changeset 3:8f3b6dcb9add

Transcribed up to section 1.9, Entropy of an Ideal Boltzmann Gas
author Dylan Holmes <ocsenave@gmail.com>
date Sun, 29 Apr 2012 02:38:22 -0500
parents afbe1fe19b36
children 299a098a30da
files org/stat-mech.org
diffstat 1 files changed, 946 insertions(+), 12 deletions(-) [+]
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     1.1 --- a/org/stat-mech.org	Sat Apr 28 23:06:48 2012 -0500
     1.2 +++ b/org/stat-mech.org	Sun Apr 29 02:38:22 2012 -0500
     1.3 @@ -91,7 +91,7 @@
     1.4  of our present review of the logic underlying classical 
     1.5  thermodynamics.
     1.6  
     1.7 -** The Primitive Thermometer. 
     1.8 +** The Primitive Thermometer
     1.9  
    1.10  The earliest stages of our
    1.11  story are necessarily speculative, since they took place long
    1.12 @@ -145,7 +145,7 @@
    1.13  present.
    1.14  
    1.15  
    1.16 -** Thermodynamic Systems.
    1.17 +** Thermodynamic Systems
    1.18  
    1.19  The \ldquo{}thermodynamic systems\rdquo{} which
    1.20  are the objects of our study may be, physically, almost any
    1.21 @@ -226,7 +226,7 @@
    1.22  of thermodynamics . The most fundamental one is a qualitative
    1.23  rather than quantitative relation, the \ldquo{}zero'th law.\rdquo{}
    1.24  
    1.25 -** Equilibrium; the \ldquo{}Zero‘th Law.\rdquo{}
    1.26 +** Equilibrium; the Zeroth Law
    1.27  
    1.28   It is a common experience
    1.29  that when objects are placed in contact with each other but
    1.30 @@ -390,7 +390,7 @@
    1.31  by specifying any two of the variables arbitrarily, whereupon the
    1.32  third, and all others we may introduce, are determined. 
    1.33  Mathematically, this is expressed by the existence of a functional
    1.34 -relationship of the form[fn:: /Edit./: The set of solutions to an equation
    1.35 +relationship of the form[fn:: The set of solutions to an equation
    1.36  like /f(X,x,t)=/ const. is called a /level set/. Here, Jaynes is
    1.37  saying that the quantities /X/, /x/, and /t/ follow a \ldquo{}functional
    1.38  rule\rdquo{}, so the set of physically allowed combinations of /X/,
    1.39 @@ -753,7 +753,7 @@
    1.40  \end{equation}
    1.41  is always satisfied. This sort of process is an old story in
    1.42  scientific investigations; although the great theoretician Boltzmann
    1.43 -is said to have remarked: \ldquo{}Elegance is for tailors \rdquo{}, it
    1.44 +is said to have remarked: \ldquo{}Elegance is for tailors\rdquo{}, it
    1.45  remains true that the attempt to reduce equations to the most
    1.46  symmetrical form has often suggested important generalizations of
    1.47  physical laws, and is a great aid to memory. Witness Maxwell's
    1.48 @@ -767,18 +767,18 @@
    1.49  interactions of the type just studied.
    1.50  
    1.51  The similarity of (1-12) to conservation laws in general may be seen
    1.52 -as follows. Let $A$ be some quantity that is conserved; the $i$th
    1.53 +as follows. Let $A$ be some quantity that is conserved; the \(i\)th
    1.54  system has an amount of it $A_i$. Now when the systems interact such
    1.55  that some $A$ is transferred between them, the amount of $A$ in the
    1.56 -$i$th system is changed by a net amount \(\Delta A_i = (A_i)_{final} -
    1.57 +\(i\)th system is changed by a net amount \(\Delta A_i = (A_i)_{final} -
    1.58  (A_i)_{initial}\); and the fact that there is no net change in the
    1.59  total amount of $A$ is expressed by the equation \(\sum_i \Delta
    1.60 -A_i = 0$. Thus, the law of conservation of matter in a chemical
    1.61 +A_i = 0\). Thus, the law of conservation of matter in a chemical
    1.62  reaction is expressed by \(\sum_i \Delta M_i = 0\), where $M_i$ is the
    1.63 -mass of the $i$th chemical component.
    1.64 +mass of the \(i\)th chemical component.
    1.65  
    1.66 -what is this new conserved quantity? Mathematically, it can be defined
    1.67 -as $Q_i = K_i\cdot M_i cdot t_i; whereupon (1-12) becomes 
    1.68 +What is this new conserved quantity? Mathematically, it can be defined
    1.69 +as $Q_i = K_i\cdot M_i \cdot t_i$; whereupon (1-12) becomes 
    1.70  
    1.71  \begin{equation}
    1.72  \sum_i \Delta Q_i = 0
    1.73 @@ -883,7 +883,7 @@
    1.74  
    1.75  # what is "the specific heat of a gas at constant pressure/volume"?
