1 j Q. nda = σ dv (3.7 A j Q = heat energy per unit time per unit area; so the product of area (da and j Q gives the rate of heat flow. On integration this gives the total rate of heat flow dq out dt that comes out of the volume V 1 j Q. nda = 1 dq out dt A V (3.8

19 3.1 reformulation of second law for open system 13 hen from Eq (3.7 t S + 1 dq out dt = V σ dv (3.9 Q out is the total amount heat coming out of the surface. herefore heat added to the volume must be dq in = dq out t S 1 dq in = σ dv (3.10 dt Multiplying both sides by dt we obtain ds 1 dq in = dt V V σ dv (3.11 he above equation is based on the assumption of local equilibrium and application of first and second laws of thermodynamics. o recover second law we note that for the closed system ds dqrev in = 0 (3.12 ds dqirr in > 0 (3.13 In order to satisfy second law of thermodynamics, the term dt V σ dv in Eq (3.1 must be positive. his quantity is a global quantity as we have integrated over volume dt σ dv 0 global condition (3.14 V Eq (3.11 can then be rewritten in the form of a sum of two inexact differentials as ds = d e S + d i S (3.15 where d e S = 1 dq in and d i S = dt V σdv. In the above equation the first term represents the external entropy due to flow as this term arises because of interaction between the system (cell and the surroundings, the second term accounts for the internal creation or production of entropy in the system itself. he relation (3.15 was derived by Prigogine in (B Open system: We are allowed to consider mass and charge transport. For the open system the entropy flux is given by ( j u 0, j i 0, j e 0. J S = 1 j u i µ i j i φ j e (3.16

20 14 reformulation of second law for open system... We may now note that small cells may be considered at any arbitrary position in the rod at any time and Eq (3.15 will remain valid for all these individual cells (assumption. hus the Prigogine equation will be true for all of these cells and hence also for the total system. his is a postulate of irreversible thermodynamics. ds i 0; (global form or σ 0 local form (3.17 as dt 0 and dv is arbitrary. (C We now look for some other consequences of the equation for entropy production. (i Considering only heat flow, we have from Eq (3.2 ( 1 σ = j u. > 0 (3.18 For isotropic substances, the angle between the heat flux vector and the gradient vector is zero. For anisotropic substances on the other hand, Eq (3.18 reveals that the angle must be less than 90 o so that we write ( 1 j u 2. > 0 (3.19 A comparison with experimental Fourier law for heat conduction gives j u = κ ( 1 κ 2 ( 2 > 0 (3.20 Hence for the coefficient of thermal conductivity κ > 0, heat must flow from high temperature to low temperature. (ii Let all the flow terms vanish. hus we have j u = j i = j e = 0; only the chemical reaction occurs so that σ = A f V ( ξ t (3.21 ( herefore A f, and ξ t must have the same sign, which means that direction of chemical reaction must be same as that of a chemical force (A f. 3.2 Mathematical details : Gauss theorem We start by defining a vector of an area as a vector d A which is normal to an area of magnitude d A. Component-wise this can be written as

21 3.2 mathematical details : gauss theorem 15 Figure 3.3 A A ' Z D D ' Y X C B C ' B ' Figure 3.4 da = idydz + jdxdz + kdxdy (3.22 Consider now a flux j = ρ v and assume for simplicity, unit density (ρ = 1 of the fluid flowing with velocity v. hen the amount of fluid coming out per unit time through the surface d A is a scalar dφ = v.d A and for the whole area φ = dφ = ν.d A (3.23 A A = (ν x dydz + ν y dxdz + ν z dxdy (3.24 A Consider now the amount of fluid per unit time coming out of a closed surface. he amount of fluid per unit time passing through ABCD (from inside in the negative direction of x dφ ABCD = V x (xdydz (3.25 he amount of fluid per unit time passing through A B C D (from inside in the positive direction of x dφ A B C D = v x(x + dxdydz (3.26 = v x (xdydz + v x(x dxdydz (3.27 x

22 16 reformulation of second law for open system... otal amount coming out (along x direction per unit time dφ x = dφ ABCD + dφ A B C D (3.28 = v x(x dxdydz (3.29 x otal amount coming out of the volume dv per unit time (taking care of the other directions y, z Now we have dφ = dφ x + dφ y + dφ z ( vx (x = + v y(y x y + v z(z dxdydz z = (. vdxdydz (3.30 φ = = = V V v dφ (. vdxdydz (. vdv (3.31 Finally consider a finite volume consisting of an infinite number of parallelepiped. Adjacent sides don t contribute to vector flux since they cancel out. Only flux through the outer surface contribute. herefore we must have from Eqs (3.23 and (3.31 (. vdv = v. da (3.32 his is the Gauss theorem. v A

