whether a process will be spontaneous, it is necessary to know the entropy change in both the

Transcription

1 93 Lecture 16 he entroy is a lovely function because it is all we need to know in order to redict whether a rocess will be sontaneous. However, it is often inconvenient to use, because to redict whether a rocess will be sontaneous, it is necessary to know the entroy change in both the system and the surroundings. It would be convenient if we could define functions that could redict sontaneity by referring to the system alone. o do this lets return to the second law. he second law says in a reversible rocess the entroy of the universe is constant; in an irreversible rocess, the entroy of the universe increases. We can exress this mathematically as d + d' 0, where d is the entroy change of the system, d' is the entroy change of the surroundings and the sum equals zero only if the rocess is reversible. herefore for sontaneous rocesses this becomes d + d' > 0. If the system and the surroundings are in thermal equilibrium at temerature, we can write d' = -dq/, where dq is the heat transferred to the system. hus we can write our condition for an irreversible rocess as d - dq/ > 0 or more generally as d - dq/ 0. his is a version of the Clausius inequality we saw earlier. Note that already this inequality refers only to the system, since the first term is the entroy change of the system and the second term contains the heat transferred to the system. But, and this is imortant, this is still the second law, so we are still redicting sontaneity, this time while referring only to the system. Now consider a constant volume system. At constant volume, we have dqv = du. If we substitute this in our inequality it becomes d - du/ 0, which with little difficulty we can rearrange to d du constant ), which becomes the condition for the sontaneity of a rocess at constant volume. o exlore some of the consequences of this simle inequality, let's consider

2 94 the additional constraint of constant energy. hen our equation becomes d 0 Constant U, ) or du, 0. d is only equal to zero for a reversible rocess, so this inequality tells us that at constant energy and volume, the entroy of the system increases for sontaneous rocesses. Now let s consider the situation where entroy is constant. he inequality now becomes du, < 0. In other words if the entroy of the system and its volume are constant, the energy of the system must decrease for a rocess to be sontaneous. his makes sense if we remember that the entroy of the universe must increase in order for a rocess to be sontaneous. If the entroy change of the system is zero, then the entroy of the surroundings must increase. If is constant so that no exansion work can be done, the only way for the entroy of the surroundings to increase is if heat flows from the system to the surroundings, which reduces the energy of the system. Hence du, 0. What haens to our original inequality, d - dq/ 0, if the heat is transferred at constant ressure? We know that dq = dh, so this yields d - dh/ 0 or d dh constant ). If we consider this inequality under conditions of constant enthaly as well it reduces to dh,p 0. In other words, at constant ressure and enthaly the entroy of the system increases for a sontaneous rocess. At constant, the inequality reduces to dh,p 0. Q: CAN ANYONE EXPLAIN WHY DH,P < 0 FOR A PONANEOU PROCE? [dtot > 0, dsys = 0, surr must be > 0, q flows to surroundings, so dh decreases] he essential content of these inequalities can be exressed by two new state functions. Note that our constant volume inequality can be exressed as du - d 0, which leads to the definition of the first of our new state functions, the Hemholtz free energy, A U -. For small changes of A its differential is da = du - d), which under conditions of constant becomes

3 95 da = du - d. Comaring this to our constant volume inequality we see that da, 0 is the constant, constant statement of the condition for sontaneity. imilarly our constant ressure inequality can be rewritten as dh - d 0, which leads to the definition of the Gibbs free energy as G = H -. Q: WHA I HE DIFFERENIAL OF G? [dg = dh - d).] Under constant this becomes dg = dh - d. Comarison of this equation with the constant inequality shows that at constant,, the condition for sontaneity is dg, 0. Let s discuss the Hemholtz free energy and the Gibbs free energy in more detail. We have shown that a rocess is sontaneous at constant volume and temerature if da < 0. In other words in a sontaneous rocess at constant volume and temerature the Hemholtz free energy of the system will decrease. When does it sto decreasing? o answer this lets consider a simle hysical roblem, a ball moving along a otential. If we ut the ball toward the to of the otential and release it, it moves sontaneously toward ositions of lower otential energy, i.e., du < 0. If we lace it at the bottom of the otential surface, and we release it, the ball will not move and du = 0, and we say the system is in equilibrium. his is not only the case for a simle otential like the one we've just considered. ake this more comlicated otential. At ositions A, B, C and D if we release a ball with no kinetic energy, it will remain stationary, du will equal 0, and the ball will be in a mechanical equilibrium.

