354 46 Advanced Discussion of the Second Law By transforming from an extensive expression for the work done, for ex- ample, δA = + Hdm H + , to the corresponding intensive quantity, e.g., δA := δA − d(Hm H )= − m H dH + , one obtains, instead of the Helmholtz free energy F (T,V,m H ,N), the Gibbs free energy F g (T,V,H,N):=F (T,V,m H (T,V,H,N),N) − m H H, 1 with dF g = −pdV + μdN − m H dH − SdT and the corresponding relations ∂F g ∂H = −m H , ∂F g ∂T = −S, etc. The second law is now dF g ≤ δA − SdT. The Gibbs free energy F g (T,V,H,N), and not Helmholtz free energy F (T,V,m H ,N), is appropriate for Hamiltonians with a Zeeman term, e.g., the Ising Hamiltonian (42.5), because the magnetic field is already taken into account in the energy values. Whereas for non-magnetic systems one can show that F (T,V,N) ≡−k B T · ln Z(T,V,N) , for magnetic systems F g (T,V,H,N) ≡−k B T · ln Z (T,V,H,N) , with Z (T,V,H,N) ≡ l e −βE l (V,H,N) . 2 46.2 On the Impossibility of Perpetual Motion of the Second Kind The following six statements can be regarded as equivalent formulations of the second law of thermodynamics: 1 Similarly, for magnetic systems one may also distinguish, e.g., be- tween a Helmholtz enthalpy I Helmholtz (T,p,m H ,N)andaGibbs enthalpy I Gibbs (T,p,H,N). 2 Here the prime is usually omitted. 46.2 On the Impossibility of Perpetual Motion of the Second Kind 355 1) It is not possible to construct an ideal heat pump in which heat can flow from a colder body to a warmer body without any work being done to accomplish this flow. This formulation is due to Robert Clausius (1857). In other words, energy will not flow spontaneously from a low temperature object to a higher temperature object. 2) An alternative statement due to William Thomson, later Lord Kelvin, is essentially that you cannot create an ideal heat engine which extracts heat and converts it all to useful work. Thus a perpetual motion machine of the second kind, which is a hypotheti- cal device undergoing a cyclic process that does nothing more than convert heat into mechanical (or other) work, clearly does not exist, since it would contradict the second law of thermodynamics. On the other hand, as we shall see later, a cyclically operating (or reciprocating) real heat pump is perfectly feasible, where an amount of heat Q 2 is absorbed at a low temperature T 2 and a greater amount of heat Q 1 = Q 2 + ΔQ is given off at a higher tem- perature T 1 (>T 2 ), but – to agree with the first law of thermodynamics – the difference ΔQ must be provided by mechanical work done on the system (hence the term “heat pump”). 3a)All Carnot heat engines 3 (irrespective of their operating substance) have the same maximum efficiency η := ΔA Q 1 = Q 1 − Q 2 Q 1 , where η ≤ T 1 − T 2 T 1 is also valid. The equal sign holds for a reversible process. The following (apparently reciprocal) statement is also valid: 3b)All Carnot heat pumps have the same maximum efficiency η := Q 1 ΔA = Q 1 Q 1 − Q 2 , with η ≤ T 1 T 1 − T 2 . The equal sign again applies for a reversible process. The efficiencies η and η are defined in a reciprocal way, and in the op- timum case also give reciprocal values. However the inequalities are not reciprocal; the irreversibility of statistical physics expresses itself here, since in both cases we have the inequality sign ≤, due to, for example, frictional losses. Figure 46.1 shows the (T,V)-diagram for a Carnot process. Depending on whether we are considering a Carnot heat engine or a Carnot heat pump the cycle (which comprises two isotherms and two adiabatics) runs either clockwise or anticlockwise. 