Chapter 11

Thermodynamics

Benson: Chap.18& Chap.19 &

Chap.20 &Chap.21

Thermodynamics deals with microscopic scale physics in terms of macroscopic quan-tities. Without knowing any detailed information of microscopic degrees of freedom,what are the macroscopic quantities we can define (through computation or measure-ment), and what are their relations?

Furthermore, although we cannot precisely describe the states of the microscopicdegrees of freedom, they may still have macroscopic physical effects. Is it possible todefine physical quantities to collectively quantify certain properties of the micro-states(at least in certain cases) such that their macroscopic effect can be quantified?

Friction is a macroscopic effective description of the collective effect of microscopic degrees of free-dom. We need the notion of friction to explain the violation of energy conservation in the macro-scopic degrees of freedom. But we may take a further step and try to keep track of the energytransferred into the microscopic degrees of freedom as “heat”.On the other hand, the notion of “heat” was historically motivated in a different way.

We will only consider thermodynamics of systems in equilibrium. Non-equilibriumthermodynamics is an interesting (and complicated topic) that will not be consideredhere.

We will consider these macroscopic quantities that are easy to measure:

1. TemperatureVolume andpressure are

typically moreimportant for

gases.

Q: (11-1) Is temperature a measure of the feeling of coldness/hotness?

Q: (11-2) Which is colder? Metal or paper?

Q: (11-3) Why is there coldness? Is it just the absence of heat?Q: (11-4) Why

is there evil? Is itjust the absence

of goodness?

2. Volume

3. Pressure

4. Quantity: mass/mole/# particles

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These are the state variables of a thermodynamic system in equilibrium, analogousto the position and velocity of a particle.

Q: (11-5) Are there other state variables? How do we know if we already have enoughstate variables or not?

When studying a class of phenomena, there are measurements available to observe the physicalsystem under study. The state variables are those representing measurement results. When thevalues of the state variables at a given time are sufficient to uniquely determine their values at alater time, we can start building a physical model, and the state variables are temporarily sufficient.It is always possible that new state variables appear because people learned to measure new featuresof a system.

Example: People did not know that an electron has a spin.

In addition to the state variables, we will also discuss macroscopic quantitiesdefined through (imagination and) calculation including

1. Heat (the transfer of microscopic energies)

2. Internal energy (the microscopic energy of a system)

3. Entropy (the amount of microscopic information omitted in a macroscopic de-

Contrary to theimpression due

to its translationǶ, entropy

isn’t a measureof randomness.

scription)

These quantities are not directly measured but (imagined to) correspond to certainmicroscopic quantities.

There is a large gap between macroscopic, thermodynamical phenomena and the behavior of micro-scopic degrees of freedom. It takes a lot of imagination to formulate principles of thermodynamics.Statistical mechanics is the advanced version of thermodynamics that explains macroscopic phe-nomena in terms of microscopic degrees of freedom via the mathematical effect of large numbers.Quantum mechanics was discovered following a study of the statistical physics problem of black-body radiation.

In the old days, people imagine that an increase in the temperature is a reflectionof the heat transferred into the substance. It was later shown that it should beidentified with a kind of energy through the experiment of Joule. (Unification ofclassical mechanics and thermodynamics.)

The key reasonwhy we can

ignoremicroscopic data

and still havedefinite relations

amongmacroscopicquantities is

large numbers.

Assumptions of this chapter:

1. All systems are (almost) at rest.

2. All systems are classical and non-relativistic.

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11.1 Ideal GasQ: (11-8) Can

we use thisequation as a

definition of thetemperature?

The equation of state of ideal gases is

PV = NkT, (11.1)

where P is the pressure, V the volume, N the number of molecules, k the Boltzmannconstant, T the temperature. The ideal gas law is equivalently written as

PV = nRT. (11.2)Q: (11-9) Whatwould go wrongif we define thetemperature by

eq.(11.1) but thegas we use doesnot behave like

ideal gas?

A real gas is approximately an ideal gas when the pressure is low and the volumeis large. In this low-temperature, large-volume limit, all gases appear to obey thesame equation of state (11.1).

The Celsius scale of temperature is defined since 1954 by absolute 0 being �273.15o C and thetriple point of water being at 0.01o C. In 2017, the Boltzmann constant was chosen to be exactlyk ⌘ 1.380649⇥ 10�23J/K by redefining the Celsius (Kelvin).

The Boltzmann constant is by definition

k ⌘ 1.380649⇥ 10�23J/K. (11.3)

The Avogadro number is by definition

NA ⌘ 6.02214076⇥ 1023 (11.4)

and the universal gas constant is

R ⌘ NAk ' 8.314J/K. (11.5)

For a real gas, the molecules are not of zero size, and the interactions amongmolecules are not of zero range, hence the ideal gas law (11.1) should be modified.An approximate description is given byVan der Waals equation of state:

✓P + a

n2

V 2

◆(V � bn) = nRT, (11.6)

where a and b are parameters depending on the nature of the molecules.

11.2 Kinetic Theory

Materials are made of molecules and atoms.

