MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.62 Lecture #2: E, A, and S: Macroscopic Propertiesfrom Microscopic {Pi} Probabilities Problem: How do we calculate a macroscopic property, which is constant intime, from a microscopic property that fluctuates in time? Example: Pressure, which is a macroscopic property that arises from themicroscopic impulses of each molecule impacting the vessel'swalls. The positions and velocities of each molecule change on a10–13s time scale (the duration of a collision)! Possible Solution: TIME AVERAGE the microscopic variable fobs is the observed macroscopic property f(q3N ,p3N ) is the instantaneous value of the sum over all ~ ~ microscopic contributions to the macroscopic property fobs = lim 1 τ∫τ f q( 3N ,p3N )dτ′ this is how a classical mechanical time average is defined τ→∞ 0 ~ ~ But this calculation is impossible because it requires knowledge of the time dependenceof a very large number, N, of q~i,p~i. Instead, we make use of ENSEMBLE THEORY, developed by J. Willard Gibbs (1839-1903) (founder of Stat. Mech.) ENSEMBLE ≡A COLLECTION OF ALL “POSSIBLE” STATES OF AN ASSEMBLY system (e.g. a molecule) → assembly of systems → ensemble In thermodynamics, the word “system” is used to specify the macroscopic object underconstruction.5.62 Spring 2008 Lecture 2, Page 2 Example: (1) Quantum – assembly consisting of 2 particles onlystate n1x n1y n1z n2x n2y n2z Constant E ensemble α 2 1 1 1 1 1 with E = 9ε0 β 1 2 1 1 1 1 E = εi i=1 N ∑ γ δ 1 1 1 1 2 1 1 2 1 1 1 1 εi = h2 8mia2 nix 2 + niy 2 + niz 2⎡⎣ ⎤⎦ ε η 1 1 1 1 1 1 1 1 2 1 1 2 h2 ε0 = 8mia2 For a 2 particle assembly there are only 6 ways E = 9ε0 can be achieved. (2) Classical – 1 particle N In general E = ∑εi i=1 pix 2 + piy 2 + piz 2 εi = 2m p2 In this case E = 2m = constant For a 2 particle assembly there are an infinite number of ways E = 9ε0 can be achieved. Note that the quantum ensemble is a set of discrete states, whereas the classical ensembleis a set of infinitely many states described by continuous variables. ENSEMBLE THEORY: SOMEHOW, WE CAN KNOW ALL POSSIBLE STATESOF AN ASSEMBLY WITHOUT WATCHING IN REAL TIME WHAT STATES THE ASSEMBLY VISITS. SO, INSTEAD OF THE INFEASIBLE TIME AVERAGE, WECOMPUTE AN AVERAGE OVER ALL FEASIBLE STATES OF AN ASSEMBLY. It is frequently feasible to list (enumerate) the states possible for the assembly without“watching in real time”. This is where combinatorics and statistics enter. revised 1/9/08 9:34 AM5.62 Spring 2008 Lecture 2, Page 3 A FUNDAMENTAL POSTULATE OF STATISTICAL MECHANICS THE ERGODIC HYPOTHESIS TIME AVERAGE ≡ ENSEMBLE AVERAGE (actually it is also an average over cells in phase space, each of volume h3N where N is the number of particles) ENSEMBLE AVERAGE Discrete case — Quantum fj ≡ a microscopic property of jth macroscopic observable quantity ≡ f= ∑Pjfj distinguishable state ofj assemblysum over distinguishable assemblystates in ensemble Pj ≡ probability that assembly is in state j. so macroscopic energy E = ∑PjE j = ensemble average energy j Continuous case — classical q3N p3N q3N p3N q3Ndp3N q3N where P ( ,p3N )dq3Ndp3N ≡ prob. of finding the assembly in the phase space   volume element dq3Ndp3N centered at  q3N ,p3N .  