Thermodynamics

Homework 6

Dustin Cassell

dcassell@physicsforums.com

December 2, 2010

Problem 6.3Consider a hypothetical atom that has just two states: a ground state withenergy zero and an excited state with energy 2 eV. Draw a graph of the par-tition function for this system as a function of temperature, and evaluate thepartition function numerically at T=300 K, 3000 K, 30,000 K, and 300,000 K.

Using mathematica to plot a graph we see the following,

Evaluating the partition function numerically,T ≈ 1 for T=300 KZ=1.00044 for T=3000 KZ=1.46132 for T=30,000Z=1.92555 for T=300,000.

Problem 6.6Estimate the probability that a hydrogen atom at room temperature is in oneof its �rst excited states (relative to the probability of being in the groundstate). Don't forget to take degeneracy into account. Then repeat the cal-culation for a hydrogen atom in the atmosphere of the star γ UMa, whose

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surface temperature is approximately 9500 K.

Since we are looking at a relative probability, we can use equation 6.11 whichstates

P (s2)

P (s1)= e−(E2−E1)/kT .

Since there are 4 degenercy states, we multiply the resulting number by 4.Below are the calculations and results,

Now using 9500 K as a temperature instead, we see that the relative proba-bility becomes P=0.000015.

Problem 6.15Suppose you have 10 atoms of weberium: 4 with energy 0 eV, 3 with energy1 eV, 2 with energy 4 eV, and 1 with energy 6 eV.

(a) Compute the average energy of all your atoms, by adding up all theirenergies and dividing by 10.

E = 4·0 eV +3·1 eV +2·4 eV +1·6 eV10

=1.7 eV/atom

(b) Compute the probability that one of your atoms chosen at random wouldhave energy E, for each of the four values of E that occur.

P(0)=.4P(1)=.3P(4)=.2P(6)=.1

(c) Compute the average energy again, using the formula E =∑

s E(s)P (s).

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E = 0 eV · 0.4 + 1 eV · 0.3 + 4 eV · 0.2 + 6 eV · 0.1 = 1.7 eV/atom

Problem 6.20This problem concerns a collection of N identical harmonic oscillators (per-haps an Einstein solid or the internal vibrations of gas molecules) at temper-ature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf,2hf, and so on.

(a) Prove by long division that

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1 + x= 1 + x + x2 + x3 + ....

For what values of x does this series have a �nite sum?

1/(1-x) has a �nite sum for values |x| < 1.

(b) Evaluate the partition function for a single harmonic oscillator. Usethe result of part (a) to simplify your answer as much as possible.

The partition function is given by equation 6.13 and states,

Z =∑

s

e−βE(s).

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Thus using the result of part a, we can simplify the expression as

Z =1

1 − e−βhf.

(c) Use formula 6.25 to �nd an expression for the average energy of a singleoscillator at temperature T. Simplify your answer as much as possible.

Equation 6.25 states,

E = − 1

Z

∂Z

∂β.

Using Mathematica for convenience we have,

(d) What is the total energy of the system of N oscillators at tempera-ture T? Your result should agree with what you found in Problem 3.25.

The total energy of the system is just ET = N · E1 = −Nfh e−fhβ

1−e−fhβ .

(e) If you haven't already done so in Problem 3.25, compute the heat capac-ity of this system and check that it has the expected limits as T → 0 andT → ∞.

Again using Mathematica we �nd,

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Problem 6.31Consider a classical �degree of freedom" that is linear rather than quadratic:E = clql for some constant c. (An example would be the kinetic energy of ahighly relativistic particle in one dimension, written in terms of its momen-tum.) Repeat the derivation of the equipartition theorem for this system,and show that the average energy is E = kT .

We start with

Z =∑q e−βcq

=1

△q

∑q

e−βcqdq

=2

△q

∞

0

e−βcqdq

=2

△q

∞

0

e−qdq

=2

△q

1

βc

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Now we let c = 2△qc

. Noting equation 6.40 which states

E = − 1

Z

∂Z

∂β

=1

β= kT.

Problem 6.36Fill in the steps between equations 6.51 and 6.52, to determine the averagespeed of the molecules in an ideal gas.

Equation 6.51 states ν =∑all ν

νD(ν)dν and equation 6.52 states ν =√

8kTπm

.

By converting equation 6.51 to an integral we use Mathematica to take careof the trival steps, and doing so yields

which is exactly equation 6.52.

Problem 6.39A particle near earth's surface traveling faster than about 11 km/s has enoughkinetic energy to completely escape from the earth, despite earth's gravita-tional pull. Molecules in the upper atmosphere that are moving faster thanthis will therefore escape if they do not su�er any collisions on the way out.

(a) The temperature of earth's upper atmosphere is actually quite high,around 1000 K. Calculate the probability of a nitrogen molecule at this tem-perature moving faster than 11 km/s, and comment on the result.

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It is next to impossible for one of these nitrogen molecules to escape.

(b) Repeat the calculation for a hydrogen molecule (H2) and for a heliumatom, and discuss the implications.

For the hydrogen molecule we see,

Similiarly for the helium atom, 1.4 × 10−12.

It is unlikely �any single� hydrogen molecule will escape, and given the Earth'satmosphere consists of only a small percent of hydrogen molecules it is aneven less likely occurance.

(c) Escape speed from the moon's surface is only about 2.4 km/s. Explainwhy the moon has no atmosphere.

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Using a simple calculation for hydrogen shows that it is very likely any givenmolecule will escape.

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