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CHAPTER 11 –Energy Levels

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CHAPTER 11 –Energy Levels CHAPTER 11 –Energy Levels In Chapter 10 we learned about energy distributions and ensembles. In chapter 11 we build on those ideas by using silimple modldels for a gas moll’lecule’s (or at’)tom’s) lddladder of energy lllevels. Salient points from Chapter 10 For independent subsystems (i.e., gas atoms or molecules), Q=qN if distinguishable (Eqn10.28); and Q=qN/N! if distinguishable (Eqn 10.30) Translational Energy Levels Use Particle‐in‐box Model (1‐dimensional) nh22 ε = n 8mL2 ∞ V = 0 V = 22 2 2 V=V=∞ V=∞ −nh/8 mLkT −nTθtr / qetr ==∑∑ e n=1 IIIIII For Ar in 1 cm box 0 L x A potential energy diagram for the one‐dimensional particle in a box is drawn here. if θ/T << 1 ∞ 2 ∞ 2 −ax 1 π −nTθtr / edx= ∫0 qedntr ≈ 2 a ∫0 1/2 ⎛⎞2π mkT qLtr = ⎜⎟2 ⎝⎠h What about a 3D box? For 3D box of sides of length a, b, and c, the energy can be written as So that qtr is a product of three one dimensional translational partition functions 3/2 3/2 ⎛⎞⎛⎞22ππmkT mkT V qqrxyz== q q q⎜⎟⎜⎟223 abc = V = tr ⎝⎠⎝⎠hhΛ For Argon T = 273 K, V = 224 x 10‐2 m3 for 1 mole, p = 1 atm 32 39 qxVxtr ==2.14 10 4.79 10 states/atom Even for the lightest molecules, H2, this approximation is good at room temperature 24 Λ = 0.714 Å for H2@T = 300 K so that qtr = 2.7 x 10 states per molecule What is the average energy of a monatomic ideal gas atom? Use the translational partition function to prove that the average energy of an ideal monatomic gas atom given by 1.5 k T. Use What about ideal gas molecules ? A potential energy surface is sketched for the ground electronic state of a diatomic molecule. Assume separability for different degrees of freedom ε j = ε j,trans + ε j,rot + ε j,vib + ε j,elec 0 n =1 -D o n =0 For their partition functions we find -De distance R o M q = exp - β (ε + ε + ε + ε ) = ∑ gi ()j,trans j,rot j,vib j,elec qtrans qrot qvib qelec j=1 The average energies become < ε > = < ε trans > + < ε rot > + < ε vib > + < ε elec > Can we find qelec? Each molecule (or atom) has its own unique ladder of electronic energy states. Thus, we have no simple model and we must evaluate the electronic partition function sum explicitly. gi −Δεε12//kT −Δ kT qggel =+01 + ge 2 +... degeneracies Fortunately, most electronic states are separated from the ground state by a large energy gap, so that we only need to Hg20 ==1, Hg 02 (considering L, S) evaluate the first term, or the NO g==2, O g 9 00 first few terms in the partition Og20 = 3 3 P0 0.03 eV function summation. OP33→ P 0.02 eV () 1 J quantum # 3 PeV2 