estatística - qualifying

1 Carnot cycle

Consider a classical ideal monoatomic gas with N atoms.

(a) Write down the expression of the internal energy.

(b) What is the work done during an isothermal expansion from initial volumeVi to final volume Vf at temperature T?

(c) What is the work done during an adiabatic expansion? Use the first law ofthermodynamics.

Assume the gas is the working fluid in a Carnot cycle. In this process, thegas is expanded at constant T = Th from V1 to V2. Then it is expandedat constant entropy from V2 to V3. At this point the temperature is Tc.It is then compressed isothermally at T = Tc to a volume V4, such that itcan be compressed at constant entropy back to V1 and Th.

(d) Compute the work done on each stage of this cycle using reservoirs at Thand Tc. Obtain the heat exchange at each stage and use the first law ofthermodynamics.

(e) What is the Carnot efficiency for this process (as a function only of tem-perature Tc and Th)?

Solution

(a) U = NkBT . Function of state only depends on T .

(b) Isothermal expansion:

W =

∫ Vf

Vi

dV P = nRT lnVfVi.

(c) Adiabatic expansion:

PV γ = const or TVγ−1 = const′, γ =5

3

Q = 0⇒ ∆U +W = 0⇒W = −∆U = −3

2nR(Tf − Ti)

(d)

1→ 2 : isothermal,∆U = 0,Qh = W = nRTh lnV2

V1

2→ 3 : adiabatic,Q = 0,W = −∆U = −3

2nR(Tc − Th)

3→ 4 : isothermal,∆U = 0,Qc = W = nRTc lnV4

V3

4→ 1 : adiabatic,Q = 0,W = −∆U = −3

2nR(Th − Tc)

Total amount of work :W = Qh − |Qc| = nR[Th lnV2

V1− Tc ln

V3

V4]

Problem 2 from Qualifying exam 2010-Spring.

(e)

η =W

Qh= 1− |Qc|

Qh= 1− Tc

Th

ln V3V4

ln V2V1

ThVγ−12 = TcV

γ−13 , ThV

γ−11 = TcV

γ−14

⇒ (V2

V1)γ−1 = (

V3

V4)γ−1 ⇒ V2

V1=V3

V4⇒ η = 1− Tc

Th

2 Lithium

The ground level of the neutral lithium atom is doubly degenerate (that is, d0= 2). The first excited level is 6-fold degenerate (d1 = 6) and is at an energy1.2 eV above the ground level.

(a) In the outer atmosphere of the Sun, which is at a temperature of about6000 K, what fraction of the neutral lithium is in the excited level? Sinceall the other levels of Li are at a much higher energy, it is safe to assumethat they are not significantly occupied.

(b) Find the average energy of Li atom at temperature T (again, consider onlythe ground state and the first excited level).

(c) Find the contribution of these levels to the specific heat per mole, CV , andsketch CV as a function of T .

Use 1mev = 11.6K.

Solution

(a) If the ground level energy is defined as zero and E is the energy of excited level:

Z =∑i

di exp(−βεi) = 2 + 6 exp(−βE)

The probability that the atom is in its excited level:

P (E) = 6 exp(−βE)/Z = 6 exp(−βE)/[2 + 6 exp(−βE)] = 3/[3 + exp(βE)]

Since E = 1.2eV , T = 6000K(∼ 0.5eV ), βE = 2.32, exp(βE) ≈ 10, we get:P (E) = 3/(3 + 10) = 0.23.

(b) The average energy per atom is:

〈ε〉 =1

Z

∂Z

∂β=

ε6e−βε

2 + 6e−βε=

3ε

eβε + 3

(c) The specific heat is:

Cv = [∂〈ε〉∂T

]V =3kB(βε)2eβε

(eβε + 3)2

limT→0

Cv = limβ→∞

Cv =3ε2

kBT 2e−ε/kBT

limT→∞

Cv = limβ→0

Cv =3ε2

16kBT 2


3 Probability of being void

A room of volume V = 10m3 is under standard conditions (classical ideal gasat pressure 1 atm and temperature T = 273K).

(a) Estimate the probability that at any instant of time a small subvolume v0of V becomes totally void, because of spontaneous statistical fluctuation,assuming v0 V .

(b) Specialize to the following three subvolumes of the room: (i) v0 = V/N ,whereN is the number of particles, (ii) 1cm3 and (iii) 1A3 (recall 1A=10−10m).

You can use the following approximate values: 1 atm=100kPa=100kN/m2, andk = 1.38× 10−23J/K for the Boltzmann constant.

Solution

(a) The probability that particle 1 is NOT in v0 is

P1 = 1− v0

V.

Therefore, the probability that all N particles are not in v0 is

PN = (1− v0

V)N

For v0/V small,

PN = exp(−Nv0

V).

(b) (i) PN = e−1 = 0.367879...for v0 = V/N.

(ii) PV = NkT implies

N = PV/Tk =105 × 10

273× 1.38× 10−23≈ 2.65× 1026.

For v0 = (0.01)3m3 = 1.0× 10−6m3 we have v0/V = 10−7 and

PN = exp(−2.65× 1019) = 0.

(iii) For v0 = 1A3 = 10−30m3 we have v0/V = 10−31 and

PN = exp(−2.65× 10−5) = 1− 2.65× 10−5 = 0.999973 ≈ 1.

Problem 5 from Qualifying exam 2011-Fall.

4 Heating in house

Does heating in winter

(a) change the energy content of the house?

(b) change the entropy content of the house?

Answer these questions in an ideal gas approximation with u = cvT for the one-particle internal energy and cv constant. Compare states of constant pressureat an initial and final temperature assuming that the initial and final states areeach in equilibrium.

Solution

(a) No. Using Ti for the initial temperature and the ideal gas equation pV = nRT ,the initial internal energy becomes

Ui = NcvTi =cvkB

pVi.

Heating at constant pressure to a temperature Tf > Ti results in a final volumeVf given by pVf = nRTf . For this volume the energy is

Uf =cvkB

pVf .

However, the house does not expand, so that the energy in the house is

Uf,house =ViVfUf = Ui.

As the volume of the house stays constant the energy inside the house is in thefinal state the same as in the initial one.

(b) For the entropy of an ideal gas at constant pressure we take the equation

S = Ncv lnT

T0+NkB ln

V

V0+ S0 = N(cv + kB) ln

T

T0+ S0,

where T > T0 and T0 is the liquefaction temperature. We neglect in the followingthe residual entropy of the liquid. The entropy in the house is then

S(T ) '(cv + kBkB

)· PVT· ln T

T0

=

(cv + kBkB

)· PVT0·[T0

TlnT

T0

].

Now, it is easy to see that the function lnx/x increases monotonically for 0 <x < e = 2.718..., decreases monotonically x > e. Therefore entropy changes byheating in the winter. If we use T0 = 78K, which is the actual boiling point ofair, even at T = 233K = −40C, T/T0 > e. So it actually decreases.


5 Bose-Einstein condensation

In this problem, we will be exploring the Bose-Einstein transition temperature,TBE , of a gas of N non-interacting spin-less Bose particles. Below TBE there isa macroscopic occupation of the ground state. Here, each of the particles hasmass m and all are enclosed in a three-dimensional volume V at temperatureT .

(a) Find an expression for the density of available single-particle states D(ε)as a function of the single-particle energy ε. Sketch D(ε)

(b) What is the allowed range of the chemical potential µ for a non-interactingBose-Einstein gas?

(c) Write down an expression for the mean occupation number of a singleparticle state, < n >, as a function of ε, T, and µ(T ), where TBE < T <∞.Sketch < n > as a function of ε.

(d) Write down an integral expression which implicitly determines µ(T ). Putthe integral into dimensionless form.

(e) Using the result from part (d), find TBE .

Solution

(a) We assume these particles are non-relativisic, and therefore have energy dispersion

ε(k) = h2k2

2m. The single-particle density of states is given by:

D(ε) = V

∫d3k

(2π)3δ(ε− h2k2

2m)

We can use a change of variables:

x =h2k2

2m→ k =

√2mx

h2

dx =h2

mkdk =

h2

m

√2mx

h2 dk =

√2h2x

m

dk =

√m

2h2xdx

D(ε) =4πV

(2π)3

∫2mxdx

h2

√m

2h2xδ(ε− x)

D(ε) =V

(2π)2

(2m

h

)3/2√ε

The plot goes like the square-root of ε. The factors in front scale the curve upor down, but do not change it’s shape.

Problem SM7 from FSU Qualifying Exam Wiki.

(b) Recall that z = eβµ. For a Bose gas, the occupation factor is:

n(ε) =1

z−1eβε − 1

The occupation factor is always greater than/equal to zero (it makes no senseto have negative occupancy). When ε = 0 the occuption factor becomes:

n(0) =1

z−1 − 1=

z

1− z ≥ 0

This inequality can only be satisfied if z < 1. Since z = eβµ, we find that µ < 0.

(c) For a Bose-Einstein gas, the occupation number is given by:

n(ε) =1

z−1eβε − 1=

1

e−µ−εkT − 1

As ε→ µ the distribution diverges, and as ε→∞ the distribution goes to zero:

(d) The particle density can be written as a sum over the occupation number forall possible energies, which in turn must be broken down into the macroscopicoccupation of the ground state plus all other energy states:

N

V=

1

V

∑k

n(ε(k)) =n(0)

V+

∫d3k

(2π)3n(ε) =

1

V

z

1− z +

∫d3k

(2π)3

1

z−1eβε − 1

Since z = eβµ, when this expression is evaluated we could solve for z and thenfind µ. At T = TBE , z → 1 (µ→ 0), so we have:

N

V=

∫d3k

(2π)3

1

eβBEh2k2

2m − 1

We can do a change of variables similar to the first part of the problem:

x =βBE h

2k2

2m→ k2 =

2mx

βBE h2

dk =1

2

(2m

βBE h2

)1/21√xdx

N

V=

4π

(2π)3

∫2mx

βBE h2

1

2

(2m

βBE h2

)1/21

x

1

ex − 1dx

N

V=

1

(2π)2

(2m

βBE h2

)3/2 ∫ √x

ex − 1dx

(e)

N

V=

1

(2π)2

(2m

βBE h2

)3/2 ∫ √x

ex − 1dx

N

V=

1

(2π)2

(2m

βBE h2

)3/2 (1.306π1/2

)(2π)2N

V

(h2

2m

)3/21

1.306π1/2= (kTBE)3/2

TBE =

((2π)2N

V

(h2

2mk

)3/21

1.306π1/2

)2/3

6 Two questions

Question One:A sealed, perfectly insulated container is partitioned into two halves. One con-tains an ideal gas at temperature T and the other, a vacuum. The partition isbroken instantly and the gas expands to fill the entire container.

(a) After the gas reaches equilibrium, is the temperature great than, less than,or equal to T? Explain.

(b) Would your answer be different if the gas were real? Explain.

Question Two:The mean free path is defined as the average distance traveled by a moleculebetween two consecutive collisions with the other molecules. Assume that theabsolute temperature of air increases by a factor of two, while the air pressureremains the same. Does the mean free path change? If yes, by how much? Ifnot, why?

Solution

Question One:

(a) The temperature is unchanged because it is an intensive quantity and it is un-changed by the size of the container.

(b) Real gases have molecules that interact, so a real gas in a large container wouldhave molecules that interact less. The increase in volume also causes a increasein the potential energy and since it is an insulated container that means thekinetic energy must decrease to keep the energy constant thus resulting in adecrease in temperature.

Question Two:

The mean free path, λ, is proportional to temperature. This makes sense if you think

about the relationship between pressure, volume, and temperature, PV ∝ T. The mean

free path is proportional to volume in this situation since the more volume there is

for the same number of particles to interact means that there will be fewer collisions.

Therefore one can think about the mean free path as being proportional to pressure, ie

λ ∝ T/V. Therefore if the temperature increase by a factor of two while the pressure

remains the same, the mean free path will increase by a factor of two as well.


7 Pressure on a solid

Consider a solid. At zero temperature, the solid has the following dependenceof pressure on the chemical potential,

P0(µ) = a(µ− µ0) + b(µ− µ0)2

Now make the temperature nonzero. According to the quantum theory of solidsconstructed by Debye, at nonzero, but small, temperature T the pressure isthe sum of the zero-temperature pressure, P0(µ), and the pressure of a gasof quasiparticles called phonons. Phonons are quanta of sound; they behavelike bosons with three possible polarizations. The energy of a phonon withmomentum p is:

ε = u|p|

where u is the speed of sound, which can be regarded as a constant in dependentof the chemical potential µ.

