Thermodynamics: Chapter 02 The Second Law of Thermodynamics: Microscopic Foundation of Thermodynamics. September 10, 2013

Transcription

1 Thermodynamics: Chapter 02 The Second Law of Thermodynamics: Microscopic Foundation of Thermodynamics September 10, 2013

2 We have talked about some basic concepts in thermodynamics, T, W, Q, C,.... Some may not be as real as others, such as heat - just another way of saying energy flow. We also tried to understand these concepts using a simply system - the Ideal Gas System. We saw the connection between the macro properties and micro-structure. In thermodynamics one key concept is energy: The thermal system stores energy, transfers energy from one system to another, convert energy (work heat or thermal energy). Energy is important also for practical reasons, we have to deal with it all the time: from refrigerator, engine, to fundamental research of the universe. The new concept of Dark Energy. Despite the difference among thermal systems, there are something that apply to all systems, such as the heat flows spon- 1

3 taneously from a hotter object to a cooler one. Closed systems always evolve toward equilibrium state (thermal, mechanical, chemical,...). What actually drives these irreversible processes? In order to understand these precisely, we need to study HOW system store energy.

4 1. Macrostate, Microstate, Multiplicity, and Large Numbers 1.1 Macrostate, Microstate, Multiplicity Macrostate: Specify the state distribution of all particle in the system: How many particles are in state-a, how many in state-b, so on so forth, Microstate: Specify the state of each particle in the system. 2

5 Multiplicity: The number of microstates corresponding to a given macrostate, Ω( M acrostate ).

6 Penny Nickel Dime 1 H H H each called microstate 2 H H T each called microstate 3 H T H each called microstate 4 T H H each called microstate all these three states for (2H, 1T) is called a macrostate 5 H T T each called microstate 6 T H T each called microstate 7 T T H each called microstate all these three states for (1H, 2T) is called a macrostate 8 T T T one microstate, also a macrostate

7 We also say (H, H, H) or (T, T, T) is better organized. (H, H, T) or (T, T, H) are more randomized. 1.2 Large Numbers and Useful Relations - a quick review Scientific notation: 13, 578, 935, 204, , 578, 935, 204, , 578, 935, 204,

8 ( ) ( ) = = Multiplication Addition Exponentiation Multiplication 13, 578, 935, 204, ( ) Logarithm Addition log(13, 578, 935, 204, ) log ( ) log log(10 13 ) = log

9 Factorial (n!): The factorial of a non-negative integer n (including 0!) is the product of all positive integers less than or equal to n. Combinatorics - Permutations and Combinations: They are two related problems, both counting possibilities to select k distinct elements from a set of n elements, where for k-permutations the order of selection is taken into account, but for k-combinations it is ignored.

10 PN k = N (N 1) (N k + 1) = N! (N k)! CN k = N (N 1) (N k+1) k! = N! k!(n k)!

11 Sept. 05 ******* Some math relations useful for dealing with large numbers: lnab = lna + lnb lna b = blna ln(a + b) b a lna + b/a d dx lnx = 1 x ln(1 + x) x when x is small Another very useful approximation is the Stirling s approxima- 3

12 tion: N! N N e N 2πN which is roughly approximated by the following when N is very large: ( ) N N N! = N N e N e lnn! ln ( N N e N ) = NlnN N 1.3 Macrostate, Microstate, Multiplicity for a system with N coins (or two-state paramagnet) (1) Number of Macrostates: = N+1. They are (0-Head, N-Tail), (1 H, N-1 T),..., (N H, 0 T)

13 (2) Number of Microstate: = 2 N : each cion has two possible states (3) Multiplicity function for macrostate (k H, N-k T) : Ω(kH, (N k)t ) = C k N = N! k! (N k)!. The probability of a Macrostate that has k Heads: P kh = Ω(kH,(N k)t ) 2 N Paramagnet: a material consists of particles that are like small magnetic needles. These needles aline parallel to any external magnetic field. Ferromagnet: (alined by internal interactions.)