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The kinetic theory was already proposed before Einstein analyzed Brownian motion to prove thatthere were atoms. Many physicists resisted this interpretation of thermodynamics in terms ofhypothetical atoms.

There are otherhidden

assumptions foryou to discover.

Assumptions:

1. Non-relativistic classical motion of molecules. (Relativistic effect and quantumeffect are ignored.)

2. Large number of identical molecules moving with “random velocities” (so thatthe gas is macroscopically uniform).

3. No interaction between molecules (and between molecules and walls) exceptbrief elastic collision (by strong repulsive force at short distance).

4. Average distance between molecules is much larger than the size of the molecules.

A gas in a box has pressure because the molecules hit on the walls very frequently.

Exercise: (11-1) Show that, with the assumptions above, the kinetic interpretationof pressure is

P =1

3⇢v2

rms. (11.7)

The parameter vrms is the root-mean-square velocity of the molecules defined by

v2rms

=1

N

NX

i=1

v2i, (11.8)

where vi is the velocity of the i-th molecule.Solution:When a molecule hits a wall normal to x̂, the change in momentum is

�px = 2mvx. (11.9)

The averaged force per molecule is

Fx =�px�t

=2mvx⇣2Lvx

⌘ =mv2xL

. (11.10)

The total averaged force isHere we assumethat there is no

preferreddirection.

Fx =m

L

X

i

(vi)2x =

m

LN

v2rms

3. (11.11)

The pressure is thus

P =F

L2= mN

v2rms

3V= ⇢

v2rms

3, (11.12)

where ⇢ ⌘ Nm/V .

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Comparing eq.(11.7) and the ideal gas law (11.1), we find that the averaged kineticenergy is

We do notinclude in Kav

the kineticenergy of the

atoms’ motionrelative to the

center of mass ofthe molecule.

Kav =1

2mv2

rms=

3

2kT. (11.13)

The absolute temperature T is thus a measure of the average (translational) kineticenergy of the molecules in the kinetic theory. This definition of T can be applied toreal gases as well.

Q: (11-10) What is the rms velocity of an ideal gas at T = 20 oC?

11.3 Advanced Topic: Boltzmann Distribution*

We talked about the averaged behavior of a large number of molecules in the kinetictheory. We would now like to discuss more detailed information — the distribution ofenergy among molecules, e.g. what is the fraction of the molecules that have energieswithin a certain range, etc.

Q: (11-11) Aftershaking

vigorously a boxof 10000

identical dice,would the

percentage ofdice landing on“1” depend on

what the dice ismade of or how

the box isshaken?

It is a remarkable fact that this distribution is independent of the details of themolecular properties, e.g. how the molecules interact with each other when they meet.The reason behind the robustness of this distribution is “large number”.

11.3.1 Effect of Large Numbers

Put N coins (say, all heads) in a box and shake the box vigorously, but not necessarilyrandomly.

Q: (11-12) What do you expect the outcome to be?Combinatorics:

The number of different arrangements with n heads and (N � n) tails is CN

n=

N !n!(N�n)! . Assuming complete randomness, the probability of n heads is

P (n) =CN

n

2N. (11.14)

For uneven coins, it is

P (n) = pn(1� p)N�nCN

n

2N(11.15)

if the probability of head is p.For large N , the number n that maximizes P (n) satisfies

0 =d

dnP (n) ' d

dn[n log(p) + (N � n) log(1� p)� n log(n)� (N � n) log(N � n)] .

(11.16)

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Here we have used Stirling’s formula

n! ' (2⇡n)1/2⇣ne

⌘n

.

The condition above impliesn

N � n=

p

1� p. (11.17)

The really remarkable part of this problem is the effect of a large N .For an even coin (p = 1/2), the ratio of the probabilities of having n = N/2 (the

expectation value) versus n = (1 ± 1%)N/2 (a 1% deviation from the expectationvalue) is

[(N/2)⇥ 1.01]![(N/2)⇥ 0.99]!

[(N/2)!]2=

8><

>:

0.6 N = 10000,

1.4⇥ 10�217151 (N = 1010),

10�1015 (N = 1020).

(11.18)

To have some feeling about very large and very small numbers, we recall that the age of the universeis about 1010 years, which is about 3⇥ 1017s or 6⇥ 1060 Planck time. a

a1 Planck time is about 5⇥ 10�44s.

Notice that it is easy to see a deviation from the expectation value with |n �(N/2)| > 1, although very hard to see |n � (N/2)| > 1

100(N/2). It is extremely hardto see a deviation by percentage from the expectation value when the number ofsamples is large.

Due to this dramatic effect of the combinatorics, shaking the box differently would hardly make anydifference unless you are Ì^.

Q: (11-13) What is the chance for all molecules in a gas to be found inside the lefthand side of a box?

11.3.2 A Simple ModelIt is not

important thatthey are

particles.

Consider a hypothetical physical system in which there are N particles (i = 1, 2, · · · )and each of them can have an energy Ei = 0, 1, 2, · · · (in a certain unit system). Whenthe total energy of the system is given, there are many different ways to distributethe total energy among the particles. We assume that there is no correlation among

Here we assumethat the particles

aredistinguishable.

the energies of different particles, except that the total energy is fixed. We considereach possible assignment of the energies as a different micro-state.

