Basic Theoretical Physics: A Concise Overview P32 ppt

41.6 Enthalpy and the Joule-Thomson Experiment; Liquefaction of Air 321 The second law can also be transformed by expressing the entropy (wholly or partially) as a function of intensive variables: e.g., S(T,p,H,N, ). The Maxwell relations can be extended to cover this case, e.g., ∂I ∂p = V − T ∂V ∂T , ∂I ∂H = T ∂H ∂T − H, or, in general ∂I ∂f j = T ∂X j ∂T − X j . (41.6) We now come to the Joule-Thomson effect, which deals with the station- ary flow of a fluid through a pipe of uniform cross-sectional area (S) 1 on the input side of a so-called throttle valve, V thr . On the output side there is also a pipe of uniform cross-section (S) 2 ,where(S) 2 =(S) 1 .Weshallsee that, as a result, p 1 = p 2 . We shall consider stationary conditions where the temperatures T on each side of the throttle valve V thr are everywhere the same. Furthermore the pipe is thermally insulated (δQ = 0). In the region of V thr itself irreversible processes involving turbulence may occur. However we are not interested in these processes themselves, only in the regions of stationary flow well away from the throttle valve. A schematic diagram is shown in Fig. 41.1. We shall now show that it is not the internal energy U (= N · u)ofthe fluid which remains constant in this stationary flow process, but the specific enthalpy, i(x):= I(T,p(x),N) N , in contrast to the Gay-Lussac experiment. Here x represents the length co- ordinate in the pipe, where the region of the throttle valve is not included. In order to prove the statement for i(x)wemustrememberthatatany time t the same number of fluid particles dN 1 =dN 2 Fig. 41.1. Joule-Thomson process. A fluid (liquid or gas) flowing in a stationary manner in a pipe passes through a throttle valve. The pressure is p 1 on the left and p 2 (<p 1 ) on the right of the valve. There are equal numbers of molecules on average in the shaded volumes left and right of the valve. In contrast to p, the temperatures are everywhere the same on both sides of the valve 322 41 The First and Second Laws of Thermodynamics pass through the cross-sections S i at given positions x i left and right of the throttle valve in a given time interval dt. Under adiabatic conditions the difference in internal energy δU := U 2 − U 1 for the two shaded sets S i · dx i in Fig. 41.1, containing the same number of particles, is identical to δA, i.e., δA = −p 1 ·S 1 · dx 1 + p 2 · S 2 · dx 2 , since on the input side (left) the fluid works against the pressure while on the output side (right) work is done on the fluid. Because of the thermal isolation, δQ =0,wethushave: dU = −p 1 dV 1 + p 2 dV 2 ≡ (−p 1 v 1 + p 2 v 2 )dN ; i.e., we have shown that u 2 + p 2 v 2 = u 1 + p 1 v 1 , or i 2 = i 1 , viz the specific enthalpy of the gas is unchanged. 2 The Joule-Thomson process has very important technical consequences, since it forms the basis for present-day cryotechnology and the Linde gas liquefaction process. In order to illustrate this, let us make a quantitative calculation of the Joule-Thomson cooling effect (or heating effect (!), as we will see below for an important exceptional case):  dT dp  i = − ∂i ∂p ∂i ∂T . (This can be described by − ∂i ∂p c (0) p , with the molecular heat capacity c (0) p := C p N ; an analogous relation applies to the Gay-Lussac process.) Using the van der Waals equation p = − a v 2 + k B T v −b , see next section, for a real gas together with the Maxwell relation ∂i ∂p = v −T ∂v ∂T and  ∂v ∂T  p = − ∂p ∂T ∂p ∂v , 2 If the temperature is different on each side of the throttle valve, we have a situ- ation corresponding to the hotly discussed topic of non-equilibrium transport phenomena (the so-called Keldysh theory). 