3 Usually more practical than Carnot machines are Stirling machines,forwhich the adiabatics of the Carnot cycle (see below) are replaced by isochores, i.e., V = constant. 356 46 Advanced Discussion of the Second Law Fig. 46.1. Carnot process in a (V, T)rep- resentation. Carnot cycle (running clock- wise: → heat engine; running anticlockwise: → heat pump) in a (V, T)-diagram with isotherms at T 1 =2andT 2 =1.5(arbi- trary units) and adiabatics T · V 5/3 =1.5 and 0.5, as for an ideal gas. Further expla- nation is given in the text We shall now calculate the value of the optimal efficiency for a Carnot cycle η opt. =1/η opt. by assuming that the working substance is an ideal gas. Firstly, there is an isothermal expansion from the upper-left point 1 to the upper-right point 1 in Fig. 46.1 at constant temperature T 1 . The heat absorbed during this expansion is Q 1 . Then we have a further expansion from 1 down to 2 , i.e., from the upper-right to the lower-right (i.e., from T 1 =2toT 2 =1.5), which occurs adiabatically according to pV κ = TV κ−1 =const., where κ = C p C v = 5 3 , down to point 2 at the lower temperature T 2 . Subsequently there is an isothermal compression to the left, i.e., from 2 → 2atT 2 ,withheat released Q 2 . Then the closed cycle W is completed with an adiabatic compression leading from 2 up to the initial point 1. It follows that W dU ≡ 0; on the other hand we have W dU = Q 1 − Q 2 − ΔA , where the last term is the work done by the system. The efficiency η := ΔA Q 1 ≡ Q 1 − Q 2 Q 1 46.2 On the Impossibility of Perpetual Motion of the Second Kind 357 can now be obtained by calculating the work done during the isothermal expansion A 1→1 = 1 1 pdV ≡ Q 1 , since for an ideal gas ΔU 1→1 =0. Furthermore 1 1 pdV = Nk B T 1 1 1 dV V , i.e. Q 1 = Nk B T 1 ln V 1 V 1 . Similarly one can show that Q 2 = Nk B T 2 ln V 2 V 2 . For the adiabatic sections of the process (1 ,T 1 ) → (2 ,T 2 )and(2,T 2 ) → (1,T 1 ) we have T 1 V κ 1 = T 2 V κ 2 and T 1 V κ 1 = T 2 V κ 2 , where κ := C P C V − 1= 2 3 . We then obtain V 1 V 1 = V 2 V 2 and η =1− T 2 ln V 2 V 2 T 1 ln V 1 V 1 , finally giving η =1− T 2 T 1 , as stated. We shall now return to the concept of entropy by discussing a fourth equivalent version of the second law: 4) The quantity S := 2 1 δQ rev T is a variable of state (where “rev” stands for “reversible”), i.e., the integral does not depend on the path. Using this definition of the state variable 358 46 Advanced Discussion of the Second Law entropy we may write: dS ≥ δQ T , with the equality symbol for reversible heat flow. We now wish to show the equivalence between 4) and 3) by considering a Carnot heat engine. Accordingto4)wehave δQ T = Q 1 T 1 − Q 2 T 2 , but from 3) for a reversible process η =1− Q 1 Q 2 ≡ 1 − T 2 T 1 , i.e., δQ rev T ≡ 0 . For an irreversible process the amount of heat given out at the lower tem- perature T 2 is greater than in the reversible case. Thus dS> δQ rev T . One may approximate general reciprocating (or cyclic) processes by in- troducing Carnot coordinates as indicated in Fig. 46.2 below: Fig. 46.2. Carnot coordinates. This (V, T) diagram shows a single curved line segment in the upper-right hand part of the diagram and two sets of four horizontal isotherms and four non-vertically curved adiabatics, respectively, forming a grid N of so-called Carnot coordinates. For an asymptotic improvement of the grid the single curve is extended to the bounding line ∂G of a two-dimensional region G, and since the internal contributions cancel each other in a pairwise manner, one may show analogously to Stokes’s integral theorem that the following is valid: H ∂G δQ rev T ≡ P Carnot processes∈N δQ rev T . Finally, instead of P ∈N one may also write the integral R G 47 Shannon’s Information Entropy What actually is entropy ? Many answers seem to be rather vague: Entropy is a measure of disorder, for example. An increase in entropy indicates loss of information, perhaps. However the fact that entropy is a quantitative mea- sure of information, which is very important in both physics and chemistry, becomes clear when addressing the following questions: –HowlargeisthenumberN of microstates or configurations of a fluid consisting of N molecules (ν =1, ,N, e.g., N ∼ 10 23 ), where each molecule is found with a probability p j in one of f different orthogonal states ψ j (j =1, ,f)? –HowlargeisthenumberN of texts of length N (= number of digits per text) which can be transmitted through a cable, where each digit originates from an alphabet {A j } j=1, ,f of length f and the various A j occur with the probabilities (= relative frequencies) p j ? – How many times N must a language student guess the next letter of a foreign-language text of length N, if he has no previous knowledge of the language and only knows that it has f letters A j with probabilities p j , j =1, ,f? These three questions are of course identical in principle, and the answer is (as shown below): N =2 N·I =e N ·S k B , where I = 1 N ldN is the so-called Shannon information entropy. Apart from a normalization factor, which arose historically, this expression is identical to the entropy used by physicists. Whereas Shannon and other information theorists used binary logarithms ld (logarithms to the base 2) physicists adopt natural logarithms (to the base e =exp(1)=2.781 ,lnx =(0.693 ) ·ldx). Then we have S k B = 1 N ln N , which apart from unessential factors is the same as I. Proof of the above relation between N and I (or S k B ) is obtained by using some “permutation gymnastics” together with the so-called Stirling approximation, i.e., as fol- lows. 360 47 Shannon’s Information Entropy The number of configurations is N = N! N 1 !N 2 ! · · N f ! because, for a length of text N there are N! permutations where exchange gives in general a new text, except when the digits are exchanged amongst each other. We then use the Stirling approximation: for N 1 ,N! ∼ = √ 2πN N e N · 1+O 1 N ; i.e., by neglecting terms which do not increase exponentially with N,wemay write N! ≈ N e N . Thus N ∼ = N e (N 1 +N 2 + +N f ) N 1 e N 1 · · N f e N f , or ln N ∼ = − f j=1 N j ·ln N j N ≡−N f j=1 p j ·ln p j , which gives N ∼ = e N· S k B , with S k B ≡− f j=1 p j · ln p j . (47.1) Using the same basic formula for S one can also calculate the thermody- namic entropy S(T ), e.g., with the Boltzmann-Gibbs probabilities p j = e −βE j Z(T ) , in agreement with the expressions Z(T )= j e −βE j ,U(T )=− dlnZ dβ ,F(T )=−k B T · ln Z(T ) , with F (T )=U(T ) − T · S(T ) , i.e. S(T ) ≡− ∂F ∂T . The relative error made in this calculation is O ln √ 2πN N , which for N 1 is negligible. 47 Shannon’s Information Entropy 361 To end this section we shall mention two further, particularly neat for- mulations of the second law: 5) A spontaneously running process in a closed system can only be reversed by doing work on the system. This is equivalent to stating: 6) Heat only flows spontaneously from a higher to a lower temperature. This last formulation due to Max Planck goes back to Robert Clausius (see above). 1 At the same time he recognized that it is not easy to prove the second law in a statistical-physical way. This is possible, but only using stochastic methods. 