1 / 44

Thermal & Kinetic Lecture 14 Permutations, Combinations, and Entropy

Thermal & Kinetic Lecture 14 Permutations, Combinations, and Entropy. Overview. Permutations and combinations. Interacting atoms and energy transfer. Distribution of energy at thermal equilibrium. Last time…. Diffusion The Einstein model of a solid Oscillators and quanta

bruno-solomon
Télécharger la présentation

Thermal & Kinetic Lecture 14 Permutations, Combinations, and Entropy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Thermal & Kinetic Lecture 14 Permutations, Combinations, and Entropy Overview Permutations and combinations Interacting atoms and energy transfer Distribution of energy at thermal equilibrium

  2. Last time… Diffusion The Einstein model of a solid Oscillators and quanta Permutations and combinations.

  3. Energy distributions Consider bringing two identical blocks together. What is the most probable distribution of energy amongst the two blocks? Most probable distribution is ‘intuitively’ that where total thermal energy is shared equally between the two blocks. However, what is the probability that the first block has more energy than the second or, indeed, ends up with all the thermal energy? Need to consider possible arrangements of energy quanta……

  4. Oscillator 2 Oscillator 1 ? The arrangement above is one possible distribution of 3 quanta amongst two oscillators. Sketch the remaining possibilities. How many possibilities in total are there? Energy distributions We have 3N independent simple harmonic oscillators (where N is the total number of atoms in the solid). Number of ways of distributing quanta of energy amongst these oscillators? Say we have 3 quanta of energy to distribute amongst 2 oscillators:

  5. Oscillator 2 Oscillator 1 Oscillator 2 Oscillator 1 3 energy quanta distributed between 2 oscillators ANS: Total of 4 possibilities (including arrangement shown on previous slide) Oscillator 2 Oscillator 1

  6. ? A CD reviewer is asked to choose her top 3 CDs from a list of 10 CDs and rank them in order of preference. How many different lists can be formed? (ABC ≠ BAC) Counting arrangements Clearly, we are not going to count by hand every arrangement of energy possible for 3N oscillators in, e.g., a mole of solid (N ~ 6 x 1023). Need to consider permutations and combinations. A permutation is an arrangement of a collection of objects where the ordering of the arrangement is important. ANS: There are 10 choices for the 1st CD, 9 for the 2nd, and 8 for the 3rd. Hence, 720 different lists. The number of permutations of r objects selected from a set of n distinct objects is denoted by nPr where nPr = n! / (n - r)!

  7. The number of combinations of r objects selected from a set of n distinct objects is denoted by nCr where Counting arrangements ? A CD club member is asked to pick 3 CDs from a list of 10 CDs. How many different choices are possible? ANS: We need to divide the previous 720 arrangements by the total number of different possible permutations of 3 choices (e.g. ABC = BAC = CAB). This is 3! permutations. Hence, 120 choices are possible.

  8. ANS: = 210 arrangements Counting arrangements ? Take a collection of 10 pool (billiard) balls, 6 of which are yellow and 4 of which are red. How many different arrangements of the coloured balls are possible (eg RRYYYYYYRR)?

  9. = 2 1 1 2 = 2 1 1 2 Counting arrangements Returning to the distribution of energy quanta amongst a collection of oscillators, we need to establish a formula for the number of possible arrangements. Consider the case of 3 quanta of energy distributed between 2 oscillators as before. We’ll adopt the same representation as Chabay and Sherwood, p. 348……. Thus, we have 4 objects arranged in a certain sequence. We need N - 1 vertical bars to separate N oscillators.

  10. This problem thus reduces to the pool ball problem except instead of red and yellow pool balls we have to arrange | and objects. = Total number of objects So, total number of arrangements of 3 quanta amongst 2 oscillators = 2 1 1 2 Number of boundaries between oscillators (= N-1) Number of quanta Counting arrangements

  11. ? How many ways can 4 quanta of energy be arranged amongst four 1D oscillators? Counting arrangements Number of ways to arrange q quanta of energy amongst N 1D oscillators:

  12. How many ways can four quanta of energy be arranged amongst four oscillators? • 21 • 42 • 256 • 35

  13. ? How many ways can 4 quanta of energy be arranged amongst four 1D oscillators? ANS: = 35 ways Counting arrangements

  14. Microstates and macrostates Each of the 35 different distributions of energy is a microstate (i.e. an individual microscopic configuration of the system, as shown above). The 35 different microstates all correspond to the same macrostate - in this case the macrostate is that the total energy of the system is 4 FUNDAMENTAL ASSUMPTION OF STATISTICAL MECHANICS Each microstate corresponding to a given macrostate is equally probable.