14 2. One thermal system with quantized microstates: Einstein s Model of a Solid Let s look at an example that is more complicated and closer to a real thermal system in reality, the Einstein Model of a Solid - First proposed by A. Einstein in A collection of micro-systems each of which can store any number of energy units. 4

15 An example of such micro-systems is the harmonic oscillator : In quantum mechanics, the size of the energy units is ω ( is Planck s Constant and ω is the angular frequency of the oscillator.) The energy levels for a oscillator is { 1 2 ω, 3 2 ω, 2 5 ω,... } A system with N atoms will have 3N oscillators (threedimensional solid, no rotation and other degree of freedom). Microstate: specify energy levels for every individual

16 oscillator in the solid. Macrostate: specify the total particles and the total thermal energy inside the solid, corresponding to N oscillators occupying total q energy units. Multiplicity: Ω(N oscillators, q energy units ) = Ω(N o, q e ). Note: Some oscillators may have zero energy unit - they are on ground state. What is Ω(N o, q e )? Let s look at a simple case Ω(5 o, 8 e ), using represents one energy unit:

17 os 1 os 2 os 3 os 4 os 5 E=0 E= The number of possible arrangements is the number of: (1) ways of choosing 8 s from 8 + (5 1) (total spots to fill with either or ), or Ω(5, 8) = C = C8 12.

18 (2) ways of choosing 4 s from 8 + (5 1), that is Ω(5, 8) = C = C4 12 In-class exercise 01: Prove C q N = CN q N. Generally, the multipiicity function can be written as: Ω(N o, q e ) = C q N+q 1 (1) Attention-1: H harmonic oscillator = p2 2m Kq2, the phase space is 2 dimensional (q, p), i.e. the Einstein solid has 2-d of freedom.

19 Attention-2: In a 3-d Einstein solid, if we have N oscillators (particles in the formular, each with one or more energy unit ω). In terms of atoms, there are total N/3 atoms. Let s take a look at an special case (not so special in reality though) - a large piece of solid (N is big!) at high temperature (q is very big, q N!).

20 Ω(N, q) = C q N+q 1 = (q + N 1)! q!(n 1)! lnω(n, q) ln(q + N)! lnq! lnn! (q + N)! q!n! (2) (q + N)ln(q + N) qlnq NlnN (3) ln(q + N) = ln[q(1 + N q )] Since we consider q N, lnq + N q (4) lnω(n, q) (q + N)(lnq + N q ) qlnq NlnN lnω(n, q) Nln ( q ) + N + N 2 N q Nln ( q N ) + N (5)

21 Exponentiating it will give: Ω(N, q) e Nln ( q N ) e N Ω(N, q) e ln ( q N )N e N Ω(N, q) Ω(N, q) ( q ) N N ( eq N ) N e N (6) If q or N increases a lit bit, Ω will increase a lot due to the large N - one important characteristics of thermal system - defined to have large number of particles.

22 In-Class Exercise 02: Calculate the multiplicity of Einstein Solid at low-temperatire limit, q N. [Problem 2.17 on p. 64]

23 3. Interacting solids and multiplicity function of two thermal systems To understand heat flow and the evolution of a thermal system (reversible or irreversible), we need to consider two systems that interact with each other with heat and/or matter exchanges. 5

24 Solid A N A, q A, U A, Ω A Solid B N B, q B, Ω B, U B Assuming system-a and system-b each has 3 parti-

25 cles: N A = N B = 3 (No particle exchange!!). Due to energy conservation, we have q total = q A + q B, assuming q total = 6. The macrostate (specified by q A and q B!) and multiplicity (Ω total ) of the combined system are: Using what we obtained about the Einstein Solid + three constraints N A = 3, N B = 3 and q A + q B = 6:

26 q A Ω A q B Ω B Ω total = Ω A Ω B Probability Total: 462 Note: When N B = 3, q B = 5, we have C = C5 7 = C 2 7 = = 21

27 Indeed, we have 462 = C 6 11! So, the combined two systems is just like one system that has N = 6 and q = 6. We can further assume that over long time, all the 462 microstates are equally probable. The so-called Fundamental Assumption of Statistical Mechanics : In an isolated thermal system at equilibrium state, all accessible microstates are equally probable. Based on this assumption, we can calculate the probability of a certain macrostate occurring in the combined system, which is given by the last column in the

28 table. The observation is, although all the accessible microstates are equally probable, the macrostates are NOT equally accessible. Now you see why different macrostates occur differently (probability of their appearance) in a system. Let s look into this more closely in actual situation, a system consists a lot of particles, (thanks to the math skills I just introduced), the sharpness of multiplicity function.

