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Entropy, probability and disorder Thermal equilibrium Experience tells us: two objects in thermal contact will attain the same temperature and keep that temperature Why? More than just energy conservation! Involves concept of entropy Entropy and disorder It is often said that entropy is a measure of disorder, and hence every system in isolation evolves to the state with “most disorder” Consider a box sliding on a floor: internal energy due to disorderly motion of the molecules kinetic energy (of the box) due to the collective, orderly, motion of all the molecules Entropy and disorder II Now the box comes to rest due to friction Temperature rise in both floor and box so the internal energy increases No more collective motion: all K.E. has been transferred into internal energy More disorder, so entropy has increased A vessel of two halves Large number of identical molecules – distribution? About 50% in left half, 50% in right half Why? Definitions Microstate: position and momentum of each molecule accurately specified Macrostate: only overall features specified Multiplicity: the number of microstates corresponding to the same macrostate Fundamental assumption Statistical Mechanics is built around this one central assumption: Every microstate is equally likely to occur This is just like throwing dice: A throw of the dice Roll one die: 1/2/3/4/5/6 all equally likely Roll a pair of dice: for each 1/2/3/4/5/6 equally likely the sum 7 is most likely, then 6 and 8, etc. Why? 6 combinations (microstates) give 7 (the macrostate): 1+6, 2+5, 3+4, 4+3, 5+2, 6+1. There are 5 combinations that give 6 or 8, etc. Four identical molecules 4 molecules ABCD 5 macrostates: Four identical molecules (2) left: A&B&C&D right: multiplicity: 1 A&B&C A&B&D A&C&D B&C&D D C B A multiplicity: 4 Four identical molecules (3) left: A&B A&C A&D B&C B&D C&D right: C&D B&D B&C A&D A&C A&B multiplicity: 6 Four identical molecules (4) #left 4 3 2 1 0 #right 0 1 2 3 4 multiplicity 1 4 6 4 1 16 probability 1/16 4/16 6/16 4/16 1/16 Ten identical molecules Multiplicity to find 10 –...– 0 molecules on left 1–10–45–120–210–252–210–120–45–10–1 Probability of210 finding 252 #left 210 = 4, 5 or 6: 1024 0.66 For large N: extremely likely that #left is very close to N/2 Generalisation Look at a gas of N molecules in a vessel with two “halves”. The total number of microstates is 2N: two possible locations for each molecule we’ve just seen the N=4 example Binomial distribution I A gas contains N molecules, N1 in the left half (“state 1”) and N2 = N – N1 in state 2 (the right half). How many microstates correspond to this situation? N1 N2 Binomial distribution II Pick the molecules one by one and place them in the left hand side: choose from N molecules for the first molecule choose from N – 1 for the second choose from N – 2 for the third, … choose from N – N1 + 1 for the N1-th molecule Binomial distribution III Number of ways of getting N1 molecules into left half: N! N ( N 1) ( N 2) ... ( N N1 1) ( N N1)! The macrostate doesn’t depend on the order of picking these molecules; there are N1! ways of picking them. Multiplicity W is mathematical N N! “combination”: W C N1 ( N N1)! N1! Verification Look at a gas with molecules A,B,C,D,E. Look at the number of ways of putting 2 molecules into the left half of the vessel. So: N = 5, N1 = 2, N – N1 = 3 Verification II The first molecule is A, B, C, D, or E. Pick the second molecule. If I first picked A then I can now pick B, C, D or E, etc: AB AC AD AE BA BC BD BE CA CB CD CE DA DB DC DE EA EB EC ED 5 4 3 2 1 5! That is 5 4 possibilities 3 2 1 3! Verification III In the end I don’t care which molecule went in first. So all pairs AB and BA, AC and CA, etc, really correspond to the same situation. We must divide by 2!