It is important toask the “right

question”.

For a given macro-state (i.e. a given total energy Etot), assuming that all micro-states are equally possible, we can count the probability of finding a particle to havea certain given energy.

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It turns out that the probability for a randomly picked particle to have n unitsof energy is P (n) / e�n/Ec for some constant Ec. (Ec is determined by Etot and N .)That is,

P (n)

P (m)=

e�n/Ec

e�m/Ec, (11.19)

or equivalently

P (n) =e�n/Ec

PE

m=0 e�m/Ec

. (11.20)

This is called the Boltzmann distribution.

Q: (11-14) In terms of P (n), what is the average number of particles of energy E = n?

Q: (11-15) How to determine Ec from Etot and N?The average energy of the N particles is always Etot/N , which should agree with

PnnP (n).Let us show how this probability can be computed by the simple example of

Etot = 6 and N = 5. In the table below, we enumerate all possible micro-states(assignment of energies to different particles) and count the number of particles witha given energy E = n.

N(0) N(1) N(2) N(3) N(4) N(5) N(6) # micro-states prob.4 0 0 0 0 0 1 5 1/423 1 0 0 0 1 0 20 4/423 0 1 0 1 0 0 20 4/422 2 0 0 1 0 0 30 6/423 0 0 2 0 0 0 10 2/422 1 1 1 0 0 0 60 12/421 3 0 1 0 0 0 20 4/422 0 3 0 0 0 0 10 2/421 2 2 0 0 0 0 30 6/420 4 1 0 0 0 0 5 1/422 1.333 0.8333 0.476 0.238 0.0952 0.0238 210 1

The numbers in the last row (2, 1.333, 0.8333, 0.476, 0.238, 0.0952, 0.0238) are theexpectation values hN(E)i of the numbers of particles with given energies (E =

0, 1, 2, 3, 4, 5, 6). The least-squares best fit of these numbers is

hN(E)i ' 2.05⇥ e�0.488E. (11.21)

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Q: (11-16) How do we know that each micro-state has the same probability?

HW: (4-2) Consider 5 particles, each can have an energy of E = 0, 1, 2, 3, · · · . Theirtotal energy Etot = 4. Make a table of all the possible distribution of energy (howmany particles have energy 0, how many 1, how many 2, etc., and how many micro-states have this energy distribution). Then count the total number of micro-states, thepossibility of each energy distribution, and the average number hN(0)i of particleswith energy E = 0, the average number hN(1)i of particles with energy E = 1,etc. Plot the diagram with hN(E)i as the vertical axis and E = 0, 1, 2, 3, · · · as thehorizontal axis. What is Ec?

11.3.3 An Argument

Consider a large number N of particles with the total energy E. Let f(Ei) be theprobability for a randomly picked particle to have the energy Ei. Let h(E 0) be theratio of the number of micro-states for (N � 2) particles to have the total energyE � E 0 over the number of micro-states for N particles to have the total energy E.Then the probability of picking two particles with energies E1 and E2 respectively is

Strictly speaking,after the first

particle is picked,the probability

for the 2ndparticle to have

energy E2 wouldbe changed. But

it is notsignificantly

change when N

is large.

h(E1 + E2) ' f(E1)f(E2). (11.22)

Notice that the left hand side depends only on the sum E1+E2. This is possible onlyif f(E) / e�E/Ec for some constant Ec.

Q: (11-17) Why does energy play a special role in thermodynamics while it is not theonly conserved quantity? Can we generalize the idea to other conserved quantitiesand define generalized notion of temperature?

11.3.4 Temperature in Kinetic Theory

The Boltzmann distribution applies to the gas in the kinetic theory. We can determineEc from the relation (11.13).

Given the Boltzmann distribution, the probability for a monoatomic molecule tohave the kinetic energy K = 1

2mv2 is proportional to e�K/Ec . The probability for a

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particle to have a velocity with its magnitude within the range (v, v + dv) isPMB is calledthe Maxwell-

Boltzmanndistribution.

PMB(v)dv = ce�E/Ecdvxdvydvz = ce�E/Ec4⇡v2dv, (11.23)

where

c =

✓m

2⇡Ec

◆3/2

(11.24)

is the normalization constant fixed by the condition that the sum of all probabilitiesequals 1 Z 1

0

PMB(v)dv = 1. (11.25)

Then vrms can be calculated asZ 1

0

v2P (v)dv =

Z 1

0

c4⇡v4e�mv2/2Ecdv = � 2c

m

d

d�

Z 1

0

4⇡v2e��mv2/2dv

�����=1/Ec

= � 2c

m

d

d�

1

c

�����=1/Ec

= � 2c

m

d

d�

✓m�

2⇡

◆�3/2������=1/Ec

=2c

m

3

2

Ec

c=

3Ec

m. (11.26)

In comparison with eq.(11.13), we see that

Ec = kT. (11.27)

The temperature thus acquires a new meaning in terms of the Boltzmann factore�E/kT .

Even though we derived eq.(11.27) from the special example of an ideal gas, weshall simply define the Boltzmann distribution by / e�E/kT .