23 L I N E A R L AW S I N P H Y S I C S A N D C H E M I S RY Linear laws in physics he differential change in entropy for an open system is postulated as a sum of two inexact differentials of entropy, one due to entropy flow and the other due to entropy production. ds = d e S + d i S (4.1 where, d e S = 1 dq in (4.2 and d i S = dt σdv (4.3 V We have shown earlier that dq in is due to entropy flow as a result of energy flow, mass flow and or charge flow. his term results from 1 A J s. da where J s is the entropy flux given by ( 1 J s = j u ( µi ( φ j wi j e i (4.4 Entropy production σ on the other hand is the sum of the scalar products of flow vectors and the corresponding gradient vectors and in addition a simple product of scalar quantities for chemical reaction. ( σ = j u. 1 ( j i. µ ( i j e. φ + A f V i ( ξ t (4.5 Positivity of entropy production demands that each term in the above expression must be separately positive, i.e., ( j u. 1 > 0, ( j i. µ i > 0 i ( j e. φ > 0, + A ( f ξ > 0 (4.6 V t 17

24 18 linear laws in physics and chemistry Experiment tells us that flow is a linear function of gradient for a wide range of physical phenomena. his implies we may write j u = λ( (Fourier law (4.7 ( j i = D i µ i > 0 (Fick s law (4.8 ( j e = κ φ (Ohm s law (4.9 hese linear laws together with the previous set of inequalities yield λ 2 ( 2 > 0 (4.10 D i ( µ i 2 > 0 (4.11 κ ( φ 2 > 0 (4.12 so that λ, D i and κ are all positive coefficients which set the direction of flow consistent with the postulate of entropy production for the system. 4.2 Linear law in chemistry What about the linear law for chemistry? Can we think of an analogous linear law for chemical reaction near equilibrium condition? dξ dt = constant A (4.13 where ξ is the advancement and A is the chemical force. o enquirer about such a linear law for chemical reaction let us begin with the following reaction scheme αa + βb k f γc + δd (4.14 k b At equilibrium concentration of A, B, C and D are n 0 A, n0 B, n0 C and n0 D. When ξ unit of reaction (advancement has taken place, the concentration of A, B, C and D changes from their equilibrium values (i.e., we follow the progress of the reaction in terms of departure from equilibrium to new values n A, n B, n C and n D. hese are given by n A = n 0 A αξ, n B = n 0 B βξ, (4.15 n C = n 0 C + γξ, n D = n 0 D + δξ, (4.16

25 4.2 linear law in chemistry 19 then the differential changes in time are dn A = αdξ dn B = βdξ (4.17 dn C = +γdξ dn D = +δdξ (4.18 Now the free energy change associated with the reaction is given by G = i G n i dn i i = A, B, C, D = i µ i dn i = [(γµ C + δµ D (αµ A + βµ B ] dξ (4.19 herefore Since G = G dξ (4.20 ξ G ξ = {(γµ C + δµ D (αµ A + βµ B } = µ (4.21 G ξ = A µ = A (4.22 We make use of the thermodynamic relation between chemical potential and concentration. o this end we note µ i = µ 0 i + RlnC i, µ 0 i refes to the standard chemical potential. For A, B, C, D and the above relation we have for µ µ = (γµ C + δµ D (αµ A + βµ B = (γµ 0 C + δµ0 D (αµ0 A + βµ0 B + (γrlnc C + δrlnc D (αrlnc A + βrlnc B ( C γ = µ 0 C + Rln Cδ D C α A Cβ B Defining the quantity written in the parenthesis as Q (4.23 µ = µ 0 + RlnQ (4.24 At equilibrium µ = 0. hen µ 0 = RlnK when Q = K. herefore µ = RlnK + RlnQ = Rln Q K (4.25

26 20 linear laws in physics and chemistry or, Q K = e µ/r (4.26 For small deviation from equilibrium µ must be small compared to K so that Q K 1 + µ R (4.27 We now turn to kinetics for computation of (dξ/dt. he rate law from mass action is dc C dt = k f C α A Cβ B k bc γ C Cδ D (4.28 Writing C C = ( n C V, number of moles of C per unit volume { 1 dn C = k f C α A V dt Cβ B 1 k } b C γ k C Cδ D ( f = k f C α A Cβ B 1 Q K (4.29 Since Q K 1 + µ R equilibrium γ V and C A C 0 A, C B C 0 B ( dξ dt ( = k f (C 0 A α (C 0 B β µ R for small deviation from (4.30 With µ = A, the chemical force ( [ ] dξ V = dt γr k f(c 0 A α (C 0 B β A (4.31 he quantity inside the bracket is a constant. he reaction (not far from equilibrium therefore follows the linear law in the spirit of Fick s law, Fourier law and Ohm s law.

