Physics 161: ThermodynamicsLecture #11

Friday 2/14/2020

Kinetic Theory of Gasesand Boltzmann�s Entropy

To this point, we have introduced thermodynamics as a framework forstudying the properties of �uids. Our �uid of choice in these lessons has been theideal gas, primarily for the reason that its equation of state is of a particularlysimple form. While occasionally we have introduced the notion of the atom,the concept that a gas consists of many independently moving particles has notbeen necessary; we have developed the laws thermodynamics without having toappeal to the microscopic details of a �uid.Thermodynamics is a subject of relationships between one quantity and

another. For example, we have the ideal gas equation of state,

PV = nRT; (ideal gas) (1)

and we have the energy,

U =3

2nRT; (monatomic ideal gas) (2)

which is a function of only the temperature. In studying the second law, welearned that these two relations are connected; and moreover must be connectedfor any �uid, to ensure that all reversible heat engines have the same e¢ ciency.The connection can be formulated through the statement that there is an ad-ditional state function S, such that the change dS is related to the change dUand the change dV ,

dU = TdS � PdVThat is as far as we can go, however. What we don�t have is something that givesus 1, or 2, from �rst principles. This is beyond the scope of thermodynamics, afact that was not lost on the founding fathers. Indeed, it was same scientists whoformulated the laws of thermodynamics who were also working on a microscopicformulation of the equation of state at the same time. For gases, this is knownas the "kinetic theory of gases".The �rst thorough (published) description of kinetic theory was by Clausius,

in 1857. In his paper "the Nature of the Motion that we call Heat", he showedhow the ideal gas law could be derived from Newton�s laws of mechanics byassuming that the atoms in a gas exert pressure when they bounce elasticallyo¤ the sides of a container. Here we will examine a simple derivation, involvingthe impulse-momentum theorem you are familiar with from Physics 160,Z

~Fdt = �(m~v)

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1 Ideal Gas Equation of State

This derivation is typically included in your freshman textbook, either as partof the text, or as one of the problems to be worked out at the end of the chapteron impulse-momentum. We repeat it here. Consider one atom of gas movingfreely inside an L � L � L cube, colliding with the faces. The atom doesn�tweigh much, so ignore the force of gravity.

Assume that the collisions with the sides of the container are elastic, so thatafter each bounce the atom comes o¤ with the same speed it had before, butmoving in the opposite direction. To calculate the pressure, we consider theaverage force exerted on one of the faces of the cube. Consider, for example,the face on the right, perpendicular to the x axis, having an area A = L2: Theatom has an incoming velocity ~v1; and an outgoing velocity ~v2; so the impluse-momentum theorem tells us thatZ

~Fdt = m~v2 �m~v1

Only the x component of the atom�s velocity changes, however, so the im-pulse is in the x - direction,Z

Fxdt = mvx;2 �mvx;1 = �2mvx;1

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the negative x direction, in fact. Since energy is conserved, vx;2 = �vx;1; andit follows that the impulse is just twice the incoming momentum. In fact, themagnitude of vx remains the same, so we can drop the initial and �nal subscripts,and write that Z

blipFxdt = �2mvx

Consider now a graph of the force Fx(t) on the atom from collisions with onlythe right-hand face of the cube, as a function of time. The force will be zeroexcept when the atom is in actual contact with the face, and then we expect theforce to rise and fall, looking like a small blip. While we don�t know the actualshape of these individual blips, but we know the area under each one from theimpluse-momentum relation.

We are interested in the time average of the force after many collisions. To�nd the average, we are asking the question, what horizontal line � �Fx can bedrawn on the graph so that the area under the line is the same as the totalarea under all of the blips? If we go out to some very long time t; the shadedarea under the horizontal line will be � �Fx � t: The area under the blips will be�n �

Rblip Fxdt; where n is the number of collisions in the time t.