29 Assuming each solid has N particles and q = q A + q B energy units, using Eq. 6 [high temperature limit, q N], the multiplicity of the combined system is Ω = ( eqa N ) N ( eq B N ) N = ( e N ) 2N (q A q B ) N (7)

30 ( e ) 2N ( q ) 2N q A =q B =q/2 Ω max = N 2 to see the offset from the peak: q A = q 2 + x, q B = q 2 x Ω = ( e N ) 2N ( q 2 ) 2 x 2 N (8) (9)

31 ln ( q 2 ( q 2 ) 2 x 2 N = Nln = Nln = N ln ( q ( q ) 2 x 2 2 ) 2 1 2x 2 q ( q ) 2 2 ( q ) 2 + ln N ln 2 ) 2 x 2 N e Nln ( q 2 )2 e N( 2x 2x q q ) x q 2

32 which gives ( q 2 ) 2 x 2 N ( q 2 ) 2N e N( 2x q ) 2 (10) Compare Eq.s [8, 9, 10], one has Ω Ω max e N( 2x q ) 2 (11)

33 Let s see how q affects the distribution: 6

34 Let s see how N affects the distribution: For a Gaussian function, 1 e of the maximum gives the half width: 1 e Ω max Ω max e N 1 e = e N 1 = N x half width = q 2 N ( ) 2xhalf 2 width q ( ) 2xhalf 2 width ( ) 2 2xhalf width q q (12) The relative width 2x half width q = 1 N, which becomes very small when N is huge (as in reality).

35 Sept. 10 ******* 4. One thermal system with continuous microstates: Ideal Gas Ideal gas system is more complicated that the Einstein s Solid: 1. Microstate is continuous : position, momentum of particles. 2. Multiplicity depends on volume, total energy, number of particles. 7

36 3. Gas can expand and be compressed, which means work may contribute in thermal process. 4. Exchange particles in addition to energy. 4.1 Macrostate, Microstate, Multiplicity for ideal gas system In classical mechanics, to fully describe the state of a particle, we need to use coordinate and momentum of

37 this particle, for example (x, p x ) for particle moving in 1-D space, (x, y, z, p x, p y, p z ) for particle moving in 3- D space. The coordinate-momentum space is called phase space. This concept can be extended to abstract phase space for generalized coordinate and momentum. You will learn this in Classical Dynamics (Level-4 course.) The oscillator has Hamiltonian (like total energy) H(q, p) = p 2 2m Kq2. Energy conservation (at moments between exchanges/collision) requires p2 2m +1 2 Kq2 = constant

38 (like elipse function x2 A 2 + y2 B 2 = constant). Over long time with energy exchange among particles, the phase space and microstate will be like this in this figure:

39 p dp p H H+dH q dq Phase space of a free par,cle (le/) and a harmonic oscillator (right) Microstate with q and p within q~q+dq and p~p+dp q A=en,on: In Ideal Gas, energy of individual par,cles may vary (H). Total energy is conserved E total =constant (isolated system).

40 For an isolated system, the macrostate of it can be described by (U, V, N). Each specific value of (U, V, N) corresponds to large amount of microstates, Ω(E, V, N), which can be written as Ω(E, V, N) = σ(e, V, N) σ 0 (13) where σ(e, V, N) = E=H(q v,p v ) dσ is the the area of the hypersphere defined by E = H(q v, p v ). Sometimes this is not an easy integration. Note: U: Total energy of the thermal system, E: energy of single particle.

41 Another question: What is actually the size in the phase space that corresponds to a microstate of a particle, the σ 0? We need to use Quantum Mechanics to find the answer. The famous Heisenberg Uncertainty Principle says: δx δp x h, (14) where h is the Planck s constant, a very small number: h = J s.

42 So, in phase space for a particle with 1-D motion, the the size corresponding to a microstate is Ω 1 = L δx L p δp x LL p h. (15) In phase space for a particle with 3-D motion, the the size corresponding to a microstate is That s for system with one particle. Ω 1 V V p h 3. (16)

43 How about two particles? Ω 2 = Ω 1 Ω 2 V 2 h 6 V p1 2 V p2 2 with constraint on momentum space by E-conservation for two particles: 1 2m (p2 1x + p2 1y + p2 1z + p2 2x + p2 2y + p2 2z ) = U p 2 1x + p2 1y + p2 1z + p2 2x + p2 2y + p2 2z = 2mU (17) Ω 2 V 2 (area of 6-D momentum hypersphere h6 of radius = 2mU) (18)

44 For two identical particles, Ω V 2 h 6 (area of momentum hypersphere.) (19) which can be extended to N identical particles: V N Ω N 1 (area of 3N-D momentum hypersphere.) N! h3n (20) The area of momentum hypersphere can be calculated. The results for d-d hypersphere is A = 2πd/2 ( d 2 1)!r(d 1) (21)

