24 How to manipulate partial derivatives

Transcription

1 24 How to manipulate partial derivatives 260 ummar he elasticit of a rubber band is understandable in terms of entropic force. he Jacobian technique allows us to handle derivatives algebraicall, helping our understanding of qualitative features (of, e.g., rubber bands). he Jacobian technique ma be full utilized, if the reader remembers (X, Y ) (A, B) = (X, Y ) (Y, X) (Y, X) = = (B, A) (B, A) (A, B), (X, Y ) (Z, W ) = (X, Y ) (A, B) (A, B) (Z, W ), and Maxwell s relation for conjugate pairs (X, x) and (Y, ): (X, x) (, Y ) = 1. Understanding a rubber band clarifies the principle of adiabatic cooling. Ke words entropic elasticit, Jacobian, Maxwell s relation, adiabatic cooling, adiabatic demagnetization. he reader should be able to: Become familiar with the Jacobian technique. Intuitivel explain rubber elasticit; get various signs of partial derivatives, and explain them intuitivel. Explain adiabatic demagnetization Rubber band experiment Let us perform simple experiments on quasistatic adiabatic processes in rubber bands. Prepare a thick rubber band (that is used to bundle, e.g., broccoli heads). We use our lips as a temperature sensor (clean the rubber band). Initiall, putting the rubber band to our lips, let us confirm it is around room temperature (slightl cool). Now, hold both ends of a small portion of the band tightl between thumbs and fingers and stretch it quickl (Fig. 24.1). hen, let us feel the temperature of

2 261 How to manipulate partial derivatives rapid change Fig stretch locall rapidl and strongl Holding the s firml between our thumbs and fingers, let us stretch a rubber band locall rapidl and strongl to realize (approximatel) an adiabatic and quasistatic process. the stretched portion with our lips. It must be warm. We have just experimentall demonstrated 1 ( ) > 0, (24.1) L where L is the length of the stretched portion of the rubber band. hat is, increasing L (stretching the band) while keeping entrop (i.e., adiabaticall) increases the temperature of the band. he thermodnamic coordinates for the rubber band are E and L, and its Gibbs relation reads or de = d + F dl, (24.2) d = 1 de F dl, (24.3) where F is the force (the component of the force parallel to the stretching direction of the rubber band) stretching the band. ince the process is adiabatic and quasistatic, is constant; Even if we rapidl pull the band, the stretching rate we can realize is ver small from the polmer point of view, so the process is (almost) quasistatic. ince the heat conduction is not a ver rapid process, during quick stretch the sstem is virtuall thermall isolated (= adiabatic). hus, is approximatel constant Polmer chain is just as kids plaing hand in hand o begin with, let us tr to understand microscopicall what we have experimentall observed macroscopicall. A rubber band is made of a bunch of polmer chains. ake a single chain that is wiggling due to thermal motion (Fig. 24.2A). tretching the chain corresponds to increasing the distance between the flags of Fig. 24.2B of the plaing kid analog. If the chain does not break, the spatial room for dancing is decreased, but since the kids must keep their entrop, the restricted dancing degrees of freedom (conformational entrop) must be compensated b some other degrees of freedom, shaking bodies (momentum-space or motional entrop). hat is, the temperature of the sstem should go up. his suggests that if the chain is stretched under constant, the entrop of the rubber band should go down: ( ) < 0. (24.4) L 1 If the reader needs a partial derivative review, read 24.6: Partial derivative review and 24.8: Remarks on the notation of partial derivatives.

