Statistical Mechanics Flashcards

Question

Dervie the E_tr

Answer 1

q_tr = (V/h³)[2πmkT]^3/2 lnq_tr = (3/2)lnT + ln[(V/h³)(2πmkT)^3/2] Then differentiate wrt T at constant V E_tr = (3/2)NkT = (3/2)nRT

Answer 2

Differentiate wrt T at constant V E_tr = (3/2)nRT C_v = (3/2)nR

Answer 3

q_elec = exp[-βε₀] x Σg_iexp[-β[εi-ε₀] Want this as specify states wrt ground state q_elec = exp(-βε₀) x q⁰_elec Usually q⁰_elec ≈ g₀ where g₀ is the degen of the ground state

Answer 4

E_elec = NkT²(dlnq_elec/dT) q_elec = g₀exp[-βε₀] where β = 1/kT E_elec = NkT² (1/kT²) = Nε₀

Answer 5

Magnetic susceptibility inversely propertional to T χ (chi) = c/T c = N(gμ_B)²/4k

Answer 6

Collection of v.many systems, each replicating @ thermo level disclosed in isothermal system of interest Fixed E, large fixed N, fixed volume, assumed knowledge of each ensemble units

Answer 7

Complete spec of ensemble when each system is assigned to a quantum state of Equally probable as isolated

Answer 8

P_i = n_i^*/N Analagous to standard definintion but not equal

Answer 9

Q = Σ exp[-βE_i] Not equal to standard

Answer 10

E = - (δlnQ/δβ)_V = kT²(δlnQ/δT)_V

Answer 11

Ideal solid (independent distinguishable systems): Q = q^N Ideal gas (independent indistuinguishable systems): Q = q^N/N!

Answer 12

S = k lnW(n`*`)

Answer 13

S = k lnW(n`*`)

Answer 14

**S** = S/N = (k/N)lnW(n`*`) Where S (non-bold) is the entropy of an enseble

Answer 15

dE = dq_rev + dw_rev (dE)_V = TdS As: E = ΣEiPi = (1/N) ΣEini`*` if heat @ constant V then Ei's fixed but population alters ni`*` -> ni`*` + dni and E-> E + (dE)_V (dE)_V = (1/N) ΣEi dni dS = (1/NT)ΣEi dni S = (k/N) lnW(n`*`)

Answer 16

S = (E/T) + klnQ

Answer 17

S = (k/N) lnW(ni`*`) = (k/N) (lnN! - Σln ni`*`!) Sterling's approx can be used as N and ni`*` large S = (k/N) (NlnN - N - Σni`*`lnni`*` + Σni`*`) = (k/N) (NlnN - Σni`*`lnni`*`) S = -k Σ(ni`*`/N) ln(ni`*`/N) as ni`*`/N = [exp(-βEi)/Q] S = -(k/Q)Σ (exp(-βEi))(-βEi - lnQ) = (βk/Q)(Σ Ei exp(-βEi)) + klnQ where βk = 1/T and (1/Q)Σ Ei exp(-βEi)) is E so S = (E/T) + klnQ

Answer 18

A = E - TS = -kT lnQ

Answer 19

P = -(δA/δV)T = kT(δlnQ/δV)T

Answer 20

H = E + PV = kT²(δlnQ/δdT)V + VkT(δlnQ/δV)T

Answer 21

G = H - TS = A + PV G = -kTlnQ + VkT(δlnQ/δV)T

Answer 22

μ = (δA/δn)_V,T = N_A(δA/δN)_V,T = -RT(δlnQ/δN)_V,T

Answer 23

S_tr = Nkln(e^5/2V/NΛ³) From S = (E/T)+Nkln(qe/N), and definititons of E_tr & q_tr

Answer 24

E_elec = Nε₀ q_elec = g₀ exp[-ε₀/kT] From S = (E/T)+Nkln(qe/N) S_elec = Nkln g₀

Answer 25

Overall entropy including elec and trs S = Nk ln[e^5/2g₀V/NΛ³T)

Answer 26

q_rot = Σg_J exp(-ε_J) ε_J = BhcJ(J+1) g_J = 2J+1 q_rot = Σ(2J+1)exp[-(θ_rot/T) J(J+1)] Where θ_rot = Bhc/k

