Contents 1 Principles of Thermodynamics 2 Thermodynamic potentials 3 Applications of thermodynamics

Transcription

1 Contents 1 Principles of Thermodynamics Introduction State variables and differential forms Equation of state Zeroth law Internal energy Work First law Second law Carnot s cycle Third law Problems Thermodynamic potentials 15.1 Fundamental equation Internal energy U and Maxwell relations Enthalpy H Free energy F Gibbs function G Grand potential Ω Thermodynamic responses Thermodynamic stability conditions Stability conditions of matter Thermodynamic potentials in electromagnetism Problems Applications of thermodynamics Classic ideal gas Free expansion of gas Mixing entropy Dilute solution, osmosis Chemical reaction Phase equilibrium Phase transitions and diagrams Coexistence Van der Waals equation of state Problems v

2 vi CONTENTS 4 Classical phase space Phase space and probability density Flow in phase space Microcanonical ensemble and entropy Problems Quantum-mechanical ensembles Density operator and entropy Density of states Energy, entropy and temperature Problems Equilibrium distributions Canonical ensemble Grand canonical ensemble Connection with thermodynamics Thermodynamic fluctuation theory Reversible minimal work Problems Ideal equilibrium systems Free spin system Classical ideal gas Diatomic ideal gas Statistics of bosons and fermions Problems Bosonic systems Bose gas and Bose condensation Black body radiation Lattice vibrations Problems Fermionic systems Conduction electrons in metals Magnetism of degenerate electron gas Problems Phase transitions Description of phase transitions Landau theory Ginzburg Landau theory Fluctuations in Landau theory Problems Index 16

3 1. Principles of Thermodynamics 1.1 Introduction Thermodynamics gives a phenomenological and very general description of matter largely independent of models of microscopic structure (which where practically nonexistent at the time of foundation of thermodynamics in the 19th century). It is based on very few basic laws plus rules of calculus. Properties of matter or concrete systems are taken from outside (experiment, statistical mechanics). System. Macrophysical entity under consideration, may interact with its environment. It is often homogeneous or consists of homogeneous phases. The usual classification according to the possibility of exchange of energy and matter between the two goes as follows: Open system: both energy and matter may be exchanged. Closed system: particle number(s) fixed, energy may be exchanged. Isolated system: no exchange of matter or energy. Thermodynamic equilibrium. State of matter without any macroscopic changes or flows. A genuine equilibrium state is unambiguously determined by externally imposed state variables like pressure, volume, electric and magnetic fields. There are no memory effects like hysteresis. Traditionally, in the description of thermodynamic equilibrium there are three different equilibria: mechanical equilibrium: no changes of form or other processes accompanied by production of macroscopic mechanical (or electromagnetic) work; chemical equilibrium: no changes in the macroscopic chemical composition of the system; thermal equilibrium: no macroscopic energy flows in a system in mechanical and chemical equilibrium. In plain words: no heat flows. In local thermodynamic equilibrium macroscopic subsets of the system are in equilibrium, but in neighbouring subsystems the equilibria are different, so that the system is not in equilibrium as a whole. Currents, heat flow etc. may occur; this is the realm of hydrodynamics. In most practically important cases local equilibrium is reached in macroscopically short time. State variables. State variables are parameters needed for characterization of the equilibrium state. Usually there is only a handful of them, 1

4 1. PRINCIPLES OF THERMODYNAMICS in many cases like the prototypical one-component gas two is enough to determine the equilibrium state, in which the rest are then functions of these parameters, state functions. State variables are either extensive or intensive, the former being proportional to the number of particles (the volume V, particle number N, internal energy U, entropy S, magnetic moment m = d 3 r M(r) etc.), whereas the latter (the temperature T, the pressure p, the chemical potential µ, magnetic field strength H) are independent of the number or particles. In thermodynamic differential forms these variables appear as conjugate pairs of extensive and intensive variables. For quantities like energy and entropy the extensiveness requires weakness of interaction energy (or correlations) between macroscopic subsystems of the original system in comparison with the "bulk" quantities prescribable to the subsystems themselves. Gravity might cause problems in this respect at very large scales. Electromagnetic interaction is usually screened in matter and thus of short range. Process. A change of state is called a process in thermodynamics. In a reversible process the direction may be inverted in the whole universe (system plus environment). These processes are always quasistatic, i.e. so slow that the state of the system is infinitesimally close to thermodynamic equilibrium. Not all quasistatic processes are reversible, however. An irreversible process is often a sudden or spontaneous change (e.g. mixing of gases, explosion), during which the system may be far from equilibrium and the description by the state variables is no sufficient. An irreversible process may occur quasistatically, though. A cyclic process (cycle) consists of repeating periods during which the system always returns to the initial state. 1. State variables and differential forms State variables are macroscopic quantities related to the equilibrium. Not all of them are independent in equilibrium, though. Once independent variables are chosen, the rest are unambiguous functions of them, say p = p(t,v,n), U = U(T,V,N), S = S(T,V,N) etc. y 1 y 1= x x Figure 1 1: Examples of processes leading from state 1 to state. When the change is infinitesimal, the rules of calculus yield the following relation between the differential of a function and the differentials of