4 96 Now let s return to the Hemholtz function. If we have a sontaneous rocess at constant and, da < 0. For examle, we could have a vessel containing air at 1.5 atm, where the surroundings are at 1.0 atm. We allow air to leak out of the system. If we let the rocess continue until it naturally stos, i.e. comes to equilibrium, then the ressures are equalized at 1 atm. WHA I DA A HE POIN WHERE WE REACH EQUILIBRIUM? o understand more about what da tells us, let s take our system now in equilibrium and ush it in the same direction, loss of ressure. Now the internal ressure is less than 1 atm. he sontaneous rocess is now for air to leak back into the system, so da < 0 for increasing ressure and by induction, da > 0 for continuing to decrease the ressure. In other words if we have a system out of equilibrium, and we use some rocess to aroach equilibrium, da < 0 for that rocess. If we ass equilibrium da > 0. In other words, before equilibrium, da < 0 and beyond equilibrium da > 0. herefore, at equilibrium, da must equal 0. his in turn imlies that when not in equilibrium da for the sontaneous rocess is < 0 but as we aroach equilibrium, da aroaches 0. In other words, A behaves in a way analogous to the otential energy for the motion of a mechanical object. Because A, G, H and U all behave in ways analogous to the otential energy, they are all called thermodynamic otentials. o clarify a final time, da < 0 imlies a sontaneous rocess at constant, and da = 0 at equilibrium for constant, and. Remember that da = du - d < 0 for a sontaneous rocess. herefore at constant and sontaneous rocesses tend toward lower energy and higher entroy. Remember, though, that in this equation, du is a measure of sontaneity only because it reflects the entroy change in the surroundings at constant,, and that the statement du - d < 0 is

5 97 exactly equivalent to the second law statement, he entroy of the universe increases in a sontaneous rocess. A has another meaning in addition to heling redict sontaneity under the aroriate conditions. If we know da for some rocess, or A for some finite rocess, we know the maximum work the system can do. herefore we can write da -dwmax and A -wmax. o show this lets start by roving our earlier assertion that a system does its maximum work under reversible conditions. We start with the Clausius inequality, d - dq/ 0, remembering that the equality holds for a reversible rocess. We can rearrange this to d dq and remembering that dq = du - dw, this becomes d du - dw. We can rearrange this to -dw -du + d which reduces to -dw -da. ince -dw, the work done by the system, is less than or equal to -da, it follows that the maximum work occurs when -dw = -da, i.e., -dwmax = -da. Remember though that -dw is only equal to -da for a reversible rocess. herefore the maximum work is only achieved for reversible rocesses. Now let s return to the Gibbs free energy. Along with the enthaly, the Gibbs free energy is the most imortant state function in chemistry. We determined earlier that if dg,p < 0 then a rocess will be sontaneous. hese conditions of constant ressure and temerature are clearly imortant for chemists, since most reactions are run under conditions of constant temerature and ressure. Using the same reasoning that we used for the Hemholtz free energy, we find that if dg < 0 at constant and P, the rocess will be sontaneous. If dg > 0 at constant and P, the reverse rocess will be sontaneous, and finally if dg = 0 the system will be under equilibrium with resect to the rocess. Again as was the case with A, as a rocess moves a system sontaneously toward equilibrium, dg moves toward 0.

6 98 An illustration of the imortance of G is the sontaneity of endothermic rocesses under some conditions. ince G = H -, if H > 0, G can be less than 0 if > 0 and is sufficiently high. Using A we found that we could redict the maximum work done by a system. his leads us to ask if there is something similar that we could learn from G. We begin with our differential of G under conditions of constant, dg = dh - d. We know that dh = du + d) = dw + dq + d). he work term can be broken down into two comonents, exansion work, which is given by -ex d, and nonexansion work, which we will simly call dwe. If we insert this in our equation for dg, and exand d) we get dg = -ex d + dwe + dq + d + d - d. For a reversible rocess, ex =, and dq = d and dwe = dwe,max, so this simlifies to dg = dwe,max + d. Under conditions of constant ressure this finally simlifies to dg = dwe,max. Note that as in the case of the Hemholtz free energy, if dg < 0, the imlication is that dwe,max < 0 and the system is caable of doing non-exansion work. If on the other hand dg > 0, then dwe,max > 0 and for a rocess to occur, work must be done on the system.