  15. ? What’s the probability of being dealt the hand of cards shown above in the order shown? Microstates and macrostates: poker hands

  16. What’s the probability of being dealt the hand of cards in the order shown? • 1/2112 • 1/8192 • 1/2,256,781 • 1/311,875,200

  17. ? What’s the probability of being dealt the hand of cards shown above in the order shown? Microstates and macrostates: poker hands

  18. As compared to the “junk” hand of cards, the probability of being dealt the Royal Flush is: • Higher • Lower • Exactly the same • Don’t know

  19. ? What’s the probability of being dealt the hand of cards shown above in the order shown? ? What’s the relevance of this to thermal equilibrium and entropy?! Microstates and macrostates: poker hands ANS: This is 1 possibility out of a total of 52!/47! possibilities. You’ll have to bear with me again for the answer......

  20. Let’s consider the smallest possible blocks – two interacting atoms. As each atom comprises three 1D oscillators this means we have six 1D oscillators in total. ? How many ways are there of distributing 4 quanta of energy amongst six 1D oscillators? Two interacting atoms: six 1D oscillators Remember, we’re still trying to find out why the total thermal energy is shared equally between the two blocks.

  21. Are many ways are there of distributing four quanta of energy amongst six 1D oscillators? • 126 • 1260 • 1,024,256 • None of these

  22. If four quanta of energy are given to one atom or the other, how many ways are there of distributing the energy quanta? • 15 • 30 • 45 • 60

  23. ? If all four quanta of energy are given to one atom or the other, how many ways are there of distributing the energy? Two interacting atoms: six 1D oscillators ANS: 4 quanta distributed amongst 3 oscillators = 15 ways x 2 = 30 ways

  24. If three quanta are given to one atom, and one quantum to the other, how many ways are there of distributing the energy? • 10 • 30 • 60 • 120

  25. ? If the four quanta are shared equally between the atoms, how many ways are there of distributing the energy? Two interacting atoms: six 1D oscillators ? If three quanta are given to one atom, and one quantum to the other how many ways are there of distributing the energy? ANS: 60 ways 3 quanta distributed amongst 3 oscillators on atom 1 : 10 ways. 1 quantum distributed amongst 3 oscillators on atom 2: 3 ways. However, could also have three quanta on atom 2, one quantum on atom 1 ANS: 36 ways. (i.e. 126 – 30 – 60, but make sure you can get the same result by counting the states as above.)

  26. Two interacting atoms: six 1D oscillators For two interacting atoms it is most probable that the thermal energy is shared equally. • We can look at this result in two ways: • if frequent observations of the two atom system are made, in 29% of the observations - i.e. 36 out of 126 – the energy will be split evenly; • for 100 identical two atom systems, at any given instant 29% will have the thermal energy split evenly between the two atoms.

  27. ? How many ways are there of distributing 10 quanta of energy amongst 10 atoms? ? How many ways are there of arranging the system so that the 10 quanta of energy are on one specific oscillator? Increasing the number of atoms…… For two atoms, 29% of the time the system will adopt a state where the energy is shared equally. So, although this is the most probable distribution, it happens < ⅓ of the time. It is almost as likely (24%) to find all the energy on one atom or the other. What happens as we add more atoms? ANS: 6.35 x 108 ANS: 1

  28. Increasing the number of atoms…… No. of arrangements increases VERY quickly for small changes in numbers of oscillators. For 300 oscillators (100 atoms) there are ~ 1.7 x 1096 ways of distributing 100 quanta of energy. 1 mole of any material contains ~ 6 x 1023 atoms. Is it possible that all the energy could be concentrated on 1 atom? Are we ever likely to see this happen? Yes! No!

  29. Module Questionnaire

  30. The aims and objectives were made clear • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  31. Handing out printed lecture notes is a good idea • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  32. Attempting questions during the lectures using the keypads aided my learning • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  33. I have no difficulty reading what is on the overhead projector • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  34. I found the module website useful • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  35. The coursework is too difficult • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  36. I find the workload expected is right for the number of credits • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  37. The subject matter is more difficult to understand than my other modules • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  38. The subject matter is interesting • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  39. The printed lecture notes are easy to follow • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  40. There is poor library support • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  41. The room is inadequate • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  42. There is the correct quantity of material • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  43. The module stimulated my interest in the subject and I would like to learn more • Strongly agree • Agree • Neutral • Disagree • Strongly Disagree

  44. THANK YOU!

More Related