Answer 27

Rotational temperature θ_rot = Bhc/k T above which sufficient # of rot states become available Can replace Σ with integral

Answer 28

T >>θ_rot where B << kT/hc Many rot states available so can integrate instead of sum

Answer 29

q_rot = T/θ_rot

Answer 30

q_rot = ∫(2J+1)exp[(-θ_rot/T)J(J+11)] dJ X = J(J+1) q_rot= ∫ exp[(-θ_rot/T)X] dX q_rot T/θ_rot

Answer 31

E_rot = NkT²(dlnq_rot/dT) = NkT C_V,rot = dE_rot/dT = Nk

Answer 32

S_rot = Nk ln(Te/θ_rot)

Answer 33

Full sum of q_rot used E_rot = (Nkθ_rot/q_rot)Σ(2J+1)J(J+1)exp[(-θ_rot/T)J(J+1)] as T << θ_rotrot = Nkθ_rot[6exp(-2θ_rot/T) + ...) Decreasing exp with T

Answer 34

E_rot = Nkθ_rot[6exp(-2θ_rot/T) + ...) So C_V,rot = 12Nk(θ_rot/T)² [exp(-2θ_rot/T) + ...) Therefore at low T, Cv -> 0

Answer 35

Explained by low and high T limit

Answer 36

q_rot = T/σθ_rot Where σ is symmetry number

Answer 37

`#` of indistinguishable forms in a molcule reached through rigid rotation σ=1 -> heteronuclear diatomics σ=2 -> homonuclear (as 180 degree)

Answer 38

neutron (n) and protons (p) are both fermions P^Ψ = -Ψ if p+n odd, fermionic nucleus but = Ψ if p+n even, bosonic nucleus as -1^(p+n)

Answer 39

|I|(uppercase i) = Sqrt[I(I+1)ℏ] I_z = m_Iℏ m_{I is 2I+1 values}

Answer 40

Can treat as 2e- problem: in triplet or singlet state

Answer 41

I_R = 1 m_I = 0, +/- 1 Wavefns are symmetric under exchange of nuclei 1<->2 α(1)α(2)>, M_I = +1 (1/sqrt2) α(1)β + β(1)α(2)>, M_I = 0 β(1)β(2)>, M_I = -1

Answer 42

I_R = 0 m_I = 0 Wavefns are antisymmetric under exchange of nuclei 1<->2 (1/sqrt2) α(1)β + β(1)α(2)>, M_I = 0

Answer 43

Ψ = Ψ_tr x Ψ_elec x Ψ_vib x Ψ_rot x Ψ_n.s Ψ_n.s = Ψ_{nuclear spin}

Answer 44

P^(Ψ_tr) = +Ψ_tr Depends only on centre of mass

Answer 45

P^Ψ_vib = +Ψ_vib As fn of only nuclear separation

Answer 46

Summarised by electronic state ^2S+1Λ^+/-_g/u Product of the inversion and reflection symmetries g x +, u x -, give symmetric g x -, u x +, gives anti-symmetric

Answer 47

P^ Ψ_elec(¹Σ⁺_g) = +Ψ_elec

Answer 48

P^ Ψ_elec(³Σ^-_g) = -Ψ_elec

Answer 49

P^Ψ_elec = (-1)^JΨ J = rot q.n., if even then symm

Answer 50

P^ Ψ_n.s = -ve if (p+n) is odd, fermionic +ve if (p+n) is even, bosonic

Answer 51

Ψ_total is -ve Ψ_tr, Ψ_vib, and Ψ_elec are +ve Ψ_n.s. is +ve when triplet or -ve when singlet. Ψ_rot is therefore -ve and J odd when triplet, and is +ve and J even when singlet

Answer 52

Ψ_total is +ve Ψ_tr, Ψ_vib, and Ψ_elec are +ve Ψ_n.s. is +ve when triplet or -ve when singlet. Ψ_rot is therefore +ve and J even when triplet, and is -ve and J odd when singlet