Setting these two areas to be the same, the average force is de�ned throughthe relation,

�Fx � t = n �Zblip

Fxdt

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SinceRblip Fxdt = �2mvx; we can replace the integral by �2mvx. Dividing by

the number of collisions, we have

�Fx �t

n= �2mvx

Now, t=n = �t is just the time between collisions, which is the same as thetime for the atom to make a round trip, from the right-face to the left-face andback. This is given by the distance divided by the speed in the x direction,�t = 2L=vx: Substituting this for �t; we �nd

�Fx �2L

vx= �2mvx

Rearranging, we see that the time-average force on the atom by the right-facein the x direction is proportional to the square of the velocity in the x direction,

�Fx =�mv2xL

and inversely proportional to the length L: Suppose now, that we consider Natoms in the box, all moving about, colliding with the faces, but not collidingwith each other. For each atom we will get exactly the same contribution tothe force, so we simply scale this result by N; replacing v2x by the average

v2x�

over all atoms.

�Fx = �Nmv2x�

LFinally, the force on the atoms by the face is the negative of the force by theatoms on the face. Dividing the latter by the area L2; we obtain the pressureon the right-face,

P =� �FxL2

= Nmv2x�

L3

Since the volume V = L3; we have an equation relating pressure to volume thatresembles the ideal gas law,

PV = Nmv2x�

(3)

Multiplying and dividing by 2; we see that the product PV is equal to twice thekinetic energy for motion in the x direction,

PV = 2N

�1

2mv2x��

(4)

Our result should now be compared with the equation of state for an ideal gas,

PV = nRT (5)

These two will be the same if the right hand sides are the same. This impliesthat temperature should be associated with kinetic energy,

nRT = 2N

�1

2mv2x��

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Dividing both sides by 2N; we have an expression for the average kinetic energyof a single atom,

1

2mv2x�=1

2

n

NRT

The ratio n=N is just the number of moles divided by the number of particles,which is the reciprocal of Avogadro�s number, A0 = 6:02� 1023:

1

2mv2x�=1

2

�R

A0

�T =

1

2kT (6)

The ratio R=A0 is denoted by the letter k; and is called Boltzmann�s constant;

k = R=A0 =8:314 J/(mole �K)6:02� 1023 moles�1

= 1:38� 10�23 J/K

Equation 6 is an important result from kinetic theory; 12kT is the averagekinetic energy of an atom in thermal equilibrium moving in the x direction. Thetotal kinetic energy of an atom moving in three dimensions is just three timesthis value,

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2mv2x�+1

2mv2y�+1

2mv2z�=3

2kT

We say that a particle in equilibrium has on average and energy of 12kT per

degree of freedom.Expressing the velocity as a vector, ~v = vxx + vy y + vz z; we can combine

the energy for the three directions, writing

1

2mv2�=3

2kT (7)

Since the atoms do not interact with one another, their total energy U is justN times the average kinetic energy per atom. Thus we obtain the result thatthe internal energy for a gas of noninteracting particles is given by,

U =3

2NkT =

3

2nRT: (8)

This implies that the heat capacity at constant volume is indeed a constant,

CV =

�@U

@T

�V

=3

2nR (9)

The agreement with what is found experimentally for a monatomic ideal gas isstriking, and suggests that the association of temperature and kinetic energythat we assumed above is correct, and that our model is correct; an ideal gas isone in which the atoms can be considered to move about independently, withoutinteractions with one another.

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2 Maxwell�s Velocity Distribution

Another important result associated with the kinetic theory of gases was shownby James C. Maxwell, in 1867. It was recognized that not all atoms in a gaswill have the same speed, and that the relation

v2�=3kT

m

was a statement about the average of the square of the velocity, the mean-squarevelocity, for a large number of atoms. Arguing that the distribution of speedsin the x; y, and z direction should obey the same relation, Maxwell pointed outthat only the exponential function could be used, and that the speed distributionin three dimensions is uniquely described by the function,

P (v) = const. � exp��mv

2

2kT

�4�v2: (10)

For small speeds, the distribution rises as v2: For large speeds, it falls o¤ expo-nentially.

To insure normalization, i.e. thatR10P (v)dv = 1; the proportionality con-

stant is given by

const. =� m

2�kT

�3=23 Boltzmann�s Entropy

In 1871, Ludwig Boltzmann showed how the basic ideas of the kinetic theory ofgases could be generalized by postulating an expression for the entropy,

S = k ln (W )

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where k is Boltzmann�s constant, and W is the number of ways that the atomsin a gas can be con�gured while still giving rise to the same macroscopic statevariables, i.e. T; V; P; U; etc. He would later show that the expression appliesto all �uids, in all states: gas, liquid, and solid.What do we mean by the number of ways W that the atoms of a system

can be con�gured? We mean the number of ways that their position and theirvelocity can be speci�ed. When you think of it, position and velocity is all thereis that needs to be speci�ed in mechanics. Once the initial positions and initialvelocities of all the particles are known, one can in principle integrate Newton�sequations of motion to predict anything in the future (or anything in the past,for that matter). It was recognized by Boltzmann that the a system in a giventhermodynamic state, say having a particular volume V and temperature T;need not have its atoms with the same position and velocity. Consider thecartoon below showing two systems, each having just four atoms with di¤erentpositions and velocities at some instant in time.