41.6 Enthalpy and the Joule-Thomson Experiment; Liquefaction of Air 323 one obtains  dT dp  i = 1 c (0) p 2a k B T (v−b) 2 v 2 − b 1 − 2a vk B T (v−b) 2 v 2 . (41.7) The first term of the van der Waals equation, − a v 2 , represents a negative internal pressure which corresponds to the long-range attractive r −6 term of the Lennard-Jones potential (see below). The parameter b in the term k B T v−b , the so-called co-volume, corresponds to a short-range hard-core repulsion ∝ r −12 : b ≈ 4πσ 3 3 . σ is the characteristic radius of the Lennard-Jones potentials, which are usu- ally written as 3 V L J. (r)=4ε ·   σ r  12 −  σ r  6  . (41.8) In the gaseous state, v  b, and (41.7) simplifies to  dT dp  i ≈ v c (0) p  T Inv. T 1 − T Inv. T  , where the so-called inversion temperature T Inv. := 2a k B v . As long as T<T Inv. , which is normally the case because the inversion tem- perature usually (but not always!) lies above room temperature, we obtain a cooling effect for a pressure drop across the throttle valve, whereas for T>T Inv. a heating effect would occur. In order to improve the efficiency of Joule-Thomson cooling, a counter- current principle is used. The cooled fluid is fed back over the fluid to be cooled in a heat exchanger and successively cooled until it eventually lique- fies. Air contains 78.08 % nitrogen (N 2 ), 20.95 % oxygen (O 2 ), 0.93 % argon (Ar) and less than 0.01 % of other noble gases (Ne, He, Kr, Xe) and hydro- gen; these are the proportions of permanent gases. A further 0.03 % consists of non-permanent components (H 2 O, CO and CO 2 , SO 2 , CH 4 , O 3 ,etc.). The liquefaction of air sets in at 90 K at normal pressure. For nitrogen the liquefaction temperature is 77 K; for hydrogen 20 K and helium 4.2K.The inversion temperature for helium is only 20 K; thus in order to liquefy helium using this method, one must first pre-cool it to a temperature T<T Inv. =20K. (He above 20 K is an example of the “exceptional case” mentioned above.) 3 see (4), Chap. 3, in [37]. 324 41 The First and Second Laws of Thermodynamics 41.7 Adiabatic Expansion of an Ideal Gas The Gay-Lussac and Joule-Thomson effects take place at constant U(T,V,N) and constant i(T,P):= U + pV N , respectively. These subsidiary conditions do not have a special name, but calorific effects which take place under conditions of thermal insulation (with no heat loss, δQ ≡ 0), i.e., where the entropy S(T,V,N) is constant, are called adiabatic. In the following we shall consider the adiabatic expansion of an ideal gas. We remind ourselves firstly that for an isothermal change of state of an ideal gas Boyle-Mariotte’s law is valid: pV = Nk B T = constant . On the other hand, for an adiabatic change, we shall show that pV κ = constant , where κ := C p C V . We may write  dT dV  S = − ∂S ∂V ∂S ∂T = ∂U ∂V + p C V , since dS = dU −δA T = C v dT T +  ∂U ∂V + p  dV T . Using the Maxwell relation ∂U ∂V + p = T ∂p ∂T , we obtain  dT dV  S = T ∂p ∂T C V . In addition, we have for an ideal gas: (C p − C V )=  ∂U ∂V + p  dV dT  p ≡ p  dV dT  p = p V T = Nk B , so that finally we obtain  dT dV  S = − C p − C V C V · T V . 41.7 Adiabatic Expansion of an Ideal Gas 325 Using the abbreviation ˜κ := C p − C V C V , we may therefore write dT T +˜κ dV V =0, i.e., d  ln T T 0 +ln  V V 0  ˜κ  =0, also ln T T 0 +ln  V V 0  ˜κ = constant , and T · V ˜κ = constant . With κ := C p C V =1+˜κ and p = Nk B T V we finally obtain: pV κ = constant . In a V,T diagram one describes the lines  dT dV  S ≡ constant  as adiabatics. For an ideal gas, since ˜κ>0, they are steeper than isotherms T = constant. In a V, p diagram the isotherms and adiabatics also form a non-trivial coordinate network with negative slope, where, since κ>1, the adiabatics show a more strongly negative slope than the isotherms. This will become important below and gives rise to Fig. 46.2. One can also treat the adiabatic expansion of a photon gas in a similar way. 42 Phase Changes, van der Waals Theory and Related Topics As we have seen through the above example of the Joule-Thomson effect, the van der Waals equation provides a useful means of introducing the topic of liquefaction. This equation will now be discussed systematically in its own right. 