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0.1 Minimum Principles and Thermodynamic Potentials The Second Law turns out to have certain very useful consequences for calculating the behavior of macroscopic systems. These are most usefully formulated as variational principles. The law states that, in any thermodynamic transformation, Rf i dQ/T ≤ Sf − Si , in an obvious notation. Equality holds for a reversible process. Now think about a system with some internal adjustable parameter, perhaps a gas with a movable partition. Let the initial configuration have the partition fixed. Then let it loose to find the equilibrium position. The total change in heat ∆Q and the total entropy change ∆S satisfy ∆Q/T ≤ ∆S, and so T ∆S ≤ ∆W + ∆U . Define A = U − T S. Then for fixed total volume ∆W = 0. If T is also fixed then ∆A = ∆U − T ∆S. The second law then states that ∆A ≤ 0. Simply stated, in an irreversible process at constant temperature and volume A decreases. When the system has found equilibrium by means of this process A will have its minimum value. It is a good exercise to show this explicitly for the partition in the ideal gas, which we will do in the problem seesions. The quantity A (also often denoted by F ) is called the Helmholtz free energy. It is one of a set of four quantities called thermodynamic potentials. A summary is best expressed in a table. Symbol Name U Internal Energy H Enthalpy A Helmholtz Free Energy G Gibbs Free Energy Definition Differential (from mechanics) T dS − P dV U + PV T dS + V dP U − TS −SdT − P dV U − TS + PV −SdT + V dP The minimum principle for G then states that for all states at a fixed T and P , the equilibrium state is that for which G is a minimum. The proof is very similar to that for A: the second law states that ∆Q ≤ T ∆S or 0 ≥ T ∆S + ∆U + P ∆V , if P is held fixed. But dG = dU − T dS + P dV , so dG ≤ 0 in an irreversible process. The Gibbs free energy is very useful because most practical experiments do take place at constant temperature and pressure. As we shall see later, it is also the most important quantity in the theory of phase transitions. The minimization principles may be understood physically by starting with the fact that, in equilibrium, a completely closed (constant energy) system will maximize its entropy. As you probably know, and as we will show later, this means that the system tends toward the most probable state, consistent with having constant energy and volume. We should think of equilibrium as the state of highest statistical weight. Conversely, the system wishes to minimize its energy if volume and entropy are fixed. The other minimum principles follow from this idea, with the sole modification that other variables are held fixed. Maximization of probability is what underlies all these ideas. The thermodynamic potentials are also important for a separate reason, this time somewhat more formal. All four of them are state functions and their differentials 1 have simple forms: dU dH dA dG 0.2 = = = = T dS − P dV T dS + V dP −SdT − P dV −SdT + V dP. (1) (2) (3) (4) Maxwell Relations The first equation implies that à and à ∂U ∂S ! ! ∂U ∂V =T (5) = −P. (6) V S The second derivatives are equal, however: ∂2U ∂2U = , ∂S∂V ∂V ∂S so à ∂T ∂V ! à S ∂P =− ∂S ! S à ∂V ∂S ! P T à ∂P ∂T ! (7) . (8) V In a similar way, we find and à ∂T ∂P ! à ∂S ∂V ! à ∂S ∂P ! T = = à ∂V =− ∂T , (9) . (10) V ! . (11) P from the differential equations for the other three potentials. These four equations are called the Maxwell relations. From them we can deduce immediately some interesting facts. For example, for most (but not all) systems V increases on heating at constant pressure: (∂V /∂T )P ≥ 0. This means that (∂S/∂P )T ≤ 0. So the entropy normally decreases on compression at constant temperature, not nearly so obvious as the first statement. A second example relates to the specific heat at constant volume, Cv . Cv = Hence à V à ∂U ∂T ! ∂ ∂T ! à ∂S ∂V ! dQ dT ∂Cv |T = T ∂V à ! = V 2 =T V à = −T T ∂S ∂T à ! . (12) V ∂2P ∂T 2 ! . V (13) The right-hand-side depends only on the equation of state of the system. Once this is known, then the volume dependence of the specific heat is known. More complicated examples of uses of the Maxwell relations abound. One of the most often encountered is the equation relating the specific heat at constant volume Cv = T (∂S/∂T )v (which is easily calculated) to the specific heat at constant pressure Cp = T (∂S/∂T )P (which is easily measured). This requires a short digression into the subject of Jacobians. Consider two functions w(y, z) and x(y, z). The area dw dx in the w − x plane is related to the area dy dz in the y − z plane by dw dx = where ∂(w, x) dy dz, ∂(y, z) ¯ ³ ´ ¯ ∂w ∂(w, x) ¯¯ ³ ∂y ´ z = ¯¯ ∂x ∂(y, z) ¯ ∂y z (14) ´ ¯¯ ¯ ¯ ¯ = J. ¯ ¯ ³ ∂w ³ ∂z ´ y ∂x ∂z y (15) J is the Jacobian of the transformation from w, x to y, z. The straight brackets indicate the determinant. Note that J = J(y, z). A special case is ∂(w, x) = ∂(y, x) à ∂w ∂y ! . (16) x In case there is yet a third set of variables s, t, the Jacobian satisfies a product rule ∂(w, x) ∂(w, x) ∂(s, t) = . ∂(y, z) ∂(s, t) ∂(y, z) ³ (17) ´ ∂S is expressed in terms of We wish to make a change of variables. Cv = T ∂T V temperature and volume. We wish to transform to temperature and pressure. Write ∂(S, V ) ∂(S, V ) ∂(T, P ) =T ∂(T, V ) ∂(T, P ) ∂(T, V ) ! à ! à ! à ! #à ! "à ∂V ∂S ∂V ∂P ∂S − = T ∂T P ∂P T ∂P T ∂T P ∂V T Cv = T = T à ∂S ∂T ! −T P à ∂S ∂P ! à T ∂V ∂T ! à P ∂P ∂V ! ´ (19) (20) T Now for any substance, we define the thermal expansion coefficient α ≡ ³ (18) 1 V ³ ∂V ∂T ´ P = fractional change in volume per degree and κT ≡ − V1 ∂V = isothermal compress∂P T ibility. Both depend only on the equation of state. Putting these results together with the Maxwell relation ! à ! à ∂V ∂S =− , (21) ∂P T ∂T P 3 we find Cv = T à ∂S ∂T ! +T P à ∂V ∂T !2 à P ∂P ∂V ! . (22) T Substituting the definitions above gives Cv = C p − T V α2 , κT (23) Cp − C v = T V α2 , κT (24) or, as is more often seen, Since κT > 0,(as we shall see below), Cp > Cv . For an ideal gas à ∂V ∂T 1 κT = − V à 1 = V and ! P ∂V ∂P 1 = V ! à =− T ∂ ∂T ! P N kT Nk 1 = = , P PV T N kT 1 ∂ N kT = 2 = 1/P. V ∂P P P V (25) (26) Hence Cp − Cv = T V T −2 P = P V /T = N k. In the dimensionless heat capacities per particle cp = Cp /N k and cv = Cv /N k, we have cp − cv = 1. For solids, it is almost always the case that the volume changes little in, say, doubling the temperature at constant pressure. (Think of the metal grill in an oven). Hence (T /V )(∂V /∂T ) P << 1 and cp ≈ cv . 0.3 The Third Law This law of nature is not as firmly based or as universal as the first two laws of thermodynamics. It states that the entropy of any system at zero temperature is zero: S(T = 0) = 0. We will discuss the conditions of validity of this law later on. For now, let us just mention some of the consequences of the law. Consider heating a substance at constant volume starting at T = 0 and ending up in some final state at temperature Tf . The final entropy is S(T ) = Z Tf 0 CV dT. T Now if CV = a + bT + cT 2 + ..., then the third law implies that a = 0. The same holds true for a process at constant pressure. All heat capacities must vanish at T = 0. This is not consistent with the relation we just deduced for ideal gases that cp − cv = 1. At very low temperatures, ideal gases cannot exist. Indeed, gases do not exist at very low temperatures, as experiment has shown. The lowest boiling point for any substance at atmospheric pressure is 4.2K for helium. Near absolute zero, only helium even remains liquid; all other substances solidify. For liquids and perfectly 4 crystalline insulating solids, it is found that CV ∼ T 3 and CP ∼ T 3 . For metals and some glassy insulators, CV ∼ T and CP ∼ T. Other consequences of the third law may be found by using the Maxwell relations. Consider an expansion process at zero temperature. S = 0 at all points, hence (∂S/∂P )T = 0. But since (∂S/∂P )T = − (∂V /∂T )P , the thermal expansion coefficient à ! 