This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 2 The Postulates of Thermodynamics
So, as science has progressed, it has been necessary to invent other forms of energy, and indeed an unfriendly critic might claim, with some reason, that the law of conservation of energy is true because we make it true by assuming the existence of forms of energy for which there is no other justiﬁcation than the desire to retain energy as a conservative quantity. – Kenneth S. Pitzer [64]. Thermodynamics is the combination of a structure plus an underlying governing equation. Before designing plays in basketball or volleyball, we ﬁrst need to lay down the rules to the game—or the structure . Once the structure is in place, we can design an inﬁnite variety of plays, and ways that the game can run. Some of these plays will be more successful than others, but all of them should ﬁt the rules. Of course, in sports you can sometimes get away with breaking the rules, but Mother Nature is not so lax. You might be able to convince your boss to fund construction of a perpetual motion machine, but the machine will never work. In this chapter we lay the foundation for the entire structure of thermodynamics. Remarkably, the structure is simple, yet powerfully predictive. The cost for such elegance and power, however, is that we must begin somewhat abstractly. We need to begin with two concepts: energy and entropy. While most of us feel comfortable and are familiar with energy, entropy might be new. However, entropy is no more abstract than energy—perhaps less so—and the approach we take allows us to become as skilled at manipulating the concept of entropy as we are at thinking 9 10 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS about energy. Therefore, in order to gain these skills we consider many examples where an underlying governing equation is speciﬁed. For example, we consider the fundamental relations that lead to the ideal gas law, the van der Waals equation of state, and more sophisticated equations of state that interrelate pressure, volume and temperature. Although the postulates are fairly simple, their meaning might be diﬃcult to grasp at ﬁrst. In fact, most students will need to revisit the postulates many times, perhaps over several years. We recommend that you think about the postulates the way you might vote in Chicago—early and often (or even after you die). Throughout most of the book, however, we will use the postulates only indirectly; in other words, we will derive some important results in this chapter, and then use these results throughout the book. It is therefore important to remember the results derived here, and summarized at the end of the chapter. 2.1 The Postulational Approach d Where does Newton’s law of motion F = dt (mv ) come from? Where did Einstein discover E = mc2 ? These equations were not derived from something—they were guessed in a ﬂash of brilliant insight. We have come to accept them for a few reasons: First, because they describe a great many experiments, and were used to predict previously unknown phenomena. Secondly, they are simple and straightforward to comprehend, although perhaps sometimes diﬃcult to implement. Thirdly, and more importantly, we accept these assertions, or postulates, of Newton and Einstein because we have never seen them violated. It is for these reasons—despite Pitzer’s ‘unfriendly critic’—that we believe that energy is always conserved. In this chapter we describe the postulates that make up the theory of thermodynamics. Just like Newton and Einstein, we must be willing to abandon our theory if experiments ever contradict the postulates. The postulates given here are not the most general possible, but are designed to be easily understood, and applicable to most systems of interest to engineers and scientists. In a few sections we brieﬂy consider generalizations. The approach in this chapter is essentially that of Callen [13]. Hence, the same restrictions apply—namely, the system must be isotropic, homogeneous, large enough to neglect surface eﬀects if we are talking about the bulk, or that we may neglect edge eﬀects if we are talking about surfaces, and no external forces. In what follows we make ﬁve fundamental postulates. The ﬁrst postulate (ﬁrst law) posits the existence of an additional form of energy called ‘internal energy’, which, along with kinetic, potential, electromagnetic and other energies, obeys a conservation principle. The second postulate assumes that every system has equilib 2.2. THE FIRST LAW: ENERGY CONSERVATION 11 rium states that are determined by a few macroscopic variables. The third postulate introduces a quantity called ‘entropy’ on which internal energy depends. The fourth postulate (the second law) and the ﬁfth (Nernst) postulate prescribe properties of entropy. The rest of thermodynamics follows from these straightforward ideas to have farreaching consequences. 2.2 The First Law: Energy Conservation What is the deﬁnition of energy? Despite using the word and the concept nearly everyday, most engineers and scientists stop short when asked this question. Nonetheless, we still ﬁnd the concept very useful. Consider a few simple thought experiments about energy. (1) How much energy does this book have if you hold it above your head before letting it drop to the ﬂoor? You might answer that it has potential energy mg l, where g  is the gravitational constant, and l is the height of the book above the ﬂoor. (2) If the book is ﬂying with speed v  while it is height l above the ﬂoor, is its energy mg l + 1 mv 2 ? (3) What if I place an ice cube on the book? We 2 note that the ice melts, and we assume (correctly) that heat was transferred from the book to the ice. What do these experiments tell us? First, we notice that the energy we ascribed to the book changed depending on the situation. That is because the situations made us think about several diﬀerent degrees of freedom or variables that we used to deﬁne the book’s energy. When we held the book over our head, we instinctively thought of position, and then calculated the potential energy of the book. When we pictured the book ﬂying, we then thought of position and velocity, and added the kinetic energy. When the ice melted on the book, we thought about concepts like temperature, heat, or maybe internal energy.1 Note that before we can talk about energy accurately, it is important to specify the system , and to specify what variables we are using . In our ﬁrst example the system is the book, and the position of the book is the variable. The other important point is the ﬁrst law, or ﬁrst postulate: Postulate I (First Law): Macroscopic systems possess an internal energy U that is subject to a conservation principle, and is extensive. An extensive variable is one that is linearly dependent on system size, and conversely, an intensive variable is one that is independent of system size.2 From
Our postulates will allow us to distinguish clearly between internal energy, heat and temperature. 2 A precise mathematical deﬁnition will be given shortly, in Postulate III.
1 12 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS a statistical mechanical, or atomistic point of view, we do not need this postulate, since we know that atoms store energy. We also know that all forms of energy are conserved. However, we choose to make this starting point clear in our framework by making it as a postulate. This internal energy is somehow transferred from the book to the ice in our thought experiment. Internal energy has properties just like other forms of energy in that it can be exchanged between diﬀerent systems, convert to and from other forms of energy, or be used to extract work. By ‘conservation principle,’ we mean that if we add up all of the forms of energy in an isolated system, the sum total of those energies is constant, although one form might have increased while another decreased, say some kinetic energy became internal energy. It is worthwhile to quote extensively from Callen [13, p.11]. The development of the principle of conservation of energy has been one of the most signiﬁcant achievements in the evolution of physics. The present form of the principle was not discovered in one magniﬁcent stroke of insight but was slowly and laboriously developed over two and a half centuries. The ﬁrst recognition of a conservation principle repeatedly failed, but in each case it was found possible to revive it by the addition of a new mathematical term—a “new kind of energy.” Thus consideration of charged systems necessitated the addition of the Coulomb interaction energy (Q1 Q2 /r ) and eventually of the energy of the electromagnetic ﬁeld. In 1905 Einstein extended the principle to the relativistic region, adding such terms as the relativistic restmass energy. In the 1930s Enrico Fermi postulated the existence of a new particle called the neutrino solely for the purpose of retaining the energy conservation principle in nuclear reactions . . . . Where is internal energy stored? Although it is not necessary to introduce atoms and molecules into the classical theory of thermodynamics, we might be bothered by this question. The answer might be in the vibrations of the atoms (kinetic energy) and the springlike forces between atoms (potential energy), or in the energies in the subatomic particles. So why do we call it ‘internal energy’ instead of kinetic + potential energy, for example? Recall our thought experiments above, where we found that the variables used to describe our system are essential to deﬁne energy. If we knew the precise positions and velocities of all the atoms in the book, we could, in principle, calculate all ∼ 1023 kinetic and potential energies of the book, and we would not need to think about internal energy. However, that approach is not only impractical, but, it turns out, is not even necessary. We can make meaningful calculations of internal energy by using just a few variables that are introduced in Postulate III. 2.3. DEFINITION OF HEAT 13 2.3 Deﬁnition of Heat If we do work on a system, then its energy must be increased. However, we often observe that after we do work on a system, its internal energy returns to its original state, although the system has done no work on its environment. For example, push the book across the table, then wait a few minutes. Although you performed work on the book, it has the same initial and ﬁnal potential, kinetic and internal energies. The reason the internal energy of a system can change without work is that energy can also be transferred in the form of heat. Since the internal energy of a system may be changed either by work, or by heat transfer, and we have postulated that energy is conserved, we can deﬁne the heat transfer to, or from, a system as dQ := dU − dW. ¯ ¯ (2.1) It is worthwhile to commit this equation to memory, or better yet: dU = dQ + ¯ dW . ¯ In this notation, work is positive when done on the system, and heat is positive when transferred to the system. We write the diﬀerentials for heat and work using d instead of d because they are imperfect diﬀerentials. ¯ Perfect diﬀerentials are not dependent upon the path taken from the ‘initial’ to the ‘ﬁnal’ states—they only depend on the initial and ﬁnal values of the independent variables. Imperfect diﬀerentials d have two important properties that distinguish ¯ them from perfect diﬀerentials d. First, imperfect diﬀerentials do depend upon the path. For example, if we set this book on the ﬂoor and push it a short distance dx away using force F and then the same distance −dx back to its original position using force −F , then the sum (perfect) diﬀerential for its position is zero—exactly the same as if we had left the book sitting there: dxtot = dx1 + dx2 = dx + (−dx) = 0. However, the work done on the book was positive in both moves, so the sum (imperfect) diﬀerential of the work is positive, even though the book ended up where it began: dWtot = F 1 · dx1 + F 2 · dx2 = F · dx + (−F ) · (−dx) = 2F · dx. ¯ Second, if we integrate a perfect diﬀerential, we obtain a diﬀerence between ﬁnal and initial states. For example, if we integrate dx from x0 to x1 , we obtain the diﬀerence ∆x = x1 − x0 . When we integrate dW , for example, we simply obtain ¯ the total work in the path W , not a diﬀerence. In order to control, and therefore measure, the internal energy of a system, we need to control both the heat ﬂow and the work done on the system. This manipulation is accomplished primarily through control of the walls of a system. The following deﬁnitions establish ideal conditions that we may approach approximately in real situations. 14 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS An adiabatic wall is one that does not allow heat to ﬂow through it. In real situations this is approximated by heavily insulated walls. On the other hand, a diathermal wall is one that permits heat to ﬂow through it freely. In addition, we often assume that the wall has a suﬃciently small mass such that its thermodynamic eﬀects are negligible. Such a wall might be approximated by one that is thin, and made of metal. When no matter is exchanged with the environment, we say the system is closed. When no mass, heat, or work is exchanged, we say the system is isolated. Example 2.3.1 A gas is placed in an insulated container that contains a frictionless piston and stir paddles attached to a falling weight (Fig.2.1). The piston is attached to a scale, so that we can calculate the pressure of the gas inside by measuring the force on the piston and dividing by its surface area. We can also measure the rate at which the weight falls. With this setup, we perform two experiments. Experiment #1: When we move the piston slowly, we ﬁnd a relationship between the pressure and volume of the form P 3 V 5 = constant. (2.2) Experiment #2: A stirrer inside the container is attached to a falling weight, which spins the stirrer; at constant volume, the pressure changes with time according to the following relation 2 mg  dl dP =− , (2.3) dt 3 V dt where m is the mass of the weight, and l is the height of the weight. Find the internal energy as a function of volume and pressure U (V, P ), relative to its value at some reference volume and pressure U0 := U [V0 , P0 ], assuming that the internal energy of the gas is a function of V and P only. Solution: Our system is the gas in the container, and our state variables