7 99 Lecture 17 Remember that our definition of the Gibbs free energy is G = H -. he Gibbs free energy is of interest to chemists because it redicts the sontaneity of reactions under conditions of constant temerature and ressure. Of equal imortance in chemistry is the condition dg, = 0, since it allows us to relate the Gibbs free energy to conditions of equilibrium. However before we can relate dg to equilibrium in ways that are useful to us, we need to further exlore the nature of G, U, H, and A and learn ways to relate them. his means in art a return to artial derivatives. Let s begin by returning to our first thermodynamic function, the energy. Our earliest definition of energy change was du = dq + dw. In chater 3 we showed that dqrev/ = d or dqrev = d. For a reversible rocess we also have dw = - d. ubstituting both of these in our equation for du yields du = d - d. his is called the fundamental relation, because it defines du strictly in terms of basic state functions, and because our other major state functions, H, G and A are all defined in terms of U. he fundamental equation is the first of four Gibbs equations we ll see, each of which exresses the differential of a thermodynamic otential in terms of its fundamental variables. he equation for U shows that U is a function of and, i.e., U = U, ). We can see this because in our fundamental relation and are the variables contained in the differentials. It can also be seen because and are the variables that are held constant when we use du to redict sontaneity. As such they are called natural variables. It is ossible to define U as a function of other variables, as we did earlier in defining U as a function of and, but as a consequence of our exloration of the second law, U allows us to redict sontaneity or equilibrium when and are held constant, and not when other variables are. Because U is a function of and, the differential du becomes

8 100 du = ) d + ) d If we comare this to our fundamental relation we see that ) = and ) = - he first equation defines the temerature as the ratio of changes in energy and entroy in a closed volume. You can see that by delving deeer into thermodynamics we are beginning to interrelate its fundamental quantities. he fact that the energy is an exact differential allows us to develo a new set of thermodynamic relations. Remember that our test for an exact differential is that if we have a function df = a dx + b dy, then a/y)x = b/x)y. Our fundamental equation is du = d + d. ince U is a state function, du is an exact differential, and we can aly the equation for df to du to get, ) = - ) his equation is one of a grou of relations called Maxwell relations. It is not immediately obvious and cannot be derived from our rules for the maniulation of artial derivatives. An alternative aroach to deriving this Maxwell relation is as follows. We've already shown that ) = and ) = - We know from our earlier discussion of artial derivatives that the mixed second derivatives,

9 U x y ) and U yx ) are equal. herefore 2 2 U )= U ) and ) = - ) Because dh, da and dg are all also erfect differentials, we can generate Maxwell relations for them as well. Let s start with the enthaly. Q: WHA I HE DEFINIION OF HE ENHALPY? [ H = U + P] Q: WHA I HE DIFFERENIAL OF H, DH? [dh = du + d)] If we substitute our fundamental equation for du, and exand d) we obtain dh = d - d + d + d = d + dp. his is the Gibbs equation for the change in enthaly. Q: HI ELL U HA H I A FUNCION OF WHICH ARIABLE? [H = H,)] We write down our test for a erfect differential, a b if df = adx + bdy then = y x and recognize that since dh is a erfect differential we can write ) = ) our second Maxwell relation. Now let s do the remaining two. Q: WHA I HE DEFINIION OF G? [G = H - = U + P - ] Q: WHA I HE DIFFERENIAL OF G, DG? [dg = du + d) - d)] = d - d + d + d - d - d and therefore, dg = dp d, the Gibbs equation for the Gibbs Free Energy. he natural variables for G are therefore and since small changes in and lead to changes in G, i.e., G = G, ). Q: WHA I HE DIFFERENIAL OF GP,)? G dg = ) d + G ) d If we comare this equation with our other equation for dg we find that G/) = - and G/), x y