1 ∂V α(T = 0) = =0 V ∂T P vanishes at T = 0. 1 1.1 Applications of Thermodynamics Adiabatic demagnetization This is a practical way to construct a refrigerator at low temperatures. The working substance is a paramagnetic salt with magnetization M subjected to an external magnetic field H (not the enthalpy). We are no longer dealing with a gas which does mechanical work. Instead, the substance does work on the external currents which produce the field. This work is dW = −HdM from electromagnetic theory. Here is the proof. A changing magnetic field creates an electric field because ~ ~ = − 1 ∂B . ∇×E c ∂t The work done by this field in a small time interval δt is δW = δt Z ~ d3 r, J~ · E and since the current density J~ is given by c ~ ∇ × H, J~ = 4π we have cδt Z ~ ·E ~ d3 r ∇×H δW = 4π ·Z ¸ Z ³ ´ ´ ³ cδt ~ · ∇×E ~ d3 r ~ ×E ~ d3 r − H = ∇· H 4π ~ 1 Z ~ ∂B = − δt H · d3 r 4π Z ∂t 1 ~ · dB ~ d3 r = − H 4π Z 1 ~ · d(H ~ + 4π m) = − H ~ d3 r 4π Z 1 ~ · dM ~, = − H 2 d3 r − H 4π 5 Figure 1: Magnetization as a function of temperature for a system of N independent spins. ~ is the total dipole moment. where m ~ is the dipole moment per unit volume and M Since H is the field associated with the external currents alone, a change in magneti~ · dM ~ . Normally H ~ and M ~ are parallel, so dW = −HdM. zation does work dW = −H Thus the first law is dU = dQ − dW = T dS + HdM . Once the first law has been determined, then everything is as for the gas except that P is replaced by −H and V by M . Our table becomes dU dH dA dG = = = = T dS + HdM T dS − M dH −SdT + HdM −SdT − M dH. (27) (28) (29) (30) The equation of state and the Gibbs free energy for a paramagnetic salt containing S = 21 spins are found experimentally to be well approximated in the range 10−2 K to 1K by the expressions: M = µB N tanh(µB H/kT ) (31) and G = −kT N log[2 cosh(µB H/kT )]. (32) Here µB is the Bohr magneton. We will derive these expressions later. The Gibbs free energy as a function of temperature for different fields looks like: The energy 6 Figure 2: Gibbs free energy as a function of temperature for various fields. crosses over from flat to downward linear at T = µB H/k. For our purposes the most important quantity is the entropy S= à ∂G ∂T ! = kN log[2 cosh µB H/kT ] − (µB H/T )N tanh(µB H/kT ). (33) H Cooling by adiabatic demagnetization proceeds by first cooling in liquid He to about 1K, in a low field (say H2 ). The system is magnetized by turning up the field, keeping the system in contact with the bath, i. e. , isothermally. Then the bath is removed so that the system is thermally isolated. Then the field is reduced to a low value (say H1 ), so that the substance is adiabatically demagnetized. The process appears graphically below. The result is that the system ends up at a temperature much lower than 1K. leaving this topic it is interesting to think about the specific heat C H = ³ Before ´ ∂S T ∂T . Looking at the slope in the graph, we see that the specific heat vanishes H at high temperatures as well as at low temperatures. The latter is a consequence of the third law, but the high-temperature behavior is a little surprising. In a gas and most other systems, the heat capacity goes to a nonzero constatn as T → ∞. C → 0 is characteristic of systems with a finite number of quantum mechanical states per particle. For spin 1/2 particles, there are precisely two states, up and down. The peak in the specific heat is called a Schottky anomaly. 7 Figure 3: Entropy as a function of temperature for different fields. Figure 4: In cooling the sysem by adiabatic demagnetization, we take it along the path shown. 