are volume and pressure. Note that the ﬁrst experiment allows us to do work on the system by decreasing its volume, or extract work from the system by increasing its volume. The second experiment can only do work on the system. The conservation of energy Eqn. (2.1) allows us to determine the changes in the internal energy by integrating the work along an appropriate path, since the system is insulated, making dQ = 0, or by calculating the loss in potential ¯ energy of the falling weight. If we consider the state of the system on a plane with volume as the x axis and pressure as the y axis, then we see that each experiment allows us to move in the plane only in a speciﬁc manner. If the system begins at a speciﬁc pressure and volume, then the second experiment allows us to increase the pressure, but 2.3. DEFINITION OF HEAT 15 Figure 2.1: Experimental apparatus described in Example 2.3.1. A P C B V
Figure 2.2: PressureVolume state plane of Example 2.3.1. 16 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS
not change the volume. The ﬁrst experiment allows us to change both the pressure and the volume, but only such that P 3 V 5 stays constant. However, these two experiments are suﬃcient to allow us to move on the state plane between any two points. For example, the points A and B in Figure 2.2 can be connected by two diﬀerent paths. First, we can draw a vertical line through point A indicating the second experiment, and a line of constant P 3 V 5 through point B, indicating the ﬁrst experiment. We can then proceed along the path using the two line segments that connect these two points. Alternatively, we could draw a vertical line through point B and a line of constant P 3 V 5 through point A and connect the points. Once we have a path connecting the points, we integrate the work required to move between them. Consider the ﬁrst path. Since the point C is connected to point A by Experiment #2 and to point B by Experiment #1, its pressure and volume must satisfy the two equations VA 35 PC VC = VC =
35 PB VB . (2.4) From which we may readily solve for PC : PC = PB VB VA
5/3 . (2.5) Similarly, for any point on the line connecting B and C, the pressure can be found as a function of volume P (V ) = PB VB V
5/3 , curve connecting B and C. (2.6) Since Experiment #1 is performed quasistatically, we can assume that the pressure in the container is that measured on the piston. Therefore, Eqn. (2.1) can be written as dU Using Eqn. (2.6) we obtain dU = −PB VB V
5/3 = = dWqs ¯ −P dV. (2.7) dV. (2.8) We can integrate Eqn. (2.8) from C to B to obtain 3 U [VB , PB ] − U [VC , PC ] = − PB VB 2 VB VA
2/3 −1 . (2.9) Note that we use the square brackets [. . .] to indicate where the quantity is being evaluated, as opposed to parentheses (. . .) to indicate a functional dependence. 2.3. DEFINITION OF HEAT
For example, we would write U (V, P ) to indicate that U depends upon V and P , and U [PA ] to indicate that we are evaluating U at PA , but arbitrary V . To integrate U along path CA, we need to ﬁnd the work done during the experiment. Recalling that the potential energy of the weight is mg l and neglecting any kinetic energy of the falling weight, an energy balance on the container plus weight yields: d(U + mg l) = 0. We can therefore write dU = = −mg l˙ dt 3 V dP, 2 (2.10) 17 where we have used Eqn. (2.3) to obtain the second line. Since the volume is constant along CA, we can integrate Eqn. (2.10) from C to A to give U [VA , PA ] − U [VC , PC ] = = 3 VA (PA − PC ) 2 3 VB VA PA − PB 2 VA 5/3 , (2.11) where we have used Eqn. (2.5). Let us call A our reference state: (PA , VA ) → (P0 , V0 ), and let B be any arbitrary state: (PB , VB ) → (P, V ). If we subtract Eqn. (2.11) from Eqn. (2.9), we obtain our desired expression for the internal energy of the system U (V, P ) − U0 = 3 (P V − P0 V0 ) 2 (2.12) Equation (2.12) is an example of an equation of state—an equation that relates several thermodynamic properties with one another. Another example is the wellknown idealgas equation of state (P v = RT ) that relates pressure, volume and temperature with one another. 2 The equation of state derived in Example 2.3.1 is used in the following example to calculate heat ﬂows in a system. Example 2.3.2 The gas from Example 2.3.1 is placed in a container with a diathermal wall and a frictionless piston. Find the amount of heat and work necessary for two diﬀerent steps: The ﬁrst step decreases the pressure from P1 to P2 at constant volume V1 . The second process increases the volume from V1 to V2 at constant pressure P2 . Assume each step is quasistatic. Solution: In the ﬁrst step, the volume is held constant, so dV = 0, and the amount of work done on the system is zero: W I = 0. Therefore, the deﬁnition for heat ﬂux Eqn. (2.1) becomes
dQ = dU, ¯ (2.13) 18 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS
which we can integrate along any path from the initial (P1 , V1 ) to the ﬁnal (P2 , V1 ) conditions to give QI
(P2 ,V1 ) =
(P1 ,V1 ) dU (P, V ) U [P2 , V1 ] − U [P1 , V1 ] 3 V1 (P2 − P1 ) . 2 = = (2.14) The third line follows from Eqn. (2.12). Since P1 > P2 , then QI < 0. To perform this step, we must extract heat from the system. In the second step, the volume changes, so work is done. If we integrate Eqn. (2.1) over the second step (P2 , V1 ) → (P2 , V2 ), we obtain QII
V2 V2 =
V1 P2 dV +
V1 V2 dU [V, P2 ] = P2
V1 dV + U [V2 , P2 ] − U [V1 , P2 ] 3 = P2 (V2 − V1 ) + P2 (V2 − V1 ) 2 5 = P2 (V2 − V1 ) . 2 (2.15) Note that the ﬁrst term on the righthand side of the third line of Eqn. (2.15) represents the work done in the second step: W II = P2 (V2 − V1 ). 2 In Example 2.6.2 we give the complete characterization of this ﬂuid, which is called a simple, ideal gas. At low densities, all gases behave ideally; later we will see that simple gases are monatomic. At higher densities, more complicated equations of state are necessary to accurately describe the behavior of liquids and gases. Several of these are given in the appendix, and we will call on them throughout the book. 2.4 Equilibrium States The density of pure water at 1 atm and 25◦ C is always 1 g/cm3 , no matter where the water comes from. This observation holds if I start with ice from Lake Mendota in the winter, or with steam from a teapot in Shanghai. We call this stable state of water an equilibrium state. Thermodynamics deals only with these equilibrium states, and not with the dynamics of the system between such states. (However, it does tell us which equilibrium states are available to a system that is not at equilibrium, and which states are not.) Thermodynamics just does not say how long, or by what path the system will attain equilibrium. These observations lead us to the second, rather sensible postulate. 2.5. ENTROPY, THE SECOND LAW, AND THE FUNDAMENTAL RELATION 19 Postulate II: There exist equilibrium states of a macroscopic system that are characterized completely by the internal energy U , the volume V , and the mole numbers of the m species N1 , N2 , . . . , Nm in the system. We have not yet said what variables we are using to describe the internal energy U , but if we know U , the volume and the mole numbers, then the equilibrium state is ﬁxed. When we do work on a system, we often wish to consider processes that occur slowly. In such a case, we assume that the system is nearly always in an equilibrium state. We call such a slow change a quasistatic process. For example, we might change the pressure in a container by changing the force on a piston slowly, perhaps by removing grains of sand that are resting on the piston one at a time. When we changed the pressure slowly in Example 2.3.1, we were changing it quasistatically. 2.5 Entropy, the Second Law, and the Fundamental Relation Since we do not have information on the positions and velocities of all of the atoms, we need to determine what set of variables is necessary to describe the internal energy of our system. A few observations can guide us: • When touching a hot stove, we notice that it transfers some internal energy to us (don’t try this experiment at home without the supervision of an adult). Intuitively we know that this mechanism of energy storage has something to do with temperature and heat ﬂow. • In order to make an air balloon smaller, we have to do work on the balloon by squeezing it. This mechanism has something to do with applied force and volume. • To place more air molecules in the balloon, we do work on the balloon by blowing, or pumping. This mechanism has something to do with the number of moles of gas in the balloon. Our observations have suggested a number of possible independent variables for the internal energy: one involving heat or temperature, a second involving volume or pressure, and a third involving mole number. We do not wish to use heat, however, since it is handled using an imperfect diﬀerential. When a system is changed quasistatically, then the work described in the second bullet above can be written as 20 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS dW = −P dV . We need to introduce a second quantity called entropy that will ¯ allow us to use only perfect diﬀerentials for changes to U from heat ﬂow. We save time and eﬀort by jumping directly to the correct answer. The justiﬁcation for these choices will have to come from the ability of the theory to explain experimentally observed phenomena. At this point it is instructive to digress momentarily and elaborate on some of the underlying mathematical ideas behind our next postulate. Generally speaking, when facing a complex problem involving multiple variables, a common strategy is to deﬁne an objective function that one tries to maximize or minimize subject to several constraints until a solution is found. As an example, a city might want to alleviate congestion by controlling the size of the streets, the location of traﬃc lights, the duration and sequence of green lights, etc. To do so, city planners might construct an objective function that they might chose to maximize or minimize. A number of choices are possible; a reasonable possibility could be to create a function that describes the average idle time per driver. Mathematically speaking, idle time would be a function of the variables listed above. For a given number of cars or “ﬂow rate” and a given street layout, one would then attempt to minimize that function by controlling the arrangement of traﬃc lights. Thermodynamics does the same thing; a function of several “natural variables”, such as the volume and the size of the system, is created and maximized or minimized (depending on the nature of that function). That function must meet several criteria, which is really what the next postulate is about. 2.5. ENTROPY, THE SECOND LAW, AND THE FUNDAMENTAL RELATION 21 Postulate III: Complete thermodynamic information is contained at equilibrium in the internal energy as a function of the quantity called entropy S , the volume V , and the mole numbers of its r constituents N1 , N2 , . . . , Nr , which are all extensive quantities . The functional form of U (S, V, N1 , N2 , . . . , Nr ) satisﬁes the following properties: • It is additive over its constituent elements (it is a ﬁrstorder, homogeneous function of its arguments, or extensive): U (λS, λV, λN1 , λN2 , . . . , λNr ) = λU (S, V, N1 , N2 , . . . , Nm ) • It is continuous and diﬀerentiable: ∂U ∂S
V ,N1 ,N2 ,...,Nr ∼ is well deﬁned everywhere • It is a monotonically increasing function of S : ∂U ∂S
V ,N1 ,N2 ,...,Nr ≥0 This postulate is true only for large systems—the socalled thermodynamic limit. Small systems, like proteins, may not satisfy the extensivity condition. We shall see later that entropy is intimately related to heat transfer and temperature, and that pressure is associated with volume changes. The properties of Postulate III can be used to make several key observations: • If we pick λ = 1/N , then we obtain for pure component systems N u(s, v ) = U (S, V, N ), (2.16) where using a small letter indicates a speciﬁc or molar property, i.e., u := U/N is the internal energy per unit mole. Thus, for pure component systems, all thermodynamic information is contained in the speciﬁc properties, and we need not know N . • The second property allows us to use the usual calculus manipulations. • The second and third properties allow us to invert the relation U = U (S , V , N1 , . . ., Nr ) to ﬁnd a unique function for S : S = S (U, V, N1 , . . . , Nr ), (2.17) 22 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS which enjoys properties similar to those enjoyed by U . Namely, we can write that ∂S > 0, and is well deﬁned. (2.18) ∂U V ,N1 ,...,Nr Property (2.18) follows from Eqn. (A.18). • Since we can invert U to ﬁnd an expression for S , we can also show that S is a homogeneous, ﬁrstorder function of its arguments S (λU, λV, λN1 , λN2 , . . . , λNr ) = λS (U, V, N1 , N2 , . . . , Nr ). (2.19) Therefore, we may deal either with S , or with U as the dependent variable. The appropriate choice is only dictated by convenience; there is no more information in U (S, V, N1 , . . . , Nr ) than there is in S (U, V, N1 , . . . , Nr ). The ﬁrst form is called the fundamental energy relation, and the second is called the fundamental entropy relation. These are sometimes called constitutive relations, since diﬀerent fundamental relations often exist to describe a single substance. For example, to describe oxygen we might sometimes use an ideal gas (at low densities), a virial relation (at moderate densities), or a PengRobinson relation (at high densities), depending on the necessary accuracy or the simplicity of the calculation. Most importantly, as we will soon see, if we know either the fundamental entropy or fundamental energy relation for a system, then we have complete thermodynamic information about that system . From either S (U, V, N1 , . . . , Nr ) or U (S, V, N1 , . . . , Nr ) we can ﬁnd the system’s P V T relation, its constantvolume heat capacity, its phase behavior, everything. This fact has farreaching ramiﬁcations. For example, in §3.6 we show that if we know the material properties of a substance (heat capacity, coeﬃcient of thermal expansion and isothermal compressibility) for a range of temperatures and pressure, we can calculate all possible thermodynamic quantities at any temperature and pressure. Or, we can also show that the change in temperature with respect to volume at constant internal energy must equal the change in pressure over temperature with respect to internal energy at constant volume for a substance. Although these relations are rigorously derived from thermodynamics, they are far from being intuitively obvious. Let’s return to the idea of ‘equilibrium states’ for a moment, and ask, How does the system ﬁnd its equilibrium state? Experimental observation suggests an essential property of all systems: each system ﬁnds one equilibrium state for given values of internal energy, volume and mole