Now imagine increasing the number of atoms from 4; to approximately 6�1023: At that level, both systems would have robust values of V and T; andwe would declare them to be identical as far as their thermodynamic stateis concerned, yet at a microscopic level they would still be in di¤erent statesbecause of having di¤erent con�gurations of the individual atoms.To calculate W; one has to count over all possible con�gurations of position

and velocity for all N atoms. It is easy to imagine counting over positionfor the atoms in a noninteracting (ideal) gas. One way to do this is to imaginecarving up the volume V into a three-dimensional grid containingM grid-points.Consider just one atom to begin with. How many di¤erent ways can you positionthe atom, locating the atom by specifying its cell in the grid? The answer is M;of course. But how many di¤erent ways can you position two identical atomson the grid? The answer is M2=2 if we do not worry about counting the casewhere both atoms are in the same cell. We must divide by a factor of 2 becausethere is only one way to con�gure two atoms at two di¤erent locations if theatoms are indistinguishable. For three atoms, the number of ways is M3=3!;and for N atoms, the number of ways is MN=N !:

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While we may not know exactly what should be the number of grid points,we can be certain that this number will increase with the size of the system. Itis surely the case that M / V: It follows, therefore, that

W / V N

N !

To count the number of ways to con�gure the velocity involves carving upvelocity-space into a similar grid. Since we are familiar with the associationbetween temperature and kinetic energy, we are not surprised to �nd that thecontribution toW coming from counting velocities is related to T:We will not gothrough the calculation here, but simply quote the result: For a noninteractingmonatomic gas (monatomic ideal gas) with N indistinguishable atoms of massm, the number of con�gurations associated with a volume V and a temperatureT goes as,

W =

�V T 3=2 � const.

�NN !

; (ideal gas)

The constant in this expression,

const. =(2�mk)

3=2

h3;

depends on Planck�s constant h = 6:26 � 10�34 J�s, curiously enough. Thisshould surprise you since Planck didn�t introduce this constant until around1902. Obviously, Boltzmann would not have had the correct value in the 1800�s,so how could the formula have been used? It turns out that, since we arenormally interested in changes in entropy, the constant drops out.Consider, for example, a change in the thermodynamic state of a monatomic

ideal gas, from (T1; V1) to (T2; V2) : For the initial state,

W1 =

�V1T

3=21 � const.

�NN !

;

and for the �nal state,

W21 =

�V2T

3=22 � const.

�NN !

:

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The change in entropy is

�S = S2 � S1= k lnW2 � k lnW1

= k ln

�W2

W1

�

= k ln

0B@�V2T

3=22

�N�V1T

3=21

�N1CA

�S = Nk ln

"�V2V1

��T2T1

�3=2#; (ideal gas)

The constants drop out because the change depends on the di¤erence of twologs, which in turn can be written as the log of the ratio of their arguments.Let us apply Boltzmann�s formula to the free expansion of an ideal gas.Example 1: Isothermal Free* Expansion of a monatomic ideal gas at tem-

perature T from a volume V to a �nal volume 2V: Calculate �S:(*Two bulbsof volume V , one initially with gas and the other initially evacuated. Open thestopcock and let the gas expand from one bulb into the other.)

�S = Nk ln

"�V2V1

��T2T1

�3=2#

= Nk ln

��2V

V

��= Nk ln 2 = nR ln 2

Example 2: Isothermal Reversible* Expansion of a monatomic ideal gasat temperature T from a volume V to a �nal volume 2V:Calculate �S:(*Putgas in a cylinder �tted with a piston. Withdraw the piston slowly until volumedoubles.)