Q: (11-18) Why? (Hint: This has to do with equilibrium at equal temperature.)

Q: (11-19) If we put two kinds of molecules (e.g. monoatomic and diatomic molecules)in the same thermally insulated box, why should the two kinds of molecules approachthe same temperature? Why should the average kinetic energy of the center-of-massmotion of the diatomic molecule agree with that of the monoatomic molecules?

11.4 Heat

11.4.1 What is “heat”?

Heat is now understood as a form of energy. In the old days, heat was an imaginaryquantity that manifests itself by making things hotter or changing phases. Heat istransferred when (but not only when) the temperature is changed.

Q: (11-20) Is there a quantity that manifests itself as kindness/cruelty?

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The conservation of heatHere we review how the notion of a conserved heat can be made concrete through observations. a

When there is no work done by/to the systems A and B, people have observed the following nontrivialrelations:

1. When TA and TB are changed to T 0A

and T 0B

, the ratio TA�T0A

T0B�TB

is always the same, independentof TA, TB, T 0

Aand T 0

B;

2. If TA�T0A

T0B�TB

= CAB when A and B are in touch, and TB�T0B

T0C�TC

= CBC when B and C are in touch,

then we have TA�T0A

T0C�TC

= CABCBC when A and C are in touch.

These relations allow us to define heat as a conserved quantity:

�Q = CA�TA = CB�TB = CC�TC , (11.28)

where CA, CB and CC are heat capacities. One can randomly pick a system R for reference(setting, e.g. CR = 1) to define heat by �Q = CR�T . Then the heat capacity of a system A is

CA ⌘ �Q

�T, (11.29)

and the specific heat is defined as the heat capacity per unit mass. (The mole specific heat is theheat capacity per mole.)

aThe discussion here relies on the assumption that the heat capacities are independent of the temperature.This is a good approximation for a large range of temperatures for many materials.

Q: (11-21) If we add heat to one of the two systems in thermal equilibrium, is itpossible that they still maintain thermal equilibrium?

Answer: Yes, the heat can become latent heat during a phase transition.There are latent

heat of fusion

and latent

heat of

vaporization.

Sometimes the heat �Q does not change the temperature. This does not meanthat heat is not conserved, but that heat may change the state of a system insteadof changing its temperature.

Example: Ice into water, water into steam, chemical reaction, etc.The conservation of heat can still be preserved by assuming fixed amount of heat

needed for a certain change of states.Heat can becreated fromwork through

friction, and heatcan be turned

into workthrough thermal

expansion.

The conservation of heat is then “broken” when Joule found that work can createheat. Therefore, the better interpretation is that heat should be identified as a kindof energy, and the conservation of heat is merely a special case of the conservation ofenergy.

The unit (cal) of heat can thus be identified with the unit of energy:1 calorie = 4.186 J.

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11.4.2 Heat Transfer

Heat transfer can occur in the following ways:

1. conductiondQ

dt= �A

dT

dx, (11.30)

where is the thermal conductivity (the thermal resistance is R = L/).

HW: (4-3) (Benson Chap. 19, Prob. 3) Hot fluid passes through a cylindricalpipe of length L with inner radius a and outer radius b. Show that the rate ofheat conduction through the swall of the pipe is

dQ

dt=

2⇡L(Ta � Tb)

log(b/a), (11.31)

where Ta and Tb are the temperatures at the inner and outer surfaces, respec-tively.

2. convection

3. radiationdQ

dt= �e�AT 4, (11.32)

where e is the emissivity, which depends on the nature of the surface. If theabsorption is given by a different formula, it would not be possible to achievethermal equilibrium at all T . Therefore, taking both emission and absorptioninto account,

dQ

dt= e�

��T 4

(body) + T 4(backgrd)

�. (11.33)

Example: Newmaterial that

cools itself.

Thermal connection may depend on the frequency of radiation. The condition ofthermal connection above in the definition of thermal equilibrium should include allfrequencies.

11.4.3 Internal Structure of Molecules

For an monoatomic ideal gas, the internal energy U = NKav =32NkT , so the molar

specific heat is Cv =32R. Hence Cp =

52R.

Equipartition of energyEach (active) degree of freedom (channel of energy) has an average energy of kT

2 .5 = 3 linearmotions + 2

rotations (Therotation along its

axis has noenergy.)

For an ideal gas composed of diatomic molecules modeled by rigid rotors, themolar specific heat is Cv =

52R and Cp =

72R.

7 = 3 linearmotions + 3

rotations + 1relative distance

= 6 kineticenergy + 1

potential energy

For an ideal gas composed of diatomic/polyatomic molecules modeled by non-rigidrotors, the molar specific heat is Cv =

72R and Cp =

92R.

The relative motion along the axis (vibration) of a diatomic molecule introduces both one degreeof freedom in kinetic energy and one degree of freedom in potential energy.

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Example: Molar Specific Heat (J/mol ·K) at 300 K and 1 atm.He Ar H2 N2 O2 Cl2 CO2 H2O (100oC)

Cv 12.5 12.5 20.4 20.8 21 25.2 28.5 27.0

Q: (11-22) Which ones are monatomic, rigid rotors and non-rigid rotors?

