4. Systems in contact with a thermal bath

Transcription

1 4. Systems in contact with a thermal bath So far, isolated systems microcanonical methods 4.1 Constant number of particles:kittelkroemer Chap. 3 Boltzmann factor Partition function canonical methods Ideal gas again! Entropy = SackurTetrode formula Chemical potential 4.2 Exchange of particles: KittelKroemer Chap. 5 Gibbs Factor Grand Partition Function or Gibbs sum grand canonical methods Phys 112 S BoltzmannGibbs 1

2 Ideal Gas Thermal Bath Consider 1 particle in thermal contact with a thermal bath made of an ideal gas with a large number M of particles We assume that the overall system is isolated and in equilibrium at temperature τ. We call U the total energy which is constant. We would like to compute the probability of having our particle in a state of energy ε. Because in equilibrium, all states of the overall system are equiprobable, this is equivalent to computing the number of states such that our particle has energy ε and the thermal bath Uε. Using the fact that for an ideal gas of energy E and particle number M, the number of states is and that g! E 3/2 M g U = 3 Res 2 M 1! exp! We obtain for large M prob! U! 3 2 M 1! 3 2 M 1 Phys 112 S BoltzmannGibbs M exp! M,!

3 Boltzmann factor System in contact with thermal bath reservoir R, U o ε Constant number of particles S,ε Configuration: 1 state of S energy of reservoir States in configuration 1! g R U o Fundamental postulate: at equilibrium Pr ob configuration! Pr ob state 1 Pr ob state 2! Pr ob state 1 Pr ob state 2 exp R U 2 1, = exp 1 o 2 1,! Prob state s = 1 Z exp s = 1 with Z = exp! s Z exp s k B T all states s Notes: exp! s =Boltzmann factor Z=partition function Probability of each state not identical! <= interaction with bath Phys 112 S BoltzmannGibbs 3 Total system =Our system S reservoir R isolated! g R U o = g R U o 1 = exp R U o 1 R U o 2 g R U o 2

4 Definition Phys 112 S BoltzmannGibbs 4 Partition Function If states are degenerate in energy, each one has to be counted separately => Z = g i exp! i with g i = multiplicity of state i i Z = all states s Example: 2 state system State 1 has energy 0 State 2 has energy ε Z = 1 exp! exp! U = = 1 exp! = C V = dq dt V,N = U T exp 1 V,N U = k b V,N exp! s Pr ob 1 = 1 1 exp!! 0 2 exp = k b k b T exp k b T k b T 1 2 exp! Pr ob 2 = 1 exp! = 1 C V Prob0 Probε 1 1 exp T T

5 Phys 112 S BoltzmannGibbs 5 Partition Function Relation with thermodynamic functions Energy U =! = 1! Z s exp! s = logz s 1 = 2 logz Entropy e s s! = p s log p s = s Z log e s Z = log Z U =, log Z, Free Energy F = U! =! log Z! = F U = 2 log Z = 2 F as implied by the thermodynamic identity Chemical potential µ =!F =!log Z!N,V!N Summation methods Direct e.g. geometric series! r i m = lim r i 1 r m1 = lim = 1 i=0 m! i=0 m! 1 r 1 r Few term approximation useful if f decreases fast! f i = f 0 f 1 f 2... Integral approximation i=0 f i! f xdx i=0 0

6 Examples Energy of a two state system Z = 1 exp! U = = 2 log Z = exp! 1 exp! = exp 1 Free particle at temperature τ Assumption: No additional degrees of freedom e.g. no spin, not a multiatom molecule Method 1: density in phase space Energy:! = p x 2 p 2 2 y p z 2M d 3 q d 3 p Z 1 =! h 3 exp = V h 3 dp x e p 2 x 2M! dp 1 y e p 2 y 2M ! dp! z e! Z 1 = V 2M 2 M = V 3 = 2,M h 2 2h 2 2h 2 Method 2: quantum state in a box: see KittelKrömer chapter 3 p 73 Same result, of course! Phys 112 S BoltzmannGibbs = VnQ with n Q = M! p z 2 2 M 3 2!!!!! =! 2 log Z 1! = 3 2!