�S = Nk ln

"�V2V1

��T2T1

�3=2#

= Nk ln

��2V

V

��= Nk ln 2 = nR ln 2

Example 3: Adiabatic Free* Expansion of a monatomic ideal gas at tem-perature T from a volume V to a �nal volume 2V:Calculate �S: (*Two bulbsof volume V , one initially with gas and the other initially evacuated. Open thestopcock and let the gas expand from one bulb into the other. The bulbs areinsulated from surroundings.)First of all, does the temperature change? From the �rst law, we have

�U = Q�W:We know that W = 0 because the two bulbs taken together forma rigid container for which the volume is constant. We know that Q = 0 because

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the system is insulated from the surroundings. This implies that �U = 0. Foran ideal gas, U is a function of only T: If �U = 0 for an ideal gas, then �Tmust also be zero. The rest of the solution follows the same pattern:

�S = Nk ln

"�V2V1

��T2T1

�3=2#

= Nk ln

��2V

V

��= Nk ln 2 = nR ln 2

Example 4: Adiabatic Reversible* Expansion of a monatomic ideal gas attemperature T from a volume V to a �nal volume 2V:Calculate �S:(*Put gasin a cylinder �tted with a piston. Withdraw the piston slowly until volumedoubles. The cylinder and piston are insulated.)Does the temperature change for this case? We know that P2V

2 = P1V

1

for an adiabatic reversible expansion of an ideal gas, where = Cp=CV = 5=3for a monatomic gas. Substituting nRT=V for P; we have

nRT2V2

V 2 =nRT1V1

V 1

The factors of nR on both sides cancel out. Collecting the T and V �s separately,we have

T2T1=

�V1V2

� �1Now use Boltzmann�s formula for a monatomic ideal gas:

�S = Nk ln

"�V2V1

��T2T1

�3=2#

= Nk ln

"�V2V1

��V1V2

�( �1)�3=2#

= Nk ln

"�V2V1

��V1V2

�( 53�1)�( 32 )#

= Nk ln

"�V2V1

��V1V2

�( 23 )�( 32 )#= Nk ln(1) = 0

This had better be zero, for the second law tells us that �S � 0 for an adiabaticprocess, with the equality taken when the process is reversible. (We usually statethis result as applying to the entire universe, but you will recall that this is thesame statement if you consider your system to be the entire universe.)

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4 W Measures Disorganization

For the ideal gas, we observe that W increases when the volume increases. Thisis because there are more ways to con�gure the atoms when the volume is larger.Similarly, W increases when the temperature increases because there are moreways to con�gure the velocities of the atoms. WhatW measures is the degree ofdisorganization in the system. The second part of the second law, the statementthat the entropy of a closed system (the universe) always increases, is thus astatement that the natural evolution of the universe is to a state of increasingdisorganization. As Maxwell noticed, this is a probabilistic statement. It is notthat nature can�t spontaneously decrease the level of disorganization, but ratherthat, once the disorganization has increased, it is exceedingly unlikely that inwill reorganize on its own because of the number of particles involved. The big-ger the number N ; the more unlikely this will be. Here is a useful analogy: Theuniverse perched precariously in some metastable state, "waiting" for somethingto happen, is analogous to Humpty-Dumpty sitting on the wall. The sponta-neous processes that occur in the universe are analogous to HD�s fall from thewall. HD smashing into pieces represents the increase in disorganization thatprevents a return to the initial state.

4.1 A Sad Ending for Boltzmann

The concept that macroscopic objects are made up of many atoms, or parts ofatoms, is common place, and to us Boltzmann�s (and Maxwell�s) view of the sta-tistical nature of the law of increasing entropy (of the universe) seems natural.By the late 1890�s, however, there was a movement afoot in the scienti�c com-munity that held that good science should be developed from the "top down",in such a manner that it does not rely on the concept of atoms or any other mi-crosopic entities.The thermodynamics purists, in particular, argued that theycould calculate anything of relevance for �uids from the �rst and second lawand empirical equations of state, without reference to microscopic interpreta-tions (such as provided by the kinetic theory of gases.) Rather than makingBoltzmann their champion, they shunned him. At one signi�cant internationalconference, the organizers refused to put him in the physical science sessions,

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and instead forced him to speak in the pure math sessions. Imagine that! Thissaddened Boltzmann in what should have been his triumphant moment, and hekilled himself in despair.

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