Example: At different temperatures, various degrees of freedom could be “frozen”or “active”, and the molecules behave like monoatomic molecules, rigid rotors or non-rigid rotors.

Exercise: (11-2) Consider ideal diatomic molecules composed of two identical parti-cles of mass m connected by a massless rigid rod. Given N such diatomic moleculesin the Boltzmann distribution at temperature T , what is the rms velocity of themolecules?

11.4.4 Phases

When a gas is compressed, eventually the distance among molecules is so short thatthe interactions among molecules cannot be ignored at all times. The macroscopicproperties of the gas is changed and we refer to it as a liquid. This is a phasetransition.

Air molecules are roughly a few (or tens of) nanometers (10�9m) apart from each other. The meanfree path is roughly 20 times the average distance. The size of a molecule is about 10�11 ⇠ 10�10m.The average distance among molecules in water is typically a few 10�10m.The nature of the interactions among molecules is different at different separation scales.

Most materials typically come in at least three phases.

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On phase transitions, more than one phase can coexist in equilibrium.The phase diagram of water is slightly different from other materials.

11.5 0-th Law of Thermodynamics: Thermal Equi-librium

By experience, when different parts of a system have different temperatures, eventu-“Different parts”

of a systemcould mean two

objects puttogether.

ally the temperature becomes uniform in space and constant in time if the system isthermally isolated from the environment. Then we say that it has reached thermalequilibrium. Thermal equilibrium is thus often defined as the situation when a sys-tem has a uniform constant temperature (the temperature is constant in space andin time).

However, strictly speaking, in some cases, no matter how long we wait, the tem-perature will never become uniform in space, although it becomes constant in time.

An example is a box of gas in a gravitational field. The gas would have differenttemperatures at different heights, since molecules with higher potential energy have

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lower kinetic energy. There is no good reason to exclude these cases from thermalequilibrium. Therefore, strictly speaking, the usual definition mentioned above isinappropriate.

A more suitable definition of thermal equilibrium is the following:Two systems

thermallyconnected canbe viewed as a

large system.The thermal

equilibrium oftwo systems is

thus also defined.

When all of the state variables of a thermally connected system are constant intime (not necessarily constant over space), the system is said to be in thermalequilibrium.

Q: (11-23) Why do we demand the condition of “thermal contact” in the definitionof thermal equilibrium?

Wikipedia:“A system is said to be in thermal equilibrium with itself if the temperature within the system isspatially and temporally uniform.”Halliday-Resnick-Walker:A and B are in intimate contact inside an insulator... “In fact, we suppose that every measurableproperty of body A and of body B has assumed a stable, unchanging a value. Then we say thatthe two bodies are in thermal equilibrium with each other.”Katz:“Two objects are in thermal equilibrium if they are in thermal contact and their temperatures donot change.”Benson:“When all its state variables are constant in time, the system is said to be in thermal equilibrium.”

Q: (11-24) Is there something inappropriate in these different definitions of thermal equilibrium?

aDoes “unchanging” refer to both space and time?

Thermal equilibrium should be defined with a time scale much longer than thetime scale of heat transfer among all degrees of freedom.

The phrases “thermally connected” and “in thermal contact” refer to the possibilityof heat (thermal energy) transfer. When heat transfer is prohibited by a division, thedivision is called a “thermal insulator”.

Q: (11-25) Why do systems in equilibrium have the same temperature?

0-th law of thermodynamicsTwo bodies in thermal equilibrium with a third are also in thermal equilibriumwith each other.

Equivalencerelations havethe following

properties. (1)reflexive, (2)

symmetric, (3)transitive.

Q: (11-26) Is thermal equilibrium an equivalence relation?

Q: (11-27) How do we verify this law?

Q: (11-28) Why should the 0th law hold? What kind of physics might break the 0th

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law?

11.6 1st Law of Thermodynamics:Energy Conservation

Due to energy conservation, the change in the internal energy of the system equalsthe heat flowing into the system minus the work done by the system:

Both �Q and�W depend on

the process.�U = �Q��W. (11.34)

This is the first law of thermodynamics.

Q: (11-29) What is the definition of U? How do we measure U?An important point that U is uniquely determined by the physical state of the

system. Without this assumption, the 1st law would be useless.Luckily, one can often define the internal energy U as a function of the state

variables. For example, for an ideal gas, U = U(P, V, T ). Whenever this gas has theIn general, P , V

and T arefunctions of

space and time.

same pressure P , volume V and temperature T , its internal energy is the same.If U is not approximately determined by macroscopic state variables, i.e., if U

is different for different macroscopically indistinguishable micro-states, the 1st lawwould be useless.

Q: (11-30) Whyis eq.(11.35)

correct?

For compressible fluids (gases),

�W =

ZPdV. (11.35)

When there is no work involved, i.e. �W = 0, and when there is no phasetransition,

�U = �Q = C�T. (11.36)

In general, the heat capacity can change with T , so that

�Q =

ZT+�T

T

C(T 0)dT 0. (11.37)

Q: (11-31) How to verify (or falsify) the 1st law through experiments when thefunction U is not given?