8 Figure 5: Typical interparticle potential 1.2 Gas-liquid transition We will first derive the van der Waals equation of state for a gas. As a thought experiment, start with an ideal gas and turn on an interaction V (r) between particles separated by a distance r. A reasonable model for V (r) has a short-range (range r0 ) repulsion and a long-range attractive part. The depth of the attractive well is Vmin . How must the ideal gas equation of state P = N kT , V (34) which assumes that the particles do not interact, be modified ? First, the particles are excluded from a volume b ∼ r03 around each other. Since V in the ideal gas is the volume in which the particles are free to move, we should replace V by V − N b in 9 Figure 6: Isotherms of the van der Waals equation of state, including unpysical sections where the compressibility is negative. the equation of state: N kT , (35) V − Nb Second, the effect of the attractive part of the interaction is to reduce the pessure. This arises because a molecule near the wall is attracted back to its fellows and strikes the wall with a smaller velocity than if it were free, as in the ideal gas. The number of molecules striking the wall per unit time is proportional to N/V. The force on any one of them is also proportional to N/V. Hence we must subtract aN 2 /V 2 from the right-hand side. The van der Waals equation of state is P = P = N kT N2 −a 2. V − Nb V (36) Our very rough derivation does not give a reasonable way to calculate a and b - let us first treat them as parameters characterizing a given substance. Note that nothing restricts us to gases in this argument. The van der Waals equation might well apply to liquids as well. That would be the regime of high density: N/V ∼ 1/b. Let us plot some isotherms. 10 At high temperatures kT >> aN/V and low densities N/V << 1/b, this reduces to the ideal gas hyperbola. At lower temperatures, T < Tc , the curves have a peculiar | < 0, which form. There is a range of volumes where ion where κT = − V1 ∂V ∂P T means that the system should expand when pressure is applied. This is obviously impossible, and the equation of state must be incorrect in this regime. Actually we can show rigorously from the second law that the equation of state can not hold everywhere. Plot Avdw (V ) at a fixed T < Tc , where Avdw is what you get from the vander Waals equation of state. Let (V1 , V2R) be the range of volumes where ∂A κT < 0. Then, ∂V |T = −P implies that A(V ) = − P dV , where the integration takes place along an isotherm. In the range (V1 , V2 ), Avdw (V ) has downward curvature. Now consider the point at Vm = V1 + V2 . We have that Avdw (Vm ) > Avdw (V1 )/2 + Avdw (V2 )/2 = Avdw (V1 /2) + Avdw (V2 /2). (The last equation follows because A is an extensive quantity. Thus the free energy of the state at Vm could be reduced by replacing it by dividing the system into two parts with volumes V1 /2 and V2 /2 in their corresponding states. Experimentally, this is exactly what happens. Let us do an experiment at constant temperature T < Tc , pressing on a gas. It contracts along the van der Waals curve, then suddenly begins to move along a horizontal line, i. e. , P = constant. The system is highly compressible: The isothermal compressibility is infinite here: 1 κT = − V à ∂V ∂P ! →∞ (37) T along this line. What is happening at this point is liquification. The system is part gas, part liquid - the two phases coexist. That is the separation of phases that we deduced from the second law above. As we continue to apply pressure, we reach a phase of low compressibility: (κT very small), which is finally the completely liquid phase. The second law actually tells us at what pressure Pe the phase coexistence sets in. Referring back to the A(V ) plot, we see that the derivatives at V1 and V2 are equal and related to the A(V1 ) and A(V2 ) by A2 − A 1 = V2 − V 1 à ∂A(V1 ) ∂V ! = T 11 à ∂A(V2 ) ∂V ! = −Pe . T (38) Figure 7: Path followed by the system has a horizontal section. However A2 − A1 = − VV12 P dV, where P(V) is taken from the equation of state. R Comparing, we have Pe (V2 − V1 ) = − VV12 P dV . Geometrically, this means that Pe is such that the