    1.76  # changed t for temperature below from capital T to lowercase t.
    1.77 -Another failure of the conservation law (1-13) was noted in 1842 by
    1.78 +Another failure of the conservation law (1-13) was [[http://web.lemoyne.edu/~giunta/mayer.html][noted in 1842]] by
    1.79  R. Mayer, a German physician, who pointed out that the data already
    1.80  available showed that the specific heat of a gas at constant pressure, 
    1.81  C_p, was greater than at constant volume $C_v$. He surmised that the
    1.82 @@ -905,6 +905,940 @@
    1.83  conservation law broke down.
    1.84  
    1.85  ** The First Law
    1.86 +Corresponding to the partially valid law of \ldquo{}conservation of
    1.87 +heat\rdquo{}, there had long been known another partially valid
    1.88 +conservation law in mechanics. The principle of conservation of
    1.89 +mechanical energy had been given by Leibnitz in 1693 in noting that,
    1.90 +according to the laws of Newtonian mechanics, one could define
    1.91 +potential and kinetic energy so that in mechanical processes they were
    1.92 +interconverted into each other, the total energy remaining
    1.93 +constant. But this too was not universally valid---the mechanical
    1.94 +energy was conserved only in the absence of frictional forces. In
    1.95 +processes involving friction, the mechanical energy seemed to
    1.96 +disappear.
    1.97 +
    1.98 +So we had a law of conservation of heat, which broke down whenever
    1.99 +mechanical work was done; and a law of conservation of mechanical
   1.100 +energy, which broke down when frictional forces were present. If, as
   1.101 +Mayer had suggested, heat was itself a form of energy, then one had
   1.102 +the possibility of accounting for both of these failures in a new law
   1.103 +of conservation of /total/ (mechanical + heat) energy. On one hand,
   1.104 +the difference $C_p-C_v$ of heat capacities of gases would be
   1.105 +accounted for by the mechanical work done in expansion; on the other
   1.106 +hand, the disappearance of mechanical energy would be accounted for by
   1.107 +the heat produced by friction.
   1.108 +
   1.109 +But to establish this requires more than just suggesting the idea and
   1.110 +illustrating its application in one or two cases --- if this is really
   1.111 +a new conservation law adequate to replace the two old ones, it must
   1.112 +be shown to be valid for /all/ substances and /all/ kinds of
   1.113 +interaction. For example, if one calorie of heat corresponded to $E$
   1.114 +ergs of mechanical energy in the gas experiments, but to a different
   1.115 +amoun $E^\prime$ in heat produced by friction, then there would be no
   1.116 +universal conservation law. This \ldquo{}first law\rdquo{} of
   1.117 +thermodynamics must therefore take the form:
   1.118 +#+begin_quote
   1.119 +There exists a /universal/ mechanical equivalent of heat, so that the
   1.120 +total (mechanical energy) + (heat energy) remeains constant in all
   1.121 +physical processes.
   1.122 +#+end_quote
   1.123 +
   1.124 +It was James Prescott Joule who provided the [[http://www.chemteam.info/Chem-History/Joule-Heat-1845.html][first experimental data]]
   1.125 +indicating this universality, and providing the first accurate
   1.126 +numerical value of this mechanical equivalent. The calorie had been
   1.127 +defined as the amount of heat required to raise the temperature of one
   1.128 +gram of water by one degree Centigrade (more precisely, to raise it
   1.129 +from 14.5 to 15.5$^\circ C$). Joule measured the heating of a number
   1.130 +of different liquids due to mechanical stirring and electrical
   1.131 +heating, and established that, within the experimental accuracy (about
   1.132 +one percent) a /calorie/ of heat always corresponded to the same
   1.133 +amount of energy. Modern measurements give this numerical value as: 1
   1.134 +calorie = 4.184 \times 10^7 ergs = 4.184 joules.
   1.135 +# capitalize Joules? I think the convention is to spell them out in lowercase.
   1.136 +
   1.137 +The circumstances of this important work are worth noting. Joule was
   1.138 +in frail health as a child, and was educated by private tutors,
   1.139 +including the chemist, John Dalton, who had formulated the atomic
   1.140 +hypothesis in the early nineteenth century. In 1839, when Joule was
   1.141 +nineteen, his father (a wealthy brewer) built a private laboratory for
   1.142 +him in Manchester, England; and the good use he made of it is shown by
   1.143 +the fact that, within a few months of the opening of this laboratory
   1.144 +(1840), he had completed his first important piece of work, at the
   1.145 +age of twenty. This was his establishment of the law of \ldquo{}Joule
   1.146 +heating,\rdquo{} $P=I^2 R$, due to the electric current in a
   1.147 +resistor. He then used this effect to determine the universality and
   1.148 +numerical value of the mechanical equivalent of heat, reported
   1.149 +in 1843. His mechanical stirring experiments reported in 1849 yielded
   1.150 +the value 1 calorie = 4.154 \times 10^7 ergs, amount 0.7% too low;
   1.151 +this determination was not improved upon for several decades.