(a) From the information above, compute the pressure at low temperatures,P (T, µ). Note that the chemical potential of phonons is equal to zero.

(b) The solid is kept at zero pressure. Assuming the temperature is small, findthe chemical potential as a function of temperature.

(c) Find the number density of atoms as a function of temperature, fixing thepressure to zero. Find the thermal expansion coefficient at zero pressure.

(d) Using general arguments, including thermodynamic stability, show that thecoefficients a and b appearing in the expression for P0 are both positive,as is the thermal expansion coefficient computer in part (c).

Solution

(a) Pressure is related to the grand potential, Σ, by the the relation PV = −Σ, whereΣ = −kT ln(Z) and Z is the grand partition function Z =

∑∞N=0 z

NQN . Forbosons the grand partition function is given by:

Z =∏k

1

1− e−β(εk−u)→∏k

1

1− e−βεk

Σ = −kT lnZ = kT∑k

ln(1− e−βεk )

We can turn the sum on k into an integral using the following relation:

1

V

∑k

=

∫d3k

(2π)3

Σ = kTV

∫d3k

(2π)3ln(1− e−βεk )


εk = u|p| = uhk

Σ =kTV 4π

(2π)3

∫k2dk ln(1− e−βuhk)

Make the substitution η = βuhk to turn the integral into a unitless quantity:

Σ =kTV 4π

(2π)3

(1

βhu

)3 ∫η2dη ln(1− e−η)

Σ = (kT )4V4π

(2π)3

(1

hu

)3 ∫η2dη ln(1− e−η)

We can group all of the constants together can call it γ and that simplifies theequation to:

Σ = T 4V γ

−PV = Σ = (T )4V γ

P = −γT 4

So the total pressure is:

Ptot = P0 + P = a(µ− µ0) + b(µ− µ0)2 − γT 4

(b) If the total pressure is kept at zero, we can write:

0 = a(µ− µ0) + b(µ− µ0)2 − γT 4

Solving this quadratic equation give us:

(µ− µ0) =−a±

√a2 + 4bγT

2b

µ =−a±

√a2 + 4bγT

2b+ µ0

This can be simplified slightly by expanding the square-root, since we are as-suming T is small (we’re not given any information about a or b, so we mustjust assume they’re both much larger than T):

µ =−a± a

√1 + 4bγT

a2

2b+ µ0

µ =−a± a(1 + 2bγT

a2 )

2b+ µ0

We have the two solutions:

µ =( 2bγT

a)

2b+ µ0 =

γT

a+ µ0

µ =−2a− ( 2bγT

a)

2b+ µ0 = −

a+ ( bγTa

)

b+ µ0

We will only consider the first solution because µ cannot be less than µ0 andwe are assuming that all of the constants are positive (γ is positive and theexplanation for the sign of a and b *should* be provided in part (d).)

(c) The number density is given by n = NV

. The mean occupation number N can befound by:

N = −(dΣ

dµ

)Another expression for N is:

N =∑k

n(ε) =∑k

1

z−1eβε − 1=∑k

1

eβ(ε−µ) − 1

Substituting in our expression for µ at zero pressure and turning the sum intoan integral:

N = V

∫d3k

(2π)3

1

eβ(ε− γTa−µ0) − 1

= V

∫d3k

(2π)3

1

eεkT− γka− µ0kT − 1

N = V

∫d3k

(2π)3

1

e−γka e( ε

kT− µ0kT

) − 1= V

∫d3k

(2π)3

1

e−γka eβ(ε−µ0) − 1

η = β(ε− µ0) = β(uhk − µ0) = βuhk − βµ0 → k =η + βµ0

βuh

dη = βuhdk → dk =dη

βuh

N =4πV

(2π)3

∫ (η + βµ0

βuh

)2dη

βuh

1

e−γka eη − 1

N =4πV

(2πβuh)3

∫ (η2 + 2ηβµ0 + β2µ2

0

)dη

1

e−γka eη − 1

N =4πV

(2πβuh)3

(∫η2dη

1

e−γka eη − 1

+ 2βµ0

∫ηdη

1

e−γka eη − 1

+ β2µ20

∫dη

1

e−γka eη − 1

)Since each of these integrals is a dimensionless constant, call them A, B, C, wehave:

N =4πV

(2πβuh)3

(A+ 2βµ0B + β2µ2

0C)

N =4πV k3

(2πuh)3

(T 3A+

T 2

k2µ0B +

T

k2µ2

0C

)So, finally, we have the number density as a function of temperature:

N

V=

4πk3

(2πuh)3

(T 3A+

T 2

k2µ0B +

T

k2µ2

0C

)

The coefficient of thermal expansion describes how the size of an object changeswith temperature:

α =1

V

(dV

dT

)V =

N(2πu)3

4πk3

(T 3A+

T 2

k2µ0B +

T

k2µ2

0C

)−1

dV

dT= −N(2πu)3

4πk3

(T 3A+

T 2

k2µ0B +

T

k2µ2

0C

)−2(3T 2A+

T4µ0B

k+

1

k2µ2

0C

)

(d) We don’t have much of an explanation for this. Looking at the simplified expres-sion for µ at zero pressure suggests a must be positive because the minimum ofµ is µ0.

8 Chain-like molecule

A chain-like molecule is composed of N rigid straight line segments connectedby hinges that can only be in two states, namely at an angle θ = 0 or π. Thefigure shows a typical configuration (but with the θ = π angles slightly smallerthan 180 for graphical clarity). Each rigid line segment carries a magneticmoment µ aligned with the segment, and the magnetic field points in the x-direction, B = Bex. Since all line segments point in the x-direction, all magneticmoments point either along or opposite to the magnetic field.

(a) Calculate the entropy S(N,T) of the chain.

(b) Calculate the average distance between the end points of the chain.

(c) Calculate the root-mean square fluctuation in the endpoint separation.

(d) Discuss your result in part (c) in the limit of very high temperatures fromthe perspective of random walks. In case you did not succeed in solvingpart (c), you should still be able to predict how ∆D(N) depends on Nusing random walk arguments.

Solution

(a) If we have a segments at π the total quantity of possible states will be:

Ω =N !

a!(N − a)!(1)

Knowing that the energy of a magnetic moment µ is given by E = −µ · B, theenergy of the system will be:

E = −(N − 2a)µB (2)

=⇒ a =E

2µB+N

2(3)

From the relation for the entropy S = kB ln Ω and from (1),(3):

S = kB lnN !− kB ln

(N

2+

E

2µB

)!− kB ln

(N

2− E

2µB

)!

Using the Stirling’s approximation (lnN ! = N lnN −N +O(lnN)) we have:

S = kBN lnN − kBN − kB(N

2− E

2µB

)[ln

(N

2− E

2µB

)− 1

]−kB

(N

2+

E

2µB

)[ln

(N

2+

E

2µB

)− 1

]

S = −kBN[(

1

2+

E

2µBN

)ln

(1

2+

E

2µBN

)+

(1

2− E

2µBN

)ln

(1

2− E

2µBN

)](4)


Using the relation 1T

=(∂S∂E

)N

and the previous one for the entropy we have:

1

T=

kB2µB

ln

(µB − E

N

µB + EN

)Solving for E:

E = −NµB tanhµB

kBT(5)

From (4) and (5) we can have S(N,T )

(b) If the distance between each two consecutive points is l, then the average distacewill be:

〈D〉 = (N − 2a)l (6)

From (2) and (6):

E = −〈D〉lµB (7)

And from (5) and (7) we have:

〈D〉 =

[NµB tanh

µB

kBT

]l

This result can also be obtain from :

〈D〉 =

∑E

(N − 2a)le−βE

Z

= − l

µB

∑E

Ee−βE

Z

= − l

µB〈E〉

= − l

µB

(−NµB tanh

µB

kBT

)=

[NµB tanh

µB

kBT

]l

(c) ⟨∆D2⟩ =

⟨D2⟩− 〈D〉2

=

∑E

(N − 2a)2l2e−βE

Z−

∑E

(N − 2a)le−βE

Z

2

=l2

µ2B2

∑E

E2e−βE

Z−

l

µB

∑E

Ee−βE

Z

2

=l2

µ2B2

[⟨E2⟩− 〈E〉2]

=l2

µ2B2

(−∂E∂β

)(8)

And from (5) and (8) we have:

⟨∆D2⟩ =

l2

µ2B2

∂

∂β(NµB tanhµBβ)

=l2

µ2B2Nµ2B2Sech2

(µB

kBT

)

=⇒ ∆D(N,T ) =√NlSech

(µB

kBT

)(d) From part (c) we have:

∆D(N,T ) =√N

l

cosh(µBkBT

)For very high temperatures cosh

(µBkBT

)→ 1, therefore:

∆D(N,T )→ l√N

9 1D ideal gas

A one-dimensional monatomic ideal gas with N atoms at temperature T isconfined to a line of length L. (An ideal gas has no collisions, so the atomscan move through each other.) Because the atoms are in motion they exert aneffective (time averaged) force F on each end of the line.

(a) Find F and show that F · L is related to the average energy of the atoms

(b) Use the equipartition theorem to find an expression for the heat capacityat constant length, CL

(c) Find the equation of state (i.e., show how to write F·L in terms of thetemperature T).

Solution

(a) Method 1:Find the force by finding the change in momentum:

F =∆p

∆t

pi = mvi pf = −mvi ∆p = 2mv

∆x = v∆t→ ∆t =2L

v

F =∆p

∆t=mv2

L

Ftotal =Nmv2

LEtotal =

1

2Nmv2

F · L = Nmv2 = 2Etotal

Method 2: (Must do parts (b) and (c) first)P = F (in one dimension). From part (b): E = 1

2NkT . From part (c): F · L =

NkTF · L = nKT = 2E

(b) The equipartition theorem gives the following relationship (where f is the numberof degrees of freedom, in this case f = 1):

E =1

2fNkT =

1

2NkT

The heat capacity at constant length is given by:

CL =

(∂E

∂T

)L

=1

2Nk


(c) Method 1:We found an expression for the energy in part (A) using the kinetic energy ofthe particles. We also found an expression for the energy in part (B) using theequipartition theorem. Equating these two expressions for the energy:

E =1

2NkT =

1

2Nmv2

From part (A) we know that:

F · L = Nmv2 → mv2 =F · LN

This gives us:NkT = F · L

Method 2:

Q1 =∑k

e−βh2k2

2m = L

∫dk

2πe−

βh2k2

2m

x =βh2k2

2mdx =

βh2k

mdk

Q1 =L

2π

√m

2βh2

∫dx√xe−x =

L√π

2π

√m

2βh2

Q1 = L

(πmkT

2h2

)1/2

QN =1

N !(Q1)N

A = −kT ln(QN ) = −kT [N ln(Q1)− ln(N !)] = kT [−N ln(Q1) +N ln(N)−N ]

A = NkT [− ln(Q1) + ln(N)− 1] = NkT

[ln

(N

Q1

)− 1

]

A = NkT

[ln

(N

L

(2h2

πmkT

) 12

)− 1

]= NkT

[ln

(N

L

)+

1

2ln

(2h2

πmkT

)− 1

]P = −∂A

∂L=NkT

L→ P · L = F · L = NkT

This method is much more complicated and time consuming than method 1.However if other thermodynamic properties are asked for it may be necessaryto compute A.

10 2-level system

Consider a 2-level system with energy states ε and ε + ∆ (∆ ≥ 0). Assumethe absolute temperature is T.

(a) Compute the partition function.

(b) Determine the free energy.

(c) Derive an expression for the specific heat C(T ). What are the low-T andhigh-T limits of this expression? Sketch your result.

Solution

(a) Partition function:

Z =∑s

e−βε = e−βε + e−β(ε+∆) (9)

(b) Free energy:

F = −kT lnZ = −kT ln(e−βε + e−β(ε+∆)) (10)

where β = 1/(kT ).

(c) To calculate the specific heat, you first need the internal energy:

U = −∂ lnZ

∂β=εe−βε + (ε+ ∆)e−β(ε+∆)

e−βε + e−β(ε+∆)= ε+

∆

e∆/kT + 1(11)

then

C(T ) =∂U

∂T=

∆2e∆/kT

kT 2(1 + e∆/kT )2 (12)

Problem 12 from Qualifying exam 2007-spring.