5 1.. STATE VARIABLES AND DIFFERENTIAL FORMS 3 indepent variables: ( ) p dp = T V,N ( ) p dt + V T,N ( ) p dv + N V,T dn. This implies that in a cyclic process the net change vanishes: dp = du = = Differential and differential form. Consider the differential form df F 1 (x,y)dx + F (x,y)dy, (1.1) where F 1 and F are given functions. An example familiar from mechanics is the differential form of work exerted by a force F on a body: dw = F dr = F x dx + F y dy + F z dz. The notation df in (1.1) means that the differential form is not necessarily a differential, therefore df may depend on the integration path. If the 1 condition F 1 / y = F / x holds, then df = df(x,y) is a differential (often referred to as the exact differential in this context). Then the integral df = F() F(1) is independent of the path and F 1 1 (x,y) = F(x,y)/ x ja F (x,y) = F(x,y)/ y are coordinates of the gradient of some function F. Integrating factor. If the form df = F 1 dx + F dy is not a differential, in case of two variables a function integrating factor λ(x, y) may be found such that, in a vicinity of the point (x,y) the condition λ df λf 1 dx + λf dy = df holds, which implies (λf 1 )/ y = (λf )/ x. Both the integrating factor λ and the function f are then state variables. In case of three or more variables the integrating factor may not exist, in general. In thermodynamics, however, the integrating factor of the differential form of heat always exists, this is partially the content of the second law. Legendre transform. Legendre transform generates changes of variables between conjugate variable pairs. Consider, for simplicity, the function f(x) and define the variable conjugate to x as y df(x) dx. (1.) The Legendre transform of f is the following function of y: g(y) f(x) yx, (1.3)

6 4 1. PRINCIPLES OF THERMODYNAMICS where on the right-hand side x is expressed as a function of y from relation (1.). Direct calculation yields so that df = ydx and dg = xdy. dg(y) dy = x, (1.4) Mathematical identities. In thermodynamics, fixed variables are usually indicated explicitly when calculating partial derivatives. This is because several sets of independent variables are in wide use and infer different physical meaning for partial derivatives in different sets. Examples of useful relations for various changes of variables are listed below. Jacobi determinants. The use of Jacobi determinants (u, v) (x,y) = u u x y v v x y is often convenient when carrying out changes of variables in differential relations. This is due to the properties ( ) (u, y) u (x,y) = (u, v), x (x,y) = (u,v) (s, t) (s, t) (x,y), y valid for Jacobians of arbitrary order and easily checkable by direct calculation for determinants. Example 1.1. Consider the function of two variables F(x, y). If by some reason we want to use the pair (x, z) as independent variables, we may writw F(x, y) = F x, y(x, z). The chain rule then yields F = x z F z F + x y 1.3 Equation of state x = F y F y x x y z y, x z Equation of state expresses the relation of state variables of the system in equilibrium. It is usually written in a form involving mechanical variables and the temperature. Equation of state does not usually include internal energy or other extensive variables of dimensions of energy, and in this sense the equation of state does not give a complete thermodynamic description of the system. A few widely used equations of state are listed below. x.

7 1.3. EQUATION OF STATE 5 Classical ideal gas. The equation of state of classical ideal gas is pv = NT. (1.5) Here, p = pressure, V = volume, N = number of molecules and T = absolute temperature. Mixture of ideal gases: Equation of state remains the same pv = NT, with N = i N i. The total pressure may be expressed as p = i p i, where p i = N i T/V = partial pressure of the ith component. Virial expansion of real gas. The equation of state of the ideal gas may be amended so that the intermolecular interaction is taken into account. Denote the (particle) number density by n N/V. In the limit of small density the pressure may be expanded in powers of the density (the virial expansion p = T [ n + n B (T) + n 3 B 3 (T) + ], (1.6) where the virial coefficients B n depend on the temperature only. Curie s law. Magnetic field strength H, magnetic induction B and magnetization M are related as. B = µ (H + M). Further the magnetic moment of the system shall often be denoted by m; in case of homogeneous field then m = V M. The magnetic equation of state expresses the dependence of magnetization M on the field strength H. Many paramagnetic materials (no spontaneous magnetic ordering) obey Curie s law M = C T H, (1.7) where C is a material constant proportional to the number density of paramagnetic atoms. Responses. Thermodynamic responses describe the reaction of state variables to the change of other state variables. They usually are easily measurable quantities. The equation of state determines mechanical responses like thermal expansion coefficient α p = 1 ( ) V, (1.8) V T p,n isothermal compressibility κ T = 1 V ( ) V = 1 p T,N n ( ) n p T, (1.9)

8 6 1. PRINCIPLES OF THERMODYNAMICS or isothermal susceptibility χ T = ( ) M H T (1.1) of magnetic material. Under an adiabatic (thermally isolated) change the responses are adiabatic compressibility κ S = 1 V ( ) V = 1 p S,N n ( ) n p S (1.11) and adiabatic susceptibility χ S = ( ) M. (1.1) H S 1.4 Zeroth law Zeroth law of thermodynamics is the observation that there is a quantity called temperature characterizing the thermal equilibrium and a thermometer to measure and compare temperatures. This comparison is transitive: if two bodies are separately in equilibrium with a third one they are in equilibrium with each other. 1.5 Internal energy In thermodynamics internal energy is the total energy of the system at rest. Usually the potential energy of the system in an external field is excluded. Then internal energy consists of the kinetic energy of relative motion of particles, energy of their interaction and structural energy of the particles. Due to interaction between the system and its environment care has to taken in dividing the world in the system and the environment, especially when long-ranged interactions occur. At the present state of our knowledge of the structure of matter it is quite obvious that the internal energy of the system is a state variable and thus an unambiguous function of the state. It is also clear that internal energy may determined even in systems which are not in a state of thermodynamic equilibrium. 1.6 Work Work is energy exchange between the system and environment which may be described in terms of work of macroscopic mechanics and electromagnetic theory.