10 102 =. Q: WHA I HE MAXWELL RELAION FOR G? 2 2 G )= G ) or - ) = ) Finally lets calculate the Maxwell relation for A. Q: WHA I HE DEFINIION OF A? [A = U - ] WHA I HE DIFFERENIAL OF A? [da = du - d) = d - d - d - d and therefore da = -d - d.] Q: WHA ARE HE NAURAL ARIABLE OF A? [A = A,)] Q: WHA I HE MAXWELL RELAION FOR A? [/) = /).] his yields us four Gibbs equations and four Maxwell relations. he Gibbs equations are: And the four Maxwell relations are: du = d + d dh = dp + d da = d d dg = d d ) = - ) s ) = ) ) = ) ) = - ) You can see that these last two of the Maxwell relations look articularly useful since they exress the difficult to measure /) in terms of the easy to measure /), and the difficult

11 103 to measure -/) in terms of the easy to measure /). hese Maxwell relations are imortant enough that I would exect you to be able to easily derive them. While both the Gibbs equations and Maxwell relations are very useful, and A fairly easy to derive, they are hard to remember. here is a simle mnemonic device called the thermodynamic square which makes it very simle to remember the Maxwell relations and the Gibbs U G equations for the thermodynamic otentials. o construct the H thermodynamic square we draw a square. At the to of the square we ut A, the first of the thermodynamic otentials alhabetically, and then roceed both alhabetically and clockwise with the rest of the otentials. Now in the corners we ut the variables common to the otentials on A either side. A and G are both functions of, G and H are both functions of, U and H are both functions of and U and A are U G both functions of. Finally we draw an arrow from to and from to. We can use the square two ways, to H P

12 104 remember the Gibbs equations, the differential forms for changes in the otentials, and to remember the Maxwell relations. o find the Gibbs equation for a otential, rotate the square until the otential you want to find is on the bottom of the square. In this case we'll use H. he differential form of a otential is the sum of two terms, each of which is a differential multilied by another variable. When we lace a otential on the bottom of the square, the differential indeendent) variables will be the ones flanking it. For the examle of dh, the differentials are d and d. Each of these differentials is multilied by the variable diagonally across from it. hus the two terms in dh are d and d. Now all we need are the signs of the two terms. For the signs we just look at the direction of the arrows. If the arrow connecting two variables oints u the sign on the term will be ositive, and if the arrow connecting the two variables oints down, the sign on the term will be negative. he arrow oints u so we have +d and the arrow oints u so we have +d, which finally yields, dh = d + d. o do a different thermodynamic otential we just turn the thermodynamic P H square so the otential we are interested in is on the bottom. Let s use the square to get the differential of A. First we turn the G U square so A is on the bottom. he differential terms are d and d. he two terms, d and d are both negative A because for both of these the arrows oint down. herefore da = -d - d.

13 105 Let s do one last examle, dg. First we turn the square so that the G is on U the bottom. Q: Which variables are the differentials? [d, d] Q: What are the airs of variables for the two terms? H A [P,] herefore the two terms are dp and d. Q: What is the sign for dp? [Arrow u, +] Q: What is the sign for d? P G [arrow down, -] o dg = d - d. Now let s use the thermodynamic square to find Maxwell relations. First let s A turn it right side u again. You can see that the arrow makes a triangle with, and. he artial from this corner has the U G variables in the same ositions, i.e., /). he other triangle from this orientation of the square is made with, H P and, which gives us the artial /). Once again the direction of the arrow gives the direction for each term. ince both arrows go u both terms are ositive. o the Maxwell relation for this side is /) = /). o find other Maxwell relations just turn the square on its side.

14 106 Lets look at the Maxwell relation for the U side. Once again we turn G P our square so the otential of interest is on the bottom. he first triangle is made by, and. Q: WHA I HE CORREPONDING A H PARIAL? [/)] he second triangle is made by,, and. Q: WHA I HE PARIAL? [-/).] o this second U Maxwell relation is /) = -/)

