Microstates, macrostates and

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Presentation transcript:

Microstates, macrostates and Thermal & Kinetic Lecture 11 Microstates, macrostates and thermal equilibrium LECTURE 11 OVERVIEW Recap…. Energy transfer Interacting atoms and blocks Thermal equilibrium and entropy A definition of temperature

Lecture next Thursday 11:00 – 12:00 ?

Feedback on SET responses Notes in A4 or A5 (booklet) format? Lecture notes and Powerpoint slides are both on the web – I really don’t see the point in making the handouts ‘exactly like the lecture slides’. “Print notes on A4 instead of A5, ‘coz A5 is 2 small an its well annoying putting it in your folder” [sic] Answers to all questions posed in lectures are available on the slides which may be downloaded from the website. Blackboard in B1 – the chalk barely makes a mark. “More note-taking”; “More worked examples” “Make lecture notes available for several lectures in case students miss a particular set of notes”; “Go through booklets systematically” (?) “Slow down”

Feedback on SET responses Please make original comments re. accent – “tree and a turd”, “tree molecules”, etc.. “Could be less irish” [sic]

Last time…. Arrangements, combinations and permutations Microstates and macrostates.

Counting arrangements Number of ways to arrange q quanta of energy amongst N 1D oscillators: How many ways can 4 quanta of energy be arranged amongst four 1D oscillators? ? ANS: = 35 ways

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.

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

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. 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? ? ANS: 126 If all four quanta of energy are given to one atom or the other, how many ways are there of distributing the energy? ? ANS: 4 quanta distributed amongst 3 oscillators = 15 ways x 2 = 30 ways

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 If the four quanta are shared equally between the atoms, how many ways are there of distributing the energy? ? ANS: 36 ways. (i.e. 126 – 30 – 60, but make sure you can get the same result by counting the states as above.)

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.

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? How many ways are there of distributing 10 quanta of energy amongst 10 atoms? ? ANS: 6.35 x 108 How many ways are there of arranging the system so that the 10 quanta of energy are on one specific oscillator? ? ANS: 1

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!

Thermal equilibrium Take 2 blocks, one containing 3 atoms, the other containing 6 atoms. Total energy of the two blocks: 100 quanta. The blocks are adiabatically isolated – no heat in or out.

Distribution of energy To the nearest integer, where on the x-axis is the maximum located? Why? ? ANS: 67 (100 quanta, split 2:1)

Thermal equilibrium Width of distribution (FWHM) is proportional to N . For a mole of a solid or a gas the odds are overwhelmingly in favour of having a uniform energy distribution. Largest number of microstates (by far!) are associated with the most probable distribution. Even for a small collection of atoms we’re dealing with very large numbers of possible microstates – much more convenient to use logarithms. (More fundamental reasons for appearance of logs which we’ll see later on…) Reconsider previous histograms using logarithms…………………

Thermal equilibrium and entropy To the nearest integer, where on the x-axis is the maximum of ln (W1W2) located? ? ANS: 67

? ? Thermal equilibrium and entropy S =k ln(W) If block 2 were three times the size of block 1 what value on the x-axis would correspond to the maximum of ln(W1W2)? ? ANS: 25 …and now we finally get to a definition of entropy! S =k ln(W) Entropy Boltzmann’s constant No. of accessible states What are the units of entropy? ? ANS: JK-1