9 1.6. WORK 7 There are different sign conventions. Here, the elementary work (differential form of work) dw is the work exerted to the environment by the system. In this case positive work means loss of energy by the system. In the paradigmatic SVN - system 1 the work related to the change of the volume is dw = pdv. (1.13) The work related to the surface energy of a liquid may be written as dw = σ da, (1.14) where σ = surface tension and A = free surface area. With positive surface energy σ > and the surface tension tends to decrease the area. An elastic deformation gives rise to the work form: dw = F dl, (1.15) where F = the force stretching the rod and L = the rod length. The tension is σ = F/A = force/cross-section area. According to Hooke s law σ = E(L L )/L, where E is Young s modulus and L the rest length of the rod. The general expression of the differential form of work is dw = i f i dx i = f dx, (1.16) where f i are the coordinates of the generalized force and X i the coordinates of the generalized displacement. Work in electromagnetism. Treatment of energetic quantities in a system in electromagnetic field requires considerable care in the definition of the system, because usually the introduction of polarizable or magnetizable body in an electromagnetic field changes the field everywhere, not only in the body itself. Unambiguous definition may be obtained, if the whole electromagnetic field is considered a part of the system. In this case the elementary work required for a change of fields in the form familiar from electrodynamics (with the sign corresponding to our convention) dw = d 3 r (E dd + H db) (1.17) may be interpreted as the work carried out by the system. With the aid of a special experimental setup in some cases it is possible to arrive at a situation in which the polarized body does not affect the fields E and H and the field energy outside the body may be dropped from the energy balance of the system so that dw = V (E dd + H db), (I) (1.18) 1 One-component isotropic homogeneous material with the state variables S, V, N (natural variables of the internal energy), often referred to as the pvt system as well.

10 8 1. PRINCIPLES OF THERMODYNAMICS when the volume V is small enough so that the fields may be regarded as uniform. If this is not possible, the energy corresponding to fields with the same sources (free charges and conducting currents) but without the polarizable body is nevertheless often subtracted from the energy related to the system including this body. The point here is that thermodynamics is brought about in the problem by the presence of polarizable material. Without it, the problem would be that of "pure" electrodynamics. For simplicity, consider still the case in which the fields E and H are the same both with and without the polarizable body. In uniform fields then the total energy might be written as E tot = U + V ( 1 ε E + 1 µ H ) (1.19) thus excluding the energy of the "empty space" from the internal energy of the system considered. In this case the differential form of electromagnetic work related to the change of the internal energy defined as (1.19) assumes, according to (1.18) the form dw = V (E dp + µ H dm). (II) (1.) This convention is often used in condensed matter and solid state physics. 1.7 First law The first law of thermodynamics is the law of conservation of energy. de = dq dw. (1.1) Here, de is the differential of the energy of the system. With the usual convention of thermodynamics, it may be identified by the differential of the internal energy: de = du, provided the momentum, angular momentum and the potential energy in external field of the system remain unaltered. If the particle number may change, the chemical potential µ is introduced by the definition du = dq dw + µdn. In a general form for several particle species the first law is du = dq f dx + i µ i dn i. (1.) Heat capacity. The ability of a body to receive heat is described by the heat capacity C A = Q T, (1.3) A

11 1.8. SECOND LAW 9 where the subscript A refers to fixed variables, e.g.: C V, C p. Specific heat is the heat capacity per unit mass. Heat capacities are usually easy to measure contrary to the internal energy. Cyclic process. Cyclic processes (cycles) are especially important in the theory of heat engines. In a cycle the system return to its initial state again and again after certain periodic stages. In a simple SVN system, in which du = dq pdv, the area enclosed by the curve describing the process in the (V,p) plane is pdv = W. (1.4) Since du =, the work during a cycle is equal to the difference of the amounts of heat received and delivered by the system. The thermal efficiency of a cycle is η = W/ Q +, where Q + is the amount of heat received by the system during a cycle. 1.8 Second law From the formal point of view the second law states two things: for the differential form of heat there is an integrating factor (the temperature) giving rise to the extensive state variable entropy S; in a reversible process: In an irreversible process ds = ds > dq T. (1.5) dq T. (1.6) There are several traditional equivalent formulations of the second law: (1.8a) Heat cannot be transferred from a colder heat reservoir to a warmer heat reservoir without any other changes. (Clausius) (1.8b) There is no cyclic process with the sole result of transferring the heat received to work. (Kelvin) (1.8c) Of all heat engines working between the temperatures T 1 and T the Carnot engine has the highest efficiency. (Carnot) All these statements are equivalent in the sense that each of them yields the others. Here, we shall not dwell on demonstration of this equivalence, however. The first law in a reversible process may now be cast in the form du = T ds f dx + i µ i dn i. (1.7)

12 1 1. PRINCIPLES OF THERMODYNAMICS 1.9 Carnot s cycle The notion of entropy may be approached by analyzing Carnot s cycle consisting of four reversible stages(fig. 1 ): a) isothermal T Q > b) adiabatic T T 1 Q = c) isothermal T 1 Q 1 > d) adiabatic T 1 T Q = p T d a Q Q W b Q 1 c V Q 1 T 1 Figure 1 : Carnot s cycle. The thermal efficiency of the process is η = W Q = 1 Q 1 Q. (1.8) Since the cycle is reversible, it may be also used as a heat pump. The efficiency of Carnot s cycle depends only on the temperatures T 1 and T of the heat reservoirs but not on the details of realization. T 3 Q 3 W 3 T Q W 1 T 1 Q 1 Figure 1 3: Determination of absolute temperature scale.

