Thermo HW Sols Copy

8/2/2019 Thermo HW Sols Copy

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Physics 212A

Problem Set #1Roya Zandi Winter, 2011

Due: Friday 14, 2011

1. Given that the efficiency, R, of a reversible heat engine that works betweentwo reservoirs at temperatures Th and Tc (Th > Tc) can be written in theform

R = 1b(Tc)

b(Th)

do the following

a) First, show that the efficiency cannot be greater than one.

b) Then, show that the function b(T) must be either always positive oralways negative.

c) Finally, show that if b(T) is chosen so that it is always positive, thenb(T) is a monotonically increasing function of T.

2. Suppose that a system with constant heat capacity C is taken from aninitial temperature TA to a higher temperature TB by placing it in contactwith n thermal reservoirs at n equally-spaced temperatures between TA andTB. That is, the temperature of the first reservoir is equal to TA, the temper-ature of the second reservoir is TA + (TB TA)/(n 1), the temperature ofthe third reservoir is equal to TA + 2(TB TA)/(n 1), and so on. Thermalreservoirs have infinite heat capacity. Now, suppose you return the systemto its initial temperature by placing it in contact with the thermal reservoirsagain, going in reverse. What is the net effect of this process? Show that asn , the process becomes one in which everything is back to its initialstate.

3. Three systems have identical heat capacities, C. They are initially attemperatures T, 2T and 3T. With the use of infinitesimal, reversible cyclesthey are brought to a common temperature, Tf.


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a) What is Tf?

b) How much work has been extracted in the process?

The heat capacities are all temperature independent, and the systemsperform no work during the process above. In addition, there is no heatexchange between the systems and any other entity.

V

P

A B

C

D

EF

T1

T2

4. Two isotherms of one mole of a substance that can undergo a gas-liquidphase transition are shown in the figure above, which is a P-V diagram. Theabsolute temperatures are T1 and T2, respectively. The substance is madeto go through one cycle of a cyclic reversible transformation, ABCDEF, asindicated in the figure. You are given the following information.

(a) ABC and DEF are isothermal transformations.

(b) FA and CD are adiabatic transformations.

(c) In the gas phase (BCDE) the substance is an ideal gas. at A thesubstance is a pure liquid.

(d) The latent heat, L along AB is given by L = 200 cal/mole. Further-more:

T2 = 300K

T1

= 150KVA = 0.5 liters

VB = 1 liter

VC = 2.71828 liters


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Calculate the amount of work done by the substance over the course of this

cycle.

5. Chapter 1, Problem 10 in Kardar.


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Physics 212A


Due: Jan. 21, 2011

1. Consider a system in equilibrium with a thermal reservoir that keeps it ata temperature T. The system is also in mechanical contact with a pressurereservoir that maintains the pressure of the system at a constant value, P.Show that the maximum amount of work that can be extracted from this

in taking it from one equilibrium state in contact with the two reservoirs toanother equilibrium state, also in contact with the two reservoirs, is equal tothe decrease in the Gibbs free energy of this system.

2. Find the Helmholtz free energy of the van der Waals liquid-vapor system,given the equation of state

P+ aN

V

2(VNb) = NkBT

and the information that the heat capacity at constant volume of this systemis equal to 3NkB/2

3. Find the heat capacity at constant pressure of the van der Waals liquid-vapor system.

4. Given that the quantity CP is the heat capacity at constant pressure:

(a) Derive the equation CPP

T

= T

2V

T2

P

(b) Prove that CP of an ideal gas is a function ofT only.

(c) In the case of a gas obeying the equation of state

PV

N= RT+ BP


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where B is a function ofT only, show that

CPN

= TPd2B

dT2+ c

(0)P

where c(0)P

is the limit of the right hand side at very low temperatures.


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Physics 212A


Due: Jan. 28, 2010

1. Derive a relationship between the isothermal compressibility, T = 1

V(V/P)T,

and the adiabatic compressibility, S = 1

V(V/P)S that allows you to

demonstrate the inequality T > S.

2. The equation of state of a magnetic system is

M = ND tanh(H/kBT)

where D and are system-dependent constants. The heat capacity at zeromagnetic field of this system is given by

CH|H=0 = N Tn

where is a constant, and the exponent n is greater than two. Does thissystem obey the third law of thermodynamics? That is, is its entropy atT = 0 unique?

3. Rewrite the following derivatives in terms of the second derivatives of theGibbs free energy: cp, , and T.

v

s|P

s

F|v

sv |H

4. Problem 8, chapter 1 in Kardar.5. Problem 9, chapter 1 in Kardar.


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Physics 212A

Problem Set #4R. Zandi Winter, 2011

Due: February 9, 2011

1. Chapter 2, Problem 8 in Kardar

2. Consider the container of gas shown in the figure below. Originally the

Moveable piston

Figure 1: The container of gas divided in two by a freely moveable partition.All walls are thermally insulating. Initially, as shown, the volume on eachside of the piston is the same.

volume on each side of the movable piston is the same, V0. The gases oneach side are also at a common pressure, P0 and a common temperature, T0.Clearly, in mechanical equilibrium the pressures must be the same, but asall walls are thermally insulating, as is the piston, there is no a priori reasonfor the temperatures to be the same. Nevertheless, they are, at the outset.

The gas in both sides is a monatomic ideal gas.The gas in the left hand side of the container is now provided with heat,

say by a heating coil inside the left-hand side of the container. The heatis transferred to the gas slowly until the pressure in the gas on each side isequal to 27P0/8.

1


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(a) What is the final right-hand volume?

(b) What is the final right-hand temperature?

(c) What is the final left-hand temperature?

(d) How much heat has been supplied to the gas on the left?

(e) How much work has been done to the gas on the right?