3 262 How to manipulate partial derivatives A Fig Fig A: A schematic picture of a single polmer chain. Each arrow is called a monomer. B: Polmer-kid analog. he temperature represents how vigorousl kids are moving around. his also includes vibration of individual bodies. he figure is after aito, N. (1967). Polmer Phsics, oko: hokabo (he original picture was due to. akai s lecture according to aito). he entrop of a chain under constant temperature is monotonicall related to the width of the range kids can pla around easil, which becomes smaller if the distance between the flags is increased. If we understand entrop as a measure of microscopic state diversit, this is not hard to guess, since stretching restricts the conformational diversit of the chain. Can we conclude this from what we have alread observed: (24.1)? Yes, we can, but we should first learn a technique to manipulate partial derivatives efficientl Freel jointed polmer chain Before explaining powerful thermodnamic techniques, let us tr to understand the polmer chain sstem statistical-mechanicall. For simplicit, let us consider a polmer chain along the x-axis (Fig. 24.3). We assume that the chain is freel- O L he chain is expanded in the vertical direction to avoid cluttering. Monomers can take the + or x-direction. he right and left directed monomer numbers N ± (N + + N = N) can be used to compute the number of conformations. jointed. hat is, there is no energ cost to change its conformation at all (just as an ideal gas can change its configuration without an energ cost). he Hamiltonian of this freel-jointed polmer consists of the chain kinetic energ K onl, which is independent of the conformation of the chain. hus, E depends onl on, since it is (just the averaged) kinetic energ Freel-jointed polmer entrop We can easil compute the (conformational) entrop 2 of the ideal rubber band using Boltzmann s principle. Following the figure caption of Fig. 24.3, let us introduce N ± so that N + + N = N, and N + N = X L/l, where l is the monomer length. We have B x N ± = 1 (N ± X). (24.5) 2 2 First, we consider onl the conformation part; there is certainl a momentum contribution, which is not discussed until the next entr.

4 263 How to manipulate partial derivatives herefore, the (conformational portion of the) microcanonical partition function reads (note that this is exactl the same problem as the chottk defect problem to get the entrop 18.2) ( ) N w(x) =. (24.6) From this, immediatel we obtain [ N+ = Nk B N log N + N + N N log N ]. (24.7) N or N + [ N + X = Nk B log N + X + N X log N X ]. (24.8) With the aid of the Gibbs relation, we find the force (note that lx = L) F = ( ) = k B l X 2l log N + X N X. (24.9) his implies L = Nl tanh(βlf ). (24.10) ince tanh x x for small x, this implies a Hookean spring for small stretches: F = (k B /Nl 2 )L. (24.11) hat is, k B / R 2 is the spring constant, where R 2 = Nl 2 is the mean square end-to-end distance of a polmer chain Ideal rubber band Can we explain what we have experienced at the beginning of this section using this entrop? Never. (24.8) implies that if L = Xl is fixed, is fixed. his is phsicall obvious, because the set of allowed conformations is completel determined b L. It is clear that we need the contribution of thermal motion. hen, the entrop should have an extra term e dependent on E. Let us assume e is not dependent on L (the ideal rubber model 3 ): [ N + X = Nk B log N + X + N X log N X ] + e (E). (24.12) Notice that e is solel determined b thermal motion (just as the corresponding term for the ideal gas; cf. (15.16)). We can make a more detailed model of a rubber band to compute more realistic entrop, but even without such microscopic details thermodnamics can tell us man qualitative features correctl. 3 Recall that the ideal gas entrop (15.16) has the same structure.