1 Molecules in a plane

Molecules confined to rotate in a plane have quantized energies

En = (h2/2I)n2 (n = 0, 1, 2, ...)

(I is the moment of inertia) The levels with n ≥ 1 are doubly degenerate.

(a) Compute the rotational partition function in the low- and high-temperaturelimits.

(b) Do a classical calculation of the partition function and compare with theresults in part (a).

(c) Compute the specific heat per molecule in the low- and high-temperaturelimits.

Solution

(a)

En =h2

2In2 (1)

Z =∑n

gne−βEn = 1 + 2

∑n=1

e−βh2

2In2

(2)

Low temperature: β 1, sum over the first few terms,

Z = 1 + 2e−βh2

2I + 2e−4βh2

2I + ... (3)

High temperature: β 1, we can use integral to estimate the summation:

Z = 2

∫ ∞0

dxe−βh2

2Ix2

− 1 (4)

= 2× 1

2

√π2I

βh2 − 1 (5)

≈

√2Iπ

βh2 (6)

(b) Classical:

Z =1

h

∫ ∞−∞

dθdPθe−p2θ

2Iβ (7)

=2π

h

∫ ∞−∞

dPθe−p2θ

2Iβ (8)

=

√2Iπ

βh2 (9)

The classical result is the same as the high temperature approximation, whichmakes sense.

Problem 4 from UCLA 1987Fall.

(c)

cV =∂U

∂T|V , U = −∂ lnZ

∂β. (10)

For low temperature,

Z ≈ 1 + 2e−βh2

2I (11)

lnZ ≈ ln(1 + 2e−βh2

2I ) (12)

U = −∂ lnZ

∂β=h2

I

1

eβh2

2I + 2(13)

cV =∂U

∂T= (

h2

I)2 1

2kT 2

eh2

2IkT

(2 + eh2

2IkT )2(14)

≈ (h2

I)2 1

2kT 2e−

h2

2IkT . (15)

For high temperature,

lnZ =1

2ln

2Iπ

βh2 (16)

U = −∂ lnZ

∂β=

1

2β=kT

2(17)

cV =∂U

∂T=k

2(18)

2 Ideal gas

(a) Show from statistical mechanics that

P = nkT

holds exactly for a classical non-interacting gas whether it is relativisticor not.

(b) Evaluate the energy density (u) of the gas in the extreme relativistic limitand show from part (a), the relation between P and u.Hint: pressure is computed from the thermodynamic relation P = −∂F/∂V ,where F = −kT lnZ; here Z is the many-particle partition function.

Solution

(a) For a non-interacting gas,

Zn =∑s

e−βEsN =∑s

e−β∑i εi = ZN1

where

Z1 =∑s

e−βEs =1

h3

∫d3xd3pe−βE =

V

h3

∫d3pe−βE .

ThenlnZ1 = lnV + function independent on V,

so

P =N

β

∂ lnZ1

∂V=

N

βV= nkT.

(b) In the extreme relativistic limit,ε = pc,

the partition function is then

Z1 =∑

e−βε =1

h3

∫d3xd3pe−βε

=V

h34π

∫ ∞0

dpp2e−βpc =8πV

(hβc)3,

the internal energy is

U = N∂ lnZ1

∂(−β)=

3N

β= 3NkT,

the relation between P and u is

P = nkT =NkT

V=

1

3

U

V=

1

3u.


3 Heat engine

A heat engine operates by extracting energy from a container of gas consistingof N atoms with constant specific heat C per atom, initially at temperature T .The engine releases heat into the atmosphere which is at temperature T0 < T .Find the maximum work which can be extracted from the gas.

Solution

The maximum work occurs when the total entropy change is zero, ∆Stotal = 0, i.e.∫ T0

T

NCdT

T+Q′

T0= 0

where the first term is the entropy change of gas in the container and the secondterm is the entropy change of the atmosphere, with Q′ being the heat released to theatmosphere. So

Q′ = T0NC lnT

T0,

then

Wmax = Q−Q′ =

∫ T

T0

CNdT −Q′ = NC(T − T0)−NCT0 lnT

T0

Problem 6 from UCLA 1989Spring.

4 Quantum oscillator

(a) What is the average energy u of a quantum oscillator of angular frequencyω in thermal equilibrium at temperature T?

(b) What is the expectation value for the fluctuations (u−u)2u2 ?

(c) In the Einstein model of a crystal each ion has 3 degrees of freedom withthe same frequency ω. Calculate the specific heat of the crystal.

Solution

(a) The partition function is

Z =

∞∑n=0

e−β(n+ 12

)hω = (eβhω

2 − e−βhω

2 )−1

u = − ∂

∂βlnZ =

hω2

(eβhω

2 + e−βhω

2 )

eβhω

2 − e−βhω

2

=hω

2coth

βhω

2

(b)(u− u)2 = u2 − u2

u2 =

∑∞n=0[(n+ 1

2)hω]2e−β(n+ 1

2)hω

Z=

∂2

∂β2Z

Zso

u2 − u2 = −∂u∂β

= − hω2

∂

∂β(coth

βhω

2) =

(hω)2

(eβhω

2 − e−βhω

2 )2

u2 − u2

u2 =4

(eβhω

2 + e−βhω

2 )2= sech2 hω

2kT.

(c) Consider N ions’ crystal,

CV = 3N∂u

∂T=

3Nk( hωkT

)2

(ehω2kT − e− hω

2kT )2


5 Gravitation

The system shown in the diagram is thermally isolated at a temperature, T ,in a gravitational field of strength, g, and has a total heat capacity, C. The ballis initially at a height, h, above its rest position, as shown. Then the ball isreleased, and a new equilibrium is reached. (Assume that Mgh kT .)

(a) What is the increase, ∆T , in the temperature of the system?

(b) What is the change in entropy of the system?

(c) What is the probability, P , that the ball will spontaneously return to itsoriginal position?

(d) Evaluate P for ∆T/T 1, Mgh = 105ergs, and T = 300K.

Solution

(a) After new equilibrium is reached,

Mgh = C∆T ⇒ ∆T =Mgh

C

The gravitational potential goes into the heat due to collisions between the balland the gas molecules.

(b)TdS = dQ = CdT

⇒ ∆S =

∫CdT

T= C ln

T + ∆T

T= C ln(1 +

Mgh

CT)

(c)

P = e−∆Sk = e−

Ck

ln(1+MghCT

) = (1

1 + MghCT

)C/k

(d) If ∆/T 1,

P = e−Ck

ln(1+ ∆TT

) ≈ e−Ck

∆TT = e−

CkMghCT = e−

MghkT

With Mgh = 105ergs, T = 300K,

⇒ P ≈ e−10−2/(1.38×10−23×300) = e−2.5×1018

≈ 10−18 ∼ 0


6 Magnetic moments

Consider a solid composed of atoms with magnetic moment µ per atom.Assume the moments interact weakly with the lattice vibrations in the solid butnot with each other. In an applied magnetic field H a moment has energy ε =−µH (ε = +µH) if it points parallel (antiparallel) to the field, and we assumethat no other orientations are possible. The system is thermally insulated,initially in equilibrium at temperature T with applied field H, and the field issuddenly switched from H to −H.

(a) Find the new temperature of the system, T ′, after equilibrium has beenreached, in terms of the given quantities and the specific heat per atom cdue to lattice vibrations. You may assume |T ′ − T | T . Is T ′ larger orsmaller than T?

(b) Give a numerical estimate as for which initial temperature range the changein temperature in this process will be larger than 1% for H = 105 gauss.Assume µ = µB (Bohr magneton). µB = 0.927× 10−20 ergs/gauss, kB =1.38× 10−16 ergs/K.

Solution

(a) For temperature T , applied field H, the number of atoms whose magnetic mo-ments are parallel to H is (equilibrium state)

N1 = NeβµH

eβµH + e−βµH

For temperature T ′, applied field −H, the number of atoms whose magneticmoments are parallel to −H is (equilibrium state)

N ′1 = Ne−β

′µH

eβ′µH + e−β′µH

The energy for getting the atoms to equilibrium is

(N1 −N ′1)2µH

Since the system is thermally isolated,

(N1 −N ′1)2µH = NC(T − T ′)

⇒ 2µH(eβµH

eβµH + e−βµH− e−β

′µH

eβ′µH + e−β′µH) = C(T − T ′)

|T − T ′| T ⇒ β′ ∼ β

2µH(eβµH − e−βµH

eβµH + e−βµH) = −C(T − T ′) = 2µH tanhβµH

⇒ T ′ = T + 2µHtanhβµH

C

T < T ′


(b)∆T

T> 0.01

⇒ 2µH

CTtanhβµH > 0.01

⇒ 2k

CβµH tanhβµH > 0.01

Choose C ∼ 3k,⇒ βµH tanhβµH > 0.015

Let x = βµH,⇒ f(x) = x tanhx

f ′(x) =4

(ex + e−x)2x+ tanhx > 0

We need to find f(x0) = 0.015, then for x > x0, f(x0) > 0.015,

⇒ x0 tanhx0 = 0.015

Suppose x 1 (proved by result)

x20 = 0.015

x0 = 0.12

so βµH > 0.12, the initial range of T is T < µH0.12k

= 56K.

7 Bose condense

Consider a system of N noninteracting free fermions of mass m and spin 1/2in a box of volume V at zero temperature. The system is thermally insulated.Assume that suddenly the fermions turn into bosons of spin 0, with no changein their mass. Will the system Bose-condense?Hints: ∫ ∞

0

dxx1/2

ex − 1= 2.32∫ ∞

0

dxx3/2

ex − 1= 1.78

Density of states per spin for particles of mass m in box of volume V :

g(ε) =V

4π2(2m

h2)3/2ε1/2

Solution

For fermions with spin 1/2,

2

∫ EF

0

g(ε)dε = N

2V

4π2(2m

h2 )3/2 2

3E

3/2F = N

LetV

4π2(2m

h2 )3/2 = k,

then

EF =h2

2m(3π2N

V)2/3 = (

N

k

3

4)2/3

E = 2

∫ EF

0

εg(ε)dε = 2V

4π2(2m

h2 )3/2 2

5E

5/2F

=3

5N(

N

k

3

4)2/3 = 0.495

N5/3

k2/3

The system is insulated, so the total energy for bosons is also E. For Bose-condense,µ = 0,

N =

∫ ∞0

g(ε)dε1

eβε − 1

= k′∫ ∞

0

ε1/2

eβε − 1dε

= k′β−3/2 ∗ 2.32

⇒ β = (N

k′1

2.32)−2/3


E′b =

∫ ∞0

εg(ε)dε1

eβε − 1

= k′β−5/2 ∗ 1.78

= k′(N

k′1

2.32)5/3 ∗ 1.78

=N5/3

k′2/3∗ 0.4387

Let E′b = E,⇒ k′ < k ⇒ V ′ < V

That is the critical volume for Bose-condense is smaller than the volume now. So no

Bose-condensation occurs.

8 Fiber modes

A thin fiber of length L is stretched between two supports. The speed ofpropagation of transverse waves on the fiber is c for both polarizations.

(a) What is the contribution of these modes to the heat capacity of the fiberat low temperatures, assuming hc/L kT? Leave any integral in dimen-sionless form to display the temperature dependence explicitly.

(b) What is the heat capacity for hc/L kT?

Solution

The modes are kn = nπ/L, n = 1, 2, 3, .... They satisfy Bose statistics.

(a)ω = kc

E = 2∑n

hωnehωn/kT − 1

For hc/L kT ,

E ≈ 2L

π

∫ ∞0

dkhω

ehω/kT − 1

=2L

πc

∫ ∞0

dωhω

ehω/kT − 1

=2L

πch(kT

h)2

∫ ∞0

xdx

ex − 1

=2L

πhc(kT )2

∫ ∞0

xdx

ex − 1

CV =∂E

∂T=

4Lk2

πhcT

(b)

E =∑n

2hωnehωn/kT − 1

= 2∑n

πhcLn

eπhcLn/kT − 1


If hcL kT ,

E ≈ 2

∞∑n=1

nπhc

Le−

nπhcL

/kT

= 2πhc

Le−πhc/LkT

CV =∂E

∂T=

2πhc

L

πhc

Lk

1

T 2e−πhc/LkT

= (πhc

L)2 2

kT 2e−πhc/LkT

9 Long bar

A long rectangular bar at temperature T exerts a tension, t(ξ, T ), when it isextended an amount ξ beyond its natural length.