15 107 Lecture 18 Let s see how we might use these Maxwell relations. Remember we said that for an ideal gas the energy is indeendent of volume at constant temerature. his is equivalent to saying that ) = 0 for an ideal gas. Let s rove this, but instead of evaluating this directly for an ideal gas, let s find an equation for ) which is true for any substance, and then aly that to the ideal gas law. uch an equation is called a hermodynamic Equation of tate. In develoing a thermodynamic equation of state, we want to exress ) in terms of our observables,,,,, n, α, κ, and C. Before we start lets write down our definitions for,, and. U = = - U = G ) ) ) = - G ) Let s try relating /) to temerature. o do this we need to use a version of the chain rule of artial differentiation. f x ) = f x ) + f y ) y x ) z y x z If we comare ) and our definition of with this equation we can see that if we set ) equal to f x, U = f, = x, and = z. ince we re relating this to temerature, z has to equal y, since y is the only unassigned variable in our equation. his means that we can write ) = ) + ) ) Comaring this with our thermodynamic definitions of and we see that

16 108 ) = -+ ) Unfortunately, /) is inconvenient to measure, and also means that to calculate the change of internal energy with volume we have to kee track of four variables,,, and. It would be convenient if we could find a more useful equivalent of /). Examining our list of Maxwell relations we see that /) = /). o this equation now becomes ) = -+ ) an equation which relates the change of energy with volume to easily measured observable quantities. Earlier we had argued that /) = 0 for an ideal gas. his conclusion was based on discussions of transfer of energy from kinetic energy, the energy of heat, to otential energy, as the molecules were searated against an external force. It was a sensible argument, but by no means a roof. With our result above we can rove this assertion. We begin with ) = -+ ) ince for an ideal gas = nr/, this becomes nr ) = -+ )= -+ = 0 his simle result shows one of the strengths of thermodynamics. It is able to obtain results that can either be alied generally, as in our thermodynamic equation, or be alied to secific models, as we have just done for the ideal gas law. Let s use this equation to calculate the value of ) for a van der Waals gas. Remember that for a van der Waals gas

17 109 = nr - nb - an 2 ubstituting into our equation for /) yields 2 ) = -+ nr ) = -+ - nb = - nr - nb + an 2 - nr - nb = an so we see that for a gas with any kind of real roerties, the energy will change with volume at constant temerature. We see also, that since the term which reresents the effect of the attractive otential is the one which makes /) non zero, that it is indeed the resence of attractive forces that makes energy change when gasses are exanded under constant temerature conditions. Now that we've increased our arsenal for maniulating artial derivatives, let s return to the Gibbs Free Energy. ince we often want to control a chemical reaction by changing the ambient temerature, and the Gibbs Free Energy redicts sontaneity under chemically useful conditions, let s look at the way G changes with temerature. Remember that G dg = ) d + G ) d = -d +d and therefore G/) = -. Our definition of G was G = H - which can be rearranged to yield G - H)/ = - and therefore G/) = G - H)/. If we collect the G's on one side of the equation we get G/) - G/ = -H/. his exression is difficult to integrate. However, if we recognize that because of the roduct rule, G ) - G = G ) his becomes G )= - H 2

18 110 his is known as the Gibbs-Hemholtz equation and relates the enthaly to the temerature deendence of G/. his is not a critically imortant equation, but the form can be useful for some tyes of calculations. Note that the forms G ) = G - H = - are of equal imortance. he Gibbs-Hemholtz equation is articularly useful in calculating the change of G for a rocess as the temerature changes. In this case the equation becomes G )= - H 2 Let s look now at how the Gibbs function varies with ressure. We begin with our differential of G, dg = d - d. If we change ressure at constant from 1 to 2 we get dg = d. and therefore, 2 G f Gi 1 = + d For liquids and solids, is almost constant so this can be aroximated as G = G + f i or for molar quantities, G = G +. f i For gases, m is deendent on ressure so we cannot take it out of the integral. For an ideal gas we have R f G f = Gi+ d= Gi+R ln. his equation is imortant in showing the ressure deendence of the Gibbs Function, but is articularly imortant in the way it allows us to begin to relate G to chemical equilibria. i