2 Work done on a closed system, δA > 0, always leads to the release of heat from the system, δQ < 0, since dU(≡ δA + δQ)=0. The entropy must therefore have increased along the “ first leg of the cycle ”, since dS = 0. The requirement that we are dealing with a closed system, i.e., dU = 0, is thus unnecessarily special, since one can always modify an open system, e.g., with heat input, δQ > 0, to be a closed one by including the heat source. In this sense, closed modified systems are the “ most general ” type of system, and we shall see in the following sections in what ways they mayplayanimportantpart. 1 We should like to thank Rainer H¨ollinger for pointing this out. 2 see, e.g., the script Quantenstatistik by U.K. 48 Canonical Ensembles in Phenomenological Thermodynamics 48.1 Closed Systems and Microcanonical Ensembles The starting point for the following considerations is a closed system, corre- sponding to the so-called microcanonical ensemble. The appropriate function of state is S(U, V, N). The relevant extremum principle is S ! = max.,sothat ΔS > 0 until equilibrium is reached. Furthermore dS = δQ rev T = dU − δA T = dU + pdV −μdN T ;thus 1 T = ∂S ∂U V,N , p T = ∂S ∂V U,N and μ T = − ∂S ∂N U,V . The probabilities p j are given by p j = 1 N , if U − δU < E j ≤ U otherwise p j = 0. Here, δU is, e.g., an instrumental uncertainty, and N (= N(U − δU,U)) is the total number of states with U − δU < E j ≤ U. We shall see below that the value of δU U is not significant unless it is extremely small in magnitude. 48.2 The Entropy of an Ideal Gas from the Microcanonical Ensemble Inserting the relation p j ≡ 1 N for U − δU < E j ≤ U, 364 48 Canonical Ensembles in Phenomenological Thermodynamics otherwise ≡ 0, into the relation for S gives S = k B ·ln N , with N = N(U −δU,U)= V N h 3N N! · ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ U−δU< p 2 1 + +p 2 N 2m ≤U d 3N p ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ . The braced multidimensional integral in p-space is the difference in volume between two spherical shells in R 3N (p) with radii R 1 (U):= √ 2mU and R 2 (U − δU):= 2m ·(U − δU) . This gives N = V N h 3N N! Ω(3N )(2m) 3N/2 · U 3N/2 − (U −δU) 3N/2 , where Ω(d) is the volume of a unit sphere in d-dimensional space. Here it is important to note that the whole second term, (U − δU) 3N/2 , can be neglected compared to the first one, U 3N/2 , except for extremely small δU, since normally we have U − δU U 3N/2 1 , because N is so large 1 , i.e., one is dealing with an exponentially small ratio. We therefore have N≡N(U)= V N Ω(3N )(2mU) 3N/2 h 3N N! and S = k B ln N = k B N · ln V V 0 + 3 2 ln U U 0 +ln s 0 V 0 U 3 2 0 . Here, V 0 and U 0 are volume and energy units (the exact value is not signifi- cant; so they can be arbitrarily chosen). The additional factor appearing in the formula, the “atomic entropy constant” s 0 := ln s 0 V 0 U 3 2 0 is obtained from the requirement that (for N 1) N · ln[s 0 V 0 U 3 2 0 ] ! =ln (2m) 3N/2 Ω(3N ) h 3N N! V N 0 U 3N 2 0 . 1 E.g., one should replace U by 1 and (U − δU) by 0.9 und study the sequence (U − δU) N for N =1, 2, 3, 4, . operating (or reciprocating) real heat pump is perfectly feasible, where an amount of heat Q 2 is absorbed at a low temperature T 2 and a greater amount of heat Q 1 = Q 2 + ΔQ is given off at a. T 1 =2andT 2 =1.5(arbi- trary units) and adiabatics T · V 5/3 =1.5 and 0.5, as for an ideal gas. Further expla- nation is given in the text We shall now calculate the value of the optimal efficiency for a Carnot cycle η opt. =1/η opt. by. following (apparently reciprocal) statement is also valid: 3b)All Carnot heat pumps have the same maximum efficiency η := Q 1 A = Q 1 Q 1 − Q 2 , with η ≤ T 1 T 1 − T 2 . The equal sign again applies