(a) What is the thermodynamic relation between dS, T, dE, t(ξ, T ) and dξ,where S is the entropy and E is the internal energy of the bar?

(b) What is the relationship between (∂S/∂ξ)T and (∂t/∂T )ξ for a quasistaticprocess?

(c) Let t(ξ, T ) = bξ(1− γT ), where b and γ are constants. What is the changein internal energy when the bar is stretched from ξ = 0 to ξ = ξ0 atconstant temperature?

(d) What is the sign and magnitude of the heat, Q, necessary to maintain thebar at constant temperature during this process?

Solution

(a) When the bar is extended the work it does is

dW = −tdξ

SodE = dQ− dW = TdS + tdξ for reversible process.

(b) For quasistatic (reversible) process, define

F = E − TS

thendF = −SdT + tdξ

⇒ (∂S

∂ξ)|T = −(

∂t

∂T)|ξ

(c) S can be expressed by S = S(T, ξ).

dE = TdS + tdξ = T [(∂S

∂T)ξdT + (

∂S

∂ξ)T dξ] + tdξ

When dT = 0,

dE = (T (∂S

∂ξ)T + t)dξ = (−T (

∂t

∂T)ξ + t)dξ

dE = [T (γbξ) + bξ(1− γT )]dξ = bξdξ

⇒ ∆E =

∫ ξ0

0

bξdξ =bξ2

0

2


(d) We only consider the heat transfer. It’s positive if it is from outside to bar.

∆Q = ∆E −∆W = ∆E −∫ ξ0

0

tdξ

=

∫ ξ0

0

bξdξ −∫ ξ0

0

tdξ

= γbT

∫ ξ0

0

ξdξ =1

2γbTξ2

0

So the sign is positive in our convention.

10 Bose statistics

Assume in a system of volume V at temperature T there are excitationswhich obey Bose statistics with ε = bp3/2, where ε is the energy and p is themomentum. The number of excitations is not conserved.

(a) Derive an expression for the temperature dependence of the heat capacity,C, of the system. Express your answer in terms of Planck’s constant andT , leaving any integral in dimensionless form.

(b) What would be the temperature dependence of C, if the system were in aspace of arbitrary dimension D?

Solution

(a) Since the number of excitations is not conserved (like photon),

N(ε) =ρ(ε)

eεkT − 1

where N(ε)dε is the average number of excitations of energy between ε andε+ dε, and ρ(ε) is the density of states.

ρ(ε)dε =V

(2π)3d3~k (19)

=V

(2π)3h3 d3~p (20)

=V

(2π)3h3 4πp2dp (21)

=V

(2π)3h3

4π

3d(p3) (22)

=V

(2π)3h3

4π

3d(ε

b)2 (23)

=V

(2π)3h3

4π

3

2ε

b2dε (24)

The total energy is

U =

∫ ∞0

N(ε)εdε (25)

=

∫ ∞0

V(2π)3h3

4π3

2ε2

b2dε

eεkT − 1

(26)

=V (kT )3

3π2h3b2

∫ ∞0

x2dx

ex − 1(27)

=2V (kT )3

3π2h3b2(28)

C =∂U

∂T=

2V k3T 2

π2h3b2(29)


(b) If in arbitrary space dimension D,

ρ(ε)dε ∝ dD~k ∝ d(pD) ∝ dε2D3 (30)

U ∝∫

ερ(ε)

eεkT − 1

dε ∝∫

εdε2D3

eεkT − 1

∝ (kT )2D3

+1 (31)

C ∝ ∂U

∂T∝ T

2D3 (32)

1 Bose statistics

Assume in a system of volume V at temperature T there are excitationswhich obey Bose statistics with ε = bp3/2, where ε is the energy and p is themomentum. The number of excitations is not conserved.

(a) Derive an expression for the temperature dependence of the heat capacity,C, of the system. Express your answer in terms of Planck’s constant andT , leaving any integral in dimensionless form.

(b) What would be the temperature dependence of C, if the system were in aspace of arbitrary dimension D?

Solution

(a) Since the number of excitations is not conserved (like photon),

N(ε) =ρ(ε)

eεkT − 1

where N(ε)dε is the average number of excitations of energy between ε andε+ dε, and ρ(ε) is the density of states.

ρ(ε)dε =V

(2π)3d3~k (1)

=V

(2π)3h3 d3~p (2)

=V

(2π)3h3 4πp2dp (3)

=V

(2π)3h3

4π

3d(p3) (4)

=V

(2π)3h3

4π

3d(ε

b)2 (5)

=V

(2π)3h3

4π

3

2ε

b2dε (6)

The total energy is

U =

∫ ∞0

N(ε)εdε (7)

=

∫ ∞0

V(2π)3h3

4π3

2ε2

b2dε

eεkT − 1

(8)

=V (kT )3

3π2h3b2

∫ ∞0

x2dx

ex − 1(9)

=2V (kT )3

3π2h3b2(10)

C =∂U

∂T=

2V k3T 2

π2h3b2(11)


(b) If in arbitrary space dimension D,

ρ(ε)dε ∝ dD~k ∝ d(pD) ∝ dε2D3 (12)

U ∝∫

ερ(ε)

eεkT − 1

dε ∝∫

εdε2D3

eεkT − 1

∝ (kT )2D3

+1 (13)

C ∝ ∂U

∂T∝ T

2D3 (14)

2 Thermodynamic engine

A peculiar thermodynamic engine operates according to the following diagram:All lines in the above thermodynamic path are straight lines. AB is horizontal

and BC is vertical.

(a) Sketch what this diagram would look like in the p− V plane.

(b) Suppose the gas used in the engine is an ideal gas with specific heatCV =const per mole. Calculate the equation for the segment AC in thep − V plane. [This means obtain a function p(V ) which describes thesegment AC. Your answer may involve constants like SA, SC , VA, VC andn (number of moles).]

Solution

(a) See the diagram.

(b)dE = TdS − PdVdE = nCV dT

P =nRT

V


⇒ nCV dT = TdS − nRT

VdV

⇒ dS =nCVT

dT +nR

VdV

S(V, T ) = S(V0, T0) + nCV lnT

T0+ nR ln

V

V0

along AC:

S =VCSA − VASCVC − VA

− SA − SCVC − VA

V

VCSA − VASCVC − VA


V = SA + nCV lnT

TA+ nR ln

V

VA

This can be solved for T = T (V ),

nCV lnT

TA=VCSA − VASCVC − VA


V − SA − nR lnV

VA

T = TA exp[1

nCV(VCSA − VASCVC − VA


V − SA − nR lnV

VA)]

P =nRT

V=nRTAV

(VAV

)RCV exp[

1

nCV(VCSA − VASCVC − VA


V − SA)]

3 Free expansion

Consider N atoms of an ideal gas in thermal equilibrium at temperature Tconfined by a barrier to a fraction f of the volume of an adiabatic enclosure.The barrier ruptures at some instant, and the gas expands to fill the wholevolume, and eventually reaches a new equilibrium state.

(a) What is the new temperature of the gas?

(b) What is the change in entropy of the gas?

Solution

(a) For ideal gas, when the barrier ruptures, the expansion is free, i.e. ∆W = 0,

⇒ ∆U = ∆Q−∆W = 0

⇒ dU = 0 = (∂U

∂V)T,NdV + (

∂U

∂T)V,NdT

⇒ dT

dV|U =

−( ∂U∂V

)T,N

( ∂U∂T

)V,N=−(T ( ∂P

∂T)V − P )

( ∂U∂T

)V,N

P =NkT

V⇒ ∂P

∂T=Nk

V⇒ T (

∂P

∂T)V − P = 0

(dT

dV)U = 0

So the temperature does not change from the initial temperature.

(b)

dU = 0 = TdS − PdV ⇒ dS

dV=P

T

dS =P

TdV

∆S =

∫ V

fV

P

TdV =

1

T

∫ V

fV

NkT

VdV = Nk ln

1

f> 0


4 Ideal gas with a surface

Consider an ideal gas of classical monatomic particles in the presence of asurface. The surface consists of N adsorption sites, each of which can accom-modate at most one atom. The absorbed atoms are bound to the sites withenergy −W , where W > 0. The gas is at density n and temperature T and isin equilibrium with the surface. What fraction of adsorption sites is occupied?

Solution

The grand partition function of the gas is

lnZ =∑l

eβ(µ−εl)

=V

h3

∫e−βp

2/2md3peβµ

= eβµ V

Λ3

where

Λ = (h2

2πmkT)1/2

is the thermal wavelength.

〈N〉 =1

β

∂

∂µlnZ

= eβµV

Λ3

⇒ eβµ = Λ(T )3n

For the sites,

Z =

N∑l=0

eβ(µ+W )lClN

= (1 + eβ(µ+W ))N

lnZ = N ln(1 + eβµeβW )

〈N〉 =1

β

∂

∂µlnZ

=N

β

βeβµeβW

1 + eβµeβW

= N(1

e−βµe−βW + 1)


5 Einstein model

The Einstein model of the lattice vibrations of a solid consisting of N atomsrepresents the solid by 3N identical one-dimensional quantum harmonic oscil-lators, each with frequency ω0. Answer the following questions related to thismodel:

(a) Find the mean energy of the system as a function of the temperature, T ,of the solid.

(b) Find the heat capacity of the system, and evaluate it in the limit kBT hω0. Discuss this result in terms of the equipartition theorem.

(c) Find the general expression relating the pressure of a system to its Helmholtzfree energy.

(d) To model anharmonic effects in the solid, one assumes that the frequencyω0 is a function of the volume, V . Find the pressure in the Einstein solidas a function of (∂ω0/∂V )T .

Solution

εn = hω0(n+1

2), n = 0, 1, ...

The partition function is

Z = [

∞∑n=0

e−βhω0(n+ 12

)]3N = limM→∞

[e−βhω0

2 (1− xM

1− x )]3N

where x = e−βhω0 . Thus

lnZ = 3N [−βhω0

2− ln(1− e−βhω0)]

E = −∂ lnZ

∂β= 3N [

hω0

2+

hω0

eβhω0 − 1]

C =dE

dT= 3NkB(

hω0

kBT)2 e−βhω0

(eβhω0 − 1)2

At high temperatures βhω0 1, thus

C ≈ 3NkB(βhω0)2

(βhω0)2= 3NkB

From equipartition, for each 1D oscillator,

ε =1

2kB +

1

2kBT = kBT

thus =3NkBT and C = 3NkB .F = E − TS

dQ = TdS = dE + pdV


thusdF = d(E − TS) = −SdT − pdV

(∂F

∂V)T = −P

F = −kT lnZ,

so

P = 3Nh(∂ω0

∂V)T [

1

2+

1

eβhω0 − 1]

6 Bosonic excitations

The bosonic low-energy excitations of a two-dimensional system of dimensionsL× L are described by the wave equation

ρ∂2u

∂t2+ C∇4u = 0

where ∇4 = (∇2)2. (The scalar u(r, t) might represent height fluctuations nor-mal to the two-dimensional plane.)

(a) Solve for the dispersion relation ω(k).

(b) Compute the density of states g(ω).

(c) Compute, to within a numerical constant, the low-temperature specificheat C(T ). (You may assume that kBT is much greater than the spacingbetween neighboring quantized energy levels.)

Solution

(a) The wave equation is

ρ∂2u

∂t2+ C∇2∇2u = 0

so substituting u(r, t) = u0ei(~k·~r−ωt) gives −ρω2 + Ck4 = 0, i.e.

ω =

√C

ρk2

where k = |~k|.(b) The density of states is given by

g(ω)dω =L2

4π22πkdk =

L2

4π

√ρ

Cdω

which gives

g(ω) =L2

4π

√ρ

C,

a constant.