Answer 53

Ψ_total is +ve Ψ_tr & Ψ_vibare +ve Ψ_elec is ve Ψ_n.s. is +ve Ψ_rot is therefore -ve and J odd

Answer 54

H₂ or ¹⁹F₂

Answer 55

H₂ or ¹⁹F₂

Answer 56

D₂, ¹⁴N₂

Answer 57

³Σ_g^- ¹⁶O₂, ¹⁸O₂

Answer 58

ortho and para H₂ Can only access only odd or even J-state, but odd is 3x more likely as triplet ortho - odd J para - even J

Answer 59

Spacing increases with J so can do J=0->1, which increases degen and causes hump Occurs at θ_rot

Answer 60

@ v.low T then (3/2)R from translations then (5/2)R when rot and trans, and hump due to degen @ high T then (7/2)R when all previous and vib

Answer 61

Low T same as theorectical, (3/2)R No hump but goes to (5/2)R Then increases more linearly to populate vibration states

Answer 62

@ high T follows n.s. degen ratio @ low T then all molcules occupt J=0 so para

Answer 63

ε_v = (v+1/2)ℏw = (v+1/2)hν v (q.n. not wavelength as in 2nd) = 0,1,2..

Answer 64

q_vib = Σ_vg_vexp(-βε_v) = Σ_v=0^∞exp(-βε_v) = Σ_v=0^∞exp[-(v+1/2)ℏw/kT] Evaluate instead of ∫ q_vib = exp[-ℏw/2kT] Σ_v=0^∞exp[-vℏw/kT] Can use Maclurian series to give: q_vib = exp(-θ_v/2T)/[1-exp(-θ_v/T)] = 1/[exp(θ_v/2T)-exp(-θ_v/2T)]

Answer 65

1+x+x²+... = 1/(1-x) Series in derivation has x = exp(-ℏw/RT)

Answer 66

θ_vib = ℏw/k = ℏν/k Temperature at which ℏw = kT, so energy spacing comparable to thermal E ℏw/kT = θ_vib/T

Answer 67

q_vib = 1/[exp(θ_v/2T)-exp(-θ_v/2T)] T<<θ_vib so exp(-θ_v/2T) -> 0 so @ limit then **q_vib = exp[-θ_v/2T]**

Answer 68

E = Nk_BT²(dlnq/dT)_V & q_vib = exp[-θ_v/2T] (dlnq/dT)_V =θ_v/2T² Energy per molecule E/N = (1/2)kθ_v = (1/2)ℏw = (1/2)h**v** This is the ZPE of harmonic oscillator

Answer 69

C_vib = dε_vib/dT = 0 As is independent of T

Answer 70

q_vib = 1/[exp(θ_v/2T)-exp(-θ_v/2T)] @ T>>θ_v, the exponents are small so can expand e^x ≈ 1+x+x²/2!+... Results in q_vib = T/θ_v

Answer 71

q_vib = T/θ_vib E = NkT²(dlnq/dT) E/N = kT² x 1/T = kT Equipartition result

Answer 72

E/N = kT Cv = k per molecule or R per mol

Answer 73

Large mass means weaker bonds and smaller force constant Results in high population of vib states θvib = ℏw/k, lower when weaker bonds

Answer 74

linear: 3N-5 vib modes non-linear: 3N-6 vib modes

Answer 75

q_vib,total = q_vib¹ x q_vib² x ... ε_{vib, total} = ε_vib¹ + ε_vib² + ...

Answer 76

E levels on reagent/product filled irrespective of if on either side Just fills lower one

Answer 77

ΔG = -NkTln(K)=-NkTln(q/N) for ideal gas q must be written in terms of ZPE @ T=0 q = q⁰ x exp(-ε⁰/kT) G = -NkTln(q^θ/N) used as want standard molar of free energy G = -NkT[ln(q^θ,0/N) + ln(exp(-ε⁰/kT)) = Nε⁰ - RTln(q^θ,0/N) ΔG = ΔE₀ - RTΣ_Jv_Jln(q^θ,0_m,J/N) ΔG = ΔE₀ - RTln[Π_J(q^θ,0/N)^Vj] = - RTln[exp(-ΔE₀/RT) Π_J(q^θ,0/N)^Vj]