number. This observation means that if you have two nonequilibrium systems that have diﬀerent phases and diﬀerent conditions but the same energy, volume and mole numbers, these two systems will 2.5. ENTROPY, THE SECOND LAW, AND THE FUNDAMENTAL RELATION 23 reach the same equilibrium point. In other words, many nonequilibrium systems will all converge on the same ﬁnal equilibrium condition. Clearly, the equilibrium point must be something special. Similar to other natural systems, we ﬁnd that the following postulate explains our observations: Postulate IV: The unconstrained variables of an isolated system arrange themselves such as to maximize entropy within the constrained equilibrium states. Consider two examples of systems with unconstrained variables. • A ﬂask contains a solution of species A and B , and the covering of the ﬂask is a membrane that is permeable to species A, but not permeable to B (Figure 2.3). The ﬂask is placed in a large vat that also contains some mixture of A and B . The ﬂask and membrane are rigid, but allow heat to pass through. For this system, V and NB are constrained for each of the two subsystems (the ﬂask and the vat). Therefore, the amount of A and the internal energy in the ﬂask will change until S for the composite system is maximized. Figure 2.3: A ﬂask whose opening is covered by a semipermeable membrane sits in a large vat. The ﬂask and the vat contain diﬀerent concentrations of species A and B . The membrane allows A to pass through, but not B . Maybe A is water, and B is a large protein, for example. • A rigid cylinder is divided into two compartments by a movable partition whose position is momentarily ﬁxed by a constraining pin. See Figure 2.4. The cylinder is insulated, and is impermeable to either mass or energy ﬂux. We remove the pin and allow the movable wall to reposition itself. The movable wall will position itself such as to make the entropy of the total system maximum, given that the total volume, internal energy and mole numbers of 24 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS the system are ﬁxed. In this example, the system is both compartments of the cylinder, and each compartment would be a subsystem. Note that the change in the total entropy of the system would be either positive or zero, but never negative. Figure 2.4: A rigid, hollow cylinder is insulated from its environment. The cylinder is divided by a
partition, and each side contains some gas. When the pin is removed the position of the inner partition is no longer constrained, and it moves. That last statement is actually general: the total entropy change of an isolated system (such as the universe) is always nonnegative. Entropy is always increasing, although energy is always constant in the universe. Therefore, there is always plenty of energy around, but the entropy might not be low enough to get the energy to do what you want. There is considerable ambiguity in identifying precisely what the second law of thermodynamics is in the literature—whether it is embodied in some or all of the properties in Postulate III, or just in Postulate IV, for example. Rudolf Clausius, who apparently ﬁrst coined the phrase entropie , meaning “transformation” in Greek, wrote the second law as Die Entropie der Welt strebt einem Maximum zu.3 Following his lead, we prefer to call Postulate IV the second law of thermodynamics. For completeness, we now state the ﬁnal postulate of thermodynamics [58]. Postulate V (The Nernst Postulate): The entropy is zero when and only when the substance is a crystal with ∂ U V ,N1 ,...,Nm = 0. ∂S In the following section we will see that the ﬁfth postulate states that the entropy is zero when the absolute temperature is zero. Although important for fundamental questions, or in statistical mechanics, we will not much use the ﬁfth postulate in this book. In fact, despite the postulate’s importance, we will often use fundamental relations that violate the Nernst Law. Why? Because a fundamental relation is usually only valid over some range of thermodynamic conditions. So, if we use a fundamental relation for, say nitrogen, in the regions where it is liquid, gas, or supercritical, but never where it is crystalline, it need not satisfy the Nernst postulate. An aside about entropy and statistical mechanics.
3 The entropy of the universe strives towards a maximum [17]. 2.5. ENTROPY, THE SECOND LAW, AND THE FUNDAMENTAL RELATION 25 Entropy always seems strange at ﬁrst sight—it usually involves symbols such as ‘<’ or ‘>’ rather than ‘=’, and it is not conserved. Also, the term statistical mechanics sounds rather intimidating. However, the ideas in statistical mechanics are rather straightforward, if their implementation is often diﬃcult. Although a quantitative understanding of statistical mechanics is not necessary to use most of this book, it is sometimes still enlightening to be aware of its ideas. A fundamental idea of statistical mechanics—so important that it is on the gravestone of Ludwig von Boltzmann, is the deﬁnition of entropy. That is correct—in statistical mechanics, entropy is not a fundamental quantity, but is actually deﬁned . There is no deﬁnition for energy, so entropy is actually less abstract than energy! Boltzmann’s deﬁnition for entropy is stunningly simple4 S (U, V, N ) = kB log Ω(U, V, N ). (2.20) The ﬁrst term is called, appropriately enough, the Boltzmann constant, and is equal to the ideal gas constant divided by Avogadro’s number. The next term Ω is just a number. It is the number of ways that the molecules in a system can arrange themselves still keeping the energy ﬁxed at U , the volume ﬁxed at V , and the mole numbers at N . Imagine a solid crystal of atoms regularly arrayed on a lattice. The positions of the atoms are moreorless ﬁxed. If we swap the atoms about, we still have the same microscopic state. So, the only possible microscopic arrangements arise from how the energy is distributed. Each atom can vibrate in its position. Or, several atoms can vibrate together. The greater the vibration, the more energy an atom has. Maybe one atom is vibrating with all the energy of the crystal, or maybe each atom vibrates the same as all the others. If we could count up all the ways that the energy could distribute itself, we could ﬁnd the entropy. Statistical mechanics deals with calculating such large numbers. Even without tackling these sorts of problems, though, we can still think qualitatively about entropy. An ideal gas consists of molecules ﬂying about in a container. The molecules are very dilute, so they rarely see each other. They are independent. The energy is either stored in the kinetic energy of the molecules, or in the vibrations in the interatomic bonds. If the molecules are monatomic, they have only kinetic energy. The possible arrangements of the molecules, then, is not just how energy is distributed, but
4 Here and throughout the text ‘log’ refers to the natural logarithm, and ‘log 10 ’, is log in base 10. 26 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS where the molecules are in the container. If we keep the energy constant, how can we lower the entropy? Well, if we decrease the volume, the molecule has fewer locations allowed. Hence, Ω goes down. Therefore, the entropy of an ideal gas decreases with decreasing volume. A third example, rubber is made up of crosslinked polymer chains. Between the chemical crosslinks, the chain can take many diﬀerent conformations. If we stretch out the rubber, the crosslinks become separated in space. This stretching deformation decreases the number of ways that the chains can arrange themselves. Hence, stretching rubber decreases its entropy. For a rubber band, the length is more important than the volume, so we substitute L for V in our list of independent variables. Considering Postulate IV, can you explain why the stretched rubber snaps back when its length is no longer constrained, or why a gas under pressure expands? The reader might ﬁnd it useful to keep these qualitative ideas in mind when trying to understand the thermodynamic behavior of material. Example 2.5.1 Van der Waals has postulated the following fundamental entropy relation for a system. Does it satisfy the postulates? S = N s0 + NR log U /N + aN/V c V /N − b , u0 + a/v0 v0 − b simple van der Waals ﬂuid. (2.21) Solution: Internal energy U does play a role. If we also stipulate that U is
conserved, then the ﬁrst postulate is satisﬁed. Postulate II is satisﬁed if we take a, b, R, u0 , v0 and s0 to be constants. Clearly, there is only one mole number, or a single species. In order to check Postulate III, we could invert Eqn. (2.21) to ﬁnd an explicit expression for U . Alternatively, we can check the equivalent criteria for S . We choose the latter. We ﬁrst check that S is a homogeneous, ﬁrstorder function of its arguments: S (λU, λV, λN ) λU aλN λV + −b λN λV λN c −λNR log [(u0 + a/v0 ) (v0 − b)] = λ { N s0 + c U /N + aN/V V /N − b NR log u0 + a/v0 v0 − b = λS (U, V, N ), = λN s0 + λNR log
c 2.6. DEFINITIONS OF T , P , AND µ
which proves that the ﬁrst property of Postulate III is satisﬁed. We check the second and third properties of Postulate III by diﬀerentiating Eqn. (2.21) with respect to U ∂S cR = (2.22) ∂U V ,N U/N + aN/V The second and third properties of Postulate III are satisﬁed only if cR > 0 and a > 0. The proposed fundamental relation says nothing about the fourth postulate, which is presumably satisﬁed. The ﬁfth postulate applied to this example requires that as U/N +aN/V → 0, then S = 0. However, Eqn. (2.21) shows that the entropy goes to negative inﬁnity in the limit. Hence, the proposed relation cannot be valid for very small values of S . In particular, for suﬃciently small values of U and large values of V , the entropy is predicted to become negative, which is unphysical. Therefore, this fundamental relation is valid in a limited range of thermodynamic states. However, as long as we stay in the region where entropy is physical, we may use this equation. In other words, at moderate values for the entropy, our system might obey this model. However, as the entropy is lowered, our system can no longer obey this model, since it violates the Nernst postulate. 2 27 2.6 Deﬁnitions of Temperature, Pressure and Chemical Potential The postulates have not yet deﬁned the important thermodynamic quantities of temperature or pressure, which might seem surprising. However, these quantities follow as deﬁnitions based on the fundamental quantities introduced in the postulates. It is important that these deﬁnitions ﬁt our observations about temperature and pressure. In the following two sections we show that the deﬁnitions given in this section lead to two important intuitive predictions: if two systems with diﬀerent temperatures are placed in thermal contact, thermodynamic equilibrium will be attained when the hightemperature body gives enough heat to the lowtemperature body until they are at the same temperature; two subsystems at diﬀerent pressures in mechanical contact reach equilibrium when the highpressure subsystem expands at the expense of the lowpressure body until both subsystems are at equal pressures. We come to the deﬁnitions of T and P by considering changes in U . For example, we have already seen in Chapter 1 that work can be done on a system to change its internal energy by changing its volume. Similarly, we can change U by adding heat or moles. Starting with our fundamental energy relation U (S, V, N1 , . . . , Nr ), and using the deﬁnition for the diﬀerential Eqn. (A.3), we can make the mathematical 28 observation dU = CHAPTER 2. THE POSTULATES OF THERMODYNAMICS ∂U ∂S dS +
V ,{Ni } ∂U ∂V r dV +
S,{Ni } i ∂U ∂Ni dNi , S,V,{Nj =i } (2.23) where, for convenience, we use {Ni } to mean N1 , . . ., Nr , and {Nj =i } means N1 , . . ., Ni−1 , Ni+1 , . . ., Nr . We already know that the (reversible) work done on a system is −P dV . We also know from experience that volume changes are driven by pressure diﬀerences between subsystems. Similarly, from experience we know that heat transfers are driven by temperature diﬀerences. And ﬁnally, the third term on the right side of Eqn. (2.23) suggests that there is some other property that should drive changes in mole numbers, or mass ﬂuxes. It turns out that these observations can be predicted by the postulates when we make the following deﬁnitions for pressure P , temperature T and chemical potential of species i µi . P T := − := ∂U ∂V ∂U ∂S ∂U ∂Ni (2.24)
S,N1 ,...,Nr (2.25)
V ,N1 ,...,Nr µi := (2.26)
S,V,Nj =i Note that the third property of Postulate III, and the deﬁnition for temperature Eqn. (2.25) require that T ≥ 0. Also, we now see that Postulate V requires that the entropy be zero when the temperature is absolute zero. The following two sections show how these deﬁnitions ﬁt our intuitive understanding of these quantities. Putting Eqs. (2.24) through (2.26) into the original diﬀerential equation (2.23), leads to the oftused diﬀerential expression for U dU = T dS − P dV + µi dNi
i (2.27) This equation should be committed to memory, since it is straightforward to write down the deﬁnitions of pressure, temperature and chemical potential from it (Try it!). The ﬁrst term on the right hand side of Eqn. (2.27) is associated with heat ﬂuxes or irreversible parts of work, the second term is the reversible work on the system, 2.6. DEFINITIONS OF T , P , AND µ 29 and the third term is associated with ‘chemical’ work. Note that Eqn. (2.27), unlike the deﬁnition for heat ﬂow Eqn. (2.1), contains only perfect diﬀerentials. If we are given a fundamental energy relation, then we can ﬁnd the temperature, pressure and chemical potentials as functions of entropy, volume and mole numbers. Such expressions are called equations of state of a system. Unlike the fundamental entropy or energy relations, these equations do not contain complete thermodynamic information, but rather are derivatives (this point is further illustrated in §3.1). A relation between U and T is sometimes called a thermal equation of state, and a relation between P and V is sometimes called a mechanical equation of state, although these distinctions are not clearcut. All three quantities (T, P, µi ) are intensive properties, meaning that they are independent of system size. If we, for example, double each of the extensive properties (S, V, N1 , . . . , Nr ), then T, P and µi remain unchanged. Alternative but equivalent deﬁnitions for temperature, pressure and chemical potential can be made for the fundamental entropy relation. Beginning with the fundamental entropy relation S (U, V, N1 , . . . , Nr ) we can write dS = ∂S ∂U dU +
V ,{Ni } ∂S ∂V r dV +
U,{Ni } i ∂S ∂Ni dNi .
