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Understanding the Second Law of Thermodynamics: The Role of Probability in Natural Processes

Explore the fundamental concept of the second law of thermodynamics by delving into the significance of probability in natural irreversible processes. This guide covers the flow of heat, two-state systems, multiplicities, and the Einstein model of a solid.

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Understanding the Second Law of Thermodynamics: The Role of Probability in Natural Processes

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  1. Second Law -Probability In a nutshell… Natural irreversible processes are

  2. Second Law -Probability In a nutshell… Natural irreversible processes are overwhelmingly probable.

  3. Second Law -Probability In a nutshell… Natural irreversible processes are overwhelmingly probable. We want to answer: what is T ?

  4. Second Law -Probability In a nutshell… Natural irreversible processes are overwhelmingly probable. We want to answer: what is T ? why does heat flow from hot cold

  5. Second Law -Probability In a nutshell… Natural irreversible processes are overwhelmingly probable. We want to answer: what is T ? why does heat flow from hot cold We will use combinatorics

  6. Two State Systems • There are many two state systems magnetism, excitation – deexcitation

  7. Two State Systems • There are many two state systems magnetism, excitation – deexcitation • This is easily modeled possible outcomes

  8. Two State Systems • There are many two state systems magnetism, excitation – deexcitation • This is easily modeled possible outcomes • Terminology • MICROSTATE every possible outcome

  9. Two State Systems • There are many two state systems magnetism, excitation – deexcitation • This is easily modeled possible outcomes • Terminology • MICROSTATE every possible outcome • MACROSTATE equivalent outcomes

  10. Two State Systems • There are many two state systems magnetism, excitation – deexcitation • This is easily modeled possible outcomes • Terminology • MICROSTATE every possible outcome • MACROSTATE equivalent outcomes • MULTIPLICITY number of microstates in a macrostate Ω(n) if there are n microstates

  11. Probability • For a two state system (heads and tails)

  12. Probability • For a two state system (heads and tails) • Probability of n heads = Ω(n)/Ω(all)

  13. Probability • For a two state system (heads and tails) • Probability of n heads = Ω(n)/Ω(all) • The multiplicity can be found by Binomial • Coefficient

  14. Probability • For a two state system (heads and tails) • Probability of n heads = Ω(n)/Ω(all) • The multiplicity can be found by Binomial • Coefficient ( Nn ) = N! • n!(N-n)!

  15. Probability • For a two state system (heads and tails) • Probability of n heads = Ω(n)/Ω(all) • The multiplicity can be found by Binomial • Coefficient ( Nn ) = N! • n!(N-n)! • Lets apply this to a two-state paramagnet

  16. PARAMAGNETISM • A two-state paramagnet showing spin up and spin down dipoles.

  17. Probability The multiplicity of an up state is: • Ω (Nup) = N! Nup! Ndn! N = Nup + Ndn The probability for an up state is Probability = Ω (Nup) = 1/ 2Nup! Ndn! ΣΩ(N)

  18. Einstein Model of a Solid • Consider a system composed of a collection of microscopic systems that can store any number of energy units.

  19. Einstein Model of a Solid • Consider a system composed of a collection of microscopic systems that can store any number of energy units. • PE = ½ ks x2

  20. Einstein Model of a Solid • Consider a system composed of a collection of microscopic systems that can store any number of energy units. • PE = ½ ks x2 Size of E = hf

  21. Einstein Model of a Solid • Consider a system composed of a collection of microscopic systems that can store any number of energy units. • PE = ½ ks x2 Separation of E = hf • f = 1 / (2π) √(ks/m)

  22. Einstein Model of a Solid • Consider a system composed of a collection of microscopic systems that can store any number of energy units. • PE = ½ ks x2 Separation of E = hf • f = 1 / (2π) √(ks/m) This model represents vibrational modes of diatomic and polyatomic gas molecules and also a solid.

  23. Einstein Model of a Solid • Consider N classical oscillators having frequency f, spaced in energy E = hf

  24. Einstein Model of a Solid • As an example let’s look at Einstein solid containing three oscillators N = 3

  25. Einstein Model of a Solid • As an example let’s look at Einstein solid containing three oscillators N = 3 • Also there are three units of energy q = 3

  26. Einstein Model of a Solid • As an example let’s look at Einstein solid containing three oscillators N = 3 • Also there are three units of energy q = 3 What is the multiplicity of Ω (0) = ?

  27. Einstein Model of a Solid • As an example let’s look at Einstein solid containing three oscillators N = 3 • Also there are three units of energy q = 3 What is the multiplicity of Ω (0) = ? Here N=3, q=0

  28. Einstein Model of a Solid • As an example let’s look at Einstein solid containing three oscillators N = 3 • Also there are three units of energy q = 3 What is the multiplicity of Ω (0) = ? Here N=3, q=0 Ω (N’ ,q) = (q + N -1 q ) = Ω(3,0)

  29. Einstein Model of a Solid • As an example let’s look at Einstein solid containing three oscillators N = 3 • Also there are three units of energy q = 3 What is the multiplicity of Ω (0) = ? Here N=3, q=0 Ω (N’ ,q) = (q + N -1 q ) = Ω(3,0) = (3-1)! / (0!2!) = 1

  30. Einstein Model of a Solid • As an example let’s look at Einstein solid containing three oscillators N = 3 • Also there are three units of energy q = 3 What is the multiplicity of Ω (0) = ? Here N=3, q=0 N’ = q+N-1 Ω (N’ ,q) = (q + N -1 q ) = Ω(3,0) = (3-1)! / (0!2!) = 1 Argument to show N’ = q+N-1 is in text p 55

  31. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2

  32. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) =

  33. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!)

  34. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6

  35. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3

  36. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10

  37. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10 Ω (0) = 1,

  38. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10 Ω (0) = 1, Ω (1) = 3,

  39. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10 Ω (0) = 1, Ω (1) = 3, Ω (2) = 6,

  40. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10 Ω (0) = 1, Ω (1) = 3, Ω (2) = 6, Ω (3) = 10

  41. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10 Ω (0) = 1, Ω (1) = 3, Ω (2) = 6, Ω (3) = 10, Ω(Total) =

  42. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10 Ω (0) = 1, Ω (1) = 3, Ω (2) = 6, Ω (3) = 10, Ω(Total) = ΣΩ(q)

  43. Einstein Model of a Solid Multiplicity of Ω (2) = ? N=3, q=2 Ω (4,2) = (q + N -1 q ) = (2+3-1)! / (2!(3-1)!) = 4!/(2!2!) = 6 Going through similar treatment for q = 3 Ω(q) = Ω (3) = 10 Ω (0) = 1, Ω (1) = 3, Ω (2) = 6, Ω (3) = 10, Ω(Total) = ΣΩ(q) =20

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