19 111 Before we can develo relationshis involving equilibria, we have to introduce a new concet, that of the chemical otential, µ. he concet of the chemical otential, µ, is develoed in the following way. Remember that in our earliest discussions of energy we talked about extensive roerties, roerties whose values deend on the amount of material. Energy is one examle of an extensive variable. his should make basic sense. For examle, 2 gallons of gasoline will take you twice as far as 1 gallon of gasoline, and therefore will do twice as much work. U to now we've only considered thermodynamic rocesses in which mole number is constant, in other words, closed systems. For these rocesses we've said that U = U,). If we allow mass transfer, i.e., if we consider oen systems, the mole number affects the energy as well. In this case we have to write U = U,,n). If we had a system with two chemical comonents, then U will deend on the mole number of each of them and U = U,,n1,n2). For a system of j comonents this becomes U = U,,n1,n2,...,nj). When we wanted to describe the change in energy for a system with no mass transfer, we wrote down the differential, du = ) d + ) d where we have already shown that /) = and /) = -. For a one comonent system in which we consider mass transfer this becomes du = ) d + ) d + n ),n,n, We call /n),, the change in energy with mole number, the chemical otential, µ. We can thus write du = d - d + µdn. For a j comonent system this becomes dn du = d - d + µ 1dn 1+ µ 2 dn v+...+ µ j dn j,

20 112 where µ j = n j Remember that G is defined by G = H - = U + P -. ubstituting our exlicit equation for du we get dg = d + d + µ dn his means that we can now define the chemical otential in terms of the free energy: j j j G µ i = ni, n, j i. his is the most common definition of the chemical otential. As we will discuss in more detail later, for a given extensive quantity Y, artial derivatives of the form Y i Y = ni are called artial molar quantities. hus the chemical otential of the ith comonent of a mixture is the artial molar free energy of that comonent, i.e. µ i = G i. Qualitatively, the chemical otential is the change in free energy of a system when a mole of the substance is added to a mixture of a given comosition at a given temerature and ressure. hus the chemical otential is a function of the comosition, temerature and ressure, i.e., µ = µ, n, ), i i j where nj reresents the mole number of each of the comonents in the mixture. An interesting line of reasoning that follows from these definitions is that if the chemical

21 113 otential is the artial molar free energy of a comonent in a mixture, that we could figure out the free energy G for a given comonent by multilying the chemical otential of the comonent by its mole number, i.e., G = n µ = n G, i i i i i and the free energy G for the whole mixture by adding u the free energies of all the comonents, i µ i i. i i G = G = n his is an exciting result. In a system in which mass can be transferred, the Gibbs free energy is the sum of the roducts of the chemical otentials of the comonents of the system with their mole numbers. Let s look at this definition for a one comonent system again. We have G µ = n,. We can write G = ng. herefore for a one comonent system we have ng) µ = = G, and n we see that for a one comonent system the chemical otential is equal to the molar Gibbs Free Energy. As is our habit, let s begin our investigation of the chemical otential by considering the case of an ideal gas. First though, we'll use our result on the ressure deendence of the free energy to relate G, the tandard Gibbs function, to G at other ressures. For ideal gases and other gases, we can use, easily measured, as a measure of n since for an ideal gas we have nr nr Bn Cn = and for a real gas n and are related by = , while for mixtures,

22 114 of real gases and n are related by n = = n i i X n i j= 1 j. he G's usually available in reference books are standard molar Gibbs free energies,. If we want to find G at some other ressure, we just use the ideal gas results we've obtained for 0 G the ressure deendence of G, 0 G ) = G + R ln. But for a one comonent system, G 0 is µ so this becomes ) 0 µ = µ + R ln. his relation of µ to G leads us to the one comonent 0 version of the Gibbs-Duhem equation. ince dg = d d and G dg = d = d d, n n n we can write d µ = d d n n.

23 115 Lecture 19 Having develoed a substantial arsenal of thermodynamic functions, I'd like to return to the subject of thermodynamic equations of state. When I was an undergraduate, deriving these equations was the greatest ain of anything I was asked to do in P-Chem. When I got to grad school though, I was taught a simle sequence of stes that made most of these derivations simle. I'd like to share this with you to sare you some of the frustration that I went through. A key concet is that most of the things we might want to measure or calculate in thermodynamics can be or must be exressed as a artial derivative. he algorithm I'm going to teach you is for reduction of any artial derivative to a function of,,,, α, κ and C, all of which are either easy to measure or commonly tabulated. he algorithm consists of the following sequence of stes. te 1) Eliminate the otentials from the artial derivative i.e. U, H, G, A) thermodynamic square. Examles a) Bring to numerator b) Eliminate by using Gibbs equations the differentials of H, U, etc.) from the A) Eliminate the otential from the artial derivative /H) We begin by bringing H to the numerator using the inverter. H ) 1 = H ) Now we eliminate the otential by substituting dh = d + dp