5 264 How to manipulate partial derivatives o this end, we must be able to manipulate thermodnamic quantities and derivatives efficientl. We begin with a review of partial differentiation Partial derivative review 4 Consider a two-variable function f = f(x, ). 5 Partial derivatives are defined as 6 f f f(x + δx, ) f(x, ) x(x, ) = lim, δx 0 δx (24.13) f f f(x, + δ) f(x, ) (x, ) = lim. δ 0 δ (24.14) Partial differentiation is extremel trick in general, however. For example, even if f/ and f/ exist at a point, f can be discontinuous at the same point (e.g., f can jump along a certain line going through the point). E is once continuousl differentiable with respect to and work coordinates, so it is convenient to introduce a stronger concept of differentiabilit. Let f be a function of several variables x = (x 1,, x n ). We could understand f as a function of vector x. We wish to stud its linear response to the change x x + δx: δf(x) = f(x + δx) f(x) = Df(x)δx + o[δx], (24.15) where o denotes higher order terms that vanish faster than δx, 7 when the limit δx 0 is taken. Here, Df(x) is a linear operator: if applied on a vector a = (a 1,, a n ), Df(x)a = n f a i. (24.16) i If such a linear map Df(x) is well-defined, we sa that f is (totall) differentiable (or strongl differentiable). If there are onl two variables, we ma write i=1 Df(x, )(dx, d) = f f dx + d. (24.17) When we sa a function is differentiable in this section, it is alwas in this strong sense Maxwell s relations Let us closel look at f(x + δx, + δ) f(x, ). here are two was to go from 4 Except at phase transitions thermodnamic functions are differentiable practicall as man times as we wish. 5 Whenever a function is mentioned, its domain should be stated explicitl, but usuall we will not do so in this book. 6 In case we wish to save space, we use the notation f/ = xf = f x. 7 denotes the norm of a vector; usuall, (a 1,, a n) = a a2 n.

6 265 How to manipulate partial derivatives (x, ) to (x + δx, + δ), δx first or δ first: f(x + δx, + δ) f(x, ) = f(x + δx, + δ) f(x + δx, ) + f(x + δx, ) f(x, ) = f (x + δx, )δ + f x (x, )δx, (24.18) f(x + δx, + δ) f(x, ) = f(x + δx, + δ) f(x, + δ) + f(x, + δ) f(x, ) he difference between these two formulas is 8 = f x (x, + δ)δx + f (x, )δ. (24.19) [f (x+δx, ) f (x, )]δ [f x (x, +δ) f x (x, )]δx = [f x (x, ) f x (x, )]δxδ. (24.20) his must vanish if the surface defined b f (i.e., the surface z = f(x, ) in 3-space) is at least twice differentiable: 9 f x = f x. (24.21) For example, for a rubber band de = d + F dl, so ( ) ( ) F =. (24.22) L uch relations are called Maxwell s relations in thermodnamics Remarks on the notation of partial derivatives When a multivariate function is written in mathematics, its independent variables are specified, and is written as f = f(x, ). herefore, if we write a partial derivative f/, it implies that the independent variable(s) other than x is fixed, so usuall there is no specification of what variable(s) to fix as ( ) f f =. (24.23) he function f ma depend on more directl through a function α = α(x, ), so it ma be more convenient to understand the quantit f as a function of x and α as f = g(x, α(x, )). Accordingl, the partial derivative of f with respect to x is better considered under the condition that fixes α. In mathematics, in such a case a new function smbol g = g(x, α) is introduced and then we compute g/. In thermodnamics, this g/ is almost alwas denoted as ( f/) α. hat is, f is understood as the smbol for a phsical quantit, and not as a function smbol (as discussed further below). Usuall, there is no accompaning specification of independent variables. hus, for example, the relevant independent variables for depend on the situations. On the left-hand side of (24.22) is a function of L and, but the temperature change due to the increase in force under constant length ( / F ) L is conceivable. In this case is understood as a function of L and F. F ma be, for example, a function of L and, as α above. In these two cases, s are different functions mathematicall, because the set of independent variables are distinct, but 8 Notation: x( f) = x(f ) = f x. 9 Young s theorem If f x and f x are continuous, or if f is twice differentiable, then f x = f x. [If f x is not continuous, we have a counterexample.] L

