Lecture 26 Statistical Thermodynamics - Nc State …Lecture 26 Statistical Thermodynamics NC State...

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Chemistry 431 Lecture 26 Statistical Thermodynamics NC State University

Transcript of Lecture 26 Statistical Thermodynamics - Nc State …Lecture 26 Statistical Thermodynamics NC State...

Page 1: Lecture 26 Statistical Thermodynamics - Nc State …Lecture 26 Statistical Thermodynamics NC State University. The Boltzmann factor ΔE If we examine the energy spacing ΔE for any

Chemistry 431

Lecture 26

Statistical Thermodynamics

NC State University

Page 2: Lecture 26 Statistical Thermodynamics - Nc State …Lecture 26 Statistical Thermodynamics NC State University. The Boltzmann factor ΔE If we examine the energy spacing ΔE for any

The Boltzmann factor

ΔE

If we examine the energy spacing ΔE for any two levelswe determine the relative population of those levels usingtemperature (RT) as our energy yardstick. The relativepopulation is given by the Boltzmann factor, exp{-ΔE/RT}.

N1

N2 N2

N1= e– ΔE/RT

The units of ΔE must bekJ/mol to be consistent

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Analogy with equilibrium constant

ΔGo

You have already learned that the relationship betweentwo species in equilibrium is given by K = exp{-ΔGo/RT}.

[A]

[B][B][A] = e– ΔGo/RT

The units of ΔGo must bekJ/mol to be consistent

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The Boltzmann factor

Δε

If we examine the energy spacing Δε of an individualmolecule we can apply the same reasoning. The relative population is still given by the Boltzmannfactor, however, the constant R (kJ/mol) is replaced by k(J/molecule) so that the expression becomes exp{-Δε/kT}.

N1

N2 N2

N1= e– Δε/kT

The units of Δε must beJ to be consistent

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We return to the concept of a partition function. Now, we have seen a number of quantum systems where there multiple energy levels and where there is degeneracy.In general, gj is the degeneracy, εj is the energy:

We assume that the energy of the lowest energy level, the ground state is ε0 = 0.

Examples:A.) Two level system (example spin ½ system)B.) Infinite energy ladder (vibrations, photons in blackbody)C.) Spherical harmonic (examples: rotation, ang. mom.)D.) Gaussian (particle in a box, on a circle etc.)

∑ −=j

jegq jβε

Statistical Thermodynamics

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Two Level System• Assume that degeneracy

g0=g1=1 (i.e. single state is found at each level).

• q = 1+e−βε

• Note that as T → 0, q → 1 and as T → ∞, q → 2.

• The ratio of the population in the two is states e−βε

where ε is the energy difference between the two states.

ε

0

1

P1 = e– ε/kT

1 + e– ε/kT

P0 = 11 + e– ε/kT

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Statistical averaging over translational energy levels

Quantum mechanics must agree with classicalphysics (mechanics) at high temperature or when the average quantum number becomes verylarge. This is the case for translational energylevels since the spacing of those levels is verysmall compared to thermal energy, kT.

Here, we consider how to average over theenergy levels given by the vibrational, rotational,and particle-in-a-box solutions.

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The vibrational partition functionThe vibrational energy levels are evenly spacedwith a separation of hν or hω.  If we take the zero‐‐point level as our “zero of energy” then εv = vhω.

qv = e– βvhcνΣv = 1

= e– βhcν vΣv = 1

= xvΣv = 1

v

S = x + x2 + x3 + x4 + ....S/x = 1 + x + x2 + x3 + ....S + 1 = 1 + x + x2 + x3 + ....S + 1 = S/x

S = 11 – x

∴ 11 – x = x + x2 + x3 + x4 + ....

x = e– βhcν = e– ε/kT

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The vibrational partition functionTherefore, we can express the vibrational partitionfunction as

qv = e– βvhcνΣv = 1

= e– βhcν v= 1

1 – e– βhcνΣv = 1

This is the partition function that we found for aninfinite ladder of energy levels.  This answer is exact.

