Finally, we produce an expression for the entropy of an ideal classical solid in terms of its macrostate internal energy E and particle number N . According to the occupation number method, S c N
( )
kk(
EEEE N Z)
where β= 1 / kT , and, according to our model of the classical, ideal solid E = 3 NkT andZ1 1
(
HvHHHH)
3 . Together these yieldS E E
NH o l
c N Nk
, N n .
( ) ( )
⎡ ⎛⎝⎛⎛⎝⎝ ⎞⎠⎞⎞⎠⎠⎣⎢
⎡⎡
⎣⎣⎢⎢
⎣⎣⎣⎣
⎤
⎦⎥
⎤⎤
⎦⎦⎥⎥
Nk ⎦⎦⎦⎦
N N Nk N
N 3 Hν (3.56)
Because this entropy does not depend upon the volume V , the pressure of an classical ideal solid vanishes – an unrealistic consequence of this overly simple model.
Example 3.2 Average kinetic energy
Problem: What is the average kinetic energy of a particle of classical, mon-atomic, ideal gas? And what is the root mean squared speed of a particle of this gas?
Solution : Since all the energy of a classical, monatomic, ideal gas is con-tained in the kinetic energy of its particles, the answers to these questions fol-low almost immediately from the monatomic, ideal classical gas equation of state E =3 NkT /2. The average kinetic energy per particle, E / N =3 kT /2, is a func-tion of the thermodynamic temperature T alone. This proporfunc-tionality between the average kinetic energy and the thermodynamic temperature obtains for sys-tems that observe the equipartition theorem.
If m is the mass of each gas particle, then the root mean squared velocity of the particles v2 is determined by
3.7 Boltzmann’s tomb 71
m v2 kT
2 3
= 2 .
Therefore,
v kT
m
2 3
= .
The hotter the gas, the faster its particles; the more massive the particle, the more slowly it moves.
3.7 Boltzmann’s tomb
It is perhaps surprising that the classical way of doing statistical mechanics has so little to do with mechanics per se , that is, with the science of motion.
Nowhere do the methods of classical statistical mechanics require F = ma . Nowhere do they require solutions to equations of motion. Particles move, but move probabilistically according to certain rules: the fundamental postulate, conservation of energy, and conservation of particle number. That Boltzmann fashioned these ideas into a statistical method is, in retrospect, prescient. For, as we shall see, Boltzmann’s approach to entropy is almost perfectly congruent with the quantum physics that emerged after his death.
Recall that the relation of relative entropy to macrostate multiplicity, S cc+klnΩ , follows from assuming that: (1) the entropy of an isolated system is a function S
( )
Ω of macrostate multiplicity Ω ; (2) the entropies of isolated subsystems are additive; and (3) their macrostate multiplicities are sta-tistically independent. The most direct way of calculating the multiplicity Ω is, as in Sections 3.1 and 3.2 , to discretize the phase space available to a system of given energy E , volume V , and number N of distinguishable particles and count the number of cells.In practice Boltzmann preferred the relatively indirect, yet powerful, occu-pation number method of determining the entropy of a system composed of N independent, distinguishable particles. According to this method we (1) dis-tribute the particles in a one-particle, 6-dimensional phase space that is divided into J cells; (2) dei ne a macrostate in which the N particles of the system are partitioned into the numbers of particles, n 1 , n 2 , …, n J , that occupy its J cells;
(3) form the multiplicity Ω of this macrostate N !/( n 1 ! n 2 ! … n J !); and (4) deter-mine the macrostate that maximizes its entropy S
[
= +]
subject to theconstraints N = n 1 + n 2 + λ + n J and E nn1 11++n2 22ε + +nJεJ . Both the direct and the occupation number methods produce a relative entropy rather than an absolute entropy.
What then is one to make of the formula for absolute entropy, S = k log W , found on Boltzmann’s tomb in the Vienna Zentralfriedhof (Central Cemetery)?
Evidently, those who dedicated Boltzmann’s tomb in 1933 believed that S = k log W encapsulated Boltzmann’s contribution to physics. But this equa-tion appears nowhere in Boltzmann’s writings, originates with Max Planck, and incorporates physics of which Boltzmann could not have been aware. The last of these three statements may be controversial and, consequently, needs justii cation.