7 Maxwell Distribution KK p. 392 Applies to classical gases in equilibrium with no potential Start from Boltzmann distribution 1 Prob 1 particle to be in state of energy ε= exp! Z 1 with Z 1 = d3 q d 3! p => h 3 exp = V M 3 2 2h 2 = VnQ Probability of finding this particle in volume dv and between p r and p r dp r = 1 exp! Vn Q For N particles, the volume density between r p and r p d r p = N V Change of variable to velocities: Probability of finding particle in v, vdv, dω Maxwell distribution Particle density distribution in velocity space KK chap. 14 p. 392 Note: 1 Could get the factor in front directly as normalization 2 Note that no h factor! p = Mv 3 One can easily compute Equipartition! Phys 112 S BoltzmannGibbs 7 1 exp! n Q d 3 p h 3 = n exp! n Q d 3 p 1 = n h 3 2 M 3 f vdvd! = n M 2 Mv 2 exp 2 v 2 dvd! 2 Mv 2 2 = 3 2! = 3 2 k BT 3/2 dvd 3 p h Number of states exp! d 3 p

8 Partition Function for N systems Weak interaction limit 2 sub systems in weak interaction: <=>States of subsystems are not disturbed by interaction just their population <=>! 12 n 1,n 2 =! 1 n 1! 2 n 2 with n 1, n 2 still meaningfull => Partition functions Distinguishable systems Suppose first that we can distinguish each of the individual components of the system: e.g. adsorption sites on a solid, spins in a lattice For 2 subsystems Z = exp! s 1 s2 Z 1 2 = Z 1Z 2 s 1 s 2 and for N systems = Z 1 N Z N Phys 112 S BoltzmannGibbs 8

9 Examples Free energy of a paramagnetic system In a magnetic field B,! = mb if m is the magnetic moment. For 1 spin Z 1 = exp mb exp mb = 2 cosh mb For N spins Z=Z 1 N since they are distinguishable U = 2 log Z Z = 2 N cosh N mb = NmB tanh mb Magnetization M = U VB mb = nm tanh M M! m B mb! Phys 112 S BoltzmannGibbs 9

10 Indistinguishable Systems Indistinguishable:in quantum mechanics, free particles are indistinguishable We should count permutations of states among particles only once s 1! s 1! s 2 e.g. for 2 particles Z 1Z 2 = e e overcounts s 1, s 2 s 2, s 1 for s 1 s 2 we have to divide by 2 s 2 More generally if we have N identical systems, the Nth power of Z 1 can be expanded as N! a 1... a s...a m N = n i n 1!...n s!...n m! a n 1..a n s...a n m 1 s m If we have a large number of possible states, most of the terms are of the form N!a i a j...a k Therefore if we have N indistinguishable systems N terms corresponding to N different states which are singly occupied. If the multiple occupation of states is negligible we are indeed overcounting by N!! 1 N! Z 1 N Z N This is a low occupation number approximation due to Gibbs no significant probability to have 2 particles in the same state: we will see how to treat the case of degenerate quantum gases Phys 112 S BoltzmannGibbs 10

11 N atoms Indistinguishable Ideal Gas Z N = 1 N! Z 1 1 N = N! n QV N log N!! N log N N log Z N = N log V n Q N! log Z N = N log n Q n N with n = N N V = concentration Interpretation: n Q as quantum concentration de Broglie wave length! = h Mv h 3M 1 2 Mv2 3 2 n! 1 Q 3! 3M h M 2! 2h 2 Phys 112 S BoltzmannGibbs 11

12 Ideal gas laws Energy U =! 2 log Z = 3! 2 N! = 3 2 Nk BT equipartition of energy 1/2τ per degree of freedom Free energy Pressure F =! log Z =!N log n Q!N n p =! F V = N V pv = N = Nk BT Entropy: Sackur Tetrode! = S = F k B = N log n Q 5., n 0 2/ identical to our expression counting states if extra degrees of freedom, additional terms! Well verified by experiment T dq Involves quantum mechanics S =! 0 Chemical potential µ =!F!N = Phys 112 S BoltzmannGibbs 12 T! N log n QV N N!N = log n n Q

13 Chemical Potential Physical Picture For an ideal gas without an external potential µ ==! log n n Q Concentration! Constancy of chemical potential = constancy of concentration <= Diffusion Wall partially transparent to particles Initial state Final state Phys 112 S BoltzmannGibbs 13