Given two states A and B of a system, the internal energies UA and UB have adifference UB � UA which is independent of how the state A is changed to the stateB. Even though we might not know what is the value of UB � UA, the combination�Q��W must be the same for all processes from A to B.

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For an ideal gas, an equilibrium state of the gas is given by P and V (or P and T , or V and T ).On the P � V diagram, the states A and B are two points, and there are infinitely many possiblepaths connecting A and B. �Q��W must be the same along all paths.

Q: (11-32) How to determine �Q in a process?

Exercise: (11-3) For an ideal gas with initial pressure P0 and volume V0, what isthe heat �Q involved to change its state to P1 = 2P0, V1 = 2V0, with the ratio P/V

kept constant in the process?

11.7 2nd Law of Thermodynamics

11.7.1 Thermal Processes

A thermal reservoir or a thermal heat bath is a thermodynamic system with aheat capacity that is large enough so that when it is in thermal contact with anothersystem of interest or the environment, its temperature remains effectively constant.

A quasi-static process is a process through which all state variables changeinfinitely slowly so that the system is always arbitrarily close to thermal equilibrium.

In a reversible process, when the process is reversed, all heat/work inputs andoutputs are interchanged, and all state variables go through the same trajectory inthe opposite order.

A reversible process must satisfy the following criteria:

1. quasi-static

2. no friction

3. any transfer of heat occurs at constant temperature or is associated with aninfinitesimal temperature difference.

A irreversible process (e.g. a spontaneous process) typically involves energylost (e.g. in radiation) and an increase in entropy.

During a thermal process, we are concerned with how physical variables (P, V, T, U,Q,W, · · · )change. While there are infinitely many possible thermal processes, some of them havespecific names:

1. Isovolumetric: V = constant,so �W = 0.

2. Isobaric: P = constant,so �W = P�V .

3. Isothermal: T = constant,so �Q = �W .

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4. Adiabatic: �Q = 0.

Exercise: (11-4) Prove that Cp = Cv +R for an ideal gas.Solution:For an ideal gas, the state variables P, V, T are related via PV = nRT , where n = 1 for onemole of the ideal gas. Hence, only two of the three variables P, V, T are independent. Wecan think of U as a function of, say, V and T .

Under an isovolumetric process, dV = 0, so dW = 0, and so

dU(V, T ) = dQ� dW = dQ = CvdT, (11.38)

whiledU(V, T ) =

@U

@VdV +

@U

@TdT =

@U

@TdT. (11.39)

so we findCv =

@U

@T. (11.40)

Similarly, under an isobaric process,

dU(V, T ) = dQ� PdV = CpdT � Pd

RT

P

�= CpdT � nRdT = (Cp �R)dT, (11.41)

anddU(V, T ) =

@U

@VdV +

@U

@TdT =

@U

@V

R

P+ Cv

�dT. (11.42)

The two equations above imply that

Cp = Cv +R+@U

@V

R

P. (11.43)

Hence, we must assume that U is independent of V (@U/@V = 0) in order to prove thatCp = Cv + R. The reason why U is independent of V for the ideal gas is that the idealgas is assumed to have no potential energy among the molecules. The internal energy U

U is independent

of V for ideal

gases.

is simply the sum of the kinetic energies of all the molecules. This assumption is a goodapproximation for real gases when the volume is large (pressure is low) so that the averagedistance between two molecules is extremely large.

Exercise: (11-5) Prove that for ideal gases,

PV � = constant (11.44)

under adiabatic processes, where � ⌘ Cp/Cv.Solution:For an adiabatic process, dQ = 0. So the first law implies that

0 = dU + dW = nCvdT + PdV = nCvdPV

nR+ PdV

=

✓Cv

R+ 1

◆PdV +

nCv

RV dP = Cv(�PdV + V dP )

= CvV1��d(PV �). (11.45)

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A heat engine is a device that converts heat into mechanical work, e.g. steamengine, generator, etc. It is a system that exchanges heat and work with externalsystems.

The thermal efficiency of a heat engine is

✏ =W

QH

. (11.46)

For the simplicityof discussions,

we often assumethat a heat

engine operatesbetween twotemperatures

with heatexchanges of QH

and QC .

The coefficient of performance (COP) is

COP =|QC |W

or|QH |W

(11.47)

for a refrigerator or a heater.

11.7.2 Statements of the 2nd Law

Kelvin-Planck Statement

“Cycle” meansthat the initial

state equals thefinal state.

(Kelvin 1851)It is impossible for a heat engine that operators in a cycle to converts its heat inputcompletely into work.

Clausius Statement

(Clausius 1850)It is impossible for a cyclic device to transfer heat continuously from a cold body toa hot body without the input of work or other effect on the environment.

Equivalence of the Statements

Consider two heat baths at temperatures TH and TC (TH > TC). Let us name the heatengine forbidden by Kelvin-Planck as KP and the refrigerator forbidden by Clausiusas C.

Consider a cyclic heat engine A that draws heat QH from TH and dumps heat QC

to TC with output work W = QH � QC over a cycle. Its reverse operation serves asa refrigerator B that draws heat QC from TC and dumps QH to TH with input workW = QH �QC .