27 H E R M O D Y N A M I C C R O S S E F F E C S 5 We have seen that entropy production σ is always positive for an irreversible process. More generally we write ( σ = j u. 1 ( j i. µ i j e. i ( φ + A f V ( j u. 1 can be written component wise as follows; ( j u. 1 = J ux x ( 1 + J uy y ( 1 + J uz z ( ξ t ( 1 So the entropy production for more general case is given by (5.1 (5.2 σ = i J i X i (5.3 Experimental ( finding is that the flux J i is related to force X i as J u = L uu X u where X u = 1. So far we have assumed that the flow of charge occurs only due to a gradient of electric field and the flow of heat is due to the gradient of temperature. hus we write J u = L uu X u and J e = L ee X e (5.4 So each flow is caused by the corresponding gradient. herefore in general, one may postulate J i = L ii X i. his is consistent with σ = i J ix i > 0. A question can now be posed as follows: Is it possible to realize a flux J i due to a thermodynamic force X j? Can there be any nonzero L ij? Experiments tells us that thermodynamic phenomena like Seebeck effect and Peltier effect suggest that charge may flow due to a thermal gradient and heat may flow due to a potential gradient. hese effects are the thermodynamic cross effects or thermoelectric phenomena. 5.1 Experiments on thermoelectric effects Consider M and N two dissimilar metal stripes connected in parallel and the two junctions are kept at different temperatures and +. 21

28 22 thermodynamic cross effects X Y + P Figure X Y Figure 5.2 he circuit has its loose ends at X and Y which are connected to a potentiometer. he thermal gradient gives rise to heat conduction as usual. But in addition it is found that an emf φ is set up between X and Y. he value of φ depends on and nature of the material M and N. φ must be measured under zero current condition using the potentiometer. his effect is known as Seebeck effect, the experimental setup is shown in the Fig 5.1. We consider an identical arrangement but here the reservoirs are kept at the same temperature initially (i.e., =0. X and Y are connected to the poles of a battery. It is observed that a temperature difference is created at the two metal junctions. he direction of the current determines the hot and the cold junctions. he effect is known as Peltier effect (Fig 5.2. hus in Eq (5.4 we must consider the cross terms also and the equations for heat and charge fluxes must allow them; J u = L uu X u + L ue X e (5.5 J e = L eu X u + L ee X e (5.6 (vector sign is omitted by considering simple product between components and hence we have the general form of fluxes as, J i = j L ij X j (5.7 At this point one may consider the possibility of appearance of the cross terms between the heat flow or the charge flow with the chemical

29 5.2 heat conduction in anisotropic crystals: 23 reaction. Let us assume that the cross effects between heat flow and chemical reaction as follows; ( ( 1 A J u = L uu + L uc x ( ( 1 A J c = L cu + L cc x (5.8 (5.9 Here J c is the reaction rate. Consider isothermal condition so that we have J u = L uc ( A. his implies a scalar cause ( A produces a vector effect. his is forbidden by symmetry. herefore (Curie law: a scaler cause can not produce a vector effect by symmetry we must have L uc = L cu = 0. Hence a coupling between a vector and a scalar process is impossible. he relation as given by Eq (5.7 is known as phenomenological linear law and it can be formally derived as follows. and J i = J i (X i, X j ( ( Ji Ji = J i (0, 0 + X i + X j +... X i (0,0 X j (0,0 ( ( Jj Jj J j = J j (0, 0 + X i + X j +... X i (0,0 X j (0,0 (5.10 (5.11 he higher order terms are neglected since we are confining ourselves to near equilibrium condition. At equilibrium there is no flux, so that Hence we can write, where ( Jj X i (0,0 = L ji. J i (0, 0 = J j (0, 0 = 0 (5.12 J i = L ii X i + L ij X j (5.13 J j = L ji X i + L jj X j ( Heat conduction in anisotropic crystals: he entropy production (σ due to heat flow only is given by ( 1 σ = J u. (5.15

30 24 thermodynamic cross effects where J u is the heat flux and ( 1 is identified as the thermodynamic force, which is the cause of the heat flow. Again from linear law: J u = const. = const 2 ( 1. = κ. (experimental Fourier law (5.16 On comparison κ = const. In anisotropic crystals all directions are not 2 equivalent and heat flow along any particular direction (x or y or z is not only due to the temperature gradient along one direction but also due to the gradients along other directions. So the fluxes along x, y and z direction are given explicitly as follows; J = L. (linear law J x = L xx x + L xy y + L xz z J y = L yx x + L yy y + L yz z J z = L zx x + L zy y + L zz z (5.17 he above relations are purely experimental. Experiment also tells us that the cross coefficients are equal, i.e., L xy = L yx, L yz = L zy etc., when gradients are small. One may anticipate that above equalities are due to existing symmetry in the crystal; but it is observed even in systems other than crystals also. So only symmetry argument is not sufficient to justify the equations. 5.3 Cross effects in chemical reactions (Onsager 1931 Consider the following reaction scheme between three species A, B and C o establish some kind of equilibrium, the reverse processes must also be considered. here are thus three reactions involving equilibrium. n a, n b and n c are the number of moles at any instant of time and At equilibrium, n a + n b + n c = n (conservation law (5.18 µ 0 A = µ0 B = µ0 C (5.19