0 For qvib we can find a closed for expression in the harmonic oscillator approximation Vibration model ⎛⎞1 ευV =+⎜⎟vhv, =0,1,2,... ⎝⎠2 1 kmm υμ==, 12, where k = force constant 2πμ mm12+ ∞ Start zero from ground vibrational level. where k = Boltzmann constant 11 O2: T = 298 K, qvib = 1.0005 qvib ==−hkTυ/ 11−−xe T = 1000 K, qvib = 1.11 Finding the average vibrational energy of a diatomic The average energy is g iven by 11 q == vib 11−−xe−hkTυ/ −θvib −θvib T T θvib k e h ν e < ε vib > = = ()1 - exp(-θvib /T) ()1 - exp(-θvib /T) O2: ‐1 T = 298 K, <εvib > = 0. 814 cm ‐1 T = 1000 K, <εvib > = 183 cm In the high temperature limit, we find that qvib Æ kT/hnu and <εvib> = kT. For polyatomics 3n-6(5) 3n-6(5) ⎛ 1 ⎞ 3n-6(5) ⎛ h ν i ⎞ ε vib = ε i and q = ⎜ ⎟ ∑ vib ∏ ⎜ ⎟ < ε vib > = ⎜ ⎟ i=1 i=1 1 - exp(- θ /T) ∑ ⎜ ⎟ ⎝ i ⎠ i=1 ⎝ () exp(θ i /T) - 1 ⎠ Populations in vibrational energy levels of diatomics Pm = f m = exp (−m θ vib / T) (1 − exp (−θ vib / T ) ) hν / cm‐1 θ / K vib 19F35Cl 1Σ+ 786 1131 1 + 19F79Br Σ 671 965 0 10 127 35 I Cl 1Σ+ 384 552 1 + 10-1 127I79Br Σ 269 387 10-2 f m 10-3 10-4 FCl ICl FBr IBr 10-5 10 -6 012345678910 m This figure shows the fractional population of vibrational levels for FCl (○), FBr (▲), ICl (□), and IBr (●). Can we find qrot for rotational energy levels ? angular momentum Rigid rotor model: A = quantum # I = moment of inertia for T >> θrot 2 σ = symmetry factor (accounts for ∞ −ε /kT TIkT8π qed=+(21AA) A == overcounting of states) rot ∫0 σθθ σ h2 rot AB− σ =1 H AA− σ = 2 OKq2 , 298 , rot = 72 HO σ = 2 C 2 CH σ =12 H H 4 CH σ =12 H 66 For nonlinear molecules 2 3/2 π IIIabc⎛⎞8π kT qrot = ⎜⎟2 σ ⎝⎠h Populations in rotational energy levels of diatomics ⎛ θ ⎞ (2J + 1) exp ⎜ - J (J + 1) rot ⎟ nJ ⎝ T ⎠ f = = 0.04 FCl J N qrot 0.03 FBr f ICl J 0.02 0.01 IBr 0.00 0 20406080100120140 J The population distribution of rotational energy levels is shown here for four interhalogen molecules. Let’s put the pieces together and build the total partition function qqqqq= tr rot vib el q 1 Consider a linear diatomic with el = 3/2 ⎛⎞281ππmkT⎛⎞2 IkT ⎛ ⎞ qV= ⎜⎟22/⎜⎟ ⎜−hkTυ ⎟ ⎝⎠hhe⎝⎠σ ⎝1− ⎠ Ideal gas (indistinguishable particles) ⎡⎤q N An ⎢⎥ ⎛⎞qeqN ⎛⎞ ⎣⎦N! F=− kTnQAA =− kTn =− NkTn A using Stirling’s ⎜⎟ ⎜⎟ =−NnqNnNNAA + ⎝⎠N! ⎝⎠N approximation =−NnqNnNAA() + Nne A () ⎛⎞eq = NnA ⎜⎟ ⎝⎠N Now we can find thermodynamic relations 1) Use to find the ideal gas equation of state or pV= NkT 2) Use to find the internal energy N 3(2) 3n-6(5) 3 1 ⎛ Nhν i ⎞ U = < E > = < ε total > = NkT + NkT + ⎜ ⎟ - N D0 ∑ ∑ ∑ θ i /T i=1 2 n=1 2 i=1 ⎝ e - 1⎠ 3) For the constant volume heat capacity we find ⎛ ∂ < E > ⎞ 3 3(2) 1 3n-6(5) CV = ⎜ ⎟ = N