13 1.9. CARNOT S CYCLE 11 Absolute temperature. An absolute temperature scale may be determined with the aid of a serial connection of Carnot s cycles as in Fig The efficiency depends only on the reservoir temperatures, therefore 1 η = Q out Q in = f(t max,t min ). (1.9) From relations f(t 3,T ) = Q / Q 3, f(t,t 1 ) = Q 1 / Q, f(t 3,T 1 ) = Q 1 / Q 3 the functional identity follows f(t 3,T )f(t,t 1 ) = f(t 3,T 1 ), which has to hold for all T i. The simplest choice is f(t,t 1 ) = T 1 T (1.3) which defines the thermodynamic (absolute) temperature scale up to the choice of the unit. For the efficiency of Carnot s cycle this yields η = 1 T min T max. (1.31) Consider now a cyclic (quasistatic) process divided to a large number of subprocesses with temperatures T i and the amount of heat received Q i. Imagine that these portions of heat are transferred by Carnot engines working between the system a huge heat reservoir at the temperature T > T i, so that the ith engine receives the heat Q i from the reservoir. Calculate now the work done in one cycle by the system and all the Carnot engines. In one cycle the work done by the system equals the heat received: W system = i Q i. (1.3) The work of ith Carnot engine is W Ci = Q i Q i, so that the total work is equal to the heat received by our combined system from the heat reservoir: W total = i Q i + i ( Q i Q i ) i Q i = Q, (1.33) which cannot be positive according to Kelvins statement of the second law, since the combined system consisting of the original cycle and the auxiliary Carnot machines did not give any heat to a heat reservoir at a temperature lower than T. Now Q i /T = Q i /T i. Therefore, replacing the sum over subprocesses by a contour integral in the state variable space, we arrive at the Clausius inequality dq T. (1.34)

14 1 1. PRINCIPLES OF THERMODYNAMICS For any reversible process this is an equality, which means that the integrand is a differential of some state variable. This state variable is the entropy S and dq T = ds. If a finite portion of our process is reversible, say from state 1 to state, the corresponding part of the contour integral in Clausius s inequality (1.34) yields the difference between the values of entropy in these states: S S 1 = and Clausius s inequality takes the form 1 dq T, (1.35) S S 1 1 dq T. (1.36) In particular, in a thermally isolated system the entropy cannot decrease. The second law seems to be in contradiction with the time-reversal invariance of the basic microscopic laws of physics, since it establishes a preferred direction of processes. The origin of this time-reversal symmetry breaking in macroscopic physics remains unclear. 1.1 Third law The third law thermodynamics, Nernst s law, states that the entropy of an equilibrium system vanishes, when the temperature approaches the absolute zero: lim S =. (1.37) T In classical thermodynamics the conjecture is that this limit exists, the particular value is explained in quantum statistical physics Problems Problem 1.1. Show that x y z y z x z = 1 (1.38) x y

15 1.11. PROBLEMS 13 and that for any function F x = y z F y F x z z. (1.39) Problem 1.. Which of the following differential forms are differentials? Find the integrating factor for those differential forms which are not differentials. (a) d u = x 4 y dx + y dy. (b) d u = (1x + 6y)dx + 6x dy, (c) d u = 1y dx + 18xy dy. Problem 1.3. Define the Legendre g(y) transform of the function f(x) as g(y) = f(x) xy, y = df dx, where on the right-hand side x is assumed to be expressed as a function of y from the condition y = f. (a) Show that d f d g = 1. dx dy (b) Construct the Legendre transform of the function f = 1 x. (c) Construct the Legendre transform of the function f = ax ln x b, where a and b are positive constants. Problem 1.4. Thermal expansivity α and isothermal compressibility κ of matter are defined as α = 1 V V T p ; κ = 1 V V p T. Show that α κ = p T T p ; α p κ = T V. Problem 1.5. Calculate the virial coefficients B, B 3 and B 4 of Clausius matter. Clausius equation of state is p + an (V bn) = NT, T(V + cn) where a, b and c are positive experimental constants and N the total number of particles. Can you determine all the virial coefficients for this matter?

16 14 1. PRINCIPLES OF THERMODYNAMICS Problem 1.6. Consider a spherical capacitor with external radius b and internal radius a charged to an initial charge Q. The capacitor is halffilled by a dielectric substance of permittivity ε in such a way that the dielectric fills the space between the plates to one side of a cross-section plane dividing the spheres in two halves, while to the other side of the plane the capacitor is empty. Express the differential form of work in terms of electric induction D and electric field E. Calculate the work exerted on the capacitor, when the charge is increased by an infinitesimal amount δq. Proceed by subtracting the differential form of work required to increase the charge by δq from Q in an empty capacitor. Express the result in terms of the polarization vector P. Problem 1.7. Consider the same capacitor but now with an initial potential difference φ between the plates. Express the differential form of work in terms of electric induction D and electric field E. Calculate the work exerted on the capacitor, when the potential difference between the plates is increased by an infinitesimal amount δφ. Proceed by subtracting the differential form of work required to increase the potential difference by δφ from φ in an empty capacitor. Express the result in terms of the polarization vector P. Problem 1.8. Experimentally it has been found that a rubber band obeys: F L T = a T L, L L F T L = a 1 L 3 L, L L where F is the tension and the constant a and the rest length of the band L are parameters. (a) Calculate ( L/ T) F and give a physical interpretation. (b) Show that df = LFdL + TFdT is a differential. (c) Determine the equation of state F = F(L, T) of the band. Problem 1.9. In a perfect gas the internal energy obeys the relation du = C V dt. Find the equation of state for such a gas in a process, in which the heat capacity C is a constant (polytropic process). Problem 1.1. Stirling s cycle consists of two isotherms at T 1 and T and two isochores (processes with constant volume) at V 1 and V. Calculate the coefficient of thermal efficiency of Stirling s cycle working on the ideal gas. Compare with the thermal efficiency of Carnot s cycle.