31 5.3 cross effects in chemical reactions (onsager A k A B k B A B k C A k A C k C B k B C C Figure 5.3 (superscript 0 denotes equilibrium, while there is slight departure from equilibrium one may write, n A = a + n 0 A (5.20 n B = b + n 0 B (5.21 n C = c + n 0 C (5.22 a, b, c are the small shifts from the equilibrium values n 0 A, n0 B, n0 C so that a << n 0 A, b << n0 B and c << n0 C. From basic thermodynamics we can write ds = de dg (5.23 Here dg = dg r (reaction and de = 0 as there is no energy change and no heat, mass or charge is flowing; only chemical reaction is taking place. herefore ds = dg r (5.24 or, ds/m = dg r /m (5.25 or, ds = dg r /m (5.26 = dg r /ρ V (ds is entropy change per unit mass; ρ = density or, Vd(ρs = dg r (5.27 Now free energy change due to slight deviation from equilibrium is given by dg r = µ A dn A + µ B dn B + µ C dn C = µ A a + µ B b + µ C c (as, dn A = a, dn B = b, dn C = c(5.28 From Eqs (5.27 and (5.28 we may write V d(ρs = µ A ȧ µ B ḃ µ C ċ dt or, Vσ = µ A ȧ µ B ḃ µ C ċ (5.29

32 26 thermodynamic cross effects where, d(ρs dt Eq (5.22 we have = σ is entropy production. With the help of the Eq (5.18 to ṅ A + ṅ B + ṅ C = ȧ + ḃ + ċ = 0 (5.30 At equilibrium µ 0 A = µ0 B = µ0 C, so we have µ 0 Aȧ + µ0 Bḃ + µ0 Cċ = µ0 A (ȧ + ḃ + ċ = 0 (5.31 Since µ 0 Aȧ + µ0 Bḃ + µ0 Cċ = 0, it can be added to the right hand side Eq (5.29 and on simplification we get, σ = (µ0 A µ A ȧ V + (µ0 B µ B ḃ V + (µ0 C µ C ċ V (5.32 where (µ0 A µ A is the measure of the deviation from equilibrium and it is identified as the thermodynamic force, ȧ V is the rate of the shift from equilibrium and is identified as flux. hus Eq (5.32 is sum of the product of two quantities, force and flux. Now using the kinetic scheme for the cyclic reaction we may write the following equations dn A dt dn B dt dn C dt At equilibrium dn A dt = (k AB + k AC n A + k BA n B + k CA n C (5.33 = k AB n A (k BA + k BC n B + k CB n C (5.34 = k AC n A + k BC n B (k CA + k CB n C (5.35 = dn B dt = dn C dt = 0, we have (k AB + k AC n 0 A + k BAn 0 B + k CAn 0 C = 0 (5.36 k AB n 0 A (k BA + k BC n 0 B + k CBn 0 C = 0 (5.37 k AC n 0 A + k BCn 0 B (k CA + k CB n 0 C = 0 (5.38 With the help of Eq (5.33 we obtain dn A dt = d(n0 A + a dt = (k AB + k AC (n 0 A + a + k BA(n 0 B + a + k CA(n 0 C + a = (k AB + k AC a + k BA b + k CA c [ ] + (k AB + k AC n 0 A + k BAn 0 B + k CAn 0 C (5.39 or, da dt = (k AB + k AC a + k BA b + k CA c (5.40

33 5.3 cross effects in chemical reactions (onsager Similarly we can also write db dt = k ABa (k BA + k BC b + k CB c (5.41 dc dt = k ACa + k BC b (k CA + k CB c (5.42 Compare Eqs ( and Eqs ( to note that the fluctuation/ deviation follow the same kinetics as their macroscopic counterpart. Now from thermodynamics µ A = µ 0 A + Rln ( n A n 0 A = µ 0 A + Rln (a + n 0 A = µ 0 A + Rln ( µ 0 A + R a n 0 A n 0 A 1 + a n 0 A (5.43 So we can write from the above equation ( µ A µ 0 A n 0 A R = a (5.44 In the similar way we have ( ( µ B µ 0 B n 0B R = b and µ C µ 0 C n 0 C R = c (5.45 Now using the Eqs (5.40, (5.44 and (5.45 flux J A is given by, ( ( J A = 1 da V dt = (k AB + k AC n 0 A µ A µ 0 A + k BAn 0 B µ B µ 0 B RV RV ( + k ACn 0 C µ C µ 0 C (5.46 RV Now the term ( µi µ 0 i represents a chemical force. hus the flux [as given by Eq (5.46] is found to be the linear function of force. he flux for B is given by, J B = 1 V db dt = k ABn 0 A RV + k CBn 0 C RV ( ( µ A µ 0 A (k AB + k BC n 0 B RV ( µ C µ 0 C µ B µ 0 B (5.47