   1.152 +
   1.153 +The first law of thermodynamics may then be stated mathematically as
   1.154 +follows:  
   1.155 +
   1.156 +#+begin_quote
   1.157 +There exists a state function (i.e., a definite function of the
   1.158 +thermodynamic state) $U$, representing the total energy of any system,
   1.159 +such that in any process in which we change from one equilibrium to
   1.160 +another, the net change in $U$ is given by the difference of the heat
   1.161 +$Q$ supplied to the system, and the mechanical work $W$ done by the
   1.162 +system.
   1.163 +#+end_quote
   1.164 +On an infinitesimal change of state, this becomes
   1.165 +
   1.166 +\begin{equation}
   1.167 +dU = dQ - dW.
   1.168 +\end{equation}
   1.169 +
   1.170 +For a system of two degrees of freedom, defined by pressure $P$,
   1.171 +volume $V$, and temperature $t$, we have $dW = PdV$. Then if we regard
   1.172 +$U$ as a function $U(V,t)$ of volume and temperature, the fact that
   1.173 +$U$ is a state function means that $dU$ must be an exact differential;
   1.174 +i.e., the integral
   1.175 +
   1.176 +\begin{equation}
   1.177 +\int_1^2 dU = U(V_2,t_2) - U(V_1,t_1)
   1.178 +\end{equation}
   1.179 +between any two thermodynamic states must be independent of the
   1.180 +path. Equivalently, the integral $\oint dU$ over any closed cyclic
   1.181 +path  (for example, integrate from state 1 to state 2 along path A,
   1.182 +then back to state 1 by a different path B) must be zero. From (1-15),
   1.183 +this gives for any cyclic integral,
   1.184 +
   1.185 +\begin{equation}
   1.186 +\oint dQ = \oint P dV
   1.187 +\end{equation}
   1.188 +
   1.189 +another form of the first law, which states that in any process in
   1.190 +which the system ends in the same thermodynamic state as the initial
   1.191 +one, the total heat absorbed by the system must be equal to the total
   1.192 +work done.
   1.193 +
   1.194 +Although the equations (1-15)-(1-17) are rather trivial
   1.195 +mathematically, it is important to avoid later conclusions that we
   1.196 +understand their exact meaning. In the first place, we have to
   1.197 +understand that we are now measuring heat energy and mechanical energy
   1.198 +in the same units; i.e. if we measured $Q$ in calories and $W$ in
   1.199 +ergs, then (1-15)  would of course not be correct. It does
   1.200 +not matter whether we apply Joule's mechanical equivalent of heat
   1.201 +to express $Q$ in ergs, or whether we apply it in the opposite way
   1.202 +to express $U$ and $W$ in calories; each procedure will be useful in
   1.203 +various problems. We can develop the general equations of
   1.204 +thermodynamics 
   1.205 +without committing ourselves to any particular units,
   1.206 +but of course all terms in a given equation must be expressed
   1.207 +in the same units.
   1.208 +
   1.209 +Secondly, we have already stressed that the theory being
   1.210 +developed must, strictly speaking, be a theory only of 
   1.211 +equilibrium states, since otherwise we have no operational definition
   1.212 +of temperature . When we integrate over any \ldquo{}path\rdquo{} in the $(V-t)$
   1.213 +plane, therefore, it must be understood that the path of 
   1.214 +integration is, strictly speaking, just a /locus of equilibrium
   1.215 +states/; nonequilibrium states cannot be represented by points
   1.216 +in the $(V-t)$ plane.
   1.217 +
   1.218 +But then, what is the relation between path of equilibrium
   1.219 +states appearing in our equations, and the sequence of conditions
   1.220 +produced experimentally when we change the state of a system in
   1.221 +the laboratory? With any change of state (heating, compression,
   1.222 +etc.) proceeding at a finite rate we do not have equilibrium in
   1.223 +termediate states; and so there is no corresponding \ldquo{}path\rdquo{} in
   1.224 +the $(V-t)$ plane ; only the initial and final equilibrium states
   1.225 +correspond to definite points. But if we carry out the change
   1.226 +of state more and more slowly, the physical states produced are
   1.227 +nearer and nearer to equilibrium state. Therefore, we interpret
   1.228 +a path of integration in the $(V-t)$ plane, not as representing
   1.229 +the intermediate states of any real experiment carried out at
   1.230 +a finite rate, but as the /limit/ of this sequence of states, in
   1.231 +the limit where the change of state takes place arbitrarily
   1.232 +slowly.