7 266 How to manipulate partial derivatives in the context of thermodnamics, we respect the identit of as the same phsical observable and use the same smbol consistentl. Needless to sa, the partial derivative depends on what to fix (= what to be the set of independent variables, mathematicall speaking), so the set of fixed variables must be explicitl indicated. Good examples are C V = ( / ) V and C P = ( / ) P (see 24.11) Jacobian expression of partial derivatives 10 o manipulate man partial derivatives, it is ver convenient to use the so-called Jacobian technique. his technique can greatl reduce the insight and skill required in thermodnamics, especiall with the Jacobian version of Maxwell s relation (24.39). he Jacobian for two functions X and Y of two independent variables x, is defined b the following determinant: ( (X, Y ) (x, ) X ) ( ) X ( ) ( ) ( ) ( ) x ( Y ) ( ) X Y X Y Y =. (24.24) x x In particular, we observe x ( ) (X, ) X (x, ) =, (24.25) which is the ke observation of this technique. Obviousl, 11 (X, Y ) = 1. (24.26) (X, Y ) here are onl two or three formulas the reader must learn b heart (the are ver eas to memorize). One is straightforwardl obtained from the definition of determinants: exchanging rows or columns switches the sign: (X, Y ) (x, ) = (X, Y ) (Y, X) X) = = (Y, (, x) (, x) (x, ). (24.27) Chain rule in terms of Jacobians If we assume that X and Y are differentiable functions of a and b, and that a and b are differentiable functions of x and, we have the following multiplicative relation: (X, Y ) (a, b) (a, b) (x, ) = (X, Y ) (x, ). (24.28) 10 In this and in the next entr, upper case (resp., lower case) letters do not necessaril mean extensive (resp., intensive) quantities, but an variables or functions sufficientl differentiable. 11 In this case, we regard X and Y are independent variables. What is independent, what is dependent In the Jacobian expression, all the letters appearing upstairs are regarded dependent variables of the variables appearing downstairs. As noted in 24.8 in thermodnamics independent variables are often implicit, and we must explicitl indicate them when we compute partial derivatives. As seen here, if the Jacobian scheme is used, the independent variables of the context are alwas explicit. his is wh the Jacobian technique can be mechanical.

8 267 How to manipulate partial derivatives his is a disguised chain rule: ( ) ( ) X X = a b ( ) a + ( ) X b a etc. Confirm (24.28) (use det(ab) = (deta)(detb)). ( ) b, (24.29) he technical significance of (24.28) must be obvious: calculus becomes algebra. We ma regard (A, B) just as an ordinar number: formall 12 we can do as follows. 13 First, split the fraction and then throw in the identical factors we wish to introduce as (X, Y ) (x, ) = (X, Y ) (x, ) = (X, Y ) (A, B) (A, B) (x, ). (24.30) From (24.28) we get at once / (X, Y ) (A, B) (A, B) = 1 (X, Y ). (24.31) In particular, we have 14 ( ) X Y /( ) = 1 X Using these relations, we can easil demonstrate ( ) ( ) / ( ) X = X as follows: (X, x) (, x) (24.28) = x (X, x) (, X) (, X) (, x) X (24.27) hen, use (24.31). A concrete example of (24.33) is ( ) ( ) P V = V Y. (24.32) (24.33) (x, X) (X, ) = (, X) (x, ). (24.34) P /( ) V P, (24.35) which relates thermal expansivit and isothermal compressibilit. For a rubber band ( ) L = (L, F ) (, F ) = (L, F ) ( ) ( ) (, F ) L (, F ) (, F ) =, (24.36) F F F which reads ( ) L = F ( ) L F / ( ) = F ( ) / L C F. (24.37) F 12 Formal Here, formall (or formal derivation ) means onl demonstrable as mechanical transformation of formulas without an mathematical justification of the procedures. his is the standard usage of formal in mathematics (and throughout this book). 13 In pragmatic thermodnamics, we can be maximall formal and seldom make an mistake. 14 Mathematicall properl speaking, on the left-hand side we regard X and Y as functions of x and, and Y = Y (x, ) is fixed. In contrast, on the right-hand side x and are understood as a function of X and Y, and Y is being kept constant.