area under the actual experimental path of the system on the P − V diagram is equal to the area under the equation of state. This is known as the R Maxwell equal area construction. The Maxwell construction gives us the locus of points at which coexistence begins on the P − V diagram. Note that there is a coexistence region which separates liquid and gas but it is also possible to go around this region of the P − V diagram. Along such a path, we go continuously from gas to liquid. A gas and a liquid are not truly qualitatively different. Rather, they have quantitatively very different densities. On a P − T diagram, on the other hand, the coexistence region is a curve. Remember that P was constant for coexistence at a fixed temperature. The general rule is that coexistence region is a curve on a plot of intensive variables, and a region of finite area if one of the variables 12 Figure 8: The Maxwell construction is extensive (like V ). The endpoint (Tc , Pc ) is called the critical point. Notice that the volume changes discontinuously as the system goes across the line. When a first derivative of the free energy such as V = (∂G/∂P )T changes discontinuously at a phase transition, it is described as a ”first-order” transition. 1.3 Clausius-Clapeyron Equation There is a general equation for the phase boundary. Let us suppose that we have an expression for the free energy of the liquid A` (V, T ) and an expression Ag (V, T ) for the gas. At the boundary, we know that P` = Pg because the system is in mechanical equilbrium and T` = Tg because the system is in thermal equilibrium. Also from the Maxwell construction A` − A g = V` − V g à ∂A` ∂V ! = T à ∂Ag ∂V ! = −P` = −Pg (39) T Hence A` − Ag = −P (V` − Vg ) or A` + P V` = Ag + P Vg . By definition, this is G` (P, T ) = Gg (P, T ), i. e. , along the phase boundary the Gibbs free energy of the two phases are equal. The equilibrium phase has the lesser free energy of the two. The whole phase diagram for a simple substance might have several phases. In the generic case, two phases are separated by a line, and three phases meet at a point, because the equilibrium of two phases is determined by one equation in two unknowns: G` (Tt , Pt ) = Gg (Tt , Pt ), while the equilibrium of three phases is determined by two equations in two unknowns: Gs (Tt , Pt ) = G` (Tt , Pt ) = Gg (Tt , Pt ),where Gs is the 13 Figure 9: Phase diagram in terms of the intensive variables P and T. The coexistence region becomes a line. Gibbs free energy of the solid and (Tt , Pt ) is called the triple point. The equations determine Tt and Pt . An important aspect of the phase diagram is that the solidliquid line doesn’t end, unlike the liquid-gas line. A solid and a liquid are qualitatively different in a sense we will make precise later on. It is not possible to go continuously from one to another. The entropy also discontinuous as we cross the phase boundary. In fact there is a relation between the discontinuity in entropy ∆S, the discontinuity in the volume ∆V , and the slope of the phase boundary on a P − T diagram, given by dP/dT . We have that G` = Gg along the line. G` and Gg are themselves continuous, as we saw above. They can be differentiated along the boundary, to get the change in entropy of the individual phases: à ∂G` ∂T ! V à ∂G` dT + ∂P ! dP = T à ∂Gg ∂T ! V à ∂Gg dT + ∂P ! dP, (40) T or so −S` + V` (dP/dT ) = −Sg + Vg (dP/dT ), (41) S` − S g ∆S dP = = . dT V` − V g ∆V (42) This is more often seen in terms of the latent heat per mole ` = T ∆S/Nm and the change in molar volume ∆v = ∆V /Nm (where Nm is the number of moles of the 14 substance): ` dP = . (43) dT T ∆v This applies to any first-order transition, not only liquid-gas. Heat is given off by the system as it melts (` > 0). When dP/dT > 0, then the system exapnds on melting, which is the usual case. Ordinary ice is an exception. It contracts on melting, and dP/dT < 0 for the solid-liquid phase transition boundary. 15