U,V,{Nj =i } (2.28) If we solve the diﬀerential of U Eqn. (2.27) for dS we ﬁnd dS = 1 P dU + dV − T T
r i µi dNi . T (2.29) When we compare Eqn. (2.29) to Eqn. (2.28) we ﬁnd the relations in the entropy formulation that are equivalent to Eqs. (2.24) through (2.26) P T 1 T µi T = = =− ∂S ∂V ∂S ∂U ∂S ∂Ni (2.30)
U,{Ni } (2.31)
V ,{Ni } .
U,V,{Nj =i } (2.32) These last relations can also be found by simple algebra from the diﬀerential for U , Eqn. (2.27). Given a fundamental energy relation U (S, V, {Ni }) one can use the deﬁnitions Eqs. (2.24) through (2.26) to ﬁnd equations of state for pressure (the mechanical equation of state), temperature (the thermal equation of state), or chemical potential (the chemical equation of state) as functions of S, V and {Ni }. 30 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS Alternatively, given the fundamental entropy relation S (U, V, {Ni }) we can use the equivalent deﬁnitions Eqs. (2.30) through (2.32) to ﬁnd equations of state for P, T and µi as functions of U, V and {Ni }. Example 2.6.1 Find the thermal and mechanical equations of state for the fundamental entropy relation of Example 2.5.1. Also ﬁnd the chemical potential as a function of temperature and volume. Solution: Since we are given an entropy relation, it is more convenient to use the alternative deﬁnitions Eqs. (2.30) through (2.32) than those based on internal energy. From Eqn. (2.22) of Example 2.5.1, and Eqn. (2.31) we have already found temperature
1 cR = . T U/N + aN/V If we solve for U explicitly, we obtain U = cNRT − aN 2 , V (2.34) (2.33) which is a thermal equation of state. If we use Eqn. (2.30) we ﬁnd P = = = T ∂S ∂V U,N acN 2 R T RT − V /N − b V 2 U/N + aN/V RT aN 2 − 2 , van der Waals equation of state. V /N − b V (2.35) In going from the ﬁrst to the second line, we have used the proposed fundamental entropy relation Eqn. (2.21). In going to the third line, we used the thermal equation of state Eqn. (2.34). The resulting mechanical equation of state Eqn. (2.35) is called the van der Waals equation of state for a ﬂuid. To ﬁnd the chemical potential, we use the alternative deﬁnition based on entropy, Eqn. (2.32) µ= = −T ∂S ∂N U,V −RT log N RT V /N − b = −RT log U /N + aN/V u0 + a/v0 V − 2 − T s0 N c T v−b T0 v0 − b c V /N − b v0 − b − cN RT U/N + aN/V − U a − N2 V + + cRT + RT v − T s0 , v−b (2.36) 2.6. DEFINITIONS OF T , P , AND µ
where we have used the thermal equation of state to eliminate internal energy in the second line. These results are useful in the next example. 2 31 Some important notes about these results are • Simple ideal gas behavior can be recovered from these results by setting a = b = 0. • For an ideal gas, the internal energy depends only on temperature, and not on volume. However, for the van der Waals ﬂuid, the internal energy decreases with decreasing volume at ﬁxed temperature (and mole number). Hence, the parameter a represents attractive forces between molecules, which are neglected in the ideal gas. • The speciﬁc volume v is required to be greater than b. Hence, b represents the repulsive forces between molecules that keep the ﬂuid from becoming inﬁnitely dense. • We cannot simultaneously use the van der Waals mechanical equation of state and assume that internal energy is independent of volume. The mechanical equation of state and the thermal equation of state come from a single fundamental relation. • We call these equations of state because each one is derived from a fundamental relation, but the fundamental relation cannot be derived from a single equation of state. We can say that the fundamental relation has more information than does an equation of state.5 Example 2.6.2 If a singlecomponent, simple ideal gas is expanded isentropically (at constant entropy) until the pressure is halved, what happens to the temperature? Solution: An ideal gas is one that obeys the mechanical equation of state
P V = NRT, ideal gas equation of state, (2.37) and the internal energy can be written as a function of temperature only U = U (T ). If the internal energy is a linear function of temperature, then we call it a simple ideal gas U = cNRT, simple ideal gas, thermal equation of state, (2.38) 5 Later we will see (in Chapter 4) that a fundamental relation for a pure species can be derived from two equations of state. 32 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS
where c is a constant. We note that the van der Waals ﬂuid of Examples 2.5.1 and 2.6.1 reduces to a simple ideal gas when we set a = b = 0. Hence, the fundamental entropy relation for a simple ideal gas can be found from Eqn. (2.21) to be S = N s0 + NR log U N u0
c V , N v0 simple ideal gas. (2.39) In this problem, we are changing T and P while keeping S constant. Therefore, it is useful to ﬁnd S = S (T, P ). We can obtain this relation by eliminating U and V from Eqn. (2.39) using Eqs. (2.37) and (2.38) S = N s0 + NR log T T0
c+1 P0 , P simple ideal gas, (2.40) where T0 := u0 /cR and P0 := RT0 /v0 . From this expression, we see that, in order to keep the process isentropic, we must keep T c+1 /P constant. Hence, if the pressure is halved, then the ﬁnal temperature Tf is Tf = Ti , 1/(c+1) 2 (2.41) where Ti is the initial temperature. 2 Example 2.6.3 Find the work necessary to complete this process, if it is done quasistatically. How much heat is transferred? Solution: If it is done quasistatically, then the work is
dWqs = −P dV. ¯ (2.42) In order to integrate this equation, we need to determine how the pressure is changing with the volume. Since the gas is ideal, we can write P= NRT . V However, the temperature is also changing with volume. From Example 2.6.2, we determined that an isentropic change in pressure requires that T c+1 /P be held constant. Hence, 1 P c+1 T = Ti . Pi Inserting this equation into the ideal gas law yields P= NRTi V
c+1 c 1 Pi
1/c . 2.6. DEFINITIONS OF T , P , AND µ
If we insert this expression into our diﬀerential equation for quasistatic work, Eqn. (2.42), we obtain dWqs = − ¯ NRTi V
c+1 c 33 1 Pi
1/c dV. We may now integrate this equation from the initial volume Vi to the ﬁnal volume; we obtain after some simpliﬁcation Wqs = cNRTi But, NRTi Pf 1 Vi = = c. Vf Pi NRTf 2 c+1 Hence, Wqs = cNRTi 1 2 c+1
1 Vi Vf 1/c −1 . (2.43) −1 . (2.44) Because Wqs is negative, we ﬁnd that the gas does work on its surroundings during the process. The process might be accomplished by gradually decreasing the pressure in the environment around the gas allowing it to expand very slowly. Or, we might remove the weight on a piston very gradually. We can ﬁnd the energy transferred during the process from the conservation of energy, or from the deﬁnition of Q, Eqn. (2.1) Qqs = Uf − Ui − Wqs = cNR (Tf − Ti ) − cNRTi = 0. 2 1
1 c+1 −1 (2.45) Therefore, an adiabatic, quasistatic process for an ideal gas is isentropic. Had the process been done rapidly, then the required work would have been higher. However, the change in internal energy would be unchanged, so it would have been necessary to remove heat from the system in order to keep it isentropic. 2 An isentropic process might seem strange at ﬁrst sight; how can one control entropy? However, we can prove that the ﬁnal result of Example 2.6.3 is actually very general: an adiabatic, isomolar, quasistatic process is isentropic . For a constantmolar process, the diﬀerential for internal energy, Eqn. (2.27), can be written dU = T dS − P dV, closed system. 34 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS If the process is done quasistatically, then energy conservation, Eqn. (2.1) allows us to write this equation dQqs + dWqs = T dS − P dV, ¯ ¯ Since dWqs = −P dV , we ﬁnd that ¯ dQqs = T dS, ¯ isomolar, (2.46) closed system. which proves that an adiabatic, quasistatic, isomolar process is isentropic for any material. In fact, some equivalent approaches to thermodynamics treat temperature as a fundamental quantity and deﬁne entropy by this relation. We prefer Callen’s way, because it allows a much easier connection later to statistical mechanics. Example 2.6.4 What is the chemical potential of a simple ideal gas as a function of temperature and pressure? Solution: We can ﬁnd the chemical potential from the fundamental relation,
Eqn. (2.39), using the alternative deﬁnition for µ µ T = ∂S ∂N U,V = −s0 − R log = −s0 − R log V U − (c + 1)R u0 N v0 N c T v + (c + 1)R. T0 v0 c (2.47) To complete the solution, we multiply each side by T . However, we still need to eliminate v in favor of T and P , which we can do using the ideal gas mechanical equation of state v = RT /P µ= = −s0 T − RT log µ◦ (T ) + RT log P. T T0
c RT + (c + 1)RT P v0 (2.48) We have split the result in this way to emphasize the dependence of µ on pressure. Note that the chemical potential of an ideal gas goes to negative inﬁnity as the pressure goes to zero. 2 Example 2.6.5 Find the change in temperature for the adiabatic compression of a simple van der Waals gas. Use this result to ﬁnd the work required for such a compression. 2.6. DEFINITIONS OF T , P , AND µ 35 Solution: If we assume that the process is quasistatic, then the work is
related to the equilibrium pressure by dWqs = −P dV. (2.49) If we use the van der Waals equation of state, Eqn. (2.35), to replace pressure, we see that we cannot integrate this equation, because the temperature is not constant during the integration. However, we can solve this problem using the thermal equation of state for a simple van der Waals ﬂuid. If we take the diﬀerential of each side of Eqn. (4.58), we obtain dU = cNRdT + aN 2 dV. V2 (2.50) When we insert Eqn. (2.50) into Eqn. (2.49) (using dU = dW for an adiabatic ¯ process), and use the mechanical equation of state, Eqn. (2.35), we obtain cNRdT = − RT dV, V /N − b simple van der Waals, adiabatic. (2.51) Note the cancelation that occurred to obtain this result. If we divide each side by NRT, we can then integrate each side from the initial thermodynamic state (T0 , V0 ) dT T0 T T c log T0 c T
T = = = dV V0 V − bN V − bN − log V0 − bN − T0 V0 − bN V − bN V 1/c , van der Waals, adiabatic. (2.52) The second line is obtained by performing the integrations, and the third line from taking the exponential of each side. Now that we have the temperature as a function of volume for the adiabatic compression of a simple van der Waals ﬂuid, we can ﬁnd the work necessary, just as we did in Example 2.6.3. The details are left as an exercise. 2 Example 2.6.6 We use a container of ideal gas as a crude refrigerator in the following way. A simple ideal gas held in a container with a piston (to manipulate volume), and a diathermal wall can be used as a refrigerator. We also have insulation available to make the diathermal wall adiabatic at will. The idea is straightforward: when the gas is compressed isothermally, it expels heat; when it is allowed to expand, it takes in heat from its surroundings. By expanding the gas in one environment (the refrigerator), and compressing it in another (outside), heat can be transferred from 36 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS Figure 2.5: Thermodynamic path for the idealgas refrigeration cycle studied in Example 2.6.6 a cold space to a warm one. The gas is cycled through compression and expansion and is called the refrigerant. The work to compress the gas is the eﬀective cost of refrigerating. Find the amount of heat that can be removed from an environment at 100◦ C and expelled to an environment at 50◦ C per mole of gas per cycle. Solution: The path used in the refrigeration cycle is sketch in Figure 2.5.