9 268 How to manipulate partial derivatives Here, C F is the heat capacit under constant force. It is explained in the following entr Expression of heat capacities he relation between heat and entrop (Clausius equalit) tells us d Q = d, so if we differentiate this with respect to under constant F, it must be the heat capacit under constant F. Generall speaking, the heat capacit under constant X (which can be extensive or intensive) alwas has the following expression: ( ) C X =. (24.38) he stabilit of the equilibrium state implies C X 0 (usuall strictl positive). Imagine the contrar. If we inject heat into a sstem, its temperature goes down, so it sucks more heat from the surrounding world, and further reduces its temperature. hat is, such a sstem becomes a bottomless heat sink. We will learn more consequences of the stabilit of equilibrium states later (ection 25) (Unified) Maxwell s relation All the Maxwell s relations can be unified in the following form 15 X (X, x) = 1, (24.39) (Y, ) where (x, X) and (, Y ) are conjugate pairs. his is the third equalit the reader should memorize. Do not forget that ( P, V ) (not (P, V )) is the conjugate pair. Let us demonstrate this. From + xdx + dy + Maxwell s relation reads ( ) ( ) =. (24.40) Y X hat is, his implies (mere a/b = c/d a/c = b/d) Perhaps, the following formula ma be better: X Y (x, X) (Y, X) = (, Y ) (X, Y ). (24.41) (x, X) (Y, X) = = 1. (24.42) (, Y ) (X, Y ) (X, x) = 1. (24.43) (, Y ) For example, (24.22) can be obtained as follows: ( ) (, ) (, ) (L, F ) = = L (L, ) (L, F ) (L, ) = (L, F ) ( ) F (L, ) =. (24.44) L 15 If the reader knows differential forms, this must be trivial: since de is exact, d 2 E = d( + xdx + dy + ) = 0. If we change onl X and Y, dx dx = d dy. he ratio of the infinitesimal areas on both sides is the Jacobian. hus, the relation is simpl due to thermodnamic quantities being state variables (do not misunderstand that the formula is due to conservation laws).

10 269 How to manipulate partial derivatives Rubber band thermodnamics Equipped with the Jacobian machiner, let us stud the rubber band in more detail. he rubber band is elastic because of the thermal motion of the polmer chains. hat is, resistance to reducing entrop is the cause of elastic bouncing. It is clearl an example of the entropic force (ection 7): obstructing the natural behavior of molecules causes a resistance force. hus, such elasticit is called the entropic elasticit. 16 An important feature is that the elastic force increases with under constant length (which is easil understood from the kid picture Fig. 24.2): ( ) F > 0. (24.45) Is this related to what we have observed (24.1)? Follow the manipulation below (as a practice): ( ) (, ) (, ) (L, F ) 0 < = = L (L, ) (L, F ) (L, ) = (L, F ) (24.46) (L, ) = (L, F ) / ( ) (, L) (F, L) (, L) F = (, L) (L, ) (, L) (, L) =. (24.47) C L Or ( ) 0 < L L (, ) (, ) (, L) ) = = = (, (24.48) (L, ) (, L) (L, ) (, L) C L ( ) (F, L) (, ) F = =. (24.49) (, L) (F, L) C L C L hat is, basicall, we do not need an foresight if we recognize the starting point derivative and the derivative we wish to produce. From our microscopic imagination visualized in Fig. 24.2, we guessed ( ) < 0. (24.50) L Let us derive this from (24.1). ( ) = (, ) L (L, ) = (, ) ( ) (L, ) (L, ) (L, ) = L L C L L < 0. (24.51) Adiabatic cooling with rubber band If a tightl stretched rubber band is equilibrated with room temperature and then suddenl (=adiabaticall) relaxed, it will cool (as we can easil feel experimentall). his is not surprising since ( ) > 0. (24.52) L 16 In contrast, the usual elasticit is called energetic elasticit, which is caused b opposing increase in energ.