   1.233 +
   1.234 +An arbitrarily slow process, so that we remain arbitrarily
   1.235 +near to equilibrium at all times, has another important property.
   1.236 +If heat is flowing at an arbitrarily small rate, the temperature
   1.237 +difference producing it must be arbitrarily small, and therefore
   1.238 +an arbitrarily small temperature change would be able to reverse
   1.239 +the direction of heat flow. If the Volume is changing very
   1.240 +slowly, the pressure difference responsible for it must be very
   1.241 +small; so a small change in pressure would be able to reverse
   1.242 +the direction of motion. In other words, a process carried out
   1.243 +arbitrarily slowly is /reversible/; if a system is arbitrarily
   1.244 +close to equilibrium, then an arbitrarily small change in its
   1.245 +environment can reverse the direction of the process.
   1.246 +Recognizing this, we can then say that the paths of integra
   1.247 +tion in our equations are to be interpreted physically as
   1.248 +/reversible paths/ . In practice, some systems (such as gases)
   1.249 +come to equilibrium so rapidly that rather fast changes of
   1.250 +state (on the time scale of our own perceptions) may be quite
   1.251 +good approximations to reversible changes; thus the change of
   1.252 +state of water vapor in a steam engine may be considered 
   1.253 +reversible to a useful engineering approximation.
   1.254 +
   1.255 +
   1.256 +** Intensive and Extensive Parameters
   1.257 +
   1.258 +The literature of thermodynamics has long recognized a distinction between two
   1.259 +kinds of quantities that may be used to define the thermodynamic
   1.260 +state. If we imagine a given system as composed of smaller 
   1.261 +subsystems, we usually find that some of the thermodynamic variables
   1.262 +have the same values in each subsystem, while others are additive,
   1.263 +the total amount being the sum of the values of each subsystem.
   1.264 +These are called /intensive/ and /extensive/ variables, respectively.
   1.265 +According to this definition, evidently, the mass of a system is
   1.266 +always an extensive quantity, and at equilibrium the temperature
   1.267 +is an intensive ‘quantity. Likewise, the energy will be extensive
   1.268 +provided that the interaction energy between the subsystems can
   1.269 +be neglected.
   1.270 +
   1.271 +It is important to note, however, that in general the terms
   1.272 +\ldquo{}intensive\rdquo{} and \ldquo{}extensive\rdquo{} 
   1.273 +so defined cannot be regarded as
   1.274 +establishing a real physical distinction between the variables.
   1.275 +This distinction is, like the notion of number of degrees of
   1.276 +freedom, in part an anthropomorphic one, because it may depend
   1.277 +on the particular kind of subdivision we choose to imagine. For
   1.278 +example, a volume of air may be imagined to consist of a number
   1.279 +of smaller contiguous volume elements. With this subdivision,
   1.280 +the pressure is the same in all subsystems, and is therefore in
   1.281 +tensive; while the volume is additive and therefore extensive .
   1.282 +But we may equally well regard the volume of air as composed of
   1.283 +its constituent nitrogen and oxygen subsystems (or we could re
   1.284 +gard pure hydrogen as composed of two subsystems, in which the
   1.285 +molecules have odd and even rotational quantum numbers 
   1.286 +respectively, etc.) . With this kind of subdivision the volume is the
   1.287 +same in all subsystems, while the pressure is the sum of the
   1.288 +partial pressures of its constituents; and it appears that the
   1.289 +roles of \ldquo{}intensive\rdquo{} and \ldquo{}extensive\rdquo{}
   1.290 + have been interchanged. Note that this ambiguity cannot be removed by requiring
   1.291 +that we consider only spatial subdivisions, such that each sub
   1.292 +system has the same local composi tion . For, consider a s tressed
   1.293 +elastic solid, such as a stretched rubber band. If we imagine
   1.294 +the rubber band as divided, conceptually, into small subsystems
   1.295 +by passing planes through it normal to its axis, then the tension
   1.296 +is the same in all subsystems, while the elongation is additive.
   1.297 +But if the dividing planes are parallel to the axis, the elonga
   1.298 +tion is the same in all subsystems, while the tension is 
   1.299 +additive; once again, the roles of \ldquo{}extensive\rdquo{} and
   1.300 +\ldquo{}intensive\rdquo{} are
   1.301 +interchanged merely by imagining a different kind of subdivision.
