Chapter 10 boltzmann distribution law bdl
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Chapter 10: Boltzmann Distribution Law (BDL). From Microscopic Energy Levels to Energy Probability Distributions to Macroscopic Properties. I. Probability Distribution. Microstate: a specific configuration of atoms (Fig 10.1 has 5 microstates)

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Chapter 10: Boltzmann Distribution Law (BDL)

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Chapter 10 boltzmann distribution law bdl

Chapter 10: Boltzmann Distribution Law (BDL)

From Microscopic Energy Levels to Energy Probability Distributions

to Macroscopic Properties


I probability distribution

I. Probability Distribution

  • Microstate: a specific configuration of atoms (Fig 10.1 has 5 microstates)

  • Macrostate: a collection of microstates at the same energy (Fig 10.1 has 2 macrostates); they differ in a property determined by finding <Property>.

  • Recall: <Prop> requires the probability distribution (i.e. weighting term) for the system.


Ii bdl

II. BDL

  • Consider a system of N atoms with allowed energy levels E1, E2, E3, …Ej…

  • These energies are independent of T.

  • What is the set of equilibrium probabilities for an atom having a particular energy? p1, p2, …pj,…, i.e. what is the probability distribution?


Chapter 10 boltzmann distribution law bdl

BDL

  • Let T, V, N be constant  dF = 0 is the condition for equilibrium.

  • dF = dU –TdS – SdT = dU – TdS = 0

  • Find dU and dS; plug into dF and minimize

  • Σpj = 1  αΣdpj = 0 is the constraint


Chapter 10 boltzmann distribution law bdl

BDL

  • U = <E> = Σ pjEj

  • dU = d<E> =Σ(pjdEj + Ejdpj) but dEj = 0 since Ej(V, N) and V and N are constant andEj does not depend on T.

  • dU = d<E> = Σ(Ejdpj)

  • S = -k Σpjℓn pj 

  • dS = -k Σ(1 + ℓn pj)dpj


Chapter 10 boltzmann distribution law bdl

BDL

  • dF = dU – TdS = 0Eqn 10.6

    = Σ[Ej + kT(1 +ℓn pj*) + α] dpj* = 0

  • Solve for pj* = exp(- Ej /kT) exp(- α/kT – 1)

  • Σpj* = 1 = exp(- α/kT – 1) Σ exp(- Ej /kT)

  • BDL = pj* = [exp(- Ej /kT)]/Σ exp(- Ej /kT) = [exp(- Ej /kT)]/QEqn 10.9(Prob 6)

  • Denominator = Q = partition functionEqn 10.10


Iii applications of bdl

III. Applications of BDL

  • p(z) = pressure of atm α N(z) α exp (-mgz/kT)

  • p(vx) = Eqn 10.15 = 1-dimensional velocity distribution = √m/(2πkT) exp (-mvx2/2kT)

    <vx2> = kT/m for 1-di

    <Ek> = kT/2 for 1-di; kT for 2-di; 3kT/2 for 3-di

  • p(v) = [m/(2πkT)]3/2 exp (-mv2/2kT) for 3-di

    = Eqn 10.17


Iv q partition function

IV. Q = Partition Function

  • Q = Σ exp(- Ej /kT)

  • Tells us how particles are distributed or partitioned into the accessible states. (Prob 3)

  • As T increases, higher energy states are populated and Q  number of accessible states. This is also true for energy levels that are very close together and easily populated. (Fig 10.c)


Chapter 10 boltzmann distribution law bdl

Q

  • An inverse statement can be made: As T decreases, higher energy states are depopulated and with the lowest state being the only one occupied. In this case, Q  1. This is also true for energy levels that are very far together and only the j = 1 level is populated. (Fig 10.5a)


Chapter 10 boltzmann distribution law bdl

Q

  • The ratio Ej/kT determines if we are in the high T (ratio is low) range or low T (ratio is high) range. (Fig 10.6)

  • If the ℓthenergy level is W(Eℓ)-fold degenerate, then Q = ΣW(Eℓ)exp(- Eℓ/kT)

  • Then pℓ* = W(Eℓ)exp(- Eℓ/kT)/Q


Chapter 10 boltzmann distribution law bdl

Q

  • A system of two independent and distinguishable particles A and B has Q = qA qB; in general Q = qN

  • A system of N independent and indistinguishable particles has Q = qN/N!

  • (Prob 5, 8)


V from partition function to thermodynamic properties

V. From Partition Function to Thermodynamic Properties

  • Recall the role of Ψ ( energy, angular momentum, position, … i.e. properties) in QM. To some extent, the role of Q ( U, S, G, H… i.e. thermo. prop.s) is similar.

  • Table 10.1Prob 11

  • Ensemble: collection of all possible microstates. Canonical (constant T, V,N), Isobaric-isothermal (T,p,N), Grand canonical (T,V,μ), microcanonical (U,V<N)


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