NOTE:To calculate an ensemble average, you need values for Pj (or P (q3N , p3N )) PROBLEM: How do we determine Pj? SOLUTION: Minimize the Helmholtz free energy, A = E – TS, holding the natural variables of A, (N, T, and V), constant. f = ∫ ⋅⋅⋅∫ P ( )f ( )d, , revised 1/9/08 9:34 AM5.62 Spring 2008 Lecture 2, Page 4 DETERMINATION OF Pj Our ensemble is a CANONICAL ENSEMBLE ≡ ensemble subject to constraints thatN,V,T are constant.A closed, thermodynamically stable system. Condition for thermodynamic stability (equilibrium) for N, V, T constant is AN,V,T ≡ MINIMUM The states of the assembly present in the ensemble, as given by {Pj}, must minimize A. Must write A in terms of {Pj}. A = E – TS and E = ∑PjE j = ∑(Γj Γ)Ej j j where Γj is the number of replicas of the j-th assembly in the ensemble, Γ is the total number of assemblies in the ensemble Γj is the probability of j-th assembly in the ensembleΓso A = ∑PjEj − TS j Now connect S and {Pj}... An isolated system at equilibrium is one of maximum entropy, S - 2nd Law. If the system is perturbed, it will relax to maximum entropy, a macro property. On a microscopic scale, it relaxes by going from a less probable state to a more probable state. So, there must be a connection between entropy (a macro property) and Pj ( a micro property). That connection is assumed to be… A CRUCIALS = −k∑ Pjln Pj ASSUMPTION!j Boltzmann wrote this down in a slightly different form. No derivation. Only revised 1/9/08 9:34 AM5.62 Spring 2008 Lecture 2, Page 5 plausibility arguments. It is an assumption on which statistical mechanics is built. It works!!! So now, A = E – TS A = ∑ PjEj + kT ∑ Pj ln Pjjj A = ∑ Pj(Ej + kT ln Pj) j Finding those Pj's that make A a minimum ... Replace Pj by Pj + δPj, then A → A + δA. The “correct” set of {Pj’s} gives δA = 0, which corresponds to the minimum of A ⎤⎡ δA = δ ∑ ⎢⎢⎣ Pj(E j + kTln Pj⎥⎥⎦ (N, V, T constant) j ⎡ ⎢⎢⎣ EjδPj + PjδEj + kT ln Pj )δPj + kTPj 1( δPj ⎤ ⎥⎥⎦ ∑ ( ) δEj = 0= Pjj = ∑δPj ⎡⎣Ej + kT ln Pj +1 ⎤⎦ set to 0 for extremum( )j revised 1/9/08 9:34 AM5.62 Spring 2008 Lecture 2, Page 6 Introduce Constraint ∑ Pj = 1 = ∑ (Pj + δPj) j j This implies ∑δPj = 0 j N or δPj=1 = −∑δPj the trick! j= 2 Remove the first term from the summation: N E1 + kT ln P1 +1) Ej + kT ln Pj +1Now δA = δP1 ⎡⎣ ( ⎤⎦+ ∑δPj ⎡⎣ ( )⎤⎦ j=2 employ the trick N N ∑ E1 + kT ln P1 +1)⎤⎦+ ∑ ( )δA = − δPj ⎡⎣ ( δPj ⎡⎣Ej + kT ln Pj +1 ⎤⎦ j=2 j=2 N δA = +∑δPj ⎡⎣(Ej − E1 )+ kT ln Pj − ln P1 )⎤⎦ = 0(j=2 δPj's are completely independent of each other for arbitrary δPj; j = 2,3 ..., thus each coefficient of each δPj must separately be zero. ∴ Ej − E1 + kT ln Pj − ln P1 )= 0( Ej − E1 = ln P1 Pj )kT ( Ej −E1 e kT = P1 Pj kTe−EJ kT∴ Pj = P1eE1 * Need to normalize Pj ∑ Pj = 1 j kT∑ e−Ej kT = 1∑ Pj = P1eE1 j j Solve