   1.302 +In spite of the fundamental ambiguity of the usual definitions, 
   1.303 +the notions of extensive and intensive variables are useful, 
   1.304 +and in practice we seem to have no difficulty in deciding
   1.305 +which quantities should be considered intensive. Perhaps the
   1.306 +distinction is better characterized, not by considering 
   1.307 +subdivisions at all, but by adopting a different definition, in which
   1.308 +we recognize that some quantities have the nature of a \ldquo{}force\rdquo{}
   1.309 +or \ldquo{}potential\rdquo{}, or some other local physical property, and are
   1.310 +therefore called intensive, while others have the nature of a
   1.311 +\ldquo{}displacement\rdquo{} or a \ldquo{}quantity\rdquo{} of 
   1.312 +something (i.e. are proportional to the size of the system), 
   1.313 +and are therefore called extensive. Admittedly, this definition is somewhat vague, in a
   1.314 +way that can also lead to ambiguities ; in any event, let us agree
   1.315 +to class pressure, stress tensor, mass density, energy density,
   1.316 +particle density, temperature, chemical potential, angular
   1.317 +velocity, as intensive, while volume, mass, energy, particle
   1.318 +numbers, strain, entropy, angular momentum, will be considered
   1.319 +extensive.
   1.320 +
   1.321 +** The Kelvin Temperature Scale
   1.322 +The form of the first law,
   1.323 +$dU = dQ - dW$, expresses the net energy increment of a system as
   1.324 +the heat energy supplied to it, minus the work done by it. In
   1.325 +the simplest systems of two degrees of freedom, defined by 
   1.326 +pressure and volume as the thermodynamic variables, the work done
   1.327 +in an infinitesimal reversible change of state can be separated
   1.328 +into a product $dW = PdV$ of an intensive and an extensive quantity.
   1.329 +Furthermore, we know that the pressure $P$ is not only the 
   1.330 +intensive factor of the work; it is also the \ldquo{}potential\rdquo{} 
   1.331 +which governs mechanical equilibrium (in this case, equilibrium with respect
   1.332 +to exchange of volume) between two systems; i .e., if they are
   1.333 +separated by a flexible but impermeable membrane, the two systems
   1.334 +will exchange volume $dV_1 = -dV_2$ in a direction determined by the
   1.335 +pressure difference, until the pressures are equalized. The
   1.336 +energy exchanged in this way between the systems is a product
   1.337 +of the form
   1.338 +#+begin_quote
   1.339 +(/intensity/ of something) \times (/quantity/ of something exchanged)
   1.340 +#+end_quote
   1.341 +
   1.342 +Now if heat is merely a particular form of energy that can
   1.343 +also be exchanged between systems, the question arises whether
   1.344 +the quantity of heat energy $dQ$ exchanged in an infinitesimal 
   1.345 +reversible change of state can also be written as a product of one
   1.346 +factor which measures the \ldquo{}intensity\rdquo{} of the heat, 
   1.347 +times another that represents the \ldquo{}quantity\rdquo{}
   1.348 + of something exchanged between
   1.349 +the systems, such that the intensity factor governs the 
   1.350 +conditions of thermal equilibrium and the direction of heat exchange,
   1.351 +in the same way that pressure does for volume exchange.
   1.352 +
   1.353 +
   1.354 +But we already know that the /temperature/ is the quantity
   1.355 +that governs the heat flow (i.e., heat flows from the hotter to
   1.356 +the cooler body until the temperatures are equalized) . So the
   1.357 +intensive factor in $dQ$ must be essentially the temperature. But
   1.358 +our temperature scale is at present still arbitrary, and we can
   1.359 +hardly expect that such a factorization will be possible for all
   1.360 +calibrations of our thermometers.
   1.361 +
   1.362 +The same thing is evidently true of pressure; if instead of
   1.363 +the pressure $P$ as ordinarily defined, we worked with any mono
   1.364 +tonic increasing function $P_1 = P_1 (P)$ we would find that $P_1$ is
   1.365 +just as good as $P$ for determining the direction of volume 
   1.366 +exchange and the condition of mechanical equilibrium; but the work
   1.367 +done would not be given by $PdV$; in general, it could not even
   1.368 +be expressed in the form $P_1 \cdot dF(V)$, where $F(V)$ is some function
   1.369 +of V.
   1.370 +
   1.371 +
   1.372 +Therefore we ask: out of all the monotonic functions $t_1(t)$ 
   1.373 +corresponding to different empirical temperature scales, is
   1.374 +there one (which we denote as $T(t)$) which forms a \ldquo{}natural\rdquo{}
   1.375 +intensity factor for heat, such that in a reversible change
   1.376 +$dQ = TdS$, where $S(U,V)$ is a new function of the thermodynamic
   1.377 +state? If so, then the temperature scale $T$ will have a great
   1.378 +theoretical advantage, in that the laws of thermodynamics will
   1.379 +take an especially simple form in terms of this particular scale,
   1.380 +and the new quantity $S$, which we call the /entropy/, will be a
   1.381 +kind of \ldquo{}volume\rdquo{} factor for heat.