24 116 H ) 1 1 = = d +d ) d ) + ) B) Eliminate the otential from /)H. We begin by bringing the otential into the numerator using the Euler chain relation. H - ) = - 1 ) H = H ) H ) H ) and now roceed as in our first examle. te 2) Eliminate µ a) Bring it to the numerator b) Eliminate it using the Gibbs-Duhem equation, d µ = - n d + n d Examle: - ) n d + µ n dp ) - d n ) + n = = ),n,n,n,n te 3) Eliminate from the artial derivative a) Bring it to the numerator b) Use a Maxwell relation if ossible c) Use the chain rule to insert under, and relace by the aroriate heat caacity. Examles A. /),n. We use the Euler cyclic rule to bring to the numerator.

25 117 - ) = - 1 ) = ) ) ) We can use a Maxwell relation to obtain ) = - ) / ) Now /) can be related to C by C = /) since q = d at constant. imilarly we can show that Cv = /). o finally we obtain, ) = - ) / 1 C For our second examle, let s look at a case where we insert under. When will we have to do this? When we can't find a Maxwell relation that fits! te 4) Bring to numerator, then use a) /) = α ) = ) ) = 1 C ) b) /) = -κ Examle: /) =? Q: What do we use to move into the numerator? [Euler cyclic rule] - ) -1 -κ κ ) = = = = ) - ) ) α α And finally te 5) Get rid of Cv using C = Cv + α 2 /κ.

26 118 Now let s do a multiste examle. Let s calculate the change of temerature with ressure under conditions of constant entroy, /). ince there are neither thermodynamic otentials nor a chemical otential in this artial, we can move directly to ste three. First we use the cyclic rule: which yields = 1 s 1 = Now we bring to the numerator using the inverter. = he lower term is simly C/ so we now have = C Looking at our thermodynamic square we find a Maxwell Relation =, which when inserted into our equation yields

27 119 = C Noting that our artial derivative is equal to α, we get as our final result α = C As a second examle let s use the same method to calculate the change in energy for a real gas during an isothermal comression, du = /),nd. For this calculation we need to eliminate /),n. First we eliminate du using du = d - d. ) = - ) = ) - ) Now we eliminate. First we turn to the thermodynamic square to get /) = -/), so ) = - ) - ) = - + α κ his simle result is an examle of the ower of thermodynamics. In order to obtain our earlier result about the isothermal change of energy for an ideal gas, du = Cv d, we had to make certain assumtions about the gas, i.e., that there were no intermolecular otentials and that therefore ) = 0. But here we have a comletely general result with no reference to a model for the structure of our system and no aroximations. As a final examle lets derive the result we use in ste 5, C 2 α Cv = κ. his is a more challenging reduction of artial derivatives since our algorithm by itself is not

28 120 enough attemts to solve this roblem by using the algorithm alone simly yield the equation C Cv = C Cv. We start simly enough using the definitions of C and Cv, C H Cv =. v We now roceed by exressing H in terms of U using the definition C H = U +. ) U + Cv = ) = + = + It is at this oint that our algorithm fails. When we run into a roadblock of this sort, the solution is to try a different aroach. he only one of our tools for maniulating artial derivatives that is not used in our algorithm is the chain rule for artial differentiation. We use it here to relate the artials and. Remember that our chain rule for artial differentials is f f y 1 f y 2 = + x y x y x 1 x 1 2 y 1 2 x y1 x2. If we choose f x to be 1 x 2 and to be f y 1 y 2, this fixes all our variables, with f = U, x1 =, x2 =, y1 = and y2 =. Plugging these into the chain rule we get

29 121 = + = + ubstituting this into our equation for C-C we get C Cv = + + = + = + We now roceed as usual with our algorithm, substituting the Gibbs equation for U, and yielding d + d C Cv = + d = + + d = We next get rid of. We look at our thermodynamic square and find that we have a useful Maxwell relation =. ubstituting this in our equation yields d C C = We now move the into the numerator using the cyclic rule.

