Chapter 15. Statistical Thermodynamics

1 Boltzmann Distribution Partition Functions Molecular Energies The Canonical Ensemble Internal energy and entropy Derived functionsChapter 15. Statistical ThermodynamicsMajor ConceptsReviewDiscrete Energy levelsParticle in a boxRigid rotorHarmonic OscillatorMathProbability Lagrange MultipliersProperties of lnMicroscopic PropertiesQuantum MechanicsSpectroscopyVibrational frequenciesBond dissociationsMacroscopic PropertiesThermodynamicsHeat capacityCoefficient of expansionStatistical MechanicsStatistical ThermodynamicsStatistical Thermodynamics provides the link between the microscopic ( , molecular) properties of matter and its macroscopic ( , bulk) properties .

2 It provides a means of calculating thermodynamic properties from the Statistical relationship between temperature and on the concept that all macroscopic systems consist of a large number of statesof differing energies, and that the numbers of atoms or molecules that populate each of these states are a function of the thermodynamic temperature of the of the first applications of this concept was the development of the kinetic theory of gases and the resulting Maxwell-Boltzmann distribution of molecular velocities, which was first developed by Maxwell in 1860 on purely heuristic grounds and was based on the assumption that gas molecules in a system at thermal equilibrium had a range of velocities and, hence, performed a detailed analysis of this distribution in the 1870 s and put it on a firm Statistical foundation.

3 He eventually extended the concept of a Statistical basis for all thermodynamic properties to all macroscopic systems. 223/ 23/ 2vMv2222v4v4v22mkTRTmMfeekTRT Maxwell-Boltzmann Distribution: Statistical ThermodynamicsStatistics and EntropyMacroscopic state:-state of a system is established by specifying its T, E ,S ..Microscopic state:-state of a system is established by specifying x, p, .. of than one microstate can lead to the same : 2 particles with total E = 2 Can be achieved by microstates 1, 1 or 2, 0 or 0, 2 Configuration:-The equivalent ways to achieve a stateW (weight):-The # of configurations comprising a stateProbability of a state:-# configuration in state / total # of configurations(Assumes that the five molecules are distinguishable.)

4 Weight of a Configuration012301230123!!!!!!lnln!!!!l n! ln!!!!ln!ln!iiNWN N N NNWN N N NNN N N NNN Using Sterling s Approximation: ln N! ~ N ln N N, which is valid for N 1,lnlnlniiiWNNNN Global maximum in f when df= 0 df f x ydx f y xdy Seek a maximum in f(x,y) subject to a constraint defined by g(x,y) = 0 Since g(x,y) is constant dg = 0 and: dg g x ydx g y xdy 0 This defines dx g y g x dyanddy g x g y dx Eliminating dx or dyfrom the equation for df:df f y f x g y g x dy 0ordf f x f y g x g y dx 0 Defines undetermined multiplier f y g yor f x g x f x g x 0or xf g 0 f y g y 0 yf g 0orSame as getting unconstrainedmaximum of K f g Undetermined Multipliers (Chemist s Toolkit )Example of Undetermined MultipliersA rectangular area is to be enclosed by a fence having a total length of 16 meters, where one side of the rectangle does not need fence because it is adjacent to a river.

5 What are the dimensions of the fence that will enclose the largest possible area?riveryxxThus, the principal (area) function is:F(x,y) = xy(1)and the constraint (16 meters) is: f(x,y) = 2x + y 16 = 0(2)If F(x,y) were not constrained, , if xand ywere independent, then the derivative (slope) of Fwould be zero:(3)and(4)However, this provides only two equations to be solved for the variables xand y, whereas three equations must be satisfied, viz., Eqs. (4) and the constraint equation f(x,y) = (,) 00and0 FFxy Example of Undetermined MultipliersThe method of undetermined multipliers involves multiplying the constraint equation by another quantity, , whose value can be chosen to make xand yappear to beindependent.