45 r: radius of the momentum hypersphere. It is the magnitude of the momentum of the particle. For a system with d = 3N and r = 2mU, V N Ω N = 1 N! h 3N 1 N! h V N 3N π3n/2 2π3N/2 ( 3N 2 1)!( 2mU) (3N 1) ( 3N 2 )!( 2mU) 3N (22) And the relation with V and U is: Ω N (U, V ) = f(n)v N U 3N/2 (23)

46 4.2 Interacting ideal gas systems System-A: (N, V A, U A ). System-B: (N, V B, U B ). Heat exchange allowed: The total energy of the two system is fixed. The total multiplicity of the system is the product of Ω A and Ω B : Ω total = [f(n)] 2 (V A V B ) N (U A U B ) 3N/2 (24) V A and V B stay unchanged. 8

47 To see where the peak of Ω total is, we can do the following: U A = U total U B Ω total = C[(U total U B )U B ] 3N/2 d du B Ω total = 3N 2 C[(U total U B )U B ] 3N/2 1 [U total 2U B ] 3N 2 C[(U total U B,peak )U B,peak ] 3N/2 1 [U total 2U B,peak ] = 0 U B,peak = U total 2 This is the only solution because we know U total U B,peak and U B,peak 0. The width of the peak is width = U total 3N/2 (25)

48 Volume change allowed: The total volume of the two system is fixed. The total multiplicity of the system is the product of Ω A and Ω B : Ω total = [f(n)] 2 (V A V B ) N (U A U B ) 3N/2 (26) U A and U B stay unchanged. Similarly, the peak of Ω total and the width of the peak are V B,peak = V total 2 (27) width = V total N (28) If we let both energy and volume exchange, the multiplicity function looks like this, with a sharp peak when the number of particles increases.

49 exp(-100.0*(2.0*x/ )^ *(2.0*y/ )^2.0) Ω total U_A V_A

50 Comments: Microstates of Ideal Gas by Quantum Mechanics. E The wavefunc/on and energy of par/cles in a box E 1 E 0

51 The wavefunction and single particle energy are given by Ψ nx,n y,n z = Asin(k x x)sin(k y y)sin(k z z) = Asin n xπx sin n yπy sin n zπx L x L y L z n x, n y, n z = 1, 2, 3,... (29) ɛ nx,n y,n z = 2 2m (k2 x + k 2 z + k 2 z ) (30) Classical mechanics: particle state is given by { q, p}, one point in q p phase space. Quantum mechanics: particle state is given by {n x, n y, n z }, one point in a 3-D space!. We will learn more when we get to quantum statistics later in the Statistical Physics.

52 5. Entropy and the Second Law of Thermodynamics Using the terminology macrostate, we have observed, interestingly, that thermal systems tend to reach the equilibrium state in when the multiplicity is at its maximum (±δ) value. This can be summarized as (the Second Law of Thermodynamics) : 9

53 Any large system in equilibrium will be found in the macrostate with the greatest muitiplicity (aside from fluctuations that are normally too small to measure). If we start from a non-equilibrium system, this can be said as:

54 Multiplicity tends to increase as time goes. Isolated systems evolve from a organized system to a more randomized system! Discussion: Is there a connection between this law and the irreversible aging process of our physical body? What hidden force caused the growth from the beginning of our life?

55 5.1 Entropy and its connection with the 2nd law of thermodynamics. Now, let s introduce the concept entropy. S = klnω (31) where k is Boltzmann s constant, k = J/K = ev/k. Practically, the convenience of such definition of logarithm of multiplicity relies in:

56 (1) the actual number of S is small and easy to handle (rather than astronomical values Ω often has). (2) the S of combined system is the sum (rather than the production) of the Ss of the subsystems. In terms of entropy, the second law of thermodynamics becomes Any large system in equilibrium will be found in the macrostate with the greatest entropy (aside from fluctuations that are normally too small to measure).

57 If we start from a non-equilibrium system, this can be said as: Entropy tends to increase as time goes. 5.2 Entropy of the systems and processes (1) Einstein s Solid with N oscillators and q units of energy and q N (high temperature approximation): S = kln ( eq ) N ( N = knln eq ) N

58 S = kn [ ln ( ) ] q N + 1 (2) An Ideal Gas: Using Eq. (22), S = kln [ 1 N! V N h 3N π3n/2 (3N/2)! (2mU)3N/2 Stirling s approximation: N! 1 N! V N h 3N π3n/2 (3N/2)! (2mU)3N/2 ( e ) N 1 V N π 3N/2 N 2πN = h 3N ( 2e 3N ] ( Ne ) N 2πN ) 3N/2 1 3πN (2mU) 3N/2 ( VN ) N e N+3N/2 1 2πN π 3N/2 h 3N ( 2 3N ) 3N/2 1 3πN (2mU) 3N/2