Step is an adiabatic compression, step b is isobaric cooling, and step c is isothermal expansion. We assume all steps are performed reversibly, so our calculation will be the maximum possible cooling. We also assume that the gas is a simple one, with c = 3/2. To help us make the calculation, it is useful to make two tables, one for the thermodynamic state at each point on the diagram, 13, and another table for the work and energies. Point 1 2 3 T [◦ C ] 50 100 50 P [atm] 1 V [liter] Step a b c ∆U [J] Q [J] 0 W [J] 0 Note that we have ﬁlled in the states already known. Similarly, we have created on the right side a table for the changes in each of the steps. Note that we have assumed an adiabatic step a. The change in step c for the internal energy is zero because the internal energy for an ideal gas depends only on temperature. We now proceed to ﬁll in the table entries one by one. The reader might 2.6. DEFINITIONS OF T , P , AND µ
wish to try and complete these tables on his own ﬁrst (probably using a pencil). Because we know the temperature and pressure at point 1, we can use our P vT equation of state to ﬁnd the volume. For an ideal gas, the EOS is V1 = N RT1 P1 (1 mol)(82.06 cm3 · atm/(mol · K))(323.15 K) = 1 atm = 26.5 liter. 37 (2.53) Note that we assumed one mole of gas, and a value for the ideal gas constant available in Table D.5. From here there are a number of ways to proceed. For the simple ideal gas, a straightforward way is to ﬁnd the change in internal energy in step a ∆Ua = cN R(T2 − T1 ) 3 (1 mol)(8.314 J/(mol · K))(50 K) = 2 = 623.6 J. = cN RT2 − cN RT1 (2.54) By conservation of energy, or the deﬁnition of heat, Eqn. (2.1), we know that Qa = ∆Ua = 623.6J. Since step b has the same magnitude in temperature change, but opposite sign, we also know that ∆Ub = −623.6J. However, we still do not know the ﬁnal thermodynamic state at point 2. This state can be found either by noting that a reversible adiabatic expansion is isentropic, as we proved earlier, or by using the following derivation. dU cN RdT cdT T T2 c log T1 V2 = = = = = dQ + dW ¯ ¯ −P dV dV − V − log v1 T1 T2 V2 V1
c (2.55) To obtain the second line, we used the fact that the step is adiabatic ( dQ = 0), ¯ and that for a simple ideal gas U = cN RT . For the second line we used the ideal gas equation of state, and integrated from state 1 to state 2 in the third line. Taking the exponential of the fourth line gives us the needed expression. Therefore, we ﬁnd V2 = = (26.5 liter) 21.4 liter. 223.15 K 373.15 K
3/2 (2.56) 38 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS
Then, using the ideal gas equation of state, we can ﬁnd the pressure at point 2 to be 1.43 atm, which is also the pressure at point 3, from our isobaric assumption in step b. Now that we know temperature and pressure at point 3, we can ﬁnd the volume, V3 = N RT3 /P3 = 18.5 liter. It is straightforward to ﬁnd the work in step b, since it is isobaric, and we assume reversibility dW ¯ Wb = = = = −P dV P2 (V2 − V3 ) (1.43 atm)(21.4 − 18.5 liter) 420.2J. 8.314 J/(mol · K) 0.08206 liter · atm/(mol · K (2.57) By energy conservation, Qb = ∆Ub − Wb = −1044J. Finally, to ﬁnd the work in step c, we assume reversible work only, and use our P vT equation of state dW ¯ = −P dV = −N RT Wc dV V = N RT log(V3 /V1 ) = (1mol)(8.314J/(mol · K))(323.15K) log( 18.5liter ) 26.5liter (2.58) = −966 J. By energy conservation we know that Qc = −Wc = 966 J. Hence, we have now completed both tables, which look like Point 1 2 3 T [◦ C ] 50 100 50 P [atm] 1 1.43 1.43 V [liter] 26.5 21.4 18.5 Step a b c ∆U [J] 623.6 623.6 0 Q [J] 0 1044 966 W [J] 623.6 420.2 966 One can deﬁne the coeﬃcient of performance as the ratio of the heat extracted divided by the work needed (assuming no work is recovered during the expansion). Heat is extracted from the environment when Q is positive, which is in step b. It is necessary to perform work on the gas during steps a and b. Hence, our coeﬃcient of performance is ε = WaQcWb = 0.925. Of course, + this assumes complete reversibility. In reality, there are several reasons why this would not be attained, such as friction in the piston. We also know from derivations in Appendix C, that any heat ﬂux, such as that necessarily through the diathermal wall, will lead to entropy generation, and hence irreversibility. 2 2.7. TEMPERATURE DIFFERENCES AND HEAT FLOW 39 U (1) U (2) (1) (2) V V N (1) N (2) Figure 2.6: We consider two initially isolated systems with diﬀerent energy, volume and mole number
(U, V, N ). The partition allows heat to ﬂow between the two systems, but is rigid and impermeable. 2.7 Temperature Diﬀerences and Heat Flow It may seem odd at ﬁrst to use something abstract like entropy as a fundamental quantity, and then deﬁne an everyday quantity like temperature based on entropy. Therefore, it is important to show that the deﬁnition for temperature and the postulates indeed lead to predictions about temperature that ﬁt our experience. From experience, we know that • Temperature is intensive. • Two systems in thermal contact reach the same temperature at equilibrium. • If two systems have diﬀerent temperatures and are placed in thermal contact, then heat ﬂows from the object of higher temperature to the colder object. • If we raise the temperature of something either at constant pressure or at constant volume, we expect its energy to go up. The ﬁrst observation has already been shown in the deﬁnition. We now show that the postulates predict the second and third observations. The fourth observation is shown in §4.1. Consider two objects that are in thermal contact with each other, but are isolated from the rest of the universe. Initially the two objects are at equilibrium, but have diﬀerent extensive properties S, U, V, N (see Figure 2.6). Since the system is closed, the total internal energy of the system must be constant: dU = dU (1) + dU (2) = 0, where the superscripts indicate either object 1 or object 2. Hence, dU (1) = −dU (2) . (2.59) 40 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS From Postulate IV, we know that the equilibrium state is attained when the entropy reaches a maximum. If each subsystem is maintained at constant volume (dV (1) = dV (2) = 0) and mole number (dN (1) = dN (2) = 0), then the entropy diﬀerential Eqn. (2.29) becomes dS = dS (1) + dS (2) 1 1 = dU (1) + (2) dU (2) T (1) T 1 1 = − (1) dU (2) , isomolar, isochoric T (2) T (2.60) where we have used Eqn. (2.59) to obtain the third line. The postulates claim that the unconstrained variables—in this case U (2) —will arrange themselves such as to maximize the entropy. In order to be a maximum, the total entropy must satisfy two conditions. First, at equilibrium (∂S/∂U (2) )V (1) ,V (2) ,N (1) ,N (2) must be zero. Hence, Eqn. (2.60) leads to T (2) = T (1) at equilibrium, which is the second observation that we wished to prove. Secondly, in order for entropy to be a maximum, entropy must change from a lower to a higher value in going from the initial to the ﬁnal conﬁguration. ∂S ∂U (2) =
V (1) ,V (2) ,N (1) ,N (2) 1 T (2) − 1 T (1) , (2.61) The inequality follows from the postulate that the entropy of an isolated system must increase in going from one equilibrium state to another. Eqn. (2.61) says that ∂ if T (2) > T (1) , then ∂US < 0, and heat ﬂows from object 2 to (2) (1) (2) (1) (2)
V ,V ,N ,N object 1. On the other hand, if T (2) < T (1) , then ∂S ∂U (2) V (1) ,V (2) ,N (1) ,N (2) > 0, and heat ﬂows from object 1 to object 2. This result is shown graphically in Figure 2.7. The equilibrium state is the maximum of the curve. If the initial state lies to the left of the maximum, then U (2) will increase. From Eqn. (2.61), we see that this is when T (1) > T (2) . This prediction agrees with our third observation about the intuitive properties of temperature: heat ﬂows from hot to cold objects. Example 2.7.1 Two tanks containing a gas insulated and isolated from the environment are connected by a pipe with a valve. The valve is initially shut, so the two tanks are not equilibrated with one another. Tank 1 has volume V1 = 5l, initial pressure P1 = 3atm, and initial temperature T1 = 250K. The second tank has volume V2 = 2l, initial pressure P2 = 5atm, and initial temperature T2 = 323K. Assuming that the gas is a simple ideal gas with c = 3/2, ﬁnd the ﬁnal temperature of the tanks. 2.7. TEMPERATURE DIFFERENCES AND HEAT FLOW 41 S Ueq (2) U (2) Figure 2.7: General form of entropy as a function of internal energy of one subsystem for the setup
in Figure 2.6. Solution: The walls of the tanks are rigid, so the volume of each tank remains ﬁxed. The valve allows the gas to pass between the tanks, so the number of moles in each tank is not conserved. However, the total number of moles N = N1 + N2 is ﬁxed. The valve also allows energy to pass between the two tanks, so the energy of each tank changes, but the total energy U = U1 + U2 is ﬁxed. Constants of the process are then V1 , V2 , N , and U . Everything else (N1 , N2 , T1 , T2 , S1 , S2 , P1 , P2 ) can change in reaching the ﬁnal state. We can ﬁnd the initial mole numbers from the ideal gas equation of state, Eqn. (2.37) Pi Vi = Ni RTi , i = 1, 2, simple ideal gas. (2.62)
Hence, N1 = 0.7312mols, N2 = 0.3773mols, and N = 1.108mols. We can ﬁnd the initial internal energies from the thermal equation of state for a simple ideal gas, Eqn. (2.38) Ui = cNi RTi , i = 1, 2, simple ideal gas. (2.63) Therefore, U1 = 2.280kJ, U2 = 1.520kJ, and U = 3.800kJ. The ﬁnal state occurs when the temperatures in the two tanks are the same. Since the resulting gas still satisﬁes the thermal equation of state for a simple ideal gas, we can ﬁnd the ﬁnal temperature T straightaway T= U = cN R (3.80kJ)(1000J/kJ)
3 2 (1.108moles)(8.3144J/mol.K) = 275K. (2.64) 42