14 Internal Degrees of Freedom Optional In general, we need to add internal degrees of freedom In most cases, the kinetic and internal degrees of freedom are independent, for one particle e.g. for multiatomic molecules Rotation 10 2 ev Z = Z { Z kin { int kinetic internal Rotation energy! j = j j 1! 0 with multiplicity 2 j1 Z int = 2 j1 exp j j 1! 0 j, 2 j1 exp j j 1! 0 1/2 dj =, exp x! x 0 o dx with x = j j 1! 0 dx = 2 j 1! 0 dj, x 0 =! 0 4. Z int exp! o! o 4 for >>! 0. for >>! 0 U = 2 /log Z = C rot = k B! o / For < <! 0, keep the first terms of the discrete sum Z int 1 3exp 2! o... log Z int 3exp 2! o U = 2 /log Z 6! 0 exp 2! o! / C rot 12k o b Vibration ev similar to harmonic oscillator Z int = exp! nh 1 = n=0 1! exp! h Spin in a magnetic field etc.., h for >>h C vib = k B 2 C V exp 2! o 0 1/ j Diatomic molecules cf. KK fig 3.9 k B Rotation Slightly better approximation than 0 Vibration k B Phys 112 S BoltzmannGibbs 14 log τ

15 Exchanging Particles with an Ideal gas Let us consider again a system with an energy level ε, which can now be occupied by a variable number N of particles We put in contact with a thermal reservoir constituting on a large number MN of free particles. Energy and particles can be exchanged with the reservoir. We consider a single state with N particles and an energy Nε We can draw qualitatively the number of states of the overall system as a function of Ν. g = g Res! 1 = e 5/2, m 2U N 2h 2 3M N M N V 3/ / 0 M N Direct Counting!!!!!!!!!<= or Seckur Tetrode g Res N exp!! µ N Phys 112 S BoltzmannGibbs 15

16 Exploration Let us show that the probability of having N particles is = g exp! µ N g t prob!, N Direct calculation to take into account N dependence both in exponent and inside the bracket g = m 2U 2!h 2 3M M V 3/2, M. N 1. /N U For N << M, M 0 1!!U 2 3 m M3!!!!! 2 g 2 n Q n M. N exp. /N 3, 1 exp. 5 2 N 3/2 M. N 2U 2!h 2 3M e 5/2 M. N 1 1. N M 3/2, 5/2 M. N 2 n Q!!!! = e n 5/2 Q n M M V e 5/2 M. N = n µ /. 3 log n n N Q exp., 3 Phys 112 S BoltzmannGibbs 16

17 R, U o ε At equilibrium Gibbs factor System in contact with thermal bath reservoir S,ε Exchange of particles Notes: Z = grand partition function Absolute activity! i = exp µ i Phys 112 S BoltzmannGibbs 17 Total system =Our system reservoir Configuration: 1 state s of S of energy ε s and of particles N is energy of reservoir States in configuration 1! g R U o s, N oi N Pr ob configuration is! g R U o s, N oi N is!! Pr ob state!s with!, N is Pr ob state 1, N i 1 Pr ob state 2, N i 2 = g R U o 1, N oi N i1 = exp g R U o 2, N oi N i 2 = 1 s, N is µ i N is Z exp i s, N is with Z = exp! all states s,n is! µ i N is i R U R 1 2 i N i N i1 N i 2, µ i N is i = 1 s, N is Z exp k B T

18 Definition Grand Partition Function Z = Note: εs,ν depends on N i both through through increase of N, and the interactions In weak interaction limit ideal gas and only 1 species, Note If the N is =N i are constant, all states s and occupation N is = N s! s! s, N s energy of each level Z = exp µ i N is! s, N is i exp µ i N i Z N i the probabilities probabilities are the same as with Z! This formalism is useful only if the N is change. i N s { Z = exp! µ s s,n s Phys 112 S BoltzmannGibbs 18

19 Examples few state systems KittelKroemer Gas absorption heme in myoglobin, hemoglobin, walls Assume that each potential absorption site can absorb at most one gas molecule. What is the fraction of sites which are occupied? State 1= 0 molecule => energy 0 State 2= 1 molecule => energy ε Fraction occupied = f = Z = 1 exp µ! = 1 exp! µ 1 exp µ! 1 exp µ! How is µ determined? <= gas in equilibrium with absorption sites = same form as Fermi Dirac If ideal applies equally well to gas and dissolved gas Langmuir adsorption isotherm p f =!n exp p Q! µ =! log n p =! log pv = N! n = N gas n Q!n Q V = P! Phys 112 S BoltzmannGibbs 19