34 28 thermodynamic cross effects and similarly for J C. Indexing A 1, B 2, and C 3, define L 12 = n 0 k BA B RV L 21 = n 0 k AB A RV [from Eq (5.46] (5.48 [from Eq (5.47] (5.49 Now imposing the condition of detailed balance for the reaction scheme for one of the three equilibria shown in Fig 5.3 we write A k AB B (5.50 k BA At equilibrium n 0 B k BA = n 0 A k AB, i.e., forward and backward reactions balance in microscopic sense for this equilibrium. he same logic holds good for other two equlibria. he above relations are considered as extra thermodynamic relations. In macroscopic thermodynamics detailed balance is not required. But this condition ensures L 12 = L 21 (5.51 he condition for detailed balance thus establishes the symmetric nature of cross effects. his is called principle of microscopic reversibility. he above proof of reciprocity relation by Onsager is based on the consideration of a cyclic chemical reaction. We shall embank on a general proof of reciprocity in a later section.

35 H E R M O D Y N A M I C F L U C U A I O N S 6 he linear and reciprocal relations describe irreversible phenomena. hey are J i = j L ij X j and L ij = L ji (6.1 respectively. J i is the measure of the rate of deviation of a thermodynamic parameter, say A i from equilibrium. For example J i = d dt (A i A 0 i (6.2 here da i = A i A 0 i is the fluctuation around equilibrium. In thermodynamics we calculate the average properties of a macroscopic system and we do not bother about fluctuations. As thermodynamic systems are large, fluctuations are insignificant, but in small systems (such as a small section of rod as shown in chapter I fluctuations are not negligible and must be taken into account. o describe a nonequilibrium process one must calculate in addition to the average value, the dispersion of the relevant thermodynamic parameters. In this section we briefly discuss the basic aspects of thermodynamic fluctuations around equilibrium. o calculate the dispersion of a thermodynamic variable from its equilibrium value, let us first start from Boltzmann s definition entropy in terms of its probability of a state W S = k B ln W; or W = e S/k B (6.3 Entropy S is maximum at equilibrium, S = S 0. Now we can express entropy as a function of thermodynamic parameter A i. S = f(a i (6.4 A i is a thermodynamic parameter, such as, temperature, pressure, volume etc. If the system slightly deviates from equilibrium then entropy can be written as ( S S = S 0 + A i A i =A 0 i ( (A i A 0 i S 2 A 2 i A i =A 0 i (A i A 0 i 2 (6.5 29

36 30 thermodynamic fluctuations Here A 0 i is the equilibrium value of the thermodynamic parameter A i. At equilibrium, entropy is maximum, so that ( ( S 2 S = 0 and A i A i =A 0 i A 2 i A i =A 0 i < 0 (6.6 Hence deviation of entropy from its equilibrium value can be written as ( S = S S 0 = 1 2 S (A i A 0 i 2 2 (6.7 Now S S W 0 k = e B W 0 A 2 i = e S k B A i =A 0 i = exp (A i A [ 0 i ( 2 2k B / 2 S A 2 i A i =A 0 i ] (6.8 ( ( 2k B / 2 S is a positive quantity, since 2 S is negative itself. hus the thermodynamic parameter (A i is distributed in a A 2 i A i =A 0 A 2 i i A i =A 0 i Gaussian way around equilibrium. Let, 2σ 2 = ( 2 S 2k B (6.9 A 2 i A i =A 0 i Hence unnormalized distribution of thermodynamic parameter A i is given by [ ] W = exp (A i A 0 i 2 W 0 2σ 2 (6.10 Now using normalization condition for the above probability distribution function we get Nexp [ (A i A 0 i 2 2σ 2 So the normalization constant N is given by ] da i = 1 (6.11 N = 1 σ 2π (6.12