(c) The mean energy is

E(T ) =

∫ ∞0

dωg(ω)hω[n(ω) +1

2]

where n(ω) = (ehω/kBT − 1)−1 is the Bose occupancy factor. Note that E(T )is infinite! This is because of the zero point energy. (If we had imposed anultraviolet (short wavelength) cutoff on the density of states, as in the Debyemodel, this wouldn’t have happened.) However, the infinite zero point energy is


temperature independent and doesn’t affect the specific heat. The T -dependentpart to E(T ) is

E(T )− E0 =L2

4π

√ρ

C

∫ ∞0

dωhω1

ehω/kBT − 1

=L2

4π

√ρ

C

(kBT )2

h

∫ ∞0

dxx

ex − 1

=π2

6

L2

4π

√ρ

C

(kBT )2

h

where the integral gives the numerical constants π2/6. The specific heat is thenlinear in T :

C(T ) =∂E

∂T=π2

6

L2

4π

√ρ

C

2k2BT

h

1 Refrigerator in air conditioned room

Consider an air conditioned room with a refrigerator inside. The temperaturein the refrigerator is maintained at T1, in the room the temperature is main-tained at T2, and the outside temperature is T3. The temperatures are relatedby T1 < T2 < T3. All heat produced by the refrigerator goes into the room andshould be pumped out by the air conditioning unit to maintain the temperatureT2.

(a) A massive object is brought from outside and placed inside the refrigerator.The object has heat capacity C. Calculate the total work W done by boththe refrigerator and the air conditioner while cooling the object to T1.Assume that both the refrigerator and the air conditioning unit are idealheat machines.

(b) Now consider another experiment. The same object is brought from theoutside into the room. First it naturally cools to room temperature T2.Next it is placed in the refrigerator and cooled to T1. Will this require thesame amount of work done by both the refrigerator and the air conditioneras in part (a)? If not, calculate the total work done. Which amount ofwork is smaller?

Solution

First we recall that an ideal heat machine is based on the Carnot cycle. In the case ofrefrigeration the work must be done to move heat from cold to hot objects. Considera cycle between a cold refrigerator interior (system 1) and hot room (system 2). Thetotal entropy of the entire system (1+2) does not change so ∆S1 +∆S2 = 0. The cycleinvolves an isothermal expansion at temperature T1 and an isothermal compression atthe temperature T2. This process involves a heat transfer Q1 = T1∆S1 is taken fromrefrigerator (since Q1 < 0) and Q2 = T2∆S2 > 0 is placed in the room. The amountof work required as follows from the energy conservation is

W = Q1 +Q2 = −Q1

(T2

T1− 1

). (1)

We thus find that cooling an object with heat capacity C from T2 to T1 requires

W = C(T2 − T1)2

T1. (2)

(a) When a massive object with temperature T3 is placed in the refrigerator coolingit to T1 requires Q1 = −C(T3−T1). The minus sign indicates that the heat is tobe removed from the system 1. We already know that this removal reduces theentropy in system 1 and raises the entropy in the room by the same amount. Thismeans that heat brought by the refrigerator into the room is Q2 = −Q1T2/T1.Finally, this heat is to be pumped outside (we move −Q2) which gives to theoutside world

Q3 = Q2T3

T2= −Q1

T3

T1(3)

http://www.phys.ufl.edu/academics/downloads/jan11-partb.pdf

The amount of work to be done on the system is

W = (Q1 +Q3) = C(T3 − T1)2

T1. (4)

The work done by the refrigerator is

WR = C(T3 − T1)

(T2

T1− 1

), (5)

and the work done by the airconditioning

WAC = Q2

(T3

T2− 1

)= −Q1

T2

T1

(T3

T2− 1

)= C(T3 − T1)

T2

T1

(T3

T2− 1

). (6)

This result shows that the two ideal heat machines working sequentially areequivalent to a single ideal machine. Indeed, here we have a problem of aheat transfer from one system at the temperature T1 to another system at thetemperature T3. The minimum amount of work needed to to that is achieved inthe Carnot cycle.

(b) The situation in the second experiment is different. While the same amount ofheat |Q1| = C(T3−T1) is taken away from the massive object this heat is takenaway at different temperatures. At intermediate temperatures the efficiency ofthe Carnot cycle is higher and heat transfer takes less work. In the first stepthe work done by the AC unit is

W (1) = C(T3 − T2)

(T3

T2− 1

). (7)

In the second step simply amounts to part (a) where one has to remove C(T2−T1)amount of heat from refrigerator to outside.

W (2) = C(T2 − T1)

(T3

T1− 1

). (8)

Clearly the total amount of work is smaller because. First the C(T3 − T2)amount of heat is removed with inverse coefficient of performance (inverse ofefficiency) T3/T2 − 1 then a second amount of heat C(T2 − T1) is removed withinverse coefficient of performance (T3/T1−1) Since the first part is more efficient(meaning less work is needed) than that of (T3/T1 − 1) in part (a) the secondmethod will require less work.

The minimum amount of work needed to cool an object with heat capacity Cfrom temperature T3 to some lower temperature T1 is

W = CT3 ln

(T3

T1

)− C(T3 − T1). (9)

2 Semiconductor

A simple model of a semiconductor is to assume that it has M donor electronlevels, all at a negative energy, −∆, and a free particle electron conductionband whose minimum energy is zero. Assume that at zero temperature theconduction band is empty, and that all the donor levels are filled, with oneelectron per donor. At finite temperature T , a number N of the electrons arethermally excited from the donor states to the conduction band, leaving behindN holes in the donor levels. Assume N << M throughout and ignore electron-electron interactions. Let the volume of the semiconductor be V .(a) Determine the energy, entropy, and (Helmholtz) free energy of the elec-trons in the conduction band. In doing so, assume Maxwell-Boltzmann statis-tics and express the entropy in terms of N and the “thermal wavelength,”λth = (2πh2/mkBT )1/2, where m is the electron mass and kB is Boltzmann’sconstant. In terms of λth, what is the condition that the conduction electronsobey Maxwell-Boltzmann statistics?(b) Determine the total energy, entropy, and free energy of the M − N elec-trons and N holes in the donor levels. Assume N >> 1 and use Stirling’sapproximation, lnN ! ' NlnN −N .(c) Determine the equilibrium number of electrons excited to the conductionband as a function of M , T , and ∆, including the free energy of the electrons inthe donor levels, but neglecting the entropy of the electrons in the conductionband.(d) Determine the equilibrium number of electrons excited to the conductionband as a function of M , T , and ∆, including the free energy of the electrons inthe conduction band, but neglecting the entropy of the electrons in the donorband.(e) Determine the equilibrium number of electrons excited to the conductionband as a function of M , T , and ∆, taking into account all the terms in the freeenergy of the electrons in both the donor levels and the conduction band.

Solution

(a) We first found the partition function Q1

Q1 =

∫ ∞0

d3rd3p

h3e−β

p2

2m

The integral of d3r was given as volume V . Performing the integral results in thefollowing

Q1 = V (2πmkBT

h2)3/2

It was given that λth = ( h2

2πmkBT)12 which gives the results

Q1 =V

λ3

Problem G1P3 from 2008Comps.

We know that the particles are indistinguishable, especially at higher temperaturestherefore

QN =1

N !(V

λ3)N

We can use this result to find the free energy (Helmholtz energy) which is defined asA = −kBT lnQN

A = −NkBT [ln(V

Nλ3) + 1] (10)

The energy is then defined as

E = − d

dβlnQN

E = − d

dβlnV

λ3

E = − d

dβlnV (

2πmkBT

h2)32

Which results in the energy for the electrons in the conduction band as

E =3

2NkBT (11)

Having already found the Helmholtz free energy and the energy and knowing thatE = A+ TS, we can solve for the entropy of the electrons in the conduction band.

S =3

2NkB +NkB [ln(

V

Nλ3) + 1] (12)

The condition that the conduction electrons obey Maxwell-Boltzmann statistics isbased on the number of degrees of freedom. λ V 1/3

(b) We first start by finding the partition function with the choose function to nor-malize the equation

QN = (MchooseM −N)e(M−N)β∆

Once again we use A = kBT lnQN to find the Helmholtz energy.

A = −kBT ln(

M !

(M −N)!N !e(M−N)β∆

)A = −kBT (ln

M !

(M −N)!N !+ lne(M−N)β∆)

A = −kBT (ln(M !)− ln(M −N)!− ln(N !) + ((M −N)β∆))

Expand using Stirling’s approximation lnN ! ' NlnN −N

A = −kBT (MlnM −M − (M −N)ln(M −N) + (M −N)−NlnN +N + (M −N)β∆)

A = −kBT (MlnM −M −Mln(M −N) +Nln(M −N) +M −N −NlnN +N + (M −N)β∆)

A = −kBT (MlnM −Mln(M −N) +Nln(M −N)−NlnN + (M −N)β∆)

A = −kBT (MlnM

M −N +NlnM −NN

+ (M −N)β∆)

The ln MM−N is zero since M N and NlnM−N

Ncan be changed to NlnM

N− Nln1,

which the ln1 = 0. Therefore, the free energy is

A = −kBT (NlnM

N+ (M −N)β∆) (13)

We define the energy as

E = − d

dβlnQN

E = − d

dβ[ln

M

N+ (M −N)β∆]

The resulting energy for the donor electrons is

E = (N −M)∆ (14)

Once again, to find the entropy we use equations (4) and (5) and E = A+ TS

S = −NkBlnN

M(15)

(c) Using the Euler relations, dE = TdS−PdV +µdN and dA = −SdT −PdV +µdN ,we find that the chemical potential for the conducting plane (neglecting the entropyof the electrons in the conduction band) is µC = dE

dN. Using equation (2), we find

µC =d

dN

3

2NkBT

µC =3

2kBT

The chemical potential for the donor plane taking into consideration the free energy ofthe electrons in the donor level results in µD = dE

d(M−N)−T dS

d(M−N). Using the results

from equations (5) and (6) we solve for the chemical potential of the donor plane.

µD =d

d(M −N)(N −M)∆ + T

d

d(M −N)NkBln(

N

M)

µD = −∆ + kBT ln(M

N)− kBT

We set the chemical potentials equal and solve for N, the equilibrium number of elec-trons excited to the conduction band.

µD = µC

−∆ + kBT ln(M

N)− kBT =

3

2kBT

ln(M

N)− 1 =

∆

kBT+

3

2M

N= e

52

+∆β

The number of electrons excited to the conduction band are as follows

N = Me−( 52

+β∆) (16)

(d) Once again we use the Euler relations and equations (2) and (3) to find the chemicalpotentials of the conducting band.

µC =dE

dN− T dS

dN

µC = −kBT lnV

Nλ3

We use the Euler relations and equation () to find the chemical potentials of theconducting band.

µD =dE

d(M −N)

µD =d

d(M −N)(N −M)∆

µD = −∆

We set the chemical potentials equal and solve for N, the equilibrium number of elec-trons excited to the conduction band.

µD = µC

−∆ = −kBT lnV

Nλ3

The number of electrons excited to the conduction band are as follows

N =V

λ3e−β∆ (17)

(e) To find the equilibrium number of electrons we set the chemical potential of theconduction bands from part C and part D equal and solve for N.

µD(partc) = µC(partd)

−∆ + kBT ln(M

N)− kBT = −kBT ln

V

Nλ3

The equilibrium number of electrons taking into account all the terms in the freeenergy of the electrons in both the donor and the conduction bands are

N =

√MV

λ3e−

12

(1+β∆)

Replacing λ,

N =

√MV (

2πmkBT

h2)3/2e−

12

(1+β∆) (18)

3 Ice in water

100g of ice at −20C is put into 100g of water at 5C. Assuming no heat flowto or from external sources, Describe exactly what happens and give reasons foryour response.

Cice = 2090J / kg oC Cwater = 4186J / kg oC (19)

Lfusion = 3.33× 105J / kg Lvaporization = 2.26x106J / kg (20)

Solution

We know that at equilibrium the temperature for the two objects will be the same.Based on looking at the specific and latent heats, we have reason to suspect theequilibrium point will be somewhere along the Tf = 0 line. So we guessed that theequilibrium was 0 and that some of the water would freeze. So we’ll try it and seewhether our answer satisfies the conditions.So here’s what happens: Twater = 5C, mass 100g, temperature of water reduces to0oC and the temperature of the ice goes to some unknown temperature Ti.Heat gained by ice during this time = Heat lost by watermCi (Ti + 20) = mCw 5Solving for Ti we have the intermediate temperature in the process at which the waterwill begin to freeze in the cup. It turns out to be,Ti ≈ −10oC.Next, some percentage (α) of the water is turned into ice.

Heat lost freezing α of the water = heat gained by ice going from −10oC to 0oC(21)

αm lf = mCw 5 (22)

Solving for α we obtain:

α = 6.25% (23)

At this point the two systems are at equilibrium with a temperature of 0oC so the

mixture will remain as both solid and liquid.