11 entrop 270 How to manipulate partial derivatives Here, L is reduced under constant and so must decrease. his is the principle of adiabatic cooling (see Fig. 24.4). Fig Fig (, L 1 ) 2 1 increasing L (, L 2 ) Initiall, the sstem is at 1. Isothermall, L is increased as L 1 L 2 (follow the big arrows). his decreases the sstem entrop. Now, L is returned to the original smaller value adiabaticall and reversibl. he entrop is maintained, and the temperature decreases (adiabatic cooling) to 2. he dotted path is the one we experienced b initial rapid stretching of a rubber band with subsequent cooling (while maintaining the length). In Fig the entrop curves merge in the 0 limit. his is the third law of thermodnamics Consequentl, we cannot reach = 0 with a finite number of adiabatic cooling processes Cooling via adiabatic demagnetization Unfortunatel, we cannot use a rubber band to cool a sstem to a ver low temperature, since it becomes brittle (the chain conformational motion freezes out easil). In actual low temperature experiments, a collection of almost non-interacting magnetic dipoles (i.e., a paramagnetic material) is used. he sstem is closel related to freel jointed polmer chains as illustrated in Fig A A, which is a cop of Fig. 24.2A, corresponds to B a paramagnet, a collection of onl weakl (ideall not) interacting spins. here is, however, a fundamental difference between spins and rotating monomers; while the monomers have the kinetic energ, spins do not. hus it is possible to decouple the spin degrees of freedom from ordinar motional degrees of freedom; we can cool the spins alone. he Gibbs relation of the magnetic sstem is de = d + BdM, (24.53) B where B is the magnetic field (in z -direction), and M the magnetization (the z- component). he correspondences B F and M L are almost perfect: M is the sum of small magnetic dipole vectors, and L is also the sum of steps (monomer orientation vectors) (see Fig. 24.5, but read its caption). hus, we expect ( ) > 0 (24.54) M

12 271 How to manipulate partial derivatives and adiabatic cooling can be realized; first appl a strong magnetic field and align all the dipoles. We can do this slowl and isothermall. hen, turn off the magnetic field to make M 0 (demagnetization). impl replacing L with M in Fig. 24.4, we can understand this adiabatic demagnetization strateg to cool a sstem Ideal magnetic sstem We can imagine a collection of noninteracting magnetic dipoles (called spins) taking onl up or down (or s = ±1) values. 18 he total magnetization of the sstem reads M = µ N s i, (24.55) where µ is the ratio of the magnetic moment and the spin (the gromagnetic ratio). It is a good exercise to compute as a function of M, but, as can be guessed easil from the freel-jointed chain case (cf. (24.10)), i=1 M = µn tanh βµb, (24.56) where B is the magnetic field. he relation between M and B for small B corresponds to Hooke s law (24.11): M = (Nµ 2 /k B )B, (24.57) which is called Curie s law: the magnetic susceptibilit χ = Nµ 2 /k B. Just as in the freel-jointed polmer model we discussed (without the e term), there is no kinetic energ of spins (or of the entities carring spins), so the entrop of this model is constant under constant M. hus, just as in the ideal rubber model, without the term similar to e the model cannot explain the use of adiabatic demagnetization to cool other sstems as a refrigerating mechanism. However, if we are interested in the spins themselves, then their coupling to other degrees of freedom (the so-called spin-lattice coupling) should not be large. Under this condition, the sstem can be described b the present model, so under the adiabatic demagnetization condition M is constant, because is constant. hus, since (24.57) or (24.56) implies B/ is constant, reducing B implies decreasing. 19 If we use the magnet as a coolant to cool other sstems, we are interested in ( ) (, ) (, ) (B, M) M) = = = (B, (24.58) B (B, ) (B, M) (B, ) (B, ) = (B, ) (B, M) (B, ) (B, ) = ( ) M. (24.59) C B B 17 Here, we discussed using the magnet to cool other sstems. o do this, the spins must couple with the lattice degrees of freedom. However, if one wishes to cool the spins onl, then this coupling is not needed (as discussed in the next entr 24.16). 18 If magnetic atoms are dilute in an insulating solid, the do not interact with each other appreciabl. 19 For a rubber band, reducing F while keeping L constant in order to change is experimentall unthinkable.