k + ∑ N k + ∑ ()N k ⎝ ∂ T ⎠V 2 n=1 2 i=1 In Class Problem A. The heat capacities of Ne and Xe are the same at T = 300 K! What is the value of their constant volume heat capacity? Why are their heat capacities the same? B. The heat cappyacity of a diatomic gas is always higher than that of a monatomic gg,as, if the total mass is the same and if the conditions are similar. Why is the heat capacity of a diatomic gas higher and what range of values may it take? C. The heat capacity of molecules is temperature dependent, even for an ideal gas of molecules. Explain why this is so! The constant pressure heat capacity ⎛ ∂ H ⎞ C P = ⎜ ⎟ = C V + R ⎝ ∂ T ⎠P 40 Heat capacity data for monatomic gases and diatomic oxygen are shown here. The data were 30 taken from I. Barin, Thermochemical Data of Pure 3.5R Substances. 20 bbbb bbb b bb b bb b b b b b b 2.5R 10 0 100 300 500 700 900 1100 1300 1500 1700 1900 2100 Temperat ure (K) We can calculate the Entropy ⎛ qN ⎞ < E > < E > ⎛ ∂ lnQ ⎞ < S > = k ln⎜ ⎟ + = k lnQ + = k lnQ + k T ⎜ ⎟ ⎜ ⎟ ⎝ N! ⎠ T T ⎝ ∂ T ⎠V 3 / 2 ⎛ 2 π m k T ⎞ 3 Use q = ⎜ ⎟ V ; < E > = Nk T ⎝ h2 ⎠ 2 3 / 2 to show that ⎛ (2 π m k T ) V e ⎞ 3 < S > = Nk ln⎜ ⎟ + Nk ⎜ 3 ⎟ ⎝ N h ⎠ 2 or 3 / 2 5 ⎛⎛ 2 π m k T ⎞ V ⎞ 3 3 < S > = Nk + Nk ln⎜⎜ ⎟ ⎟ = Nk lnM + Nk lnT + Nk lnV + K′ ⎜ 2 ⎟ 2 ⎝⎝ h ⎠ N ⎠ 2 2 In Class Problem Consider the absolute entropies for the gases shown on the right. Explain the trend in the entropies for the He 126.0 monatomic series of gases. Ne 146. 2 Ar 154.7 Kr 164.0 Xe 170.0 HHg 175. 0 We can calculate the chemical potential of an ideal gas ⎛⎞∂∂F ⎡⎤ ⎛⎞eq ⎛⎞ eq 1 μ ==−⎜⎟⎢⎥NkTAA n ⎜⎟ =−+ kT n ⎜⎟ NkT ⎝⎠∂∂NNTV, ⎣⎦ ⎝⎠ NTV, ⎝⎠ NN ⎛⎞Nq⎛⎞ =+=kTAA n⎜⎟ kT− kT n ⎜⎟ ⎝⎠eq⎝⎠ N If we write ⎛⎞qkT 0 pqkT0 = then we find that μ =−kTA n⎜⎟ with int 0 ⎝⎠p 0 pint = standard state pressure In an alternative form, we find 00 0 μμμ0int=−kTAA np ⇒ = + kT np μ = standard state chemical potential Equipartition theorem −ε xkT/ ∑ε ( xe) ( ) ε = x , ∑e−ε(xkT)/ x a degree of fdfreedom x ∞ −ε xkT/ ε ()xe() dx ∫−∞ ε = ∞ edx−ε()xkT/ ∫−∞ trans 2 2 Suppose ε ( x) = cx εαn n 1 εαrot AA+1 ε = kT e ( ) 2 vib ⎛⎞1 2 kT εαv =+⎜⎟vhv x = : ⎝⎠2 2c fluctuations ε vib = kT (both KE, PE) only holds at very high T Einstein model of solids CNk= 3 N atoms ⇒ v Dulong+ Petit equipartition actual at low T, Cv → 0 (as explained by Einstein) Einstein: solid = 3N oscillators −βhv −βhv −1 ⎛⎞e q =−(1 ehv) →ε = ⎜⎟−βhv ⎝⎠1− e 0 as T → 0 2 ∂ ε ⎛⎞hv e−hv/ kT CNv ==33 Nk⎜⎟ 2 ∂T ⎝⎠kT −hv/ kT ⎛⎞hv⎛⎞1/− hv kT (1− e ) 33Nk⎜⎟= Nk as T → 0 ⎜⎟⎜⎟2 ⎝⎠kT ⎝⎠11−+()hv / kT 3 actually CTv ~ at low T (explained by Debye model).
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