37 thermodynamic fluctuations 31 Normalized probability distribution function for the thermodynamic parameter A i, becomes P(A i = W W 0 = 1 σ 2π exp [ (A i A 0 i 2 2σ 2 ] (6.13 We know that a probability distribution function of the form P(x = 1 ] [ σ 2π exp (x µ2 2σ 2 ( < x < (6.14 is a normalized Gaussian distribution where µ = mean, σ = < (x µ 2 > is the standard deviation. So in our case the variance, σ 2 =< (A i A 0 i 2 >= ( 2 S k B (6.15 A 2 i A i =A 0 i For example, fluctuation in volume A i = V, the dispersion can be calculated as follows: 2 S A 2 A 0 = 2 S i V i 2 eqm = ( S V V = ( P V [From Maxwell s relations ( S V P = ( P V ]. For ideal gas P = Nk B V ; P V eqm ( P = Nk B S V eqm (6.16 (6.17 Since equilibrium deals with average properties we must replace V eqm by < V > and from Eq (6.16 we have Now with the help of Eq (6.15 we get 2 S V 2 eqm = Nk B < V > 2 (6.18 < (V < V > 2 >= < V >2 N and the relative fluctuation in volume is given by < (V < V > 2 > < V > 2 = 1 N (6.19 (6.20

38 32 thermodynamic fluctuations If N is very large (large system, the fluctuation is negligible i.e., thermodynamics is valid for N. hus in this limit, thermodynamic can be applied in a straightforward way without bothering about fluctuation or dispersion.

39 H E R M O D Y N A M I C F L U C U A I O N S, F O R C E A N D F L U X 7 In any thermodynamic system one must consider both the average and fluctuation around the average to completely describe the system. In N limit the fluctuations become insignificant. In irreversible thermodynamics, parameters are slightly away from equilibrium in small local cells. Hence fluctuations must be considered. If the system is in equilibrium, only the average values need to be considered. Let entropy at equilibrium is maximum (S 0 and in any other time entropy is S. S = S({A i } (7.1 {A i } is a set of thermodynamic parameters. he fluctuations of the variables {A i } around their equilibrium values (A 0 i are defined as the Onsagar coordinates α i = A i A 0 i (7.2 So entropy can be also expressed as a function of {α i } S = S({α i } (7.3 If the fluctuation is small then one can write (upto second term of the expansion S = S 0 + i ( S α i + 1 ( 2 S α i α j (7.4 α i α i =0 2 α i α j α i,α j =0 At equilibrium S is maximum and so ( S α i i ij α i =0 = 0 (7.5 Hence the departure of entropy from its equilibrium value is given by S = S S 0 = 1 g ij α i α j (7.6 2 ij 33

40 34 thermodynamic fluctuations, force and flux where g ij = ( 2 S α i α j α i,α j =0 < 0. he Boltzmann distribution S = k BlnW does not need any equilibrium condition. It is valid irrespective of the system being in equilibrium or not. he criterion for equilibrium is S is maximum which makes W most probable. Since W W 0 = e S k B (7.7 S = S({α i } (7.8 S = S({α i } (7.9 he probability of finding the system with fluctuations between α 1 and α 1 + dα 1, α 2 and α 2 + dα 2,...and α n and α n + dα n is given by dw = e S k B dα 1 dα 2...dα n ( Entropy production from fluctuation theory: hermodynamic force X i is defined as Now with the help of Eq (7.6 we obtain X i = S α i (7.11 X i = j g ij α j (7.12 he entropy production is S t = i S α i j α i t = i X i J i (7.13 is identified as the flux, corre- he rate of change of fluctuation, i.e., α i t sponding to the thermodynamic force X i

41 7.2 calculation of averages < α i x i >: Calculation of averages < α i X i >: < α i X j > = = α i X j W(α 1, α 2,...α n dα 1 dα 2...dα n α i S α j e S k B dα 1 dα 2...dα n ( k = k B α i e S B dα 1 dα 2...dα n α j [ ] k = k B α i e S α i B k k B e S B α 1 dα 2...dα n (7.14 α j Since the distribution function vanishes at the boundary and α i α j = δ ij, < α i X j > = k B δ ij Wdα 1 dα 2...dα n = k B δ ij ( Properties of fluctuation: he thermodynamic parameters (variable or state functions such as temperature, pressure etc. are macroscopic quantities and are given by < E kin > = < 1 2 mv2 >; P = < F A > (7.16 Note that from mechanics involving microscopic variables the velocity v and the force F are given by v 2 = ( dx 2 ; F = m d2 x dt dt 2 (7.17 If time t replaced by t then kinetic energy (E kin and pressure (P remain invariant. hus kinetic energy and pressure (or force obey time reversal symmetry. he Newton s equation (F = mẍ and Schrödinger equation (i h ψ t = Ĥψ also obey time reversal symmetry at the microscopic level. At the macroscopic level, the time reversal symmetry makes its presence felt in the principle of detailed balance. Let us restrict our attention to the fluctuating parameters, α i (t + τ, t is large time and τ is small time interval away from t. Now the value of fluctuation at t + τ, α i (t + τ may not be equal to fluctuation at t τ, α i (t τ in general, but following averages over a large number of observations are equal < α j (tα i (t + τ > = < α j (tα i (t τ > (7.18

42 36 thermodynamic fluctuations, force and flux and also < α j (tα i (t + τ > = < α j (tα i (t τ > = < α j (t + τα i (t > (7.19 he first equality presents the time reversal property of fluctuations. he second equality indicates invariance under time translation (t t + τ. his implies that fluctuations are stationary. he properties of the time reversal symmetry is an important ingredient for establishing reciprocity relation which we treat in the next chapter.