But i rst, what does S = k log W mean? The convention that S stands for entropy had been established by Clausius in 1865 and k was Planck’s choice for what we now call Boltzmann’s constant (probably after the German word Konstant ). But the letter W , taken from the German word Wahrscheinlichkeit for probability , is problematic. Certainly the word probability misleads. For probabilities are usually normalized to lie between 0 and 1 inclusive. And if 0 W 1 , the equation S = k log W leads to negative entropy – contrary to Clausius’s convention and our expectation. To be sure, Planck envisioned other ways of normalizing probabilities and called W not the probability but rather the thermodynamic probability . Others have referred to W as the thermody-namic or statistical weight . In actual practice, Boltzmann always used W as a part of the occupation number method and always replaced W by the multino-mial coefi cient N !/( n 1 ! n 2 !. λ n J !). Thus W appears to be equivalent to what we call the multiplicit y Ω of an occupation number macrostate of a system of N distinguishable particles.
Given these identii cations, Boltzmann’s epitaph , S = k log W , is equivalent to S k lnΩ . And so we return to the question: What is the physics required to turn the equation describing the entropy of a classical system, S cc+klnΩ , into the equation inscribed on Boltzmann’s tomb? As we shall see in Section 4.5 , this missing physics is the third law of thermodynamics – a law that allows one to adopt a conventional value for the entropy in the T→ 0 limit.
But Walther Nernst (1884–1941) i rst articulated a version of the third law in 1906 – too late for Boltzmann to exploit before his death in September of that year. And neither the ideal gas nor the ideal solid models that Boltzmann stud-ied are third law compliant. Evidently the equation S = k log W summarizes not only Boltzmann’s direct contribution to physics but also his service in laying a foundation upon which a new quantum statistical physics, then only emerging at the time of his death, could be built.
Problems 73
Problems
3.1 Room temperature density
Determine the number of particles in 1 cm 3 of ideal gas when T = 320 K and the pressure is 1 atmosphere. (Consult Appendix 1: Physical constants and standard dei nitions.)
3.2 Van der Waals equations of state
The goal of much of statistical mechanics is to produce meaningful partition functions. But suppose someone gives you the ready-made single-particle partition function
Z V Nb
m e aN V
1
2 3 2
−
((
V)
⎛⎝⎜⎛⎛⎝⎝ πβ⎞⎠⎟⎞⎞⎠⎠ β ,where a and b are constants that characterize a gas system and β= 1 / kT . Find (a) the corresponding energy equation of state and (b) the entropy function S ( E , V , N ). (Note that β is a function of the extensive variables E , V , and N .) From the entropy function S E V N
(
, ,V)
i nd (c) the energy equation of state again and (d) the pressure equation of state.3.3 Ideal gas of diatomic molecules
The single-particle partition function of an ideal gas of diatomic molecules is
Z V
1 m 2 5 2
⎛
⎝⎜
⎛⎛
⎝⎝
⎞
⎠⎟
⎞⎞
⎠⎠
π
β ,
where β= 1 / kT . Find (a) the corresponding energy equation of state and the pressure equations of state. (b) Use the equipartition theorem to make a mean-ingful statement about the number of quadratic terms in the energy of the diatomic molecules in this gas.
3.4 Mixing of ideal gases
Two containers of ideal gas, one of N 2 and another of O 2 molecules, have the same volume V , pressure P , and temperature T . A valve connecting the two containers is opened and the resulting mixture comes to equilibrium. Assume
the container walls are adiabatic. What is the resulting increase ΔS in the entropy of this two-gas system? Use Eq. ( 3.49 ).
3.5 Extensivity of ideal solid
Impose the extensivity condition S
(
λEE,λλNNN)
λS E NSS E(
,)
on the entropy of an ideal classical solid given in ( 3.56 ). (Note: In much the same way that there is a family of extensive entropies of an ideal classical gas, there is a family of extensive entropies of the ideal classical solid.)3.6 Room temperature speed of N
2molecules
Use the data given in Appendix I , Physical Constants and Standard Dei nitions, to determine the root mean squared speed in meters per second of an N 2 mol-ecule at room temperature ( T = 300 K). The mass of a nitrogen molmol-ecule is 28.0 u.