4 Otto cycle

An Otto cycle engine burns gasoline and goes through a reversible processconsisting of two adiabatic, and two isovolumetric parts in the PV diagramshown in the problem. (a) Show that the efficiency of the engine can be writtenas e = 1− Tb−Tc

Ta−Td . (b) Show that the efficiency can also be written as

e = 1− V1

V2

γ−1

= 1− 1

rγ−1.

where r = V2

V1γ =

CpCv

.

Solution

(a)

e =dW

dQh(24)

But by energy conservation the work done is the heat taken from the hot reservoir anddumped to the cold reservoir. Therefore,

e =−(dQh + dQc)

dQh(25)

=−(dTh dS + dTc dS)

dQh dS(26)

giving us the classic result:

e = 1− dTcdTh

. (27)

or,

e = 1− Tb − TcTa − Td

. (28)

(b) For adiabatic processes we know that PV γ = constant, therefore,

P1 Vγ1 = P2 V

γ2 (29)

P1

P2V γ1 = V γ2 (30)

but from the ideal gas law, P1P2

= T1T2

(V1V2

)−1

, thus we have,

T1

T2

(V1

V2

)−1

V γ1 = V γ2 (31)

T1 Vγ−11 = T2 V

γ−12 . (32)

Problem G1P14-2 from 2008Comps.

But since this has to be true for the two adiabatic processes, we have two equations:

Ta Vγ−11 = Tb V

γ−12 Td V

γ−11 = Tc V

γ−12

subtracting these two equations from each other we are left with,

(Ta − Td)V γ−11 = (Tb − Tc)V γ−1

2

therefore,

Ta − TdTb − Tc

=

(V1

V2

)γ−1

Thus we may plug this into our previous result (equation 10) to obtain, e = 1− V1

V2

γ−1

5 Rotational gas

Consider the rotational motion of a dilute gas of N diatomic molecules, eachwith moment of inertia I. In this problem you are asked to consider three cases:a quantum system of diatomic molecules in which the two atoms are distin-guishable (parts A-C); a quantum system of diatomic molecules in which thetwo atoms may be considered as indistinguishable spinless bosons (part D), anda classical system of rotors (part E). In this problem we will consider rotationalmotion only. You may use the fact that a quantum rotor has energy levels

EJ = h2

2I J(J + 1).In parts A-C, use quantum statistical mechanics for the case where the twoatoms in each molecule are distinguishable.(a) Give an expression for the partition function Z as a function of temperatureT. You do not need to evaluate the expressions in a closed form. Be sure toproperly include all degeneracy factors in your expression.(b) When the temperature T is low but non-zero, derive an asymptotic expres-sion for the rotational specific heat per molecule C as a function of temperature.(c) In the limit of high temperature, the expression for Z in part (a) can be ap-proximated as an integral over a continuous variable. By evaluating the integral,give the limiting value of the specific heat per molecule C.In part (d) consider the case where the atoms in each molecule can be regardedas indistinguishable spinless Bosons.(d) Give the expression for the partition function Z in this case which replacesthe one in part (a). Explain your reasoning. You do not need to evaluate theexpression.In part (e), consider the relation to classical statistical mechanics.(e) The expression for the specific heat using quantum statistical mechanicsshould approach the classical result in the high temperature limit. Show thatyour result of part (c) does approach the correct classical limit in which thespecific heat is determined by the number of degrees of freedom for the rotationalmotion.

Solution

(a) The particles are distinguishable so the partition function is given by

Z = (Z1)N =

(∞∑J=0

gJe−EJ/kBT

)N,

where gJ is the degeneracy factor. The quantum rotor energy levels have a degeneracyof 2J + 1. Hence, we have

Z =

(∞∑J=0

(2J + 1) exp

(−h2J(J + 1)

2IkBT

))N. (33)


(b) For kBT h2

2I, we should keep only the the first two terms (J=0 and J=1) in

the partition function to get the leading order temperature dependence of the specificheat. This gives us

Z =(

1 + 3e−βh2/I)N

(34)

for the partition function. First, we calculate the internal energy

U = −∂ln(Z)

∂β= −N ∂

∂βln(

1 + 3e−βh2/I)

=3Nh2

I

e−βh2/I

3e−βh2/I + 1≈ 3Nh2

Ie−βh

2/I .

Notice that it is sufficient to keep only the leading order temperature dependence ofthe energy above. Lastly, we find the specific heat per molecule at low temperature

C =1

N

dU

dT= − 1

NkBT 2

dU

dβ= 3kB

(h2

IkBT

)2e−h

2/IkBT . (35)

(c) For kBT h2

2I, we can write

Z1 ≈∫ ∞

0

dJ(2J + 1) exp

(−h2J(J + 1)

2IkBT

). (36)

Let

x = J(J + 1)dx

dJ= 2J + 1.

Substituting these last two expressions into equation (4) we get

Z1 ≈∫ ∞

0

dx exp

(−h2x

2IkBT

)=

2IkBT

h2 . (37)

Calculating the energy we obtain

U

N= − 1

Z1

∂Z1

∂β=

1

β= kBT.

For the limiting value of the specific heat per molecule, we have then

C =1

N

dU

dT= kB. (38)

(d) The total wave function for bosons must be symmetric with respect to interchangeof identical particles. Recall that we are considering rotational motion only in thisproblem. Hence, the eigenfunctions of our Hamiltonian are the spherical harmonics.We use the following property of the spherical harmonics

PYJm(θ, φ) = (−1)JYJm(θ, φ),

where P is the parity operator. This means that we should keep only the J even statesso that we have a symmetric wave function under particle exchange. Therefore,

Z =

(∞∑

J=0,2,4,...

gJe−EJ/kBT

)N=

(∞∑

J=0,2,4,...

(2J + 1) exp

(−h2J(J + 1)

2IkBT

))N. (39)

(e) Here we use the equipartition theorem. The rigid rotor Hamiltonian has twoquadratic degrees of freedom f. Using classical theory we obtain for the specific heatper molecule

C =f

2kB =

2

2kB = kB. (40)

This agrees with the limiting value of the specific heat per molecule that we obtainedin part (c).

6 Diatomic molecules

Consider a classical gas of N diatomic molecules in thermal equilibrium witha heat bath at temperature T. The molecules move freely inside a cubic box ofvolume V. Each molecule consists of two atoms of mass m1and m2respectively.The two atoms are permanently bound together inside the molecule by intra-molecular forces with potential

U(r) = U0

( ra

)αwhere r is the relative separation of the two atoms, U0 is a positive constantwith units of energy, a is a characteristic length, and αis a positive constant.The molecules do not interact with each other.Let j = 1, 2...N label the molecules, and for the jth molecule denote by Rj ,Pj , rj , and pjthe coordinates of the center of mass, the momentum of the cen-ter of mass, the relative separation of its component atoms and their relativemomentum, The classical Hamiltonian is

E =

N∑j=1

[~P 2j

2M+~p2j

2µ+ U(rj)

]

where M = 12 (m1 +m2) and µ = m1m2/(m1 +m2) .

(a) Show that the classical partition function ZN is given by

ZN = AV N

N !

(kBT

2πh2

)3N ∫ ∞0

dr 4πr2e−U(r)/kBT

NAnd express the constant A in terms of M and µ.

(b) In the thermodynamic limit N¿ ¿1, N/V fixed, calculate the average energyin therms of N, kBT and α. [Hint: the change of variable t ≡ U(r)/kBTmay be useful. You should not need to evaluate any integrals for thispart.]

(c) Calculate the heat capacity at constant volume CV .

(d) For the case α = 2, explain how the answer to (c) could have been obtainedwithout detailed calculation.

Solution

(a)The 1-particle partition function is determined by

Z1 =

∫d3P d3p d3R d3r

h6e−βE (41)


Notice that the Hamiltonian is independent of R. Thus, integrating over therelative coordinate results in a factor of volume V. This leaves

Z1 = V

∫d3P

h3e−β

~P2j

2Md3p

h3

∫d3p

h3e−β

~p2j2µ

∫d3r e−βU(r) (42)

The first two integrals are common in statistical mechanics. They can be pre-formed easily by integrating in spherical coordinates. This results in

Z1 = V

(2πMkBT

h2

) 32(

2πµkBT

h2

) 32∫dr 4πr2e−βU(r) (43)

The partition function for N particles is justZN1N !

because the particles are non-interacting and indistinguishable. Thus,

ZN = AV N

N !

(kBT

2πh2

)3N ∫ ∞0

dr 4πr2e−U(r)/kBT

N(44)

Where

A ≡ (Mµ)3N/2 (45)

(b) Let

t ≡ U0

( ra

)αβ (46)

r = a

(t

βU0

)1/α

(47)

dr =1

βU0

a

α

(t

βU0

) 1−αα

dt (48)

Now, the N-particle partition function becomes

ZN = (Mµ)3N/2V N(kBT

2πh2

)3Na3N

αN

(kBT

U0

)3N/α∫ ∞0

dt t3−αα e−t

N

(49)This integral is just a finite constant. The average energy is given by

E = kBT2 d ln(ZN )

dT(50)

⇒ E = 3

(1 + α

α

)NkBT (51)

(c) Finding the heat capacity is trivial

CV =

(∂E

∂T

)V

(52)

CV = 3

(1 + α

α

)NkB (53)

(d) When α = 2, the Hamiltonian only depends on quadratic terms. This means thatthe equipartition theorem is valid. In the Hamiltonian there are nine indepen-dent quadratic degrees of freedom - three from the center of mass momentum,three from the relative momentum, and three from the potential energy expres-sion. The equipartition theorem states that

E =f

2NkbT (54)

where f is the number of quadratic degrees of freedom. This is 9 for our case:

E =9

2NkbT (55)

meaning that the specific heat is

CV =9

2Nkb (56)

These expressions agree exactly with equations (52) and (54).

7 Classical ideal gas

For a classical ideal gas the number of molecules having component momentabetween px and px + dpx, py and py + dpy, and pz and pz + dpz is

fN =N

(2πmkT )3/2exp

(−∑i p

2i

2mkT

)d3p

(a) Show that the total number of particles is N.

(b) Find the number of particles having speeds between v and v + dv.

(c) If the mean free path ` is independent of the speed v, the mean time betweencollisions for a particle of speed v is `/v. Calculate the mean time betweencollisions for particles of all speeds.Does this result = `/(v2)1/2?Please Note:

∫ ∞0

e−x2

dx =

√π

2

Solution

(a) Denote the total number of particles by M. We know that

M =

∫f(p)d3p, (57)

where f(p) is the number of particles with momentum between p and p+ dp.

M =

∫ ∫ ∫N

(2πmkT )3/2exp

(−(p2

x + p2y + p2

z)

2mkT

)dpxdpydpz. (58)

Doing the integration, we get

M =N

(2πmkT )3/2

(√2πmkT

)3

= N. (59)

(b) We know p = mv. Replacing the momentum in given distribution function bythis equation and using the fact that dp = mdv, we get

F (v) = N( m

2πkT

)3/2

exp

(−m(v2

x + v2y + v2

z)

2kT

)dvxdvydvz. (60)

In spherical coordinates we obtain

F (v) = 4πN( m

2πkT

)3/2

v2 exp

(−mv2

2kT

)dv. (61)


(c) To find mean free time we use

τ =

∫`

vNF (v)dv. (62)

τ = 4π`( m

2πkT

)3/2∫vdv exp

(−mv2

2kT

). (63)

This integral returns

τ = 2`

√m

2πkT. (64)

Finally, we must calculate `/(v2)1/2.

v2 =

∫v2F (v)

Ndv = 4π

( m

2πkT

)3/2∫v4dv exp

(−mv2

2kT

)=

3kT

m. (65)

Hence we have,`

(v2)1/2= `

√m

3kT. (66)

This is not the same result that we obtained for the mean time τ .

8 Van der Waals gas

The Van der Waals equation of state for one mole of a nonideal gas is (P +aV )(V − b) = RT . Show that the entropy for this gas is S = CvlnT +Rln(V −b) + const. Use this result that, for an adiabatic change in a Van der Waals gas,

T (V − b)RCv = constant.