43 R E G R E S S I O N O F F L U C U A I O N A N D W O L AW S O F I R R E V E R S I B L E H E R M O D Y N A M I C S 8 Onsager assumed that fluctuation regresses to equilibrium exactly in the same way as the corresponding macroscopic quantities (Regression hypothesis, i.e., regression of fluctuation follows a linear law viz. α i = j A ij α j (8.1 We present here a heuristic basis of this hypothesis. Consider, for example, the following reaction scheme A k f B (8.2 k b he rate equations for the above reaction scheme are dc A dt dc B dt = k f C A + k b C B (8.3 = +k f C A k b C B (8.4 C A and C B being the concentrations of A and B, respectively. At equilibrium k f k b = C0 B C 0 A = K (8.5 C 0 B and C0 A denotes the equilibrium concentrations and K is the equilibrium constant. Let us consider small fluctuation of the concentrations α A and α B from the corresponding equilibrium values. C A = C 0 A + α A; C B = C 0 B + α B (8.6 With the help of kinetic Eqs ( and detailed balance Eq (8.5 we obtain dα A dt dα B dt = k f α A + k b α B (8.7 = +k f α A k b α B (8.8 37

44 38 regression of fluctuation and two laws... or, d α A dt α B = k f k f k b k b α A α B (8.9 his is the regression kinetics of fluctuation. hus from the Eqs ( and Eqs ( it can be said that both the kinetics follow the same linear law. We now return to the previous discussion in the last chapter to note the following expression for force X i = ij g ij α j (8.10 By reverse transform one can write α j = m G jm X m (8.11 From Eqs (8.1 and (8.11 we get α i = j or, α i = m A ij G jm X m m L im X m (8.12 where the matrix product is j A ijg jm = L im. Eq (8.12 is exactly the linear law. Now from the definition of derivative we have α i = α i(t + τ α i (t τ = m L im X m (8.13 τ is neither very small, nor very large, but of some intermediate value. Multiplying both side of Eq (8.13 by α j (t we get α j (tα i (t + τ α j (tα i (t τ = m L im α j X m (8.14 Now taking ensemble average and using Eq (7.15 ( α j X m = k B δ jm we obtain α j (tα i (t + τ α j (tα i (t τ = k B L im α j X m m = k B L im δ jm m = k B L ij (8.15

45 regression of fluctuation and two laws Similarly (just interchanging i and j in Eq (8.15 we obtain α i (tα j (t + τ α i (tα j (t τ = k B L ji (8.16 From the stationary property of fluctuation [Eq (7.19] one can easily see that α i (tα j (t + τ α i (tα j (t τ = α j(tα i (t + τ α j (tα i (t (8.17 τ hus the right hand sides of the Eqs (8.15 and (8.16 must be same, i.e., L ij = L ji (8.18 his is the second law, reciprocity relation of irreversible thermodynamics.

46

47 A P P L I C A I O N O F H E L AW S O F I R R E V E R S I B L E H E R M O D Y N A M I C S, I : H E R M O E L E C R I C E F F E C S 9 he linear law and the expression for entropy production are given by J i = j L ij X j ; ds dt = σ = j J i X i (9.1 o have the reciprocal relation (L ij = L ji the above two equations are necessary but not sufficient. One must require Onsager co-ordinates i.e., α i = J i ; and X i = δ S δα i (9.2 where S = 1 g ijαi α 2 j (9.3 i,j Our object here is to use the two laws of irreversible thermodynamics to establish some relationship between transport coefficients. In this section we consider thermodynamic phenomena. (i Seebeck effect: Consider a metallic strip M (say, copper with two end points A and B connected to a dissimilar metallic strip N (say, silver. A temperature difference ( between two junction points are maintained externally and these points are connected to a potentiometer to measure potential difference between the two junction points as shown in Fig 9.1. We observe a potential difference ( φ = 0 between two points A and B under zero current condition. At the junction point A we have ds A = de (9.4 A + B C P Figure