Answer 78

ΔG = - RTln[exp(-ΔE₀/RT) Π_J(q^θ,0/N)^Vj] Generally ΔG = -RTlnK(T) so K(T) = Π_J(q^θ,0/N)^Vjexp(-ΔE₀/RT)

Answer 79

q includes all dof for each J For an ideal gas: θ applies to q_tr,m = V^θ_m/Λ³ = RT/P^θΛ³ q⁰ implies energies relative to ε₀ -> ground state @ T=0 ΔE₀ is from ground states between reactants and products v_J is the stoichiometry coefficients

Answer 80

Λ_J = h/Sqrt[2πm_JkT] = (h/Sqrt[2πmukT]) x M_J^-1/2 = Λ₀ x M_J^-1/2

Answer 81

Calc modes of motion individually before combining trs: q^θ,0_J,m,tr = V^θ_m elec: q⁰_J,m,elec = g₀ + g₁exp[-(ε₁-ε₀)/kT]+... vib, rel to ZPE not bottom of potential well: q⁰_J,m,elec = 1/[1-exp(-hv/kT)] rot, high T limit as usually : q_J,m,rot = T/σθ_rot

Answer 82

Mass number of molecule - i.e. neutrons and protons (just integer values)

Answer 83

g₀ = 2 Spin up or down -> must be retained in product/reactants

Answer 84

Rate constant of thermal ionisation K(T) = (kT/p^θ) (1/Λ³_e) exp[-I₁(atom)/RT] Found as atoms/ions have no vibrations or rotations, same mass, ideal gas assumed

Answer 85

M <-> M⁺ + e- α is degree of dissociation M: 1-α, M⁺: α, e-: α partial pressures (molefraction x pressure) M: (1-α/1+α)P, M⁺: (α/1+α)P = e- K(T) = [(P_M+/P^θ)(P_e-/P^θ)]/(P_M/P^θ) Cancel to K(T) = (α²/1-α²) (p/p^θ) If α small then can find value

Answer 86

R <-> ‡ -> P for ‡->R, then K^‡ for ‡->P, then K^‡ Overall rate = d[P]/dt = K^‡[‡]

Answer 87

K^‡ = (p^‡p^θ/p_ABp_C) Assume ideal gas eqn: K^‡ = (p^θ/RT) `([‡]/[AB][C])` [‡] = (RT/p^θ)K^‡`[AB][C]` Therefore d[P]/dt = k^‡[‡] = k_r`[AB][C]` where k_r = k^‡(RT/p^θ) K^‡ And K = [(N_Aq_m^‡)/(q_AB,m q_C,m)] exp(-ΔE₀^‡/RT)

Answer 88

v of forming products proportional to natural v of loose vib mode (assumed to be ‡) Each time oscillation occurs it gives a chance for moleucle to escape and form products k^‡ = κv where κ= transiction coefficient, prob of forming products when oscillation occuring

Answer 89

q^‡ = ~q^‡ q_vib ' where ~q^‡ is the mpf for all other modes in ‡ q_vib' is mpf for TS accounting only for E fo low-v vibration q_vib' = (1-exp[-hv/kT])^-1, relative to ZPE When hv/kT << 1 then can expand as series q_vib ' = (1-1-(hv/kT) +...)^-1 = kT/hv where v is freq of a loose vibration

Answer 90

k_r = k^‡ (RT/p^θ) K^‡ k_r = (κ-v)(RT/p^θ)(kT/hv) ~K^‡ Where ~K^‡ is eqm constant with q_vib' removed k_r = κ(RT/p^θ)(kT/hv)(N_A~q_‡/q_ABq_C) exp(-ΔE^‡/RT)

Answer 91

k_r = κ(RT/p^θ)(kT/hv)(N_A~q_‡/q_ABq_C) exp(-ΔE^‡/RT) where all q's are molar, 0, and θ

Answer 92

v cancelled - validates can use this and other models all terms can be calc or measuured so is calculable simple cases -> good results

Answer 93

q_‡ not readily accessible (shapes&modes) so must assume and estimate Strong assumption: pre-equilibrium between R&‡, and all ‡ -> P Must have a fitting parameter κ (kappa)