Solution

We start off with the partial equation for entropy.

dS =∂S

∂VdV +

∂S

∂TdT

We know that ∂E = T∂S and Cv = ∂E∂T

. Therefore,

dS =∂S

∂VdV +

CvTdT

Substitute CvTdT for Cvln(lnT ).

dS =∂S

∂VdV + Cvd(lnT )

Using the relation ∂S∂V

= ∂P∂T

we can substitute into the equation.

dS =∂P

∂TdV + Cvd(lnT )

We can use the initial equation to easily solve for ∂T∂P

= V−bR

, Therefore,

dS =R

V − bdV + Cvd(lnT )

RV−bdV can be expressed as R∂ln(V − b).

dS = dln(V − b) + Cvd(lnT )

We integrate through to find our final answer, which includes an integration constant.

S = CvlnT +Rln(V − b) + const

Now we must show that for an adiabatic process (S = constant), S = CvlnT+Rln(V −b) + const can be expressed as T (V − b)

RCv = constant. To do so, we know that any

exponent in logarithmic function can be brought down in front of the natural log.Thus we can reverse this process to produce.

constant = lnTCv + ln(V − b)R + const

Using the logarithmic multiplication rules,

Constant = ln(TCv (V − b)R)


Now we place both sides to the exponential to cancel out the natural log.

constant = TCv (V − b)R

Since the specific heat is remaining constant, we can multiply the exponents by 1Cv

resulting in the final answer of

constant = T (V − b)RCv

9 1d polymer

Let us consider a polymer consisting of N(a positive even integer) monomersconnected end-to-end and lying along the x-axis. The monomers may be re-garded as small arrows of lengths l without thickness and lie along the x-axiswith no preferred direction.One end of the polymer is fixed at the origin O.

(a) Find the number N+ of monomers pointing in the positive direction longthe x-axis, and the number N− of monomeres pointing in the negativedirection along the x-axis, when the free end x-coordinate is X = 2nl,where −N/2 ≤ n ≤ N/2.

(b) Write down the total number ΩN (X) of allowed conformations when thefree end x-coordinate is X = 2nl as in (a).

(c) Let SN (X) be the conformational entropy of our polymer whose end x-coordinate is X. Compute the entropy difference SN (X)− SN (0) to orderX2, assuming that | n |<< N .

(d) Find the required average force in pN (picoNewton = 10−12N to make theend-to-end distance of the polymer be R = 200A. Assume that N = 4×104, the monomer size l is 1A(= 1×10−10m), and the ambient temperatureis 300K. Assume that the only contribution to the forces comes fromthe conformational entropy of the polymer for which you may use thefunctional form obtained in (cA. Do not hesitate to use thermodynamicconsiderations, if microscopic calculations can be avoided.

Solution

(a) We know that the total number of monomers is given by:

N = N+ +N− (67)

We want to find the N− and N+ when X = 2nl. We have that:

X = l(N+ −N−) (68)

Problem ”problem1” from 2008Comps.

Taking equation 67 and solving for N− we get

N− = N −N+ (69)

Substituting in equation 68 and solving for N+

N+ =N

2+X

2l(70)

Substituting equation 70 in equation 69 we have that:

N− =N

2− X

2l(71)

(b) For the total number of ΩN (X) we have that ΩN (X) = N !N+!N−!

. Substituting

N+ and N− we have

ΩN (X) =N !(

N2

+ X2l

)!(N2− X

2l

)!

(72)

(c) We are going to call the entropy difference ∆S. But first we want to get S(X),for that we use the relation SN = KB ln ΩN

Substituting equation 72 we have

SN (X) = KB ln

[N !(

N2

+ X2l

)!(N2− X

2l

)!

]Using the approximation M ! '

√2πe−MMM we have

SN (X) = KB ln

√

2πe−NNN[√2πe−(N

2+X

2l)(N

2+ X

2l)(N

2+X

2l)] [√

2πe−(N2−X

2l)(N

2− X

2l)(N

2−X

2l)]

Looking at this equation we can see that it can be written as

SN (X) = KB ln

[NN

√2π(N

2+ X

2l)(N

2+X

2l)(N

2− X

2l)(N

2−X

2l)

]

SN (X) = KB

N lnN − ln

√2π − ln

[(N

2+X

2l

)(N2

+X2l

)]− ln

[(N

2− X

2l

)(N2−X

2l)]

Using some properties of logarithms

SN (X) = KB

lnNN − ln

√2π −

(N

2+X

2l

)ln

[N

2+X

2l

]−(N

2− X

2l

)ln

[N

2− X

2l

]

Factorizing N/2 in the logarithm

SN (X) = KB

N lnN − ln

√2π −

(N

2+X

2l

)ln

[N

2

(1 +

X

Nl

)]−(N

2− X

2l

)ln

[N

2

(1− X

Nl

)]

SN (X) = KB

N lnN − ln

√2π −N ln

N

2−(N

2+X

2l

)ln

(1 +

X

Nl

)−(N

2− X

2l

)ln

(1− X

Nl

)SN (X) = KB

− ln√

2π +N ln 2−(N

2+X

2l

)ln

(1 +

X

Nl

)−(N

2− X

2l

)ln

(1− X

Nl

)

We know that the series for logarithmic function is given by

ln(1 + x) = x− x2

2+x3

3− x4

4+ .... − 1 < x < 1

Expanding the ln and keeping only to second order in X2

SN (X) = KB

− ln√

2π +N ln 2−(N

2+X

2l

)(X

Nl− X2

2N2l2

)−(N

2− X

2l

)(− X

Nl− X2

2N2l2

)

SN (X) = KB

[− ln√

2π +N ln 2 +X2

2Nl2− X2

Nl2

]

SN (X) = KB

[− ln√

2π +N ln 2− X2

2Nl2

]Therefore,

∆S = −KBX2

2Nl2(73)

(d) The force is given by

F = − ∂U∂X

From thermodynamics we know that

T =

(∂U

∂S

)N,V

Therefore

F = −T(∂S

∂X

)N,E

Substituting equation 73 we get

F =TkBX

Nl2

F = .207pN

10 Noninteracting atoms

A gas of N indistinguishable classical non-interacting atoms of mass m is heldin a neutron atom trap by a potential of the form V (r) = ar. The gas is inthermal equilibrium at a temperature T.

(a) Find the partition function Z1 for a single atom trapped in this potential.Express your answer in the form Z1(T, a) = ATαa−η. Find the exponentsof α and η, and the constant A.

(b) Find the entropy of the gas in terms of N , kB , and Z1(T, a) = ATαa−η.

(c) The gas can be cooled if the potential is lowered reversibly(by decreasinga) while no heat is allowed to be exchanged with the surroundings. FindT as a function of a and the initial values To and ao.

Solution

(a) The Hamiltonian is:

mH =p2

2m+ ar (74)

Thus the partition function is,

Z1 =

∫d3p d3r

h3e−β H (75)

=1

h34π

∫ ∞0

e−βp2

2m p2dp · 4π

∫ ∞0

e−β arr2dr (76)

letting x ≡ βar, and y ≡√

β2m

p, thus,

Z1 =1

h34π

(β

2m

)− 32∫ ∞

0

e−y2

y2dy︸︷︷︸=√π4

· (βa)−3 4π

∫ ∞0

e−xx2dx︸︷︷︸=2

(77)

=8π

h3(2πmkBT )

32

(kBT

a

)3

(78)

Therefore,

Z1 =8π

h3(2πmkB)

32 k3

B · T92 a−3 . (79)

Thus, α = 92

and η = 3.

(b)

S = −(∂A

∂T

)V,N

(80)

= − ∂

∂T(−kBT lnZN ) (81)

Problem StatMech4 from 2008Comps.

but ZN = 1N !ZN1 , therefore,

= kB∂

∂T

(T ln

(1

N !ZN1

))(82)

= kB∂

∂T(T (N lnZ1 − lnN !)) (83)

using Stirling’s approximation lnN ! ≈ N lnN −N ,

= N kB

((lnZ1 − lnN + 1) + T

1

Z1

∂Z1

∂T

)(84)

but ∂Z1∂T

= αTZ1 (where α = 9

2), therefore,

S = N kB

((lnZ1 − lnN + 1) + T

1

Z1

α

TZ1

)(85)

= N kB ((lnZ1 − lnN + α) (86)

(87)

Therefore,

S = N kB

(ln

(Z1

N

)+

11

2

). (88)

(c) Since dQ = TdS = 0, and T is not equal to 0, then dS = 0. So,

So = S (89)

N kB

(ln

(Z1(To, ao)

N

)+

11

2

)= N kB

(ln

(Z1(T, a)

N

)+

11

2

)(90)

Z1(To, ao) = Z1(T, a) (91)

ATαo a−ηo = ATαa−η (92)(

T

To

)α=

(a

ao

)η(93)

Therefore,

T = To

(a

ao

) 23

. (94)

1 Molecule of four atoms

A molecule consists of four identical atoms at the corner of a tetrahedron.

(a) What is the number of (i) translational, (ii) rotational, and (iii) vibrationaldegrees of freedom for this molecule?

(b) On the basis of the equipartition principle, what are the values of Cv andγ = Cp/Cv for one mole of an ideal gas composed of these molecules?

(c) Assuming that the vibrations of the molecule are quantized, what is Cv fortemperatures such that kBT hωv? Here ωv is the lowest frequency ofvibration of the molecule.

(d) At very low T the rotations of the molecule are also quantized, i.e. Erot =(h2/2I)l(l + 1), where l = 0, 1, 2... and I is the moment of inertia of themolecule. What is the value of Cv at very low temperatures (kBT h2/2I)? Assume the gas is still ideal at low T .

Solution

(a) We have four particles which have three positions each, as they are moving inthree dimensions. This gives a total of 12 degrees of freedom. Three of theseare translations of the center of mass, and since the molecule has no continuousrotational symmetries (unlike a diatomic linear molecule, for example) thenthere are also three rotational degrees of freedom. This leaves six of the originaltwelve degrees of freedom which must be vibrational degrees of freedom. Notethat we may not have this many modes, but the sum of the number of modestimes the number of degrees of freedom within each mode will be six.

(b) The classical equipartition theorem states that for each independent quadraticvariable in the energy (Hamiltonian) we have a mean energy of kBT/2 whenthat system is at a temperature T . Note this only holds when the level spacingis small compared to kBT . The kinetic energy

T =

N∑i=1

1

2(v2x + v2

y + v2z)

has three such independent quadratic degrees of freedom for each particle, so ithas a mean of 3kBT/2 for each particle.

The same is true of the kinetic energy of rotation about the center of mass, sinceit looks like

Trot =

N∑i=1

1

2

(I1ω

21 + I2ω

22 + I3ω

23

)and so it also has a mean of 3kBT/2 for each particle.

The vibrational degrees of freedom enter the Hamiltonian with both a quadratickinetic energy term and a quadratic potential energy term. Each vibrational

Problem opathf99 from ProfDiag bank.

degree of freedom within a normal mode behaves like a one-dimensional oscillatorwith energy

H =p2

2m+

1

2kx2,

and so each vibrational degree of freedom contributes an average of 2×kBT/2 =kBT to the energy.

Putting these together, we have (recall there is one mole of the gas)

U = NA

(3

2kBT +

3

2kBT + 6kBT

)= 9NAkBT = 9RT

so thatCv = 9R.

For an ideal gas we have that Cp = Cv + nR so that for this mole of gasCp = Cv +R = 10R, so that

γ =CpCv

=10

9.

(c) At sufficiently low temperature the classical equipartition theorem no longer holds,because kBT can become comparable to the level spacing of the oscillator (orthe rigid-rotator rotational spectrum). Since part d) implies that the rotationalmotion has much smaller frequencies than the vibrational motion, we know thatfor temperatures T so that kBT hωv the vibrational oscillators are in theirground state, and so the internal energy U associated with them does not dependon T and we have

U = NA

(3

2kBT +

3

2kBT

),

and so Cv = 3R.

(d) Now even the rotational degrees of freedom are in their ground state and so theenergy associated with them does not depend on T , so all that is left are thetranslational degrees of freedom and

U = NA3

2kBT,

giving Cv = 3R/2.

2 Electron gas

Consider an ideal gas of N electrons that move freely in 2D, and are confinedto an area A.

(a) Estimate the temperature T0 below which quantum mechanical considera-tions are important.

(b) Calculate the Fermi energy assuming the system is at the temperatureT = 0K. Compare the Fermi energy to kBT0, where T0 is calculated above.

(c) (c) What is the physical interpretation of the Fermi energy for a givensystem?

Solution

(a) The area of a 2D system of N electrons is: A ∼ Nπ(∆x)2. Therefore ∆x ∼(1/π)(A/N)1/2. Using the uncertainty relation: ∆x∆p ∼ h, we can then write:∆p ∼ h/∆x.