17 . Thermodynamic potentials.1 Fundamental equation Thermodynamic potentials are extensive state variables of dimensions of energy. Their purpose is to allow for simple treatment of equilibrium for systems interacting with the environment. In thermodynamics all variables are either extensive or intensive. Mathematically this may expressed in homogeneity relations with respect to the system size. Thus, extensive variables (e.g. N, V, U, S,...) are first-order homogeneous functions, whereas intensive variables (like p, T, µ,...)are independent of the size of the system. Natural variables. are those whose differentials appear in the differential form of the first law: du = T ds pdv + µdn so that S, V ja N are natural variables of internal energy. With all intensive variables fixed, extensivity of all these variables means U(λS,λV,λN) = λu(s,v,n). (.1) Differentiating both sides with respect to the auxiliary parameter λ and putting λ = 1 thereafter we arive at the identity (Euler equation for homogeneous functions): U = S ( ) U + V S V,N ( ) ( ) U U + N. V S,N N S,V From the first law it follows that ( ) U = T, S V,N ( ) U = p, V S,N ( ) U N S,V = µ. Thus, we arrive at the fundamental equation U = TS pv + µn. (.) 15

18 16. THERMODYNAMIC POTENTIALS. Internal energy U and Maxwell relations The first law du = T ds pdv + µdn yields ( ) U T =, (.3a) S V,N ( ) U p =, (.3b) V S,N ( ) U µ =. (.3c) N S,V From the definition of heat capacity it follows that ( ) ( ) dq U C V = = dt T V,N V,N. (.4) Since U may be assumed to be single-valued smooth state variable, result of iterative differentiation does not depend on order T/ N = ( U/ S)/ N = ( U/ N)/ S = µ/ S. This procedure gives rise to Maxwell relations: ( ) ( ) T p =, (.5a) V S,N S V,N ( ) ( ) T µ =, (.5b) N S,V S V,N ( ) ( ) p µ =. (.5c) N V S,V These and similar relations for other thermodynamic potentials are often useful in expressing differential relations in terms of response functions and state variables. In an irreversible process T δs > δq = δu + δw µδn, therefore S,N δu < T δs pδv + µδn. (.6) In an irreversible process with fixed S, V and N the internal energy decreases. Thus, in equilibrium U assumes the mimimal value with S, V and N fixed (implying, of course, that something else may change). If some other work may be done in a reversible process, then U = R = W free, where the free work W free = R is the work the system may carry out in given circumstances. If the process is irreversible, then W free U (.7) even if (S,V,N) are kept fixed. Thus, the minimal work needed to bring about the change of internal energy U is R = U.

19 .3. ENTHALPY H 17.3 Enthalpy H Other thermodynamic potentials are Legendre transformsof the internal energy U(S,V,N) with respect to natural variables S, V or N. Enthalpy (or the heat function) is obtained by using p instead of V : H U + pv. (.8) The differential follows from the definition and the first law: dh = T ds + V dp + µdn. (.9) Natural variables are (S,p,N). From the definition of heat capacity it follows that ( ) ( ) dq H C p = =. (.1) dt T p,n p,n From the expression for dh three more Maxwell relations follow: ( ) ( ) T V =, (.11a) p S,N S p,n ( ) ( ) T µ =, (.11b) N S,p S p,n ( ) ( ) V µ =. (.11c) N p S,p Ia an irreversible change δq = δu + δw µdn < T δs. Substitution of δu = δ(h pv ) yields δh δ(pv ) + δw µdn < T δs, i.e S,N δh < T δs + V δp + µδn. (.1) If in the process S, p and N remain constant, spontaneous changes drive the system to the state with mimimum enthalpy. Many practically important processes (phase transitions, chemical reactions etc) take place at constant (ambient) pressure. If the conditions include also thermal isolation, the enthalpy is the natural energy quantity to use. In hydrodynamics adiabatic flow is a popular approximation. Then the specific (per unit mass) internal energy u appears in the energy equation only in the combination u+p/ρ = h, which is the specific enthalpy and thus the natural energy variable. When (S,p,N) are fixed, the portion of the energy of the system freely exchangeable for work obeys the condition W free H. (.13) The mimimum work required to bring about H is thus R = H.

20 18. THERMODYNAMIC POTENTIALS Joule Thomson process. Consider thermally isolated forced flow of gas through a throttle valve or a porous wall. Movement of pistons is devised to keep pressures p 1 > p fixed. Although the flow is far from equilibrium and not reversible, a hypothetical reversible process between the same states is useful, because state variables are process-independent. For the transfer of an infinitesip mal quantity of matter the work by the system is dw = p dv + p 1 dv 1. Initially V 1 = V i and V =. Finally V 1 = and V = V f. For constant pressures the work is 1 p W = dw = p V f p 1 V i. Figure 1: Flow through porous wall. Thermal isolation means Q =, therefore U = W. From this it follows U f + p V f = U i + p 1 V i. Thus, the quantity U + pv, i.e. the enthalpy H remains constant: the process is isenthalpic, H = H f H i =. (.14) Imagine now a reversible isenthalpic process of decreasing the pressure by infinitesimal steps. The response of the temperature to this is given by the Joule Thomson coefficient ( ) T. (.15) p H To express this coefficient in terms of already introduced quantities, use the Jacobi determinant method. It is good policy to introduce variables which are the natural variables of the thermodynamic potential appearing in this definition, because then its first derivatives are state variables. Thus, ( ) T = (T,H) p H (p,h) = (p,s) (p, H) [ ( T = 1 ) ( ) H T p S S S p p (T, H) (p,s) = 1 (T, H) T (p, S) ( ) ( ) ] T H = S p Using the Maxwell relation (.11a) rewrite ( ) ( ) T V = = T ( ) ( ) V S p S C p S T Thus, we arrive at the expression ( ) [ ( V T = T ) V p H C p T p T p p ] S p ( ) T V. (.16) p S C p = T ( ) V. C p T p = V C p (Tα p 1). (.17) The latter form follows form the definition of the thermal expansion coefficient : α p = V 1 ( V/ T) p.