2 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS Note the usefulness of this strategy to solve thermodynamics problems: ﬁnd which variables are constant. 2.8 Pressure Diﬀerences and Volume Changes In §2.7 we found that our intuition about temperature is predicted by the postulates: temperature diﬀerences lead to heat ﬂow from hot to cold objects. Similarly, we show here that pressure diﬀerences lead to changes in volume. In other words, if two objects are in mechanical contact, the one at higher pressure will expand at the expense of the volume of the other object, until the pressures are equal. Consider two containers of gas that have diﬀerent temperatures, and diﬀerent pressures. Also, we relax the constraint on volume, and allow the volumes of the two subsystems to change. See Figure 2.4. The two containers are separated by a movable, diathermal partition. Initially, the partition is insulating, and ﬁxed. Then, we allow the partition to move and transfer heat between the two compartments. Throughout the process, the composite system is closed to the universe. Since the system is closed, again the internal energy is conserved, and Eqn. (2.59) holds. Since the composite system is closed to the universe, the total volume must also be conserved, and dV (1) = −dV (2) . (2.65) Again we hold the mole numbers in each subsystem constant, so that the diﬀerential for S Eqn. (2.29) becomes dS = dS (1) + dS (2) 1 P (1) = dU (1) + (1) dV (1) + T (1) T 1 1 = − (2) dU (1) + (1) T T 1 P (2) (2) dV T (2) dV (1) , (2.66) T P (1) P (2) − (2) T (1) T dU (2) + (2) where we have used Eqs. (2.59) and (2.65) to obtain the third line. Since the variables U (1) and V (1) are unconstrained, each of the terms on the right side of the last line of Eqn. (2.66) must be zero at equilibrium. Hence, we again ﬁnd that T (1) = T (2) . However, we also ﬁnd that P (1) = P (2) at equilibrium. Consider the special case when the two compartments are at equal temperature T . Then, the change in volume for the compartments is given by the stipulations that ∆S > 0, according to the postulates. Thus, T ∂S ∂V (1) = P (1) − P (2) . (2.67) 2.9. THERMODYNAMICS IN ONE DIMENSION 43 If P (1) > P (2) , then the volume of V (1) must increase to satisfy the postulates. Likewise P (1) < P (2) requires that V (1) decrease. These results agree with our observation about pressure. Example 2.8.1 Find the ﬁnal pressure of the gas in Example 2.7.1. Solution:
We have now proven that the ﬁnal state occurs when the temperatures and pressure in the two tanks are the same. Since the resulting gas still satisﬁes the mechanical equation of state for a simple ideal gas, we can ﬁnd the ﬁnal pressure P P= 2 NRT (1.108 mols)(0.08206 l·atm/mol·K)(275 K) = = 3.572 atm. V1 + V2 7l (2.68) In Problem 2.8.A, one ﬁnds that there is a role played by chemical potential for moles analogous to that played by temperature for energy, and by pressure for volume. Hence, we ﬁnd that for two systems in thermal, mechanical, and chemical contact, equilibrium implies T (1) = T (2) , P
(1) thermal equilibrium mechanical equilibrium chemical equilibrium. (2.69) (2.70) (2.71) =P = (2) , (1) µ1 (2) µ1 , These results are completely general6 , and should be remembered. We also ﬁnd that if the chemical potential of a species is nonuniform, that species will move from the region of high chemical potential to the region of low chemical potential. Roughly speaking, the chemical potential depends strongly on concentration, so we sometimes say that moles diﬀuse to oﬀset diﬀerences in concentration. However, as stability analysis shows in Chapter 4, that approximation is not always correct. 2.9 Thermodynamics in One Dimension So far we have considered systems where the independent variables are (S, V, N ), or their conjugates. However, some important systems are only two dimensional, such
6 These are completely general for a 3D system, that is. For other systems, there may also exist an analog to equal pressure. 44 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS as adsorption on a surface, or a thin ﬁlm, and, hence, volume plays no role. It is also possible to apply thermodynamics to such systems with a slight modiﬁcation to the development we have seen so far. In this section, we consider a onedimensional system—the length of a rubber band. In later sections, we will also consider twodimensional systems, such as a catalytic surface and a thin ﬁlm. To consider the rubber band, we must use an appropriate replacement for volume, and then the quantity analogous to pressure. For the rubber band, we replace volume with the length of the rubber band: V → L, and pressure with tension: P → −T . The sign is diﬀerent, because the work done on a rubber band is +T dL, as opposed to −P dV for a ﬂuid. Hence, we can write the thermodynamic deﬁnition for tension that is consistent with the mechanical work as T := ∂U ∂L .
S (2.72) Rubber is comprised of crosslinked polymeric chains; an ideal rubber is made ˜ ˜ up of one large molecule, so that N = 1/NA , where NA is Avogadro’s number. Therefore, one cannot easily change the size of the system, and the number of moles is ﬁxed. Hence, a fundamental entropy relation for a rubber band must be of the form S = S (U, L), and extensivity is a little diﬀerent. A few simple experiments one can easily perform at home show two important trends: (1) the tension in a rubber band increases with length, and (2) the tension increases with increasing temperature.7 Hence, any reasonable fundamental relation should predict these two observations. The latter observation is curious, since it is the opposite as that seen for metal springs, where the tension decreases with temperature. In the following example we consider a fundamental entropy relation that can be derived from straightforward statistical mechanical arguments. Example 2.9.1 Treloar [93] has suggested the fundamental entropy relation for a rubber band as S = S0 − nc L0 A0 R 1 2 L L0
2 + L0 L + CL log 1 + U − U0 , CL T0 (2.73) where S0 , U0 and T0 are constants with straightforward interpretation. We have four more constants nc , CL , A0 and L0 ; the last two have straightforward meaning:
Try suspending a weight with a rubber band, and heating it with a hair dryer. However, be careful to make sure that the rubber band is well equilibrated. Stretch and heat the rubber band a bit ﬁrst. Then, keeping the weight attached, let the rubber band cool to room temperature. Mark the height of the weight, and then begin warming the rubber band with the dryer. As the temperature increases, the tension should increase, raising the height of the weight.
7 2.9. THERMODYNAMICS IN ONE DIMENSION 45 the crosssectional area and length of the rubber band at rest. What does this model predict for tension as a function of length and temperature? Does it agree with experiment? Solution: To ﬁnd tension and temperature, we derive two equations of state
from the fundamental relation. We can ﬁnd temperature in the usual way from its (alternative) deﬁnition T := = 1 ∂S ∂U L,Nc T0 1 + U − U0 , CL T0 (2.74) where we used the fundamental relation Eqn. (2.73) to obtain the second line. From this result we can ﬁnd the thermal equation of state U = U0 + CL (T − T0 ). (2.75) Curiously, we ﬁnd that the internal energy does not depend upon length at constant temperature. Hence, an isothermal stretch of the rubber band stores no energy in the rubber band. Nonetheless, the rubber band can do work on its environment by pulling up a weight, say. It does this by increasing its entropy . Actually, there is a strong analogy with compression of an ideal gas. The student should be able to prove that isothermal compression of an ideal gas does not change its internal energy, but rather decreases its entropy. To ﬁnd the tension in the rubber band, we could use the thermodynamic deﬁnition Eqn. (2.72). However, it is more convenient to use the alternative expression T = −T ∂S ∂L U,Nc = nc A0 RT L L0 − L0 L 2 (2.76) where we have again used the fundamental relation Eqn. (2.73). We ﬁnd several important features of the model from this result: • The tension in the rubber band increases nearly linearly with length. This is observed approximately experimentally whenever the extension is not too great. • Our thermal equation of state shows that lengthening the rubber band at constant temperature does not change the internal energy. 46 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS
• The tension increases linearly with temperature, again in accord with experiment.8 • The tension in the rubber band is zero only when its length is L0 , its natural rest length. • The second term in the square brackets of Eqn. (2.76) is related to incompressibility of the rubber band, and shows that a compressive force is necessary to decrease the length of the rubber band from its rest length. 2 The fundamental relation considered in the previous example captures many of the key experimental features. However, a real rubber band shows richer behavior: • The internal energy does not grow linearly with temperature. • At suﬃciently low temperatures, the rubber can crystallize, and show no strong elastic behavior. • At large extensions, the tension grows more rapidly than linear with length, and the rubber can eventually crystallize, or break at large stretches. The physical meaning of the constants is now clear. T0 is a reference temperature. The rest length of the rubber band (the length where the tension is zero) is L0 .9 Given the limitations and successes of this fundamental relation, one can draw many analogies with an ideal gas which also fails at low temperatures and low volumes. In other words, the rubber band behaves like an ideal gas, except that increasing the length of the rubber band is like decreasing an ideal gas’s volume. For example, compressing an ideal gas isothermally changes the gas’s entropy, but not its internal energy. Therefore, we choose to call Eqn. (2.73) a fundamental entropy relation for a simple, ideal rubber band.10 We have shown that this fundamental relation
We get to see in §3.7.1 the interesting result that the linear dependence in temperature is tied directly to the second note above. There we also show how isothermal stretching aﬀects the entropy, but not the energy of a real rubber band. 9 From Eqn. (2.75) we see that CL is the constantlength heat capacity. Heat capacities are rigorously deﬁned in §3.5. 10 In the language of mechanics, it is the equation that arises for an incompressible, neoHookean solid in extensional deformation.
8 2.10. SUMMARY yields T
T 47 = T ψ (L) =0 < 0. (2.77) ∂U ∂L ∂S ∂L T In words, the last two relations say that isothermal stretching of the rubber band decreases the entropy, but leaves the internal energy unchanged. In §3.7 we show that experimental observation of the ﬁrst relation leads inescapably to the second and third relations. 2.10 Summary In this chapter we introduced or used several fundamental quantities: • Volume V , area A, or length L. • Mole number N . • Internal energy U . • Entropy S . • Equilibrium states. The properties of U, S and equilibrium states are described in the four fundamental postulates that form the basis of all thermodynamics. We know that a function of the form U = U (S, V, N ) or one of the form S = S (U, V, N ) completely determines the thermodynamic properties of a substance. From these fundamental quantities, we derived • Temperature T := • Pressure P := −
∂U ∂S V ,N . ∂U ∂V S,N . ∂U ∂N S,V . • Chemical potential µ := • Heat ﬂow Q: dQ = dU − dW . ¯ ¯ 48 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS The ideas in the ﬁrst three deﬁnitions are neatly summarized in the diﬀerential for internal energy
r dU = T dS − P dV + We proved that • In an isomolar, quasistatic process µi dNi .