48 42 application of the laws of irreversible thermodynamics... (energy is released to maintain lower temperature. At the junction point B ( + ds B = de (9.5 (energy is supplied de is the measure of thermal energy. he open point C is at equilibrium temperature ( 0, we assume 0 (since C-point is closer to A than B. ds C = φde (non-mechanical work due to charge separation (9.6 Eq (9.5 can be rewritten as ds B = = = 1 de + de (1 + / de de (9.7 2 From Eqs ( the total entropy change is given by (ds A + ds B + ds C, i.e., ds = φ de de (9.8 2 Here de = dq V (heat flow at constant volume. So entropy production is given by dṡ = dq V 2 dt corresponding force terms are X Q = 2 one can write φ de dt = 2 J Q φ J e (9.9 where J Q and J e are heat flux and charge flux respectively and the and X e = φ. From linear law ( J Q = L 11 2 J e = L 21 ( 2 Under zero current condition J e = 0 L 21 ( 2 ( φ L 22 or, ( φ ( φ L 12 L 22 ( φ J e =0 = 0 = L 21 L 22 ( 1 (9.10 (9.11

49 application of the laws of irreversible thermodynamics A + B C Figure 9.2 ( φ, change of potential per unit change of temperature is called J e =0 thermoelectric power which is observed to be independent of nature of the substance. (ii Peltier effect: he experimental arrangement is same as in Fig 9.2. We maintain (with the help of an external source, battery a potential difference between two junction points A, B and = 0 condition to start with. One then observes a temperature difference between two points A and B. Now putting = 0 in Eq (9.10 we get ( φ J Q = L 12 ( φ J e = L 22 ( JQ = L 12 (9.12 L 22 J e =0 Using reciprocal relation (L 12 = L 21, Eqs (9.11 and (9.12 result in ( ( φ JQ = = L 12 (9.13 J c =0 J e L 22 =0 he above equation presents the connection between the two effects (Seebeck effect and Peltier effect. Some notable conclusions (i Reciprocity relation is valid when the time reversal symmetry is maintained. In absence of time reversal symmetry, the reciprocity relation will not be valid, for example, in presence of magnetic field. (a moving charged particle in a magnetic field (B experiences a Lorentz force F = e v. B which is devoid of any time reversal symmetry due to the fact that v(t v( t and hence reciprocity relation will not be valid. (ii Fourier law and Ohm s law are in a sense equivalent to Onsager s direct effect but not exactly same as these laws involve some spatial variation ( φ x or x

50 44 application of the laws of irreversible thermodynamics... whereas the fluctuations are defined on the basis of Onsager coordinates (α i or α i. hus L ij = L ji assumes certain microscopic basis (fluctuation in terms of Onsager coordinates and hence is not as general as first law or second law of thermodynamics whose foundation is completely independent of any microscopic basis.

51 A P P L I C A I O N I I : R A N S P O R O F H E A A N D M A S S 10 Let consider a gas in a large enclosure composed of two vessels at equilibrium, connected by means of a capillary or a membrane as shown in Fig A temperature difference is maintained between two vessels (vessel-i is at the temperature and vessel-ii at +. he system is allowed to settle upto a time it reaches a state in which the transport of matter vanishes, while the transport of energy as well as the entropy production remain non-vanishing. he state variables tend asymptotically to time independent values, or a nonequilibrium steady state. (No confusion should arise between steady state and equilibrium which is characterized by a zero entropy production. At the steady state one can observe a pressure difference, called thermomolecular pressure due to temperature difference between two vessels. For vessel-i and vessel-ii the entropy changes due to energy flow are given by, and ds II = de + = ds I = de (10.1 de (1 + / = 1 de de ( respectively. he entropy due to transport of mass is given by (for dn mole transport from vessel-i to vessel-ii ds m = dg m = 1 [µ I( dn + µ II (dn] = µ dn (10.3 where µ is the chemical potential. he total entropy change from Eqs (10.1 to 10.3 can be written as ds = ds I + ds II + ds m = µ de dn ( I II + Figure

52 46 application ii: transport of heat and mass Relation between µ and p: We know dg = Vdp Sd + i µ i dn i (10.5 From the above thermodynamic relation we have ( ( G G = µ i ; = V n i,p P,{n i } µ i P = ( G = ( G = V = V i (10.6 P n i n i P n i V is specific molar volume (for only one component, the index i is omitted. So for finite change from Eq. (10.6 we can write Now putting the value of µ into Eq (10.4 we get µ = V P (10.7 ds = V P de dn ( Here de = dq v (as no expansion work is involved since volume is fixed. Entropy production rate is given by σ = ds dt = 2 ( dqv dt V P ( dn dt = 2 J Q V P J m (10.9 where J Q and J m are heat and mass flux, respectively and the corresponding force terms are X Q = 2 write and X m = V P. Using linear law one can ( ( V P J Q = L 11 2 L 12 J m = L 21 ( 2 L 22 ( V P (10.10 For given 0, J m = 0, i.e., a stationary in which no mass flow but heat flow persists (it is not an equilibrium state since J Q 0 and entropy production has a non-zero value. L 11 ( 2 ( V P L 12 ( P J m =0 = 0 = L 21 L 22 ( 1 V (10.11