13 272 How to manipulate partial derivatives herefore, if Curie s law of the form M = a(b/ ) holds, then gives the cooling rate. δ = Q24.1[Basic problems] (1) F = x sin, and = x + z. Express ( ) F ab δb (24.60) C B and ( ) F z (24.61) in terms of x and. [sin ; sin + x cos ] (2) For a gas P V and E are functions of onl. how that actuall P V/ is a constant. [Compute ( E/ V ) = ( / V ) P = ( P/ ) V P = 0.] (3) For a general gas, find the temperature change d due to adiabatic free expansion V V + dv. [Compute ( / V ) E. ( / V ) E = [P ( P/ ) V ]/C V ] (4) If M is a function of B/, E is a function of onl. [how ( E/ B) = 0 using ( B/ ) M = B/ ; quite parallel to (2).] Q24.2 [Negative temperature is ver hot] Let us consider a two-state-spin sstem containing 1 mole of spins (cf ). Assume that under the magnetic field B, the energ gap between the up and down spin states is 600 K. uppose the initial temperature of the magnet is 500 K. 20 (1) What is the temperature of this sstem measured 21 with an ideal gas thermometer containing 10 6 moles of monatomic gas particles? (Assume the initial temperature of the thermometer is around room temperature, so its intial internal energ ma be ignored.) (2) If, instead, the original negative temperature sstem is thermall equilibrated with a 1 mole of ideal monatomic gas that is initiall 200 K, what is the equilibrium temperature? oln (1) he relation between the magnetization and the temperature can be solved as (2µB/k B = 600 K) M = µn A tanh 300. (24.62) Energ conservation tells us ( M B is the energ of the magnet in the magnetic field) 300N A k B tanh = 300k BN A tanh nk B. (24.63) In this case we ma expect that the temperature is extremel high, so 300N A tanh = 161N A 3 2 n = = K. (24.64) ince the temperature is outrageousl high, we must pa attention to relativit. hat is, 20 Is < 0 possible? We have alread discussed (see Q14.1 (2)) the impossibilit of < 0, if the spatial degrees of freedom are involved. Here, we assume that the spins and lattice vibrations (phonons) are for a sufficientl long time decoupled. 21 In this case, upon contact with the thermometer, the sstem temperature changes drasticall, so the process cannot be a temperature measuring process. Here, we simpl wish to know what happens after the drastic change.

14 273 How to manipulate partial derivatives the gas must be superrelativistic. hen, is the temperature higher or lower? (You can of course get the answer quantitativel.) (2) 161N A k B N Ak B = 300N A k B tanh N Ak B. (24.65) hat is, we must solve (numericall) hat is, = 374 K. 361 = 300 tanh (24.66) 2 Q24.3 [Ideal rubber model example] For a rubber band the tensile force F is given b ( ) L F = a L2 0, (24.67) L 0 L 2 where a and L 0 are positive constants. Also the constant length heat capacit C L is independent of. (1) Find the entrop of the band. (2) If the band is adiabaticall and reversibl stretched from the initial length L = L 0 at = 0 to the final L = 2L 0, what is the final temperature? oln. (1) he Gibbs relation reads de = d + F dl, so and hus, d = 1 de F dl (24.68) ( ) ( ) E C L = = C L. (24.69) L L d = C ( ) L L d a L2 0 L 0 L 2 dl. (24.70) his is an exact form (or perfect differential), so (, L) = ( 0, L 0 ) + C L log { L 2 L 2 ( 0 1 a + L L 0 L 1 )} L 0 (2) hus (24.71) ince the change is adiabatic and reversible, or = 0 e al 0/C L > 0. (, 2L 0 ) = ( 0, L 0 ) + C L log 0 al 0 (24.72) C L log 0 = al 0 (24.73)