First we show that if the Clausius statement is violated, that is, if C exists, thenthe Kelvin-Planck statement is also violated, i.e. KP exists.

If C exists, A and C together draw heat (QH � QC) from TH and dump heat(QC � QC) = 0 heat to TC with output work W = (QH � QC). This violates theKelvin-Planck statement.

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Next we prove that if the Kelvin-Planck statement is violated then the Clausiusstatement is also violated, that is, if KP exists, then C also exists.

If KP exists, we can have B and KP put together to draw heat QC from TC anddump heat QC to TH , while the work W = QH � QC needed by B is provided byKP .

11.8 EntropyQ: (11-33) How

muchinformation isignored when

you toss N coinsbut do not checkwhich sides they

land on?

Entropy is not really about “disorder” or “randomness”. It is a measure of the degreeof ignorance. First we have to define what is “ignored”, then we can ask how muchinformation is lost. The larger the entropy, the more information is ignored.

In thermodynamics, we ignore the information about the micro-states. Entropytells us how much microscopic information is ignored when we choose to describe asystem in terms of macroscopic properties.

For instance, if macro-state #1 corresponds to N1 micro-states, and macro-state#2 N2 micro-states, then the entropy of macro-state #1 is larger than that of macro-state #2.

Recall Stirling’sformula n! 'p

2⇡n�n

e

�n.

Since the number of information goes up exponentially, it is more convenient todefine the entropy so that it is linearly related to the logarithm of the number ofignored information.

In fact, “heat” can be understood as the energy of micro-states. Typically, thenumber of accessible mirco-states is larger when there is more energy, so the transferof heat also transfers the ignorance about micro-states. On the other hand, dependingon the status of the system, the same amount of heat may correspond to differentamounts of ignorance.

Q: (11-34) Can we find the quantitive relation between heat transfer and the transferof ignorance (entropy)?

11.8.1 Carnot EngineFor a deep

understanding ofphysics, it is veryimportant to be

familiar withexplicit

computations onconcrete

examples.

Before entropy was understood as ignorance, entropy was motivated by a detailedstudy of concrete thermodynamical systems, in particular, the Carnot engine.

A Carnot engine operates between two temperatures TH and TC through thefollowing reversible processes (called Carnot cycles):

1. isothermal(Pa, Va) ! (Pb, Vb), satisfying PV = nRT .

2. adiabatic(Pb, Vb) ! (Pc, Vc), satisfying PV � = constant.

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3. isothermal(Pc, Vc) ! (Pd, Vd), satisfying PV = nRT .

4. adiabatic(Pd, Vd) ! (Pa, Va), satisfying PV � = constant.

The relations among the quantities are

PaVa = PbVb, (11.48)

PcVc = PdVd, (11.49)

PbV�

b= PcV

�

c, (11.50)

PdV�

d= PaV

�

a. (11.51)

Multiplying all the equations above on both sides, we find

VaVc = VbVd. (11.52)

During each stage of the Carnot cycle, we compute the heat transfer to the engineand the work done by the engine:

1. isothermal Q = W :

QH = W =

ZPdV = nRTH

ZdV

V= nRTH log(Vb/Va) > 0. (11.53)

2. adiabatic Q = 0:

W =

ZPdV = PbV

�

b

ZdV

V �= PbV

�

b[V 1��

c� V 1��

b]. (11.54)

3. isothermal Q = WNote that here

QC is defined asthe heat drawn

by the engine, soit is negative.

QC = W = nRTC log(Vd/Vc) < 0. (11.55)

4. adiabaticW =

ZPdV = PdV

�

d[V 1��

a� V 1��

d]. (11.56)

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The Carnot efficiency is

✏C ⌘ QH � |QC |QH

=TH � TC

TH

. (11.57)

Carnot’s Theorems

1. All reversible engines operating between two given reservoirs (at given temper-atures) have the same efficiency.

2. No cyclic heat engine has a greater efficiency than a reversible engine operatingbetween the same two temperatures.

Exercise: (11-6) Prove Carnot’s theorem.Solution:Proof of the 1st theorem:

If two reversible engines have two different efficiencies ✏ > ✏0. We can turn the 2ndengine to its reversed process, such that the work output of the 1st engine equals the workinput of the 2nd engine.

The two engines together serve as a single engine which draws heat QH �Q0H

from TH

and dumps heat QC �Q0C= QH �Q0

Hto TC . Since ✏ > ✏0, QC �Q0

C< 0, so this violates

the Clausius statement of the 2nd law.Proof of the 2nd theorem:

Since all reversible engines have the same efficiency, we only need to prove that irre-versible engines cannot have a greater efficiency.

Consider the same setting as above, with the 1st engine replaced by the irreversibleengine. If the efficiency ✏ is greater, the Clausius statement of the 2nd law is violated.

Q: (11-35) What is so special about the Carnot engine that it has the highest possibleefficiency? Can you think of another cycle of thermal processes that would have thesame nice properties as the Carnot engines?