=2. A B = B A Binomial distribution plotted Look at N=4, 10, 1000: P(N1) 0 1 2 N1 3 40 2 4 6 N1 8 10 0 200 400 600 8001000 N1 Probability and equilibrium As time elapses, the molecules will wander all over the vessel After a certain length of time any molecule could be in either half with equal probability Given this situation it is overwhelmingly probable that very nearly half of them are in the left half of the vessel Second Law of Thermodynamics Microscopic version: If a system with many molecules is permitted to change in isolation, the system will evolve to the macrostate with largest multiplicity and will then remain in that macrostate Spot the “arrow of time”! Boltzmann’s Epitaph: S = k logW Boltzmann linked heat, temperature, and multiplicity (!) Entropy defined by S = k ln W W: multiplicity; k: Boltzmann’s constant s = “dimensionless entropy” = ln W Second Law of Thermodynamics Macroscopic version: A system evolves to attain the state with maximum entropy Spot the “arrow of time”! Question 1 Is entropy a state variable? a) Yes b) No c) Depends on the system Question 2 The total entropy of two systems, with respective entropies S1 and S2, is given by a) S = S1 + S2 b) S = S1 · S2 c) S = S1 – S2 d) S = S1 / S2 Entropy and multiplicity Motion of each molecule of a gas in a vessel can be specified by location and velocity multiplicity due to location and velocity Ignore the velocity part for the time being and look at the multiplicity due to location only Multiplicity due to location I Divide the available space up into c small cells. Put N particles inside the space: W=cN. For c=3, N=2: W=32=9 AB A A B B A B AB B A A B B A AB Multiplicity due to location II Increasing the available space is equivalent to increasing the number of cells c. The volume is proportional to the number of cells c Hence W VN “Slow” and “fast” processes Slow processes are reversible: we’re always very close to equilibrium so we can run things backwards Fast processes are irreversible: we really upset the system, get it out of equilibrium so we cannot run things backwards (without expending extra energy) Slow isothermal expansion Slow isothermal expansion of ideal gas; small volume change Wfinal V V V 1 N Winitial V V N N “velocity part” of multiplicity doesn’t change since T is constant V V Slow isothermal expansion (2) Use the First Law: N Q pV kTV V Wfinal V 1 Winitial V N Q 1 NkT Big numbers take logarithm N V V Slow isothermal expansion (3) Manipulation: N Wfinal Q ln ln 1 NkT Winitial Q Q Q N ln 1 N NkT kT NkT or k ln Wfinal k ln Winitial Q T V V Slow isothermal expansion (4) Use definition of entropy: Q k ln Wfinal k ln Winitial T Q Sfinal Sinitial S T V V valid for slow isothermal expansion Example To melt an ice cube of 20 g at 0 °C we slowly add 6700 J of heat. What is the change in entropy? In multiplicity? 24.5 J K-1; Wfinal 10770 ,000 ,000 ,000 ,000 ,000 ,000 ,000 Winitial Very fast adiabatic expansion Expand very rapidly into same volume V+V which is now empty Isothermal: same #collisions, #molecules, etc. Q Entropy change: S 0 ??? T Q NO! Entropy is a state variable and therefore S T S = same as for slow isothermal expansion Slow adiabatic expansion Same volume change, but need to push air out of the way so temperature drops Q Again we ask: S 0 ??? T YES! The “location part” of multiplicity increases as with slow isothermal expansion The “velocity part” decreases as temperature drops The two exactly cancel Constant volume process Heat is added to any (ideal or non-ideal) gas whose volume is kept constant. What is the change in entropy? dQ nCV dT dQ nCV dT ; dS T T Integrate (assuming CV is constant) T2 T2 T1 T1 S dS nCV dT T2 nCV ln T T1 Constant pressure processes Heat is added to an ideal gas under constant pressure. What is the change in entropy? T2 a) S nCV ln T1 V2 c) S nC p ln V1 p2 b) S nC p ln p1 d) 0 Entropy and isothermal processes An ideal gas expands isothermally. What is the change in entropy? Q Constant temperature so S T V2 First Law: Q W nRT ln (done previously) V1 V2 p1 Therefore S nR ln nR ln V1 p2 Entropy and equilibrium We have established a link between multiplicity and thermodynamic properties such as heat and temperature Now we see how maximum entropy corresponds to maximum probability and hence to equilibrium Equilibrium volume In general the number of microstates depends on both the volume available and the momentum (velocity) of the molecules Let’s ignore the momentum part and look at the spatial microstates only. Equilibrium volume II Say we have 3 molecules in a vessel which we split up into 6 equal parts. A partition can be placed anywhere between the