i=1 (2.27) dQrev = T dS . ¯ • The fundamental postulates say that heat ﬂows from hot to cold objects until the temperature is equal. • Compressible objects in mechanical contact change volumes until the pressures are equal, the higher pressure object expanding at the expense of the other. • Compounds diﬀuse across permeable boundaries from regions of high chemical potential to regions of low chemical potential. The last three results were written mathematically for two subsystems (1) and (2) in thermal, mechanical and chemical contact T (1) = T (2) , P (1) = P (2) , µ1 = µ1 ,
(1) (2) thermal equilibrium mechanical equilibrium chemical equilibrium. (2.69) (2.70) (2.71) All of the above results in the summary should be memorized. We also proved that experimental results imply that • Compressing an ideal gas isothermally decreases its entropy, but leaves its internal energy constant. • Stretching a rubber band isothermally decreases its entropy, but leaves its internal energy constant. For examples, we used fundamental relations for a simple ideal gas, a simple van der Waals ﬂuid, and an incompressible, neoHookean rubber band. We also performed calculations for work and heat ﬂow of expansion (or contraction), and work required for adiabatic volume (or length) changes. We showed how an ideal gas can be used as a refrigerator, and estimated the minimum amount of work required to do so. 2.11. EXERCISES 49 2.11 Exercises 2.3.A. The same experiments from Example 2.3.1 are repeated for another gas, which exhibits diﬀerent results. The ﬁrst experiment reveals that P+ a ˜ V2
c V −˜ b c+1 = constant. The second experiment (using the falling weight) reveals that dP mg dl =− , dt c(V − bN ) dt where a, b, c are constants, and N is the number of moles of the gas. Find the equation of state U = U (P, V, N ). 2.3.B. Two points on the pressurevolume plane of Example 2.3.1 are connected by a line: P V =constant. Find the work and heat ﬂow necessary to change the volume and pressure of the container in this example to follow this line from [P0 , V0 ] to [Pf , Vf ]. Find the heat and work necessary to make the system follow a straight line between the points. How do your answers compare? 2.3.C. A gas is placed in a container identical to the one in Example 2.3.1. It is reported that the equation of state for the gas is given by U (P, V ) = U0 + A P V 2 − P0 V02 . If the volume is changed slowly (quasistatically) and adiabatically, how will the pressure change? 2.3.D. The gas in Example 2.3.1 begins at a pressure of 1 atm, and a volume of 1 liter. If it is compressed adiabatically to half its original volume, how much work is done? If a riﬂe bullet of mass 4g has the same kinetic energy, how fast is it moving in miles/hour? How high must a golf ball (of mass 42g) be raised to gain the same amount of potential energy? 2.3.E. How much work does it take to compress N = 3 mols of an ideal gas adiabatically from V = 1 liter at T = 24◦ C to a volume of 10ml? The pressure of an ideal gas can be found from the temperature and volume by the idealgas equation of state P V = NRT . 2.3.F. The gas from Example 2.3.1 is now to be used as part of a refrigeration cycle. To move the ﬂuid in the cycle (look at the tubes in the back of your refrigerator to see how this ﬂuid is carried), a pump raises its pressure from approximately 20 psig at 25◦ C to 32 psig. Assuming that this pump operates adiabatically, and 50 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS that the gas is pumped at 23 ft3 /min (based on the upstream volume), estimate the minimum power (work/time) necessary to pump the ﬂuid. Note that a ﬂowing system necessarily has irreversibilities that make the necessary power greater than you would predict from a quasistatic process. 2.3.G. We repeat the experiments of Example 2.3.1 with a diﬀerent gas, and observe diﬀerent behavior. The second experiment reveals that the pressure changes with time according to dP dl = −f0 (V )mg , dt dt where f0 (V ) is some measured function of volume with dimensions of inverse volume. The ﬁrst experiment reveals [P − P1 (V )] exp −
V V0 f0 (V ′ )dV ′ cf0 (V ) ∼ constant, where c is a constant, and P1 (V ) is another measured function of volume, with dimensions of pressure. V0 is some constant reference volume. Find the equation of state U = U (P, V ). Note that Problem 2.3.A is a special case, which might provide a check for your answer. 2.4.A A substance called a glassy polymer is placed into a container that allows us to control the work done on and the heat added to the polymer. In State #1, the polymer has a certain energy U , volume V and mole number N . We measure the pressure at these conditions to be P1 , which appears to be stationary with time. Then, we perform various steps of work on the system, and heat extractions such that the volume and energy are the same as before. We again measure the pressure, and ﬁnd P2 = P1 , but P2 also appears to be stationary. Are states 1 and 2 at equilibrium? Explain. 2.5.A. Determine which of the following proposed fundamental relations satisfy the postulates of thermodynamics. 1. S = s0 N + AU 1/4 V 1/2 N 1/4 2. S=
i=1 r r Ni si,0 +
i=1 Ni ci log U N u0 V 2B N2 r +
i=1 1/3 Ni R log V v0 Ni 3. S = AU N
β 2.11. EXERCISES 4. U= 5. U= 6. U= 7. S = N s0 + NR log U A V B N C , where A, B, C, R, ci , s0 , si,0 , u0 , and v0 are constants. 51 AS 3 +B NV AV 3 NS AN 2 exp (S/NR) V 2.5.B. A simple van der Waals ﬂuid is compressed isentropically to half its volume, if the initial state is (T0 , v0 ), ﬁnd the ﬁnal temperature, the work required, and heat ﬂow. 2.5.C. One mole of oxygen at 25◦ C and 15 atm is compressed isoenergetically to 35 atm. Calculate the change in entropy using both a simple ideal gas (c = 5/2) and a simple van der Waals ﬂuid. The other parameters for the van der Waals ﬂuid can be found from the critical properties of oxygen, as shown in §B.3. Compare the two answers. Why are they diﬀerent? How does your answer ﬁt with the deﬁnition of entropy, Eqn. (2.20)? 2.5.D. Do Problem C above, but this time use the simpliﬁed RedlichKwong model, Eqn. (2.78) instead of the van der Waals model. 2.6.A. Find U (P, V, N ) for the van der Waals ﬂuid whose fundamental entropy relation is given by Eqn. (2.21). Compare your answer with the result from Problem 2.3.A. 2.6.B. Check the postulates, and ﬁnd P (T, V ) for the following fundamental entropy relation v−b 4b2 u3 N S = NR log + (2.78) 2, v0 v 3 3a log v+b − 4Ab where R, b, v0 and a are constants. The resulting relation is called the RedlichKwong equation of state . Note that the thermal equation of state resulting from Eqn. (2.78) is not particularly accurate. In Chapter 3, a more realistic thermal equation of state is derived that is compatible with the RedlichKwong mechanical equation of state. 52 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS 2.6.C. Check the postulates, and ﬁnd P (T, V ) for the following fundamental energy relation 2 − N R log vv0b − a1 ψ (v ) − s U = N a0 ψ (v ) − , (2.79) 4a2 ψ (v ) where a0 , v0 , R, b, a1 , a2 and A are constants, and √ 1 v + b + 2b √ ψ (v ) := A + √ log . 2 2b v + b − 2b The resulting relation is similar to the PengRobinson equation of state . Note that the thermal equation of state resulting from Eqn. (2.79) is not particularly accurate. In Chapter 3, a more realistic thermal equation of state is derived that is compatible with the PengRobinson mechanical equation of state. 2.6.D. Check the postulates, and ﬁnd P (T, V ) for the following fundamental energy relation N [R log (v − b) − a1 ψ (v ) − S/N ]2 , (2.80) U = N a0 ψ (v ) − 4a2 ψ (v ) where a0 , a1 , a2 , R, b and A are constants, and ψ (v ) := A + 1 log γ−β v+β−b v+γ−b The resulting relation is called Martin’s generalized cubic equation of state . Note that the thermal equation of state resulting from Eqn. (2.80) is not particularly accurate. In Chapter 3, a more realistic thermal equation of state is derived that is compatible with Martin’s generalized mechanical equation of state. 2.6.E. Check the postulates, and ﬁnd P (T, V ) for the following fundamental entropy relation V /N − b U + aN 2 /V S = N s0 + NR log u0 + a/v0 v0 b(V + N v0 − 2N b) b(V − N v0 ) 4+ (V − bN )v0 (V − bN )v0
c 1 + Nc (2.81) where a, v0 , R, b and c are constants. The resulting relation is called the CarnahanStarling generalization to the van der Waals equation of state. Note that the thermal equation of state resulting from Eqn. (2.81) has a constant heat capacity. In Chapter 3, a thermal equation of state is derived that yields arbitrary temperature dependence for the heat capacity. 2.11. EXERCISES 53 2.6.F. Check the postulates, and ﬁnd P (T, V ) for the following fundamental entropy relation S = N s0 + NR log V N v0 − N 2R V B0 + bN 2V − 2ψ2 (U, V, N )3/2 3 N ψ1 (V, N ) (2.82) where A0 , B0 , C0 , B1 , R, a, b, c, s0 , v0 , α, γ are constants, and ψ1 (V, N ) := and ψ2 (U, V, N ) := N u0 − U − 3N C0 + 3c V 1 N2 + γ 2V 2 exp −γN 2 V2 + B1 2 N 2 A0 aN 3 +2 V V αN 3 1 − 5V 3 2 . The resulting relation is called the BenedictWebbRubin equation of state , commonly used for alkanes. Note that the thermal equation of state resulting from Eqn. (2.82) is not very accurate. In Chapter 3, a more realistic thermal equation of state is derived that is compatible with the PVT equation of state that arises from Eqn. (2.82). 2.6.G Find the work and heat ﬂow necessary to compress a van der Waals ﬂuid isothermally. 2.6.H One mole of a simple ideal gas (c = 3/2) is placed into a container with a nearly frictionless piston to be used as a refrigerator. The cycle of the refrigerator has three steps. In step (a), the gas is compressed adiabatically from 50◦ C, 1atm to 150◦ C. In step (b), the gas is cooled isobarically back to 50◦ C. In step (c), the gas is expanded back to its original thermodynamic conditions. During step (c), the gas can be used to extract heat from an object at 50◦ C. Find the work necessary to operate the refrigerator, and the amount of heat that can ideally be extracted. 2.6.I. Find the equations of state for those fundamental relations in Problem 2.5.A. that satisfy the ﬁrst four thermodynamic postulates. 2.6.J. If you compress a van der Waals ﬂuid isentropically until its volume is halved, what happens to the temperature? Find the quasistatic work, and show that the process is also adiabatic. 2.6.K. From Eqn. (2.35), we see that the van der Waals equation of state oﬀers a correction to the ideal gas through the constants a and b. These constants may be found from the socalled critical properties of a ﬂuid Pc and Tc using Eqn. (4.22). From its critical properties, ﬁnd a and b for nitrogen. If you have 1 mole of nitrogen at room temperature and pressure, what does the ideal gas model predict for the volume? What does the van der Waals ﬂuid predict? What is the percentage 54 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS diﬀerence? Now do the same calculations for 5atm of pressure. For the vdW ﬂuid, you must solve a cubic equation in volume. You may ﬁnd the equations in Appendix A.7 useful. 2.6.L. For toluene, we estimate the parameters of the van der Waals equation of state to be a = 2.4876432Pa·m6 /mol2 and b = 0.000149797m3 /mol. Make a plot of speciﬁc volume versus pressure for T = 580K, and the range P < 3.78 × 106 Pa and 6.223 × 10−04 < v < 0.007m3 /mol. Note that the van der Waals equation of state is cubic in volume. Hence, you might ﬁnd the equations in §A.7 useful. What diﬃculties arise as the pressure approaches 3.78 × 106 Pa? (Hint: the range of volumes given are the physically acceptable roots to the cubic equation for a vapor, as we discuss in detail in §4.2.2.) 2.6.M. For oxygen, we estimate the parameters of the PengRobinson equation of state to be a = 0.164995Pa·m6 /mol2 and b = 1.98414 × 10−5 m3 /mol at T = 120K. If you wish to store 1000 moles of oxygen in a 1 cubic meter container, what will the pressure be? If you triple the pressure, how many moles will be stored in the same container? Note that the PengRobinson equation of state is cubic in volume. Hence, you might ﬁnd the equations in §A.7 useful. 2.6.N. From Statistical Mechanics it is known that the idealgas law holds only when the individual molecules rarely interact with one another—i.e., at low densities. However, we can also see this is true from the equations of state. For example, in the van der Waals equation of state, P= a RT − v − b v2 we see that the pressure is well approximated by RT /v only for large values of v . To make this more quantitative, we could say that va should be small relative to RT . 