42.1 Van der Waals Theory In textbooks on chemistry the van der Waals equation is usually written as  p + A V 2  ·(V − B)=n Mol R 0 T. In physics however one favors the equivalent form: p = − a v 2 + k B T v −b , (42.1) with a := A N ,v:= V N ,Nk B = n Mol R 0 ,b:= B N . The meaning of the terms have already been explained in connection with Lennard-Jones potentials (see (41.8)). Consider the following p, v-diagram with three typical isotherms (i.e., p versus v at constant T , see Fig. 42.1): a) At high temperatures, well above T c , the behavior approximates that of an ideal gas, p ≈ k B T v , i.e. with negative first derivative dp dv ; but in the second derivative, d 2 p dv 2 , 328 42 Phase Changes, van der Waals Theory and Related Topics there is already a slight depression in the region of v ≈ v c , which corre- sponds later to the critical density. This becomes more pronounced as the temperature is lowered. b) At exactly T c , for the so-called critical isotherm, which is defined by both derivatives being simultaneously zero: dp dv |T c ,p c ≡ d 2 p dv 2 |T c ,p c ≡ 0 . In addition the critical atomic density  c = v −1 c is also determined by T c and p c . c) An isotherm in the so-called coexistence region, T ≡ T 3 <T c . This region is characterized such that in an interval v (3) 1 <v<v (3) 2 the slope dp dv of the formal solution p(T 3 ,v) of (42.1) is no longer negative, but becomes positive. In this region the solution is thermodynamically unstable, since an increase in pressure would increase the volume, not cause a decrease. The bounding points v (3) i of the region of instability for i =1, 2(e.g.,the lowest point on the third curve in Fig. 42.1) define the so-called “spin- odal line”, a fictitious curve along which the above-mentioned isothermal compressibility diverges. On the other hand, more important are the coexistence lines, which are given by p i (v) for the liquid and vapor states, respectively (i =1, 2). These can be found in every relevant textbook. Fig. 42.1. Three typical solutions to van der Waals’ equation are shown, from top to bottom: p = −0.05/v 2 + T/(v − 0.05) for, (i), T = 4; (ii) T ≡ T c = 3; and, (iii), T =2 42.1 Van der Waals Theory 329 The solution for p(T 3 ,v) is stable outside the spinodal. At sufficiently small v, or sufficiently large v, respectively, i.e., for v<v liquid 1 or v>v vap or 2 , the system is in the liquid or vapor state, respectively. One obtains the bound- ing points of the transition, v liquid 1 and v vap or 2 , by means of a so-called Maxwell construction, defining the coexistence region. Here one joins both bounding points of the curve (42.1) by a straight line which defines the saturation vapor pressure p s (T 3 ) for the temperature T = T 3 : p(T 3 ,v liquid 1 )=p(T 3 ,v vap or 2 )=p s (T 3 ) . This straight line section is divided approximately in the middle by the solu- tion curve in such a way that (this is the definition of the Maxwell construc- tion) both parts above the right part and below the left part of the straight line up to the curve are exactly equal in area. The construction is described in Fig. 42.2. This means that the integral of the work done,  vdp or −  pdv, for the closed path from (v liquid 1 ,p s ) , firstly along the straight line towards (v vap or 2 ,p s ) and then back along the solution curve p(T 3 ,v) to the van der Waals’ equation (42.1), is exactly zero, which implies, as we shall see later, that our fluid Fig. 42.2. Maxwell construction. The fig- ure shows the fictitious function p(V )= −(V +1)· V · (V − 1), which corresponds qualitatively to the Maxwell theory; i.e., the l.h.s. and r.h.s. represent the marginal values of the liquid and vapor phases, respectively. The saturation-pressure line is represented by the “Maxwell segment”, i.e., the straight line p ≡ 0 between V = −1andV =+1,where according to Maxwell’s construction the ar- eas above and