   1.382 +
   1.383 +We recall that $dQ = dU + PdV$ is not an exact differential;
   1.384 +i.e., on a change from one equilibrium state to another the
   1.385 +integral
   1.386 +
   1.387 +\[\int_1^2 dQ\]
   1.388 +
   1.389 +cannot be set equal to the difference $Q_2 - Q_1$ of values of any
   1.390 +state function $Q(U,V)$, since the integral has different values
   1.391 +for different paths connecting the same initial and final states.
   1.392 +Thus there is no \ldquo{}heat function\rdquo{} $Q(U,V)$, and the notion of
   1.393 +\ldquo{}amount of heat\rdquo{} $Q$ stored in a body has no meaning 
   1.394 +(nor does the \ldquo{}amount of work\rdquo{} $W$; 
   1.395 +only the total energy is a well-defined quantity). 
   1.396 +But we want the entropy $S(U,V)$ to be a definite quantity,
   1.397 +like the energy or volume, and so $dS$ must be an exact differential.
   1.398 +On an infinitesimal reversible change from one equilibrium state
   1.399 +to another, the first law requires that it satisfy[fn:: The first
   1.400 +equality comes from our requirement that $dQ = T\,dS$. The second
   1.401 +equality comes from the fact that $dU = dQ - dW$ (the first law) and
   1.402 +that $dW = PdV$ in the case where the state has two degrees of
   1.403 +freedom, pressure and volume.]
   1.404 +
   1.405 +\begin{equation}
   1.406 +dS(U,V) = \frac{dQ}{T} = \frac{dU}{T} + \frac{P}{T}dV
   1.407 +\end{equation}
   1.408 +
   1.409 +Thus $(1/T)$ must be an /integrating factor/ which converts $dQ$ into
   1.410 +an exact differential [[fn::A differential $M(x,y)dx +
   1.411 +N(x,y)dy$ is called /exact/ if there is a scalar function
   1.412 +$\Phi(x,y)$ such that $M = \frac{\partial \Phi}{\partial x}$ and
   1.413 +$N=\frac{\partial \Phi}{\partial y}$. If there is, \Phi is called the
   1.414 +/potential function/ of the differential, Conceptually, this means
   1.415 +that M(x,y)dx + N(x,y) dy is the derivative of a scalar potential and
   1.416 +so consequently corresponds to a conservative field.
   1.417 + 
   1.418 +Even if there is no such potential function
   1.419 +\Phi for the given differential, it is possible to coerce an
   1.420 +inexact differential into an exact one by multiplying by an unknown
   1.421 +function $\mu(x,y)$ (called an /integrating factor/) and requiring the
   1.422 +resulting differential $\mu M\, dx + \mu N\, dy$ to be exact.
   1.423 +
   1.424 +To complete the analogy, here we have the differential $dQ =
   1.425 +dU + PdV$ (by the first law) which is not exact---conceptually, there
   1.426 +is no scalar potential nor conserved quantity corresponding to
   1.427 +$dQ$. We have introduced a new differential $dS = \frac{1}{T}dQ$, and we
   1.428 +are searching for the temperature scale $T(U,V)$ which makes $dS$
   1.429 +exact (i.e. which makes $S$ correspond to a conserved quantity). This means
   1.430 +that $\frac{1}{T}$ is playing the role of the integrating factor
   1.431 +\ldquo{}\mu\rdquo{} for the differential $dQ$.]]
   1.432 +
   1.433 +Now the question of the existence and properties of 
   1.434 +integrating factors is a purely mathematical one, which can be 
   1.435 +investigated independently of the properties of any particular
   1.436 +substance. Let us denote this integrating factor for the moment
   1.437 +by $w(U,V) = T^{-1}$; then the first law becomes
   1.438 +
   1.439 +\begin{equation}
   1.440 +dS(U,V) = w dU + w P dV
   1.441 +\end{equation}
   1.442 +
   1.443 +from which the derivatives are
   1.444 +
   1.445 +\begin{equation}
   1.446 +\left(\frac{\partial S}{\partial U}\right)_V = w, \qquad
   1.447 +\left(\frac{\partial S}{\partial V}\right)_U = wP. 
