Ch 11: The Statistical Mechanics of Simple Gases and Solids Models for Molecular Motion Correspondence Principle Gases and Solids

I. Classical  Quantum Mechanics Energy: E = T + V  મ = C d 2 /dx 2 + V(x) મ Ψ = EΨSchroedinger EqnEqn 11.4 મ = Hamiltonian operator = 1-di energy operator; time independent; must be Hermitian; examples of CM functions  QM operators E = value of energy; eigenvalue; real number; {E} values are the allowed energy values.

CM  QM Ψ(x) = wave function = math function for system = eigenfunction; time independent –│Ψ 2 │dx = probability of finding particle in (x, dx+dx) –│Ψ 2 │ = probability density > 0 –Ψ(x) must be finite, continuous and single-valued Given system, the goal of QM is to – write મ – apply boundary conditions –find the set of {Ψ} and {E} such that મ Ψ = EΨ. –8 steps

II. Quantum Mechanical Models Particle in the Box = model for translational energy Simple Harmonic Oscillator (SHO) = model for molecular vibrations Rigid Rotor (RR) = model for molecular rotations. Each has simplifying assumptions and yields {Ψ} and {E}.

A. Translational Motion in 1-di Consider free particle of mass, m, constrained to move in 1-di (x-axis). Let V(x) = 0 from x = -∞ to x = +∞. મ Ψ = -(h 2 /8π 2 m) Ψ”(x) = εΨ(x) We need to find a function whose 2 nd derivative yields the function back again. Ψ(x) = sin ax, cos ax, exp(± iax) ε = a 2 h 2 /8π 2 m; a = real number; ε is continuous.

Translational Motion in 1-di Now put the particle in a 1-di box where V(x) = 0 between (0, L). But outside the box (x≤0 and x≥L), V(x) = ∞. (2 regions) Outside box, Ψ(x) = 0Eqn 11.8Why? Boundary conditions: Ψ(0) = 0 = Ψ(L) મ Ψ = -(h 2 /8π 2 m) Ψ”(x) = εΨ(x) Solve for eigenfunctions and eigenvalues

Translational Motion in 1-di Inside box, Ψ n (x) = √2/L sin (nπx/L) and energies are ε n = n 2 h 2 /8mL 2 Eqn n = positive integer = quantum number {ε n } are quantized, divergent, nondegenerate. (Fig 11.4, Example 11.1) Translational temp = Θ trans = h 2 /(8mL 2 k) Θ trans /T << 1  q tr = (2πmkT/h 2 ) ½ L Note as T and L incr, q tr incr. Recall S(T,V)

Correspondence Principle As T  ∞ and/or n  ∞ and/or m  ∞, then lim QM = CM –CM is a special case of QM, not an exception or inconsistency. –As n  ∞, │Ψ 2 │ become more uniform (CM result); see Fig 11.5

Translational Motion in 3-di Consider a box of length x = a, depth y = b and height z = c with V = 0 inside box and V = ∞ outside box (boundary conditions) Separate the variables to convert one 3-di problem (Eqn 11.16) into 3 1-di problems. I.e., solve 3 1-di box problems Write મ (x, y, z) = મ x + મ y + મ z Write S-Eqn

Translational Motion in 3-di Write Ψ(x, y, z) = Ψ x Ψ y Ψ z Each is an independent sin function. Write ε = ε nx + ε ny + ε nz Eqn –Quantized energies Solve for eigenfunctions and eigenvalues. Write q tr = q x q y q z = (2πmkT/h 2 ) 3/2 V Example 11.2 (q tr = states/atom) Prob 11.3, 5, 15

B. Vibrational Motion Define system, i.e. V(x) Write મ (x) Write and solve S-Eqn Find {Ψ n (x)} = NC x HP x AS Find {ε n } Write q vib and Θ vib Example 11.3 (note error; q vib = 1)

C. Rotational Motion Define system, i.e. V(θ,φ) Write મ (θ,φ) Write and solve S-Eqn Find {Ψ ℓ (θ,φ) = Θ(θ) Φ(φ)} = NC x SH Find {ε ℓ }; note degeneracy Write q rot and Θ rot (linear and nonlinear) Example 11.4 (q rot = 72) Prob 11.10

D. Electronic Energy This refers to the various allowed electronic states: ground state [lowest energy, 1s 2, 2s for Li), first excited state (1s 2, 2p for Li*), second excited state…] q elec = Σ g i exp (-Δε i β) Table 11.3 Finally, q = π q j where j = trans, vib, rot, elec; Table 11.2

III. QM  q  Ideal Gas Properties (micro  macro) Combine Table 10.1 with partition functions for trans, vib, rot, elect as appropriate: IGL, U, C v, S (T 11.4), F, μ Note Q = q N or q N /N! Distinguish monatomic vs diatomic vs polyatomic gas. Distinguish linear vs nonlinear molecular gas

Equipartition Theorem Internal energy is distributed uniformly over each DegF. Full contributions α T. #atoms#DegFTransRotVib = N= 3NTotalUContrib atom133RT/200 diatomic263RT/2RT linearN3N3RT/2RT(3N-5)RT nonlinN3N3RT/2 (3N-6)RT

Problems 11.1, 11.11, 17

IV. Heat Capacity of Solids Dulong and Petit: As T  0K, C V  3R Expt: As T  0K, C V  0 Einstein: As T  0K, C V  T -2 exp(-hν/β) [accounts for ZPE that can take up energy] Better Expts: As T  0K, C V  T 3 Debye: As T  0K, C V  T 3 [allows coupling between vibrational modes] Prob 11.9