6 This results in a third variable being introduced into the three-equation problem. Because f(x,y) = 0, maximizing the new function F F (x,y) F(x,y) + f(x,y)(5)is equivalent to the original problem, except that now there are three variables, x, y, and , to satisfy three equations:(6)Thus Eq. 5 becomesF (x,y) = xy+ (2x + y -16)(7)Applying Eqs. (6),yielding = 4, which results in x= 4 and y= 8. Hence, the maximum area possible is A = 32 m2.''00and( , ) 0 FFf x yxy '202 Fyyx '0 Fxxy 2162216 0xy Find a maximum of fx,y e x2 y2 Subject to the constraint gx,y x 4y 17 0 From slope formula df f x ydx f y xdy 0df 2xe x2 y2 dx 2ye x2 y2 dy 0 Global maximum: x = y = 0 Need to find constrainedmaximum Find undetermined multiplierK(x,y) f(x,y) g(x,y) e x2 y2 x 4y 17 An unconstrained maximum in K must K x 2xe x2 y2 0 K y 2ye x2 y2 4 0 2xf y2f This implies 2x y24x y Original condition g = 0gx,y x 44x 17 0 Constrained maximum : x = 1.

7 Y = 4 Example of Undetermined MultipliersMost Probable Distribution = = ! 1! 2! 3!..Configurations:(permutations)Total energy:Maximum probability (and, hence, maximum entropy) occurs when each particle is in a different energy minimum energy occurs when all particles are in the lowest energy level. Thus, must find the maximum probability that is possible, consistent with a given total energy, E, and a given total number of particles, is an example of a classic problem, in which one must determine the extrema ( , maxima and/or minima) of a function, , that are consistent with constraints that may be imposed because of other functions, , energy and number of particles.

8 This problem is typically solved by using the so-called LaGrange Method of Undetermined : N = 20,000; E = 10,000; three energy levels 1=0, 2=1, 3= E requires that N2+ 2N3= 10,000; constant N requires that N1+ N2+ N3= 20,0000 < N3 < 3333; W is maximum when N3~1300. = Total number of particles:The most probable distribution is the one with greatest weight, W. Thus, must maximize lnW. Because there are two constraints (constant E and constant N), must use two undetermined multipliers: g(xi) = 0 and h(xi) = 0 so K = (f + ag +bh) Use this to approach to find most probable population:K lnW aN Njj bE Nj jj (constant N) (constant E)Want constrained maximum of lnW(equivalent to unconstrained maximum of K) Use Stirling'sApproximation for lnW:K NlnN NjlnNjj aN Njj bE Nj jj Can solve for any single population Ni(all others 0).

9 N Ni 0 Ni Ni 1 Nj Ni 0 K Ni lnNi Ni 1Ni a 1 b i 0 K Ni lnNi 1 a b i 0lnNi 1 a b iMost Probable DistributionN Njj Ae b jj A Ne b jj Ni Ne b ie b jj b 1kbTpi NiN e ikbTe jkbTj A exp 1 a Ni Ae b iIf , then A (a) can be eliminated by introducing N:Boltzmann Temperature (will prove later)Boltzmann DistributionBoltzmann DistributionlnNi 1 a b i =e =e For relative populations:Gives populations of states, not more than one state at same energy, must account for degeneracy of state, gi. = e Most Probable DistributionIn summary, the populations in the configuration of greatest weight, subject to the constraints of fixed E and N, depend on the energy of the state, according to the Boltzmann Distribution.

10 IikTikTiNeNe The denominator of this expression is denoted by qand is called the partition function, a concept that is absolutely central to the Statistical interpretation of thermodynamic properties which is being developed can be seen in the above equation, because kis a constant (Boltzmann s Constant), the thermodynamic temperature, T, is the unique factor that determines the most probable populations of the states of a system that is at thermal Probable DistributionIf comparing the relative populations of only two states, iand j, for example,iijjkTikTjkTNeeNe The Boltzmann distribution gives the relative populations of states, notenergy levels.