cells. One molecule is on the left-hand side, the other two on the right-hand side. What is the equilibrium volume? Look for maximum entropy! Equilibrium volume III Number of cells on the left c1, on the right c2. We’ll look at c1=4, c2=2: A BC A B A A C C B BC Equilibrium volume IV Left: W1= c1=4. Right: W2= (c2)2=4. s = ln 4 + ln 4 = ln 16 = 2.77 A BC A B A A C C B BC Question The dimensionless entropy of this system of 6 cells and one partition dividing it into c1 and c2 cells is a) s = ln (c1+ c2) b) s = ln (c1+ c22) c) s = ln c1+ln 2c2 d) s = ln c1+2·ln c2 Equilibrium volume V c1 1 2 3 4 5 total W 25 32 24 16 5 102 s 3.22 3.47 3.18 2.77 1.61 P(W) 0.25 0.31 0.24 0.16 0.05 1 Maximum entropy and probability s1 s2 s dimensionless entropy 4 3 0.3 2 0.2 1 0.1 0 1 2 3 c1 4 5 Probability 0.4 0.0 1 2 3 c1 4 5 Maximum probability Probability maximum coincides with entropy maximum Volume V1 = c1·dV where dV is the cell size ds 0 Most likely situation when dV1 N1 1 N1 2 1 Same density on both sides: V 2 ; V 4 2 1 1 Question Which relationship holds for the probabilities of finding a the system in a microstate corresponding to c1=2,3,4 ? a) P(2) < P(3) < P(4) b) P(2) = P(3) = P(4) c) P(2) > P(3) > P(4) Entropy and mixing Suppose we remove the partition. What is the entropy of this system? Answer: ln 63 = ln 216 = 5.38 The additional entropy of 1.91 is called “the entropy of mixing” Generalisation I Look at N1 particles occupying c1 cells on the left, N2 particles occupying c2 cells on right. Volume of each cell = dV. N1 N2 W W W c c Multiplicity: 1 2 1 2 N1 N2 V1 V2 dV dV V1 V2 V1N1 (V V1) N 2 N1 N2 (dV ) N Generalisation II V N1 (V V1) N 2 Entropy: s ln W ln 1 N (dV ) N1 ln V1 N 2 ln(V V1) N ln dV Maximum entropy for equal densities: V1 V2 N1 N2 ds N1 N2 N1 N 2 0 dV1 V1 V V1 V1 V2 Multiplicity and energy According to quantum mechanics, atoms in a crystal have energies 0, e, 2e, 3e, … (This is called the Einstein model of solids) Say we have three atoms with total energy 3e Microstates are distinguished by the different energies E1, E2, E3. The microstates 3e 2e E1=3e e 0 Energy 3e E1=2e 2e e 0 3e E1=e 2e e 0 E1=0 3e 2e e 0 E1,E2,E3 E1,E2,E3 E1,E2,E3 E1,E2,E3 Question What is the probability for any of these three atoms to have energy 0? e? 2e? 3e? a) 4/10,3/10,2/10,1/10 b) 1/4,1/4,1/4,1/4 c) 1/10,2/10,3/10,4/10 d) not sure Generalisation If there are n atoms and the total energy is qe, then the number of microstates is given by (q n 1)! W( q , n ) q!(n 1)! Works for previous example (n=3,q=3): 5! W 10 3!2! Partition I Look at 10 particles, with energy 20e. n1=3 particles on the left-hand side, n2=7 on right-hand-side What is the most likely energy distribution? Plot W as a function of q1. Partition II W against energy on left-hand side 1000000 W W1 W2 W 500000 0 0 2 4 6 8 10 12 14 16 18 20 q1 Partition III Entropy against energy on left-hand side s1 s2 s 15 s 10 5 0 0 2 4 6 8 10 12 14 16 18 20 q1 Partition IV You expect the atoms on the left to have the same energy on average as the atoms on the right Calculation/plot shows this: W maximum for n1 q1 q2 q1 (q1 q2 ) n1 n2 n1 n2 Entropy, energy, temperature I The internal energy on the left U1 = q1e. ds 0 Equilibrium/entropy maximum when dU1 Use s = s1 + s2: ds1 ds 2 0 dU1 dU1 ds 2 ds 2 dU1 ds 2 Use U2 = U – U1: dU 2 dU1 dU 2 dU1 Entropy, energy, temperature II It follows that the most likely distribution of energy corresponds to a situation where ds1 ds 2 dS1 dS2 or dU1 dU 2 dU1 dU 2 We know that in this situation T1 = T2 Clearly the two are linked! Entropy, energy, temperature III Remember: we kept V, N constant so the only way in which energy could be exchanged was through heat transfer Q U Remember: S T T due to heating only 1 S S T U due to heating only U fixed external parameters Entropy, energy, temperature IV In our example the only external parameter was the volume In general, gravitational, electric or magnetic fields, elastic energy etc. could all change The definition of temperature only holds if all of these are held fixed PS225 – Thermal Physics topics The atomic hypothesis Heat and heat transfer Kinetic theory The Boltzmann factor The First Law of Thermodynamics Specific Heat Entropy Heat engines Phase transitions