(f) What is the entropy change in the gas on the left?

(g) What is the entropy change in the gas on the right?

3. Consider the case of a box with total volume V divided into two unequalparts by an imaginary partition. Let the volume of the first part be V1,where V1/V = p = 1/2. The volume of the second part is, then, V2, whereV2/V = 1 p. Assume that the total number of molecules in an ideal gascontained in this box is N, where N 1.

(a) What is the probability that N1 of the molecules are in the first partof the box and N2 = N N1 are in the second part of the box?

(b) For what value of N1 is this probability a maximum?

(c) Call the answer to the above question Nmax. Making use of Stirlingsformula, find an expression for the probability that Nmax + moleculesare in the first part of the box and N1Nmax molecules are in thesecond part of the box, under the assumption that


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(a) Use the two equations above to show that the probabilities add up to

1:N

N1=0 PN1 = 1.(b) Use those two equations to show that the expectation value of the

number of molecules in the first portion of the box is equal to pN:N1 =

N

N1=0N1PN1 = pN.

(c) Use those two equations to obtain a simple, closed form expression for(N1 N1)

2.

3


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Physics 212A


Due: February 18, 2011

1. Problem 7, chapter 4 in Kardar.2. Consider a system that can exchange energy and volume with a largereservoir.

a) Show that the entropy of this combined system (system & reservoir)is maximized when the temperature of the system is equal to the tem-perature of the reservoir and the pressure of the system is equal to thepressure of the reservoir.

b) Assume that the reservoir is much larger than the system. Expand tosecond order in the energy and volume of the system. Show that twoinequalities must be satisfied in order that the entropy of the combinedsystem is a maximum at the extremum point.

c) Show that when you sum over all ways in which the system shares en-ergy and volume with the reservoir and then take the log, you generate

the Gibbs free energy of the system.

3. Consider, again a system that shares energy and volume with a reser-voir. Calculate the relevant partition function in the case that the systemis a monatomic ideal gas. From this, obtain the Gibbs free energy of themonatomic ideal gas.

4. Consider a set ofN noninteracting molecules in a container that consistsof two portions. One portion of the container has a volume V1 and the otherhas a volume V2. The potential energy of a molecule in the first portion of

the container is equal to 1, and the potential energy of a molecule in thesecond portion of the container is equal to 2.

a) Calculate the partition function of this system.


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b) Use this result to find the pressure in each portion of the container at

a temperature T. Are the two values of the pressure that you obtainthe same?


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Physics 212A


Due: March 2, 2011

1. Find the contribution of the diagram shown below to the virial expansion

for the grand partition function of an interacting classical gas. See Fig. 1.

i

k

Figure 1: A second-order contribution to the virial expansion of the inter-

acting gas.

Now, find the correction to the equation of state for the pressure. How does

the correction depend on the density of the gas? To find the correction, stay

at the lowest possible order in the virial expansion.

2. Here, I want you to try to work out another approximation for the grand

potential of the interacting gas. You will make use of the linked cluster

expansion, and you will calculate the diagrammatic sum shown in Fig. 2 for

the grand potential. The sum that you will end up doing is an infinite one,

but it will also be a sum that is possible to perform. Write down your result

for the grand potential.

3. Consider the case of a system that shares both volume and energy with

a bath. Find the mean square deviation of the volume occupied by the

system about the optimal volume. Show that this is simply related to the

appropriately defined compressibility. Which compressibility is it, and what

1


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+

+

+

+

+

...

Figure 2: The diagrammatic sum for the contributions of interactions to the

grand potential of the interacting gas.

is the relationship?

2


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Physics 212A

Problem Set #7R. Zandi Winter, 2011

Due: March 11, 2011

1. The two equations of state of the monatomic ideal gas can be written inthe form

PV =2

3U (1)

U = 32NkBT (2)

The above equations of state have also been derived for the classical ideal gas.I would like you to show that the equation of state (1) holds for the quantummechanical monatomic ideal gas, whether its statistics are Fermi-Dirac orBose-Einstein.

2. Problem 11 Chapter 7 in Kardar

3. Problem 16 Chapter 7 in Kardar


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Physics 212A

Midterm ExamRoya Zandi Winter, 2011

Solutions

1. Given the ideal gas equations of state

PV = NkBT

U = NBT3

where B is a positive constant,

a) What is the Helmholtz free energy of this system, up to terms inde-pendent ofV and T?

b) What is the relationship between temperature and volume in an adia-batic expansion or compression of this ideal gas?

c) What is the heat capacity at constant pressure of this gas?

The Helmholtz free energy of this system is obtained by integrating upthe two equations above. We end up with the result

F = NkBT lnV

NNB

T3

2+ DNT (1)

Ive inserted the N in the log to keep the free energy properly extensive. Ifyou forgot the factor, you will not be docked any points. Here, the constantD cannot be obtained on the basis of the information given. I got the secondterm on the right hand side of Eq. (1) by making use of the equality

U= T2

T

F

T

(2)

To find the relationship between the volume and the temperature in anadiabatic compression or expansion of this gas, take the derivative of the


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Helmholtz free energy with respect to temperature, which is minus the en-

tropy, and demand that it be a constant. This translates into the relation

kB lnV

N+

3

2BT2 = a constant (3)

As the volume increases, the temperature decreases.Finally, the heat capacity at constant pressure can be obtained from the

second derivative of the Gibbs free energy, or by other means. Performing aLegendre transformation on the Helmholtz free energy, we find

G(T,P,N) = NkBT ln

kBT

P

NB

T3

2+ DNT+ NkBT (4)

Making use of the relationship

CP = T

2G(T,P,N)

T2

P,N

(5)

we findCP = 3NBT

2 + NkB (6)

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Thermo HW Sols Copy