21 .4. FREE ENERGY F 19 In the process the pressure decreases, so that the gas is cooled, if Tα p > 1, or heated, if Tα p < 1. For the ideal gas the Joule Thomson coefficient vanishes, so that the temperature of an ideal gas remains the same. For real gases the coefficient is positive below a certain pressure-dependent inversion temperature, so that the gas is cooled. Thus, the Joule-Thomson process may be and is used for cooling and eventually liquifying gases..4 Free energy F The Legendre transform of the internal energy with respect to S yields the free energy (Helmholtz free energy): F = U S( U/ S) V,N i.e. The corresponding differential is F U TS. (.18) df = S dt pdv + µdn. (.19) The natural variables are T, V and N. The Maxwell relations are ( ) ( ) S p =, (.a) V T,N T V,N ( ) ( ) S µ =, (.b) N T,V T V,N ( ) ( ) p µ =. (.c) N V T,V As before, for an irreversible process T,N δf < S dt pδv + µδn. (.1) Thus, with fixed T, V and N, the system evolves towards the minimum of the free energy. For the free work at fixed T,V,N it follows W free F. (.) Free energy is an extremely important tool in statistical mechanics: in many cases it is the natural macroscopic quantity to calculate for a given microscopic model..5 Gibbs function G The Legendre transform of U with respect to both S and V leads to the Gibbs function (Gibbs free energy) G U TS + pv, (.3)

22 . THERMODYNAMIC POTENTIALS with the differential dg = S dt + V dp + µdn. (.4) The natural variables are T, p and N and the Maxwell relations ( ) ( ) S V =, (.5a) p T,N T p,n ( ) ( ) S µ =, (.5b) N T,p T p,n ( ) ( ) V µ =. (.5c) N p T,p With fixed T, p and N, a non-equilibrium system evolves towards the minimum of the Gibbs function: and the free work is T,N δg < S δt + V δp + µδn, (.6) W free G. (.7) The Gibbs function is a suitable thermodynamic potential for systems which change at fixed pressure and temperature (no thermal isolation). Since these parameters are perhaps most easily of all adjustable, the Gibbs potential has a wide scope of applications both in physics and chemistry. From the fundamental equation it follows that G = µn, (.8) i.e. the chemical potential is the Gibbs function per particle in a singlespecies system. Since from (.35) it follows that dg = µdn + Ndµ and, taking into account the alternative form (.4), we arrive at the Gibbs Duhem equation dµ = S N dt + V dp, (.9) N showing that the natural variables of the chemical potential are T,p..6 Grand potential Ω The grand potential is also an important quantity for calculations in statistical mechanics when the number of particles cannot be fixed. The definition is Ω U TS µn (.3) leading to the differential dω = S dt pdv N dµ, (.31)

23 .7. THERMODYNAMIC RESPONSES 1 showing that the natural variables are T, V and µ. The Maxwell relations are ( ) ( ) S p =, (.3a) V T,µ T V,µ ( ) ( ) S N =, (.3b) µ T,V T V,µ ( ) ( ) p N =. (.3c) µ V T,V In an irreversible change the inequality T,µ δω < S δt pδv N δµ, (.33) holds revealing that in a process with fixed T, V and µ the system tends to state with the minimum of Ω. The free work under these conditions is From the fundamental equation it follows that W vapaa Ω. (.34) Ω = pv (.35) revealing that knowledge of Ω is tantamount to knowing the equation of state (although in non-standard variables)..7 Thermodynamic responses Thermodynamic responses have the form of partial derivatives ( K/ A) B,C,... and reveal the effect of an infinitesimal change of a state variable (A) to some quantity (K) describing the system at equilibrium. Usually these are (the most) directly measurable quantities. Coefficient of (volumninal) thermal expansion. Definition α p = 1 ( ) V. (.36) V T p,n In terms of number density n = N/V : α p = 1 n ( ) n. (.37) T p In isotropic substance the coefficient of linear thermal expansion is one third of this, since a small change of volume is three times the change of length.

24 . THERMODYNAMIC POTENTIALS Isothermal compressibility. Reaction to pressure at constant temperature κ T = 1 ( ) V = 1 ( ) n. (.38) V p T,N n p T Adiabatic compressibility. Pressure acting in thermal isolation κ S = 1 V ( ) V = 1 p S,N n This quantity determines the speed of sound: c s = ( ) n p S,N. (.39) 1 mnκs. (.4) Here, m is the particle mass and mn = mn/v is the mass density. Isochoric heat capacity. Definition by reversible process, thus Q = T S, and the heat capacity in general C Q T = T S condition T. condition Due to the chosen scale of temperature, heat capacities are dimensionless. Specifically at constant volume we obtain ( ) S C V = T. (.41) T V,N Since under these conditions du = T ds pdv +µdn reduces to du = dq = T ds and, on the other hand, S = ( F/ T) V,N, we arrive at relations C V = ( ) ( U ) F = T T V,N T V,N Isobaric heat capacity. Analogously. (.4) C p = T ( ) S T p,n (dh) p,n = T ds = dq, and S = ( G/ T) p,n yield C p = ( ) H = T T p,n, (.43) ( ) G T p,n. (.44)

25 .7. THERMODYNAMIC RESPONSES 3 Connections. Relations between heat capacities under different conditions are due to differences in work. For C p, change variables ( ) ( ) ( ) ( ) ( ) S S(V (p,t),t) S S V = = +. T T T V T p p Free energy Maxwell ( S/ V ) T = ( p/ T) V (.a) yields ( ) ( ) p V C p = C V + T. (.45) T T The change of variables in ( p/ T) V gives rise to the result V V p C p = C V + V T α p κ T. (.46) Compressibility is positive in stable matter, therefore C p > C V. Construction of potentials. An equation of state like p = p(t, V ) and a thermal response, say C V, are required to this end. Consider, for instance, van der Waals matter: ( ) p + a N V (V Nb) = NT. The heat capacity C V is directly a partial derivative of the internal energy: ( ) U C V =. T To calculate the other one, change variables ( ) U = (U,T) V T (V,T) = (U,T) (V, S) (V, S) (V,T) ( ) [( ) ( ) S U T = T V S V S V V T ( ) ( ) ] U T. (.47) S V V S Here, derivatives of U are p and T, this is the point of introducing the natural variables of U. Thus ( ) ( ) ( ) U S p = p + T = T p, V T V T T V where the free-energy Maxwell (.a) has been used once more. This relation shows, in particular, that for the van der Waals equation of state ( ) CV = U V V T =, T i.e. C V is a function of temperature only! Then the integration is simple: [ ( ) ] p U(T,V ) = C V dt + T p dv = C V (T)dT a N T V. V p