59 = ( ) VN N ( e 5N/ ) 3N/2 2πN h 3N 1 3N (4πmU) 3N/2 3πN ) 3N/2 (4πmU) 3N/2 e 5N/2 = 1 2πN 1 3πN ( VN ) N ( 1 3Nh 2 ( ) VN N ( 4πmU ) 3N/2 3Nh e 5N/2 [ 2 (V ) N ( 4πmU ) 3N/2 N 3Nh e 5N/2] 2 [ (V ) ( ) 4πmU 3/2 N 3Nh e 5/2] 2 S kln knln S kn ln V N ( 4πmU 3Nh 2 This gives ) 3/2 + 5 (32) 2 We have arrived at the famous Sackur-Tetrode Equa-

60 tion. (Very useful. Will see it again in Chapter 3.) One application: If we change the gas from V i to V f while keeping U and N fixed, we will have S knln V f V i What causes entropy to increase in this scenario? (1) Putting heat into the system: Quasistatic isothermal expansion (U thermal = Nf 1 2kT : T remains fixed

61 U=fixed) (2) Free expansion into vacuum: U = Q + W = 0 Both (1) and (2) satisfy the condition of U and N being fixed! Discussion: Will slow (quasistatic) expansion or compression by itself (no heat flow!) change the entropy of a system? - No.

62 Note-1: U thermal = Nf 1 2 kt and T = 2 3 < ɛ kin > /k Quantum: V - smaller, U - higher. Note-2: Quantum Mechanics: When higher energy levels are occupies, for slow process, molecules will NOT change to different energy states, The energy arrangement remains the same No change in entropy. (3) Entropy of mixing and the Gibbs paradox:

63 Mixing two boxes of different gases, each starts from 1/2 of the final volume: We can use Eq. (32), the Sackur-Tetrode Equation, S A = S A,final S A,initial = Nkln V f V i = Nkln2 S B = Nkln V f V i = Nkln2 (33) The entropy of mixing is: S total = S A + S B = 2Nkln2 (34) It is a net increase!

64 In-Class Exercise 03: Calculate the entropy of mixing of the two types of gases whose relative proportion is arbitrary. [Problem 2.37 on p. 81] Gibbs paradox: We have learned the entropy of ideal gas: S = kln [ 1 N! V N h 3N 1 ] π3n/2 (3N/2)! (2mU)3N/2 The factor N! was introduced so that it works for identical particles. If we assume particles are distinguishable, it should be S = kln [ V N h 3N π3n/2 (3N/2)! (2mU)3N/2 ]

65 Using Stirling s approximation: N! repeat what we did, V N h 3N π3n/2 (3N/2)! (2mU)3N/2 V N π 3N/2 h 3N ( 2e 3N ) 3N/2 1 3πN (2mU) 3N/2 = V N e 3N/2 π 3N/2 h 3N ( 2 3N ) 3N/2 1 3πN (2mU) 3N/2 = V N e 3N/2 ( 1 1 ) 3N/2 h 3N 1 3N (4πmU) 3N/2 3πN = 1 V N ( 1 ) 3N/2 (4πmU) 3N/2 e 3N/2 3πN V N ( 4πmU S kln 3Nh 2 ) 3N/2 e 3N/2 3Nh [ 2 V N ( 4πmU 3Nh 2 ) 3N/2 e 3N/2] ( Ne ) N 2πN and

66 [ knln V [ kn lnv ( ) 4πmU 3/2 3Nh e 3/2] 2 ( ) ] 4πmU 3/2 3Nh This introduces disturbing consequence: If we insert a partition into a box of He, dividing it into half, the entropy of the final state (Ω box/2 Ω box/2 ) is less than the original state (Ω box ): S 1 = k(n/2) S 2 = k(n/2) [ ln(v/2) [ ln(v/2) ( ) ] 4πm(U/2) 3/2 3(N/2)h ( ) ] 4πm(U/2) 3/2 3(N/2)h

67 [ S 1 + S 2 = 2k(N/2) ln(v/2) [ ( S 1 + S 2 = kn ln(v/2) 4πmU < kn [ lnv ( 4πmU 3Nh 2 3Nh 2 ) ] 3/ ( ) ] 4πm(U/2) 3/2 3(N/2)h ) ] 3/ = S before inserting This is against the 2nd law! This was first pointed out by J. Willard Gibbs, and is called the Gibbs paradox. For identical particles, we have to use quantum mechanics.