below the straight line on the r.h.s. and the l.h.s. are identical 330 42 Phase Changes, van der Waals Theory and Related Topics system must possess the same value of the chemical potential μ in both the liquid and vapor states: μ 1 (p s ,T)=μ 2 (p s ,T) . If v lies between both bounding values, the value of v gives the ratio of the fluid system that is either in the liquid or the vapor state: v = x ·v liquid 1 +(1− x) · v vap or 2 , where 0 ≤ x ≤ 1 . Only this straight line (= genuine “Maxwell part”) of the coexistence curve and the condition μ 1 = μ 2 then still have a meaning, even though the van der Waals curve itself loses its validity. This is due to the increasing influence of thermal density fluctuations which occur in particular on approaching the critical point T = T c ,p= p c ,v = v c . Near this point according to the van der Waals theory the following Taylor expansion would be valid (with terms that can be neglected written as + ): −(p − p c )=A · (T − T c ) · (v −v c )+B · (v −v c ) 3 + , (42.2) since for T ≡ T c there is a saddle-point with negative slope, and for T>T c the slope is always negative, whereas for p = p s <p c the equation p ≡ p s leads to three real solutions. The coefficients A and B of the above Taylor expansion are therefore positive. 42.2 Magnetic Phase Changes; The Arrott Equation In magnetism the so-called Arrott equation, H = A · (T − T 0 ) · m h + B ·(m h ) 3 (42.3) has analogous properties to the van der Waals equation in the neighborhood of the critical point, (42.2). T 0 is the Curie temperature. As in the van der Waals equation, the coef- ficients A and B are positive. For T ≡ T 0 , i.e., on the critical isotherm, H = B · (m h ) δ , with a critical exponent δ =3.ForT  T 0 we have H = A · (T − T 0 ) · m H , i.e., so-called Curie-Weiss behavior. Finally, for T<T 0 and positive ΔH we have a linear increase in H,ifm H increases linearly from the so-called spontaneous boundary value m (0) H that arises for H =0 + , together with 42.2 Magnetic Phase Changes; The Arrott Equation 331 a discontinuous transition for negative ΔH. This corresponds to a coexis- tence region of positively and negatively magnetized domains, respectively, along a straight line section on the x-axis, which corresponds to the Maxwell straight line, i.e., for −m (0) H <m H <m (0) H . Therefore, one can also perform a Maxwell construction here, but the result, the axis H ≡ 0, is trivial, due to reasons of symmetry. However, the fact that the Maxwell line corresponds to a domain structure with H ≡ 0, and thus a corresponding droplet structure in the case of a fluid, is by no means trivial. The spontaneous magnetization m (0) H follows from (42.3) with H ≡ 0for T<T 0 , and behaves as m (0) H ∝ (T 0 − T ) β , with a critical exponent β = 1 2 . Finally, one can also show that the magnetic susceptibility χ = dm H dH  =  dH dm H  −1  , is divergent, viz for T>T 0 as χ = 1 A · (T − T 0 ) . On the other hand, for T<T 0 it behaves as χ = 1 2A ·(T 0 − T ) , i.e., generally as χ ∝|T − T 0 | γ , with γ =1 Fig. 42.3. The Arrott equation. The figure shows three typical solutions to the Arrott equation, which are discussed in the text. The equation is H = A ·(T −T 0 ) ·M + B ·M 3 and is plotted for three cases: (i), T =3,T 0 =1; (ii), T = T 0 = 1, and (iii), T = −0.5, T 0 =1. In all cases we have assumed A ≡ 1 . lines  dT dV  S ≡ constant  as adiabatics. For an ideal gas, since ˜κ>0, they are steeper than isotherms T = constant. In a V, p diagram the isotherms and adiabatics also form a non-trivial coordinate network. shall consider the adiabatic expansion of an ideal gas. We remind ourselves firstly that for an isothermal change of state of an ideal gas Boyle-Mariotte’s law is valid: pV = Nk B T = constant. adiabatic expansion of a photon gas in a similar way. 42 Phase Changes, van der Waals Theory and Related Topics As we have seen through the above example of the Joule-Thomson effect, the van der