26 4. THERMODYNAMIC POTENTIALS.8 Thermodynamic stability conditions Let the near-equilibrium system be divided to (semi)macroscopic subsystems (labeled by index α) each in a local equilibrium, but with different pressure, temperature etc. in neighbouring subsystems. Extensive quantities remain additive: S = α S α, V = α V α, U = α U α. Let N jα be the particle number of species j in the subsystem α. Then N j = α N jα for the jth species. Due to local equilibrium S α = S α (U α,v α, {N jα }). α p α T α V α Figure : System near equilibrium For a small change of S α S α = 1 T α U α + p α T α V α j µ jα T α N jα. Assume a system isolated as a whole, then U, V and N j remain constant. To simplify notation, consider two subsystems: α = A, B. Conservation laws yield U B = U A, V B = V A and N jb = N ja. Thus, S = S α α ( 1 = 1 ) ( pa U A + p ) B V A T A T B T A T B j ( µja µ ) jb N ja. T A T B At equilibrium S = identically. Since the fluctuations U A, V A and N ja are arbitrary, the equilibrium conditions: T A = T B p A = p B (.48) µ ja = µ jb follow. Thus, in equilibrium the temperature is the same everywhere, as well as the pressure (provided no external fields impose inhomogeneity) and the chemical potential for each particle species. The conditions hold also in the case system consists of different phases (constant pressure requires flat interfaces, however)..9 Stability conditions of matter In stable equilibrium the entropy must be at maximum. To analyze this, the second variation of the entropy with respect to { U α, V α and N jα } may be used.

27 .1. THERMODYNAMIC POTENTIALS IN ELECTROMAGNETISM 5 Let T, p and {µ j } be the common equilibrium values. For simplicity, assume one species. In Taylor expansion at the equilibrium point S = S + ds + 1 d(ds) the linear term vanishes. Since d X = for any independent variable X, we obtain S = S + 1 d(ds) = 1 ( 1 d α T α ) (du α + p α dv α µ α dn α ) + 1 α 1 T α (dp α dv α dµ α dn α ). Here, du α + p α dv α µ α dn α = T α ds α, so that (denote dx X) S tot S S = 1 ( T α S α p α V α + µ α N α ). (.49) T α The condition of a stable equilibrium is that this expression is negative definite. Since any subsystem α is at local equilibrium, only three of fluctuations of the quantities T α, S α, p α, V α, µ α, N α are independent, the rest must be expressed as functions of the chosen three. Let N α =. Then only two independent variables remain. Choose T α and V α and express S α as p α functions thereof. Maxwell relations allow for simplification and the result is S tot = 1 T α { CV,α T ( T α) + 1 κ T V α ( V α ) }. (.5) Another possibility V α = with T α and N α as independent variables leads to { S tot = 1 ( ) } C V,α µ T T ( T α) + ( N α ). (.51) N α T,V α α From these expressions it is readily seen that the total entropy is at maximum, when the following stability conditions hold: : C V > κ T > ( ). (.5) µ > N T,V Otherwise the equilibrium is unstable and small spontaneous disturbances give rise to growing changes which lead to another state..1 Thermodynamic potentials in electromagnetism The starting point here is the basic differential form of work (1.17) dw = d 3 r (E dd + H db),

28 6. THERMODYNAMIC POTENTIALS whose addition to previously introduced differentials gives rise to differentials of thermodynamic potentials in electromagnetism, say (relevant parameters only explicit) du = TdS + d 3 r (E dd + H db). Material parameters contained in vectors E and B like the permittivity ε and permeability µ should be expressed here as functions of the entropy S. This is inconvenient, therefore a preferable choice is the free energy, for which df = SdT + d 3 r (E dd + H db). (.53) Here, ε and µ are functions of the temperature. Thermodynamics potentials assume minimum values at equilibrium, when their natural variables are fixed. Since free charges are sources of the electric induction D and the vector potential A the source of the magnetic induction B, the free energy (.53) is the choice for problems with fixed charges of conductors and fixed vector potentials (the latter might be difficult to control in real world, though). For other cases, new potentials should be formed by suitable Legendre transforms. For instance, the potential F E = F d 3 r E D gives rise to the differential d F E = SdT d 3 r D de, (.54) which reveals that the natural variables are T and E. Thus, this potential minimizes at equilibrium when the field E (or the electric potential) is kept constant. Similarly, the potential F H = F d 3 r H B with the differential d F H = SdT d 3 r B dh, (.55) and natural variables T and H is suitable for cases with fixed currents. Combinations of these transform may appear useful as well. Unfortunately, there seems to be no standard nomenclature of the different potentials in the electromagnetic case (cf. the Helmholtz free energy and the Gibbs function of an S, V, N system). Example.1. Consider a vertical parallel-plate capacitor in contact with a liquid reservoir. Let us calculate, how high the liquid with the dielectric constant ε r rises between the vertical plates, when the capacitor is charged and disconnected from any voltage source. The potential energy of the liquid in the gravitational field is W g = 1 gρwdy, where g is the acceleration of gravity, ρ the density of the liquid, y the height of the liquid slab between the plates, d the separation of the