   1.448 +\end{equation}
   1.449 +
   1.450 +The condition that $dS$ be exact is that the cross-derivatives be
   1.451 +equal, as in (1-4):
   1.452 +
   1.453 +\begin{equation}
   1.454 +\frac{\partial^2 S}{\partial U \partial V} = \frac{\partial^2
   1.455 +S}{\partial V \partial U},
   1.456 +\end{equation}
   1.457 +
   1.458 +or 
   1.459 +
   1.460 +\begin{equation}
   1.461 +\left(\frac{\partial w}{\partial V}\right)_U = \left(\frac{\partial
   1.462 +P}{\partial U}\right)_V + P\cdot \left(\frac{\partial w}{\partial U}\right)_V.
   1.463 +\end{equation}
   1.464 +
   1.465 +Any function $w(U,V)$ satisfying this differential equation is an
   1.466 +integrating factor for $dQ$.
   1.467 +
   1.468 +But if $w(U,V)$ is one such integrating factor, which leads
   1.469 +to the new state function $S(U,V)$, it is evident that
   1.470 +$w_1(U,V) \equiv w \cdot f(S)$ is an equally good integrating factor, where
   1.471 +$f(S)$ is an arbitrary function. Use of $w_1$ will lead to a 
   1.472 +different state function 
   1.473 +
   1.474 +#what's with the variable collision?
   1.475 +\begin{equation}
   1.476 +S_1(U,V) = \int^S f(S) dS
   1.477 +\end{equation}
   1.478 +
   1.479 +The mere conversion of into an exact differential is, therefore, 
   1.480 +not enough to determine any unique entropy function $S(U,V)$.
   1.481 +However, the derivative
   1.482 +
   1.483 +\begin{equation}
   1.484 +\left(\frac{dU}{dV}\right)_S = -P
   1.485 +\end{equation}
   1.486 +
   1.487 +is evidently uniquely determined; so also, therefore, is the
   1.488 +family of lines of constant entropy, called /adiabats/, in the
   1.489 +$(U-V)$ plane. But, as (1-24) shows, the numerical value of $S$ on
   1.490 +each adiabat is still completely undetermined.
   1.491 +
   1.492 +In order to fix the relative values of $S$ on different 
   1.493 +adiabats we need to add the condition, not yet put into the equations, 
   1.494 +that the integrating factor $w(U,V) = T^{-1}$ is to define a new
   1.495 +temperature scale . In other words, we now ask: out of the
   1.496 +infinite number of different integrating factors allowed by
   1.497 +the differential equation (1-23), is it possible to find one
   1.498 +which is a function only of the empirical temperature $t$? If
   1.499 +$w=w(t)$, we can write
   1.500 +
   1.501 +\begin{equation}
   1.502 +\left(\frac{\partial w}{\partial V}\right)_U = \frac{dw}{dt}\left(\frac{\partial
   1.503 +t}{\partial V}\right)_U
   1.504 +\end{equation}
   1.505 +\begin{equation}
   1.506 +\left(\frac{\partial w}{\partial U}\right)_V = \frac{dw}{dt}\left(\frac{\partial
   1.507 +t}{\partial U}\right)_V
   1.508 +\end{equation}
   1.509 +
   1.510 +
   1.511 +and (1-23) becomes
   1.512 +\begin{equation}
   1.513 +\frac{d}{dt}\log{w} = \frac{\left(\frac{\partial P}{\partial
   1.514 +U}\right)_V}{\left(\frac{\partial t}{\partial V}\right)_U-P\left(\frac{\partial t}{\partial U}\right)_V}
   1.515 +\end{equation}
   1.516 +
   1.517 +
   1.518 +which shows that $w$ will be determined to within a multiplicative
   1.519 +factor.
   1.520 +
   1.521 +Is the temperature scale thus defined independent of the
   1.522 +empirical scale from which we started? To answer this, let
   1.523 +$t_1 = t_1(t)$ be any monotonic function which defines a different
   1.524 +empirical temperature scale. In place of (1-28), we then have
   1.525 +
   1.526 +\begin{equation}
   1.527 +\frac{d}{dt_1}\log{w} \quad=\quad \frac{\left(\frac{\partial P}{\partial
   1.528 +U}\right)_V}{\left(\frac{\partial t_1}{\partial V}\right)_U-P\left(\frac{\partial t_1}{\partial U}\right)_V}
   1.529 +\quad = \quad 
   1.530 + \frac{\left(\frac{\partial P}{\partial
   1.531 +U}\right)_V}{\frac{dt_1}{dt}\left[ \left(\frac{\partial t}{\partial
   1.532 +V}\right)_U-P\left(\frac{\partial t}{\partial U}\right)_V\right]},
   1.533 +\end{equation}
   1.534 +or
   1.535 +\begin{equation}
   1.536 +\frac{d}{dt_1}\log{w_1} = \frac{dt}{dt_1}\frac{d}{dt}\log{w}
   1.537 +\end{equation}
   1.538 +
   1.539 +which reduces to $d \log{w_1} = d \log{w}$, or
   1.540 +\begin{equation}
   1.541 +w_1 = C\cdot w
   1.542 +\end{equation}
   1.543 +
   1.544 +Therefore, integrating factors derived from whatever empirical
   1.545 +temperature scale can differ among themselves only by a 
   1.546 +multiplicative factor. For any given substance, therefore, except
   1.547 +for this factor (which corresponds just to our freedom to choose
   1.548 +the size of the units in which we measure temperature), there is
   1.549 +only /one/ temperature scale $T(t) = 1/w$ with the property that
   1.550 +$dS = dQ/T$ is an exact differential.