For high frequency modes ε >> kT and qv ≈ 1.For very low frequency modes ε << kT and qv ≈ kT/ε.

v

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The rotational partition functionThe degeneracy of rotational levels is J(J+1). The energies are given by ε = hcBJ(J+1) whereJ is the rotational quantum number and B is therotational constant.

qr = (2J + 1)e– βhcBJ(J + 1)ΣJ = 0

A large number of rotational levels populated sincethe rotational constant is of the order of 10 cm ‐1 formany molecules and kT > 10 cm‐1 for temperatureshigher than about 15 K.

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The high temperature form of therotational partition function

qr = (2J + 1)e– βhcBJ(J + 1)dJ0

The partition function can be expressed as anintegral at high temperature.

It turns out that the integral can be solved analytically by making the substitution u = J(J+1).

u = J(J + 1), du = (2J + 1)dJ

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The rotational partition function is an exponential integral

qr = e– βhcBudu0

= 1βhcB

Making the substitution u = J(J+1) the integralreduces to an easily soluble exponential integral.

The rotational constant B is:

B = h4πcI where I is the moment of inertia

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The translational partition functionThe translational partition function consists ofa sum over a very large number of states

q = e–(n2–1)βε = e–(n2–1)βεdn1

Σn = 1

This sum can be expressed as an integral overthe states n. The integral can be expressed asa Gaussian.

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Energies from particle in the box can be used to calculate energy level spacingThe difference in energy levels n in the particle-in-the-box solutions has the general form.

where ε n gives the energy of a large number of translational energy levels derived from the particle in the box solutions. We can write ε = h2/8mX2 to use a factor in the Boltzmanndistribution, so that εn = ε(n2 - 1)

εn = h2

8mX 2 n2 – 1

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The translational partition function is a Gaussian

The Gaussian in n integrates to π1/2/2.

Substitute for ε to obtain

e–n2βεndn0

= 1βε

1 / 2

e–x 2d0

= 12

πβε

1 / 2

qx = 2πmh2β

1 / 2

X

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The translational partition function in three dimensions X,Y, and Z

The volume is V = XYZ

This expression for the translational partitionfunction derived from the particle in the boxis the same as that derived classically from the integral over all velocities.

q = 2πmh2β

3/2

XYZ= 2πmh2β

3/2

V

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The translational partition function can be expressed in terms of a

thermal wavelength Λ

where the thermal wavelength is defined as:

q = 2πmkTh2

3/2

V = VΛ3

Λ = h2

2πmkT

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Average energy

Relative to a reference energy U0 we have:

⎟⎟⎟

⎜⎜⎜

∂∂−=− βln

0QUU

q = e– εnβΣn = 1

<ε> =εne– εnβΣ

n = 1

e– εnβΣn = 1

∞ =– ∂q

∂βq = – ∂ln q

∂β

Q = qN

N! ⇒ ln Q = N ln q – N ln N

U = – N∂ln q∂β = – ∂ln Q

∂β

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Average vibrational energyThe partition function is:

Q = 11 – e– βhν

N

ln Q = N ln 11 – e– βhν

= – N ln 1 – e– βhν

∂ N ln 1 – e– βhν

∂β = Nhνe– βhν

1 – e– βhν

U – U0 = Nhν e– βhν

1 – e– βhν

Note that the high temperature limit is: U – U0 = NkT

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Connection with Planck law for blackbody radiation

Recall that the assumption made by Planck is thatthere is an infinite energy ladder of equally spacedlevels due to the standing waves of radiation insidea blackbody. Thus the average energy is exactlywhat we just derived, except for a small rearrange-ment.U – U0 = Nhν e– βhν

1 – e– βhν= Nhν

1eβhν – 1

We have seen that U – U0 = NkT at high temperature

U – U0 = Nhcλ

1ehc/λkT – 1

is the thermal energy, which can

replace NkT in the Rayleigh–Jeans law to make the Planck law.

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Average rotational energyThe partition function is:

Q = qrN= 1

βhcBN

ln Q = – N ln βhcB

∂ N ln βhcB∂β = N

βhcB(hcB) = Nβ

U – U0 = NkT

Note that the high temperature limit is: U – U0 = NkT

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Average translational energyThe partition function is:

Q = qtransN = 2πm

βh2

3/2

VN

ln Q = – 3N2

2πmβh2 – N ln V

∂ – 3N2 ln 2πm

βh2

∂β = 3Nβh2

4πm2πmβ2h2 = 3N

U – U0 = 3NkT2

Note that the high temperature limit is: U – U0 = 3NkT/2