29 .11. PROBLEMS 7 plates and w the width of the plates. The energy of the electric field between the plates is W Q = Q C = Q d wε [h y + ε, ry] where Q is the charge of the capacitor and h the height of the plates. Minimization of the free energy F with respect to y leads to the thirdorder equation y y + = h ε 1 Q ρw gε, which only has one real solution most conveniently obtained by some symbolic calculation programme like Maple or Mathematica..11 Problems Problem.1. Calculate the value of the expression T S T S p V V p V p p V. Problem.. (a) During a thermally isolated free expansion of a gas no work is carried out and no heat exchanged, thus the internal energy of the gas remains constant (the expansion is not necessarily a quasistatic process, however). Show that the Joule coefficient for a free expansion of a gas is T V U,N = 1 C V p Tαp. κ T (b) Is the van der Waals gas heated or cooled in the free expansion? Hint: it is more convenient not to calculate α p and κ T separately. The van der Waals equation of state is p + a N (V Nb) = NT. V Problem.3. Find out in which systems the heat capacity C V does not depend on the volume of the system. Problem.4. The free energy of a crystal in which the ions have just two quantum states is F = NT ln 1 + e ǫ/t, where ǫ is a constant. (a) Find the entropy S of the system as a function of the internal energy U and the number of particles N, i.e. the fundamental relation S = S(U, N).

30 8. THERMODYNAMIC POTENTIALS (b) Calculate the heat capacity C ǫ as a function of temperature T. Plot it and find the position of the peak, known as the Schottky anomaly. Such a peak is characteristic of a system in which atoms have a few low-lying closely spaced energy levels, and at low temperatures may dominate all other contributions to the heat capacity of the solid. Problem.5. Show (N is kept fixed) that the internal energy of the Clausius gas is U(T, V ) = Nu 1(T) an T(V + Nc), where u 1(T) is a function of temperature only, whose explicit form cannot be determined thermodynamically. However, there is another relation (show this as well) U(T, V ) = C V (T, V ), dt, establishing a connection between u 1 and the isochoric heat capacity of the gas. The equation of state of Clausius s gas is p + an (V bn) = NT. T(V + cn) Problem.6. For a unit volume of dielectric at constant density find the difference c E c D between the heat capacities of a homogeneous isotropic dielectric at constant electric field strength E and electric induction D. Problem.7. Show without resorting to the connection between C p and C V that in a stable thermodynamic equilibrium C p > and κ S > Problem.8. By minimizing a suitable thermodynamic potential, find how high dielectric liquid rises between the vertical plates of a parallelplate capacitor connected to a voltage source with the constant electromotive force E. Express your answer in terms of the electric field in the capacitor rather than the emf.

31 3. Applications of thermodynamics 3.1 Classic ideal gas For a complete thermodynamic description of a system, knowledge of the equation of state and some thermodynamic potential (energy function) is required. From the equation of state mechanical responses may be inferred and vice versa: mechanical responses suffice to reconstruct the equation of state. To determine internal energy or some other thermodynamic potential a thermal response is needed, however. The equation of state of the perfect gas pv = NT immediately yields coefficient of thermal expansion α p = 1 V and isothermal compressibility κ T = 1 V ( ) V = N T p,n V p = 1 T (3.1) ( ) V = NT p T,N V p = 1 p. (3.) It is en empirical observation that in conditions in which the equation of state of a real gas coincides with that of the ideal gas (low pressure, high temperature) the heat capacity is constant. Denote C V = 1 fn. (3.3) The quantity 1 f is the specific heat capacity, (specific heat), i.e. heat capacity per molecule. In chemistry heat capacity per mole and in hydrodynamics per mass unit are preferred. The factor f is the number of the effective degrees of freedom, whose classic value depends on the number of modes of translational, rotational and vibrational motion of the molecule: monatomic molecule f = 3 3 translations diatomic molecule f = 5 3 transl. + rotations polyatomic molecule f = 6 3 transl. + 3 rot. Each active vibrational mode adds two effective degrees of freedom (for both kinetic and potential energy of the corresponding mode of harmonic oscillation). In what follows f is assumed constant. 9

32 3 3. APPLICATIONS OF THERMODYNAMICS The differential of the entropy ( ) ( ) S S ds = dt + dv T V V T = 1 ( ) p T C V dt + dv (3.4) T is readily integrated in the (T,V ) plane (N is fixed) with the use of the heat capacity and mechanical responses (see Fig. 3 1) Assuming constant C V and using the equation of state we obtain S = S + T T dt C V T + V V V dv N V = S + C V ln T T + N ln V V. The integration constants S and V as extensive quantities may be written as S = Ns, V = Nv, where s and v are the the specific entropy and volume at the reference point. The entropy of the ideal gas is thus [ ( ) ] f/ T V S = Ns + N ln. (3.5) T Nv The specific entropy at the reference point s shall be defined later with the aid of statistical mechanics. The expression (3.5) for entropy does not vanish in the limit T but even diverges in contradiction with the III law. For real V gases, however, lowering the temperature leads either to phase transitions to liquid or solid or the appearance of quantum corrections. All thermodynamic information may be calculated starting from the known entropy. The internal energy may also be calculated in the same fashion as before for the van der Waals gas by integration in the (T,V ) plane: V T Figure 3 1: Integration path for entropy. du ( S(T,V ),V ) = T ds(t,v ) pdv ( ) ( ) S S = T dt + T dv pdv T V V T [ ( ) ] S = C V dt + T p dv. V T From the Maxwell (.a) and the equation of state it follows that ( S/ V ) T = ( p/ T) V = N/V = p/t, so that the coefficient of dv vanishes. The equation of state renders the heat capacity C V independent of volume as well. Thus, the internal energy of the ideal gas is independent of the volume and may be written as U = U + 1 f(t T )N. T