To estimate the temperature below which quantum effects become important,we will use:

E ∼ kBT0 ∼ (∆p)2/2m ∼ h2/(2m(∆x)2) ∼ h2Nπ2/(2mA). (1)

Therefore, as a crude approximation: T0 ∼ h2Nπ2/(2mAkB).

(b) Calculate the 2D Fermi energy to compare to the energy scale in part (a):

N = 2A

kF∫0

d2k

(2π)2 =2A

kF∫0

2πkdk

(2π)2 =4πA

(2π)2

kF∫0

kdk =A

2πkF

2 (2)

The factor of 2 is to account for the spin of the electron. The Fermi energy istherefore:

EF =h2kF

2

2m=

h2

2m

(2πN

A

)=πh2N

mA

The Fermi temperature T0 is then:

T0 =π

kB

(h2N

mA

)Note that this is close to the crude result estimated in part (a).

(c) The Fermi energy is a measure of the energy scale where quantum effects becomeimportant for describing the properties of a system of Fermions. For most metalsat room temperature, for example, the Fermi energy is very large, and thereforeFermi-Dirac statistics are needed to describe the properties of the system asopposed to a more classical Maxwell-Boltzmann approach.


3 Thermal and free expansion

One mole of an ideal gas undergoes a reversible thermal expansion from volumeV to volume 2V .

(a) What is the change in entropy of the gas?

(b) What is the change in entropy of the universe?

Now suppose the same expansion takes place as a free expansion.

(c) What is the change in entropy of the gas?

(d) What is the change in entropy of the universe?

Solution

(a) Since the process is isothermal, the internal energy does not change. Thus thework done during the expansion must be equal to the heat absorbed by the gas.

∆Q = W =

∫ 2V

V

PdV =

∫ 2V

V

RT

VdV = RT ln 2. (3)

The entropy change of the gas is then

S =∆Q

T= R ln 2. (4)

(b) The entropy change of the heat bath the gas is coupled to when expanding is∆Sbath = −∆Q/T = −R ln 2, and so the total entropy change of the universe iszero,

∆Su = ∆Sgas + ∆Sbath = 0, (5)

as it must be for a reversible process.

(c) We recall that free expansion is an irreversible process in which a gas expandsinto an insulated evacuated chamber. While for real gases the temperature maychange in the process, for ideal gases free expansion takes place at a constanttemperature. This is easy to see: in an ideal gas the molecules do not interactand their average kinetic energy remains 3/2 kT (for monoatomic gas). For afree expansion there is again no change in the internal energy of the gas, so theinitial and final states of the gas are the same as in part (a). The change inentropy of the gas is thus again

∆Sgas = R ln 2. (6)

(d) For a free expansion the state of the bath does not change and so ∆Sbath = 0.Thus for free expansion the total change in entropy of the universe is

∆Su = ∆Sgas + ∆Sbath = R ln 2 (7)

reflecting that free expansion is not reversible.


4 Entropy change

A certain amount of water of heat capacity C is at a temperature of 0C. It isplaced in contact with a heat reservoir at 100C and the two come into thermalequilibrium.

(a) What is the entropy change of the universe?

(b) The process is now divided into two stages: first the water is brought intocontact with a heat reservoir at 50C and comes into thermal equilibrium;then it is placed in contact with the heat reservoir at 100C. What is theentropy change of the universe?

(c) If we were to continue this subdivision into an infinite number of heat baths,what would the entropy change of the universe be?

Solution

(a) The entropy change of the water is

∆Sw =

∫ Tf

Ti

dQ

T=

∫ Tf

Ti

CdT

T= C ln

TiTf

(8)

where Ti = 273K and Tf = 373K, so

∆Sw = C ln373

273. (9)

The entropy change of the reservoir is

∆Sr = −∆Q

TR= −C TR − Ti

TR= −C(1− Ti

TR) (10)

where Ti = 273K and TR = 373K. Thus the total entropy change of the universeis

∆S = (ln373

273+

273

373− 1)C = 0.044C (11)

(b) and after the second stage

∆S2 = (ln373

323+

323

373− 1)C = 0.010C. (12)

Thus the total entropy change of the universe is

∆S = ∆S1 + ∆S2 = 0.023C (13)

which is less than that found in part (a)(as we should expect).

(c) In this limit all heat transfer would be adiabatic and ∆S = 0.

Problem 4 from Qualifying exam 2007-spring.

5 Spins in magnet

A magnet consists of N >> 1 noninteracting spins which can be in any ofthree states: σ =1, 0, and -1. Each of these spins is described by a Hamiltonian:

E0σ2 − µ0Hσ (14)

where H is the applied magnetic field.

(a) Write the partition function for this system.

(b) Calculate the Helmholtz free energy

(c) Calculate the entropy per spin, S/N in the limit as T →∞.

(d) Does the specific heat per spin vanish, or tend to a constant as T → ∞?Explain your answer.

Solution

(a)

Z =∑

exp(−β(E0σ2 − µ0Hσ)) =

(e−β(E0−µ0H) + e−β(E0+µ0H) + 1

)N(15)

(b)

F = −kT lnZ = −NkT ln(e−β(E0−µ0H) + e−β(E0+µ0H) + 1

)(16)

(c) As T →∞, the probability of each state being occupied is roughly equal. There-fore, S/N = k ln 3. One can also solve this more explicitly by calculatingS = −dF/dT , and then the limit.

(d) C = TdS/dT (constant V ). Since the entropy/spin approaches a constant as thetemperature approaches infinity, this means that dS/dT per spin is zero, andthe specific heat must be zero.


6 Magnons

Spin waves (magnons) in a ferromagnetic solid have a dispersion relation ofthe form:

ω = A|k|2

Treating the magnons as elementary excitations, calculate the temperature de-pendence of the heat capacity of the spin system at low temperatures. Youmight find the following integral to be useful:

∞∫0

x3/2dx

ex − 1= finiteconstant

Solution

The magnons are integer spin particles, but the number is not conserved. Therefore,we will use the Planck distribution to describe them:

E =V

(2π)3

∞∫0

hω

eβhω − 1(4πk2dk) =

V 4π

(2π)3

∞∫0

hω

eβhω − 1k2dk

Change of variable: let x = βAhk2. We will then have that: dx = 2βAhkdk. Ourintegral then becomes:

E = V 4π(2π)3

hA∞∫0

(x

βAh

)21

ex−1dx

2βAh1k

= V 4π(2π)3

hA∞∫0

(x

βAh

)21

ex−1dx

2βAh

(x

βAh

)−1/2

→ E ∼ 1

β5/2

∞∫0

x3/2

ex−1dx ∼ T 5/2

The heat capacity is therefore:

C =dE

dT∼ d

dTT 5/2 ∼ T 3/2

.


7 Hurricane

Some people believe that opening the windows of a house during a hurricanecould reduce the risk of having its roof blown off. Examine the possibilitiesdiscussed below.

(a) Hurricane is a massive area of low pressure. The pressure in the center ofthe storm is below 920 mbar1. What would be the vertical component ofthe force on the roof of a 10 by 10 meter house if the pressure inside thehouse is 1013 mbar? Would this force lift a 10 ton roof? Would openingof the windows help in this case and how realistic is this assumption?

(b) A typical wind speed of the category five storm is 70 m/s. Imagine thatthe house has opened windows and the flow of air is passing below andabove the cylindrically shaped roof, see Fig. 1. What is the lifting forceon the roof in this situation 2.

Figure 1: Airflow through and above the house.

Solution

(a) The pressure difference is 9.33 kPa. Which corresponds to a lifting force of 933kN. A force required to lift a 10 ton roof is F = mg =100 kN. Thus, this pressuredifference will result in the roof being blown off. However, the assumption thatthe pressure inside stays unchanged is unrealistic: hurricane is a relatively slowmoving storm, it takes hours for the pressure to drop and even a small airexchange would equilibrate the inside-outside static pressure.

192 kPa2You may use the following constants: molar mass of air is m = 29 g/mol, gas constant

R = Nak = 8.3 J/(K mol)

(b) We can estimate the pressure below and above the roof using the Bernoulli’s law

p2 − p1 =ρ

2(v2

1 − v22), (17)

here we use subscript 1 for outside of the house and 2 for the inside. To estimatethe density one can use the ideal gas law

ρ =Pm

RT= 1.3kg/m3, (18)

or to recall that a molar volume is 22 dm3/mol. The distance over the roofof the house is π/2 times longer than the distance directly through the housewhich means that the speed of the flow is v1 = v2π/2, therefore

p2 − p1 =ρv2

2

2

(π2

4− 1

)≈ 4.6kPa. (19)

The resulting in this case force 460 kN is sufficient to lift the roof. Unlike inpart (a), this scenario is much more realistic.

8 Harmonic oscillator

Consider a one-dimensional quantum harmonic oscillator with energy levels

En = (n+ 1/2)hω

(a) What is the average energy of this oscillator at temperature T?

(b) What is the specific heat of this oscillator?

(c) Give expressions for the specific heat appropriate for high temperature(kBT hω) and low temperature (kBT hω) limits.

Solution

(a) The average energy is

〈E〉 =

∑nEne

−βEn∑n e−βEn

= − d

dβlnZ

where Z is the partition function

Z =∑n

e−βhω(n+1/2) =e−

12βhω

1− eβhω =1

2 sinhβhω/2

Thus

〈E〉 =d

dβln(2 sinhβhω/2) =

hω

2cothβhω/2

(b)

C(T ) =d〈E〉dT

=1

kBT 2

1

sinh2 βhω/2

(c) For kBT hω we can make the approximation sinh2 βhω/2 ' (βhω/2)2 and soin this limit

C(T ) ' kBFor kBT hω then sinh2 βhω/2 ' 1

4eβhω and

C(T ) ' kB(hω

kBT

)2

e−hω/kBT


9 Rubber band

A simple model of a rubber band is a one-dimensional (horizontal) chain con-sisting of N (N 1) linked segments, as shown in the diagram. Each segmenthas just two possible states: horizontal with length a and vertical, contributingnothing to the length. The segments are linked such that they cannot comeapart. The chain is in thermal contact with a reservoir at temperature T .

(a) If there is no energy difference between the two states, what is the averagelength of the chain?

(b) The chain is now fixed at one end and a weight hung from the other end,supplying a force F = Mg as shown below. Determine the average lengthof the chain at any temperature T . Find the length in the limits of lowand high T .

(c) In which temperature limit is the extension proportional to F (Hooke’slaw)? Calculate the constant of proportionality.

Solution

Z =∑σj

e−βEj (20)

(a)

Ej = ε0; lj =1

2(1− σj)a (21)

L0 =∑j

〈lj〉 =N

2a (22)


(b)

Ej = ε0 −Mga

2(1− σj) (23)

〈lj〉 =

∑σj=±1 e

βMg a2

(1−σj) 12(1− σj)a∑

σj=±1 eβMg a

2(1−σj)

(24)

=2(a

2cosh Mga

2T+ a

2sinh Mga

2T)

2 cosh Mga2T

(25)

=a

2(1 + tanh

Mga

2T) (26)

L =

N∑j=1

〈lj〉 =Na

2(1 + tanh

Mga

2T) = L0(1 + tanh

Mga

2T) (27)

(c) Hooke’s law is valid for Mga2T 1⇒ T 1

2kBMga.

10 Water molecule

A water molecule can vibrate in the ” flexing ” mode in which the hydrogenatoms move towards and away from each other without stretching the HO bond.The oscillation of this mode is approximately harmonic with a frequency of5.0× 1013Hz. What is the probability of a water molecule in its flexing groundstate and in the first excited state? Assume that the water is in equilibrium atroom temperature. The Boltzmann constant k is 1.38× 10−23J/K and Plancksconstant h is 6.63× 10−34J ∗ s.

Solution

The energy levels of a harmonic oscillator are given by En = (n+1/2)hf , where f is the

frequency and n = 0, 1, 2, 3, .... So the partition function is Z =∑n e−(n+1/2)βhf =

e− 1

2βhf

1−e−βhf = 0.018. At T = 300K, hf/kT = 8.01. The probability of being in first

excited state is P1 = e−(3/2)hf/kT /Z = 3.33× 10−4. The probability in being ground

state is e−(1/2)hf/kT /Z = .9997.


estatística - qualifying

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