11.8.2 Definition of Entropy

For the Carnot engine, or any reversible engine,

|QH ||QC |

=TH

TC

, (11.58)

henceQH

TH

+QC

TC

= 0. (11.59)

More generally, consider infinitely many infinitesimal steps of reversible processes,we have X �Q

T= 0, (11.60)

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and taking the limit, we have IdQ

T= 0 (11.61)

for reversible processes.In the space of states, two reversible processes from state A to state B must have

the same value forRdQ/T . Mathematically, this means that we can define a certain

Like thepotential energy,

this quantityinvolves an

arbitrary additiveconstant. Here

this arbitraryconstant

corresponds tothe possibility ofanother level of

unknownmicro-states.

physical quantity for each state, analogous to the potential energy.This quantity is called the entropy. The difference in entropy between two states

is defined by�S =

ZdQ

T(11.62)

with the integral along any reversible process. Picking any reference state, we candefine the entropy of all states that can be continuously connected to this referencestate through reversible processes.

The equation above is also expressed as

dS =dQ

T. (11.63)

Q: (11-36) If we connect two states by two different thermal processes; one is reversibleand the other irreversible. Which process has a larger value for the integral

RdQ

T?

Q: (11-37) What is the change in entropy �S when a gas with (P, V ) is changed to(P/2, 2V ) (at the same temperature)?

Q: (11-38) What is �S when 1 gram of ice melts into water at the same temperature?

Q: (11-39) What is �S for an isobaric process for a gas from T to T 0?

Exercise: (11-7) What is �S for a generic reversible process for a gas?Solution:

dQ = dU + dW = nCvdT + PdV

) dS =dQ

T= nCv

dT

T+

nR

VdV = nd[Cv log(T ) +R log(V )]

) �S = n[Cv log(T0/T ) +R log(V 0/V )] (11.64)

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Very roughly, we can understand eq.(11.64) as S = k log(N ), where N is the number of micro-states,as follows.Let there be N particles (N = nN0). The position of each particle can be anywhere in a volume V .Suppose we discretize the volume by a minimal volume �V , the total number of possible spatialdistributions of all particles is (V/�V )N .The average kinetic energy of a particles is p2/2m = 3kT/2, so the average magnitude of momentumof a particle is p =

p3mkT . The momentum of each particle is thus distributed around a sphere

of this radius in the 3-dimensional momentum space. Roughly speaking, the volume of the denselydistributed region has a volume / p3 = (3mkT )3/2, and the total number of possible momentumdistributions of all particles is / ((3mkT )3/2/�Vp)N .Multiplying the numbers of spatial and momentum distributions, the total number N of micro-statesof these N particles is a constant times V NT 3N/2. Thus,

S ⌘ k log(N ) = kN

✓3

2log(T ) + log(V )

◆+ const. (11.65)

This leads to eq.(11.64).

Q: (11-40) What is �S when a gas of volume V is released in a larger container ofThis is calledadiabatic free

expansion.

volume V 0? Assume that the gas is thermally isolated.

HW: (4-4&5 with the credit for 2 problems) Consider a monoatomic ideal gas in aninitial state with volume V0, pressure P0 and temperature T0. Consider the followingprocesses:(1) The volume is increased to V1 under an isobaric process.(2) Then the pressure is increased to P1 under an isothermal process.(3) The volume is changed back to V0 through an adiabatic process.Questions:(a) How much work is done by the gas in each process?(b) How much heat is transferred to the system in each process?(c) What is the final temperature after the last process?(d) How is the entropy of the system changed in each process?Give all of your answers in terms of P0, V0, P1 and V1. You will learn

much more inour courses on

Thermodynamics

and Statistical

Mechanics.

We skip it here, but the notion of entropy can be extended to other systems andthe definition (11.63) still holds.

We cannot explain it here, but this definition of entropy agrees with the countingof the ignorance about micro-states.

11.8.3 3rd Statement of 2nd Law

In addition to the Kelvin-Planck and the Clausius statements, the 2nd law is alsostated as

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The entropy of an isolated system cannot decrease.

This statement agrees with the Clausius statement, but is more informative.

11.8.4 Other Definitions of Entropy

For a given macro-state with N possible micro-states, the entropy is

S = �kX

i

Pi log(Pi), (11.66)

where Pi is the probability of the i-th micro-state.Q: (11-41) Forfixed N , is the

case thatPi = 1/N for alli the case with

the largestentropy?

Assuming that all micro-states are equally probable,

S = k logN, (11.67)

where N is the number of micro-states.The entropy defined in eq.(11.63) can be explained this way in statistical mechan-

ics.If somehow it is known that only one micro-state is possible, (say, P1 = 1 and

Pi 6=1 = 0),S = 0. (11.68)

Exercise: (11-8) For N coins, what is the macro-state with the largest entropy? (Amacro-state is specified by the number of heads/tails.)

Exercise: (11-9) For N coins, what is the entropy if nothing about the heads or tailsis known?

Q: (11-42) How is the entropy changed when the macro-states are changed by shakinga box of coins?

Exercise: (11-10) There are N particles and N boxes. What is the entropy for themacro-state with all particles in a given box? What is the entropy for the macro-statewith all particles in different boxes?

This kind of notion of entropy is clearly also useful in information sciences.

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