2 v Suppose that we arbitrarily choose an accuracy of 10% as a ‘good’ approximation by an equation of state, then we could require that va / RT < 0.10 for the idealgas 2 v law to hold. Make plots of vmin , the minimum value of speciﬁc volume allowing ideal gas to be a good assumption, as functions of temperature (50–400K). Use the van der Waals, RedlichKwong, and PengRobinson equations of state for estimates of nonideal behavior, and assume that the gas is nitrogen. Two of these equations of state are known to be more accurate. Which is the outlier (i.e., the bad equation of state)? Note that the parameters for the equation of state can be estimated using the critical properties of the ﬂuid, as shown in the appendix. 2.6.O. Launching potatoes with an air gun appears to be a popular pastime in rural parts of the U.S. From a thermodynamics point of view, one uses the work available 2.11. EXERCISES 55 Figure 2.8: Schematic of a device that uses compressed air to launch a potato. The air is held in a rigid container with adiabatic walls. The friction and inertia of all devices but the potato are neglected. The analysis of this device is described in Problem 2.6.O. from expanding air to give kinetic energy to a potato. Here we wish to estimate the minimum compression of air necessary to launch a 500 gram potato a distance of 100 meters. For the sake of analysis, we consider a simple setup as shown in Figure 2.8. A real air gun does not work in the way illustrated, but the analysis is similar. To calculate the minimum work, we make the following simplifying assumptions. First, the air is assumed to be described by a simple ideal gas, with c = 3/2. Hence, we can use the analysis in Examples 2.6.2 and 3. Secondly, we neglect all irreversible losses, such as the sliding of the piston, nonquasistatic expansion, and the air drag of the potato as it ﬂies through the air. Assuming that the air begins at room temperature, and that there is 0.1 mole of gas, ﬁnd the initial volume and pressure, and the nal temperature and volume of the air, assuming that the process is adiabatic. What do you expect to happen to any humidity that might be in the air after the expansion? 2.6.P. Finish Example 2.6.5. In other words, ﬁnd the work necessary to compress a simple van der Waals gas adiabatically. 2.6.Q Using the van der Waals equation of state, calculate the amount of work necessary to compress one mole of nitrogen from 25ml to 3ml at 25◦ C. How much does the internal energy change in the process? 2.6.R What does the van der Waals equation of state predict for the pressure of 3 mols of oxygen at room temperature when it has a volume of 10 mls? Is the ideal 56 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS gas law a good approximation for this value? 2.6.S Using the PengRobinson model from Problem 2.6.C, predict how the pressure would change for the adiabatic, reversible compression of the ﬂuid. 2.6.T In Problem 2.3.F you calculate the minimum power to compress an ideal gas. Now we consider the adiabatic compression of a gas that is too dense to be ideal, and use the van der Waals model to estimate the work. Find the minimum power to pump hydrogen from 13 atm to 15 atm assuming that the pump operates adiabatically, and pumped at 23 ft3 /min (based on the upstream volume), estimate the minimum power (work/time) necessary to pump the ﬂuid. For a diatomic gas (like hydrogen) at suﬃciently high temperatures, we can assume that the gas is simple, and c = 5/2. The other parameters are found from the expressions in §B.3 and the critical properties of the ﬂuid. The critical properties for some ﬂuids are found in Table D.3 and many more can be found on the NIST Web Book. You might use the following steps. • Begin by working in speciﬁc quantities only. Find the initial pressure, speciﬁc volume and temperature. From the thermal equation of state, ﬁnd the speciﬁc internal energy. • Find the ﬁnal state of the system assuming irreversible, adiabatic compression. Example 2.6.5 might be useful here. • How is the work (per mole) related to the change in speciﬁc internal energy? • Now ﬁnd the moles per minute pumped through the system, in order to ﬁnd the minimum work per time required of the pump. 2.6.U Estimate the minimum amount of work necessary to use nitrogen as a refrigerant in the manner shown in Example 2.6.6. However, in this case, the initial pressure is 25 atm, so that we cannot assume that the gas is ideal. Instead use the simple van der Waals model, with c = 5/2. The other parameters in the model can be found from the critical properties, as given in §B.3. 2.6.V What is the change in chemical potential predicted by a simple van der Waals ﬂuid to compress one mole of carbon dioxide isothermally from 15 to 25 bar at 25◦ C? 2.8.A. Consider an isolated system that is separated into two compartments by a rigid, diathermal wall that is permeable to species 1, but not to other species in the system. Show that the postulates indicate that at equilibrium the chemical potential of species 1 must be the same in each compartment. Also show that the moles of species 1 move from the compartment of large chemical potential to the compartment of low chemical potential. 2.11. EXERCISES 57 2.8.B. Find the ﬁnal temperature of the gas in Example 2.7.1 if it is described by a simple van der Waals ﬂuid. 2.8.C. A tank of 1 liter is connected by a pipe ﬁtted with a valve to another tank of 2 liters. Both tanks are insulated and contain Argon, but the ﬁrst tank is at a pressure of 5 bar, and temperature of 220K, whereas the second tank is at a pressure of 10 bar and a temperature of 250K. What are the ﬁnal temperature and pressure when the valve is opened? Assume that the gas is described by a PengRobinson mechanical equation of state, given in §B.6. Also assume that the thermal equation of state is √ a0 v + (1 + 2)b 3 √ u = u0 + √ log + R (T − T0 ) 2 2 2b v + (1 − 2)b 2.8.D. Assuming that a simple ideal gas with c = 5/2 is used as the refrigerant, design a refrigerator to keep a temperature of 10◦ C with an outside temperature of 22◦ C. The cycle of the refrigerant has four steps. In the ﬁrst step, the piston is used to expand the gas isothermally at the lower temperature. Secondly, the gas is compressed adiabatically until it reaches the higher temperature. Thirdly, it is compressed further at the higher temperature isothermally. Finally, in the fourth step, the gas is expanded adiabatically back to its original thermodynamic conditions. Calculate the amount of work required at each step, and how much heat is transferred. How much force would you require in your design and how much distance to obtain the necessary work? You might ﬁnd it useful to sketch the process on a pressure/volume diagram. If the refrigerator leaks 1 Joule/hour to the surroundings, how many times per hour must you cycle the refrigerant for your design? A real refrigerator is only slightly more complicated than this conceptually. First, the refrigerant is cycled by pumping it around a tube, so the process is continuous. Secondly, the refrigerant undergoes phase changes between liquid and gas, which give greater heat transfers. The details for design of a real system is given in Chapter 5. 2.8.E. Two tanks are isolated from the environment and connected by a valve. Initially, each tank has oxygen at volumes 10 and 25 liters, and pressures 40atm and 27atm, respectively. Each is at temperature 298K. At these pressures, we cannot assume that the gases are ideal. Instead, we use the simple vdW ﬂuid (with c = 5/2), and determine the a and b parameters from the critical properties of O2 (see §B.3 and Table table:critical properties, or use the NIST Web Book). Determine the ﬁnal temperature and pressure of the tanks after the valve is opened, and equilibrium is reached. 2.9.A. Does the fundamental relation in Example 2.9.1 satisfy the postulates of thermodynamics? 58 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS 2.9.B. The following fundamental entropy relation is proposed for a rubber band U − U0 A0 L0 nc R S = S0 + CL log 1 + + log 1 − CL T0 2 L − L0 L1 − L0
2 . (2.83) What does this fundamental relation predict for T (T, L)? What physical interpretation does L1 have? Is this fundamental relation in closer agreement with experiment? 2.9.C. What is the tension as a function of length for a reversible, adiabatic stretch of the rubber band from Example 2.9.1, if the initial length is L0 , and the initial temperature is T0 ? 2.9.D. A rubber band can also be used as a refrigerator, as in Problem 2.6.H. By following the cycle drawn in Figure 2.9 in a clockwise fashion, heat can be extracted at the lower temperature from A to B, and expelled at a higher temperature from C to D. Calculate the amount of work necessary if the refrigerator is operated at high temperature TH and low temperature TC . Assume that LA = 3L0 and LB = 2L0 . Also ﬁnd the amount of heat extracted during AB and CD for each cycle. 2.9.E. Recent experiments have measured the tension in a single strand of DNA in water by attaching one end of the strand to a glass slide, and the other end to a 3µm magnetic bead. By applying a known magnetic ﬁeld and measuring the extension of the strand under an optical microscope, a plot of endtoend length as a function of tension can be generated [11]. Using straightforward statistical mechanical ideas, fundamental relations can be found for a single DNA molecule. The Gaussian chain model in simpliﬁed form is U (S, L, NK ) = U0 + CL T0 exp kB L2 S − S0 − −1 , CL 2CL NK a2 K (2.84) where U0 , T0 , and S0 are constants with straightforward interpretation, kB is Boltzmann’s constant (equal to the ideal gas constant divided by Avogadro’s number), CL is the constantlength heat capacity of the strand, L is the endtoend length of the DNA strand, NK is the number of ‘segments’ in the strand, and aK is the length of a segment. A somewhat more sophisticated model is called the wormlike chain, with simpliﬁed fundamental relation U (S, L, Np ) = U0 + CL T0 S − S0 kB Np exp − CL 2CL −1 . L Np lp
2 − L + 2Np lp (2.85) Np lp 4(Np lp − L) 2.11. EXERCISES 59 Find expressions for tension as a function of length for these two fundamental relations. Make plots of length versus logarithm of the force, and compare with the data in Table 2.1, assuming that (1) they were taken at room temperature, (2) the contour length (Np lp ) is 32.7µm, and (3) that the persistence length lp is 53nm. These latter properties can be found from other experiments unrelated to the elasticity. Note that aK = 2lp and NK = Np /2. L(µm) 8.0 9.3 10.0 10.8 16.4 17.9 17.8 19.0 18.3 21.2 22.0 22.7 23.0 F (fN) 30.04 29.01 35.15 39.04 82.71 84.16 103.78 103.78 117.27 181.45 178.31 223.75 248.46 L(µm) 24.1 24.3 25.1 25.8 26.3 27.0 27.0 27.4 27.5 27.7 28.7 28.9 29.4 F (fN) 301.07 442.08 301.07 412.25 535.69 535.69 733.50 936.62 1195.98 828.86 1609.29 1195.98 2242.39 L(µm) 29.5 29.5 29.7 30.3 30.3 30.6 31.2 31.6 31.6 31.8 32.4 33.0 33.0 F (fN) 1449.23 1850.54 2446.95 3920.71 3920.71 4587.87 5463.12 7746.39 15040.12 10793.82 25394.48 56696.27 67512.45 Table 2.1: Force as a function of stretching as measured by [11] for stretching a strand of DNA at 298 K. 2.9.F. Isothermally compress an ideal gas to 1/2 its initial volume, and calculate the changes in energy and entropy. Can the resulting system do work? How does this relate to the stretching of a rubber band? What is the analogy with Eqs. (2.77)? 2.9.G. A recent paper [95] recommends a new fundamental relation for the extension of a single polymer S = S0 − 2− NK kB 2 4 NK 1+ 2 10 + 2 3NK 27NK L NK aK
2 L NK aK 2 − U − U0 , CL T0 (2.86) log 1 − + CL log 1 + where S0 , T0 , kB , NK , aK , and CL are constants with the same meaning as Problem 2.9.E. How does this expression compare with the data in Table 2.1? 60 CHAPTER 2. THE POSTULATES OF THERMODYNAMICS 12 D
10 C 8
T L0 N RT0 6 A B
adiabats 4 y 2 2 2.5 3 3.5 Dimensionless Length, L/L0 4 Figure 2.9: Tension as a function of stretching ratio for a rubber band as modeled by Treloar. The solid lines are isothermal, and the dashed lines are isentropic. The rubber band may be used as a refrigerator by going clockwise around the cycle shown: A → B → C → D → A. ...
View
Full
Document
 Spring '08
 LIZ

Click to edit the document details