Classical Statistical Mechanics in the Canonical Ensemble: Application to the Classical Ideal Gas

Canonical Ensemble in Classical Statistical Mechanics As we’ve seen, classical phase space for a system with f degrees of freedom is f generalized coordinates & f generalized momenta (qi,pi). The classical mechanics problem is done in the Hamiltonian formulation with a Hamiltonian energy function H(q,p). There may also be a few constants of motion (conserved quantities): energy, particle number, volume, ... 2

The Canonical Distribution in Classical Statistical Mechanics The Partition Function has the form: Z ≡ ∫∫∫d3r1d3r2…d3rN d3p1d3p2…d3pN e(-E/kT) A 6N Dimensional Integral! This assumes that we’ve already solved the classical mechanics problem for each particle in the system so that we know the total energy E for the N particles as a function of all positions ri & momenta pi. E  E(r1,r2,r3,…rN,p1,p2,p3,…pN) 3

P(E) ≡ e[- E/(kBT)]/Z CLASSICAL Statistical Mechanics: Let A ≡ any measurable, macroscopic quantity. The thermodynamic average of A ≡ <A>. This is what is measured. Use probability theory to calculate <A> : P(E) ≡ e[- E/(kBT)]/Z <A>≡ ∫∫∫(A)d3r1d3r2…d3rN d3p1d3p2…d3pNP(E) Another 6N Dimensional Integral!

Relationship of Z to Macroscopic Parameters Summary of the Canonical Ensemble (Derivations are in the book! Results are general & apply whether it’s a classical or a quantum system!) Mean Energy: Ē  E = -∂(lnZ)/∂β <(ΔE)2> = [∂2(lnZ)/∂β2] β = 1/(kBT), kB = Boltzmann’s constant. Entropy: S = kBβĒ + kBlnZ An important, frequently used result! 5

Summary of the Canonical Ensemble: Helmholtz Free Energy: F = E – TS = – (kBT)lnZ Note that this gives: Z = exp[-F/(kBT)] dF = S dT – PdV, so S = – (∂F/∂T)V, P = – (∂F/∂V)T Gibbs Free Energy: G = F + PV = PV – kBT lnZ. Enthalpy: H = E + PV = PV – ∂(lnZ)/∂β 6

Summary of the Canonical Ensemble: Mean Energy: Ē = – ∂(lnZ)/∂ = - (1/Z)(∂Z/∂) Mean Squared Energy: <E2> = (rprEr2)/(rpr) = (1/Z)(∂2Z/∂2) nth Moment: <En> = (rprErn)/(rpr) = (-1)n(1/Z)(∂nZ/∂n) Mean Square Deviation: <(ΔE)2> = <E2> - (Ē)2 = ∂2lnZ/∂2 = -∂Ē/∂. Constant Volume Heat Capacity CV = (∂Ē/∂T)V = (∂Ē/∂)(d/dT) = - kBT2∂Ē/ ∂ 7

The Classical Ideal Gas Z ≡ ∫∫∫d3r1d3r2…d3rN d3p1d3p2…d3pN e(-E/kT) So, in Classical Statistical Mechanics, the Canonical Probability Distribution is: P(E) = [e-E/(kT)]/Z Z ≡ ∫∫∫d3r1d3r2…d3rN d3p1d3p2…d3pN e(-E/kT) This is the tool we will use in what follows. As we.ve seen, from the partition function Z all thermodynamic properties can be calculated: pressure, energy, entropy…. 8

Canonical Ensemble in Classical Statistical Mechanics. Consider an Ideal Gas from the point of view of microscopic physics. It is the simplest macroscopic system. Therefore, its useful use it to introduce the use of the Canonical Ensemble in Classical Statistical Mechanics. The ideal gas Equation of State is PV= nRT n is the number of moles of gas. 9

We’ll do Classical Statistical Mechanics, but very briefly, lets consider the simple Quantum Mechanics of an ideal gas & the take the classical limit. From the microscopic perspective, an ideal gas is a system of N non interacting Particles of mass m confined in a volume V = abc. (a, b, c are the box’s sides) Since there is no interaction, each molecule can be considered a “Particle in a Box” as in elementary quantum mechanics. 10

where nx, ny, nx = integers Since there is no interaction, each molecule can be considered a “Particle in a Box” as in elementary quantum mechanics. The energy levels for such a system have the form: where nx, ny, nx = integers 11

Z = (q)N (2) (1) The energy levels for each molecule in the Ideal Gas have the form: l (nx, ny, nx = integers) (1) The Ideal Gas molecules are non interacting, so the gas Partition Function has the simple form: Z = (q)N (2) where q  One Particle Partition Function 12

Ideal Gas Partition Function: Z = (q)N (2) q  One Particle Partition Function Using (2) in the Canonical Ensemble formalism gives the expressions on the right for: mean energy E, equation of state P & entropy S: 13

The Partition Function for the 1- dimensional particle in a “box” under the assumption that the energy levels are so closely spaced that the sum becomes an integral over phase space can be written: (3) For the 3 – dimensional particle in a “box”, th 3 dimensions are independent so that the Partition Function can be written as the product of 3 terms like equation (3). That is: (4) 14

Now, using the Canonical Ensemble expressions from before: Mean Energy & the Equation of State can be obtained (per mole): To obtain, for one mole of gas:

 “Gibbs’ Paradox” The Entropy can also be obtained: (5) As first discussed by Gibbs, Entropy in Eq. (5) is NOT CORRECT! Specifically, its dependence on particle number N is wrong!  “Gibbs’ Paradox” in the first part of Ch. 7!