20 Ionized impurities in a semiconductor Donor= tends to have 1 more electron e.g. P in Si State 1= ionized D : no electron => energy= 0 State 2=neutral 1 extra electron spin up => energy =ε d State 3= neutral 1 extra electron spin down => energy =ε d Fraction ionized f = 1 1 2{ exp!! µ d 2 spin!states when!donor is!neutral Z =1 2exp d!! µ d = energy of electron on donor site KK use ionization energy I= d Acceptor= tends to have 1 fewer electron e.g. Al in Si State 1= ionized A : one electron completing the bond=> energy= ε a State 2= 1 neutral: missing a bond electron spin up => energy =0 State 3= 1 neutral: missing a bond electron spin down => energy =0 Fraction ionized f = µ determined by electron concentration in semiconductor = Fermi level Phys 112 S BoltzmannGibbs 20 exp! a! µ exp!! µ a 2{ 2 spin!states when!acceptor is!neutral a = energy of electron on acceptor site = 1 1 2exp! µ! a

21 Grand Partition Function Relation to thermodynamic functions Mean number of particles Entropy N i =!! N is p s,nis = s! = p s,nis log p s,nis s N is N is N is Z exp i!! = log Z s N is = log Z! µ i N is s, N is µ i N is i = log Z µ i Energy Free Energy Ω defined as U =! = 2 log Z µ i N i = 2 i µ i i F = U! =! log Z µ i N i G = F pv! F µ i N i i = log Z i µ i log Z Phys 112 S BoltzmannGibbs 21

22 Difference of Concentration and Potential Let us consider two isolated systems in interaction with each other. µ int 1 > µ int 2 If concentration 1 > concentration 2 But if the systems are in equilibrium, they have to have the same total chemical potential This means that the only way to maintain the difference of concentration is to give some potential energy to system 2 with respect to system 1 > µ int 2! 1 < µ int 1 diffusion drift! 2 Recall!!U = U { N! K { µ = In practice, if you attempt to prevent an evolution of the concentration, you will have to generate a difference of potential energy. Example: Barometric pressure equation Let us consider a horizontal slab of gas in the constant gravitational potential of the earth. Assume that atmosphere is isothermal and in equilibrium => constant total chemical potential kinetic Potential U,V, N N = U K! { N =µ int In equilibrium, µ 1 = µ 2 µ int 1! 1 = µ int 2! 2 µ z = µ 0 mgz =! log n z mgz = constant n Q! n z = n o e mgz Phys 112 S BoltzmannGibbs 22

23 Three different ways µ=constant Thermodynamic equilibrium between slabs at various z At constant temperature µ total =C!!!!! log n z Single particle occupancy of levels at different altitude n z! prob K, z! exp z K = n mgz 0 exp Hydrodynamic equilibrium! p dpda pda! p dpda! mgn zdadz = 0 1 n z pda!mgn zdadz pv = N p z = n z 1 n n z z =! mg Generalizes to case when the temperature is not constant At a deeper level the constancy of the total chemical potential in a system in equilibrium reflects the balance between the drift current generated by the external potential and the diffusion current due to the random thermal velocity. cf. Chapter 10 of the notes Phys 112 S BoltzmannGibbs 23 n Q = z C! n z = n 0 exp mgz, n z p z =!mg

24 Other Examples Equilibrium across a membrane: Osmotic diffusion Concentrations across the membrane want to be equal: but the quantity of material can be very different e.g reverse osmotic diffusion for desalinization Ions : interplay of chemistry and electricity When the cell needs more molecules, trapping sites are generated. Process continues till most of the sites are occupied.example Lungs If these molecules are ions this generates a potential difference with outside. Process continues till the electric potential difference stops the diffusion! log n in n out = q V V in out!!!qv in < qv out! n in > n out This would limit the concentration of ions! In order to circumvent the problem,the membrane recognizes the ions and lets them through.this is not thermal equilibrium! K Battery KK p.129 Example of mix of chemistry and electricity Consider an electrolyte AB: ions A B In the middle of the cell, equilibrium between positive and negative ions. B A B A But on electrodes, selective depletion repulsion r A! N AN e! B P e! BP B A E V B A B A x z V =! E r B A.dr A B N P Stops r!. r A neutralization E = E z Case where the electrodes are not connected z = N P N o Phys 112 S BoltzmannGibbs 24 P z