30 122 1 C C = = Finally, using the relations α = and κ =, we get C C = α κ 2

31 123 Lecture 20 Let s return now to our discussion of the chemical otential. We showed that for a erfect gas, µ µ 0 = +. However, if, as we want, we relate µ to real chemical systems, we will R ln 0 have to deal with real gases. We could deal with this situation by going back to our definition of µ for a one comonent system, µ = G, and recalculate its ressure deendence by inserting the virial equation, f G f Gi i - = d and integrating to yield R 2 = 1+ B +C ) 1 C G f =G +R +B +C)d=G +R ln +B- )+ - )) 0 which gives us for µ for a real gas, 0 0 C 2 02 µ = µ +R ln +B- )+ - )...) his equation is far more accurate for real gases than our ideal gas version, but it has the disadvantage that it is awkward to handle with all its terms. hese considerations led G.N. Lewis to suggest the following equation for the ressure deendence of µ for real gases: µ µ 0 = +, R ln f 0 where f is called the fugacity and has units of ressure. Remember that for an ideal gas, the term R ln / 0 comes from integration of d. Relacing the ressure with the fugacity is a

32 124 reflection that the ressure deendence of the volume for a real gas is different than that for an ideal gas. Fugacity is related to ressure by the definition f = γ, where γ is called the fugacity coefficient, is a function of and, and is different for each substance. As with any real gas roerty, we want the fugacity to reduce to the ressure in the limit of low ressure, since real gases behave like ideal gases in the limit of low ressure, i.e., lim f 0 = ince f = γ, we can also say, limγ 1. = 0 It is imortant to note that like the virial coefficients that we use to calculate the,, deendence of real gases, the fugacity coefficient is an emirical quantity, which has to be measured or calculated from other emirical quantities. o see how we can calculate this lets insert our definition of fugacity into our equation for µ as a function of. 0 f 0 γ 0 µ = µ + R ln = µ + R ln = µ + R ln + R lnγ It is interesting to note here that in this last form, our first two terms are the same as our exression for the chemical otential of an ideal gas. Remember that an ideal gas is distinguished from a real gas because its intermolecular forces are negligible. ince our exression for µ for a real gas is identical to the exression for an ideal gas excet for the term R ln γ, R ln γ must contain all the information about the effect of the intermolecular forces of a substance on the chemical otential. o calculate the value of γ, begin with dµ = dg = d constant ). hen

33 ' 125 f f ' f md = µ µ ' = R ln R ln = R ln 0 0 f ' For a erfect gas the difference between µ at two different ressures is ' d = µ µ = R ln 0 0 0' m ' where here the suerscrit o indicates an ideal gas. If we subtract the ideal gas value from the real gas value we will get which rearranges to ' o f m- m)d = R[ ln - ln ] f f ' 1 ln )= - )d o m m f ' R. ' We can simlify this equation by taking its limit as ' 0 and noting that lim f ' = ' ' 0 herefore in the limit of low ressures, our equation becomes f 1 o ln )= m- m)d R 0 But f = γ so this becomes 1 o lnγ = m- m)d R 0 For an ideal gas m = R/. Now remember we defined a roerty of gases called the comression factor Z, where Z = m/r. For ideal gases Z = 1, but for real gases we can write m = RZ/. herefore we can write 1 RZ R Z -1 R 0 0 lnγ = - )d = )d.

34 126 Note that if we return to our virial descrition of a real gas we have so Z = R = 1+ B + C +... ), m lnγ = B +C +...)d = B + C +... his result is consistent with our understanding of ln γ as the comonent of the fugacity corresonding to the intermolecular forces. Now let s look at the imlications of this result. Our equation for the chemical otential of a real gas is 0 µ = µ + R ln + R ln γ 0 he first two terms corresond to the chemical otential of an ideal gas so this third term with ln γ will tell us how the real gas differs from an ideal gas. ince our equation for γ is in terms of the comression factor lets remind ourselves how the comression factor varies with ressure. At ressures close to 0, the comression factor is close to the ideal gas value of one. As we increase the ressure, the attractive forces come into lay and z dros below one. For most gases, at ressures somewhere between 200 and 400 atm, the reulsive forces begin to dominate and z becomes greater than one. We see from this that at ressures less than 200 atm, since Z<1, R ln γ is less than one and therefore µreal < µideal. If we have a gas at high ressure, where the reulsive forces dominate and Z > 1, ln γ > 0 and therefore at high ressures µreal > µideal.