   1.551 +
   1.552 +To find a feasible way of realizing this temperature scale
   1.553 +experimentally, multiply numerator and denominator of the right
   1.554 +hand side of (1-28) by the heat capacity at constant volume,
   1.555 +$C_V^\prime = (\partial U/\partial t) V$, the prime denoting that 
   1.556 +it is in terms of the empirical temperature scale $t$. 
   1.557 +Integrating between any two states denoted 1 and 2, we have
   1.558 +
   1.559 +\begin{equation}
   1.560 +\frac{T_1}{T_2} = \exp\left\{\int_{t_1}^{t_2}
   1.561 +\frac{\left(\frac{\partial P}{\partial t}\right)_V dt}{P - C_V^\prime
   1.562 +\left(\frac{\partial t}{\partial V}\right)_U} \right\}
   1.563 +\end{equation}
   1.564 +
   1.565 +If the quantities on the right-hand side have been determined
   1.566 +experimentally, then a numerical integration yields the ratio
   1.567 +of Kelvin temperatures of the two states.
   1.568 +
   1.569 +This process is particularly simple if we choose for our
   1.570 +system a volume of gas with the property found in Joule's famous
   1.571 +expansion experiment; when the gas expands freely into a vacuum
   1.572 +(i.e., without doing work, or $U = \text{const.}$), there is no change in
   1.573 +temperature. Real gases when sufficiently far from their condensation 
   1.574 +points are found to obey this rule very accurately.
   1.575 +But then 
   1.576 +
   1.577 +\begin{equation}
   1.578 +\left(\frac{dt}{dV}\right)_U = 0
   1.579 +\end{equation}
   1.580 +
   1.581 +and on a change of state in which we heat this gas at constant
   1.582 +volume, (1-31) collapses to
   1.583 +
   1.584 +\begin{equation}
   1.585 +\frac{T_1}{T_2} = \exp\left\{\int_{t_1}^{t_2}
   1.586 +\frac{1}{P}\left(\frac{\partial P}{\partial t}\right)_V dt\right\} = \frac{P_2}{P_1}.
   1.587 +\end{equation}
   1.588 +
   1.589 +Therefore, with a constant-volume ideal gas thermometer, (or more
   1.590 +generally, a thermometer using any substance obeying (1-32) and
   1.591 +held at constant volume), the measured pressure is directly 
   1.592 +proportional to the Kelvin temperature.
   1.593 +
   1.594 +For an imperfect gas, if we have measured  $(\partial t /\partial
   1.595 +V)_U$ and $C_V^\prime$, Eq. (1-31) determines the necessary
   1.596 +corrections to (1-33). However, an alternative form of (1-31), in
   1.597 +which the roles of pressure and volume are interchanged, proves to be
   1.598 +more convenient for experimental determinations. To derive it, introduce the
   1.599 +enthalpy function
   1.600 +
   1.601 +\begin{equation}H = U + PV\end{equation}
   1.602 +
   1.603 +with the property
   1.604 +
   1.605 +\begin{equation}
   1.606 +dH = dQ + VdP
   1.607 +\end{equation}
   1.608 +
   1.609 +Equation (1-19) then becomes
   1.610 +
   1.611 +\begin{equation}
   1.612 +dS = \frac{dH}{T} - \frac{V}{T}dP.
   1.613 +\end{equation}
   1.614 +
   1.615 +Repeating the steps (1-20) to (1-31) of the above derivation
   1.616 +starting from (1-36) instead of from (1-19), we arrive at
   1.617 +
   1.618 +\begin{equation}
   1.619 +\frac{T_2}{T_1} = \exp\left\{\int_{t_1}^{t_2}
   1.620 +\frac{\left(\frac{dV}{dt}\right)_P dt}{V + C_P^\prime
   1.621 +\left(\frac{\partial t}{\partial P}\right)_H}\right\}
   1.622 +\end{equation}
   1.623 +
   1.624 +or
   1.625 +
   1.626 +\begin{equation}
   1.627 +\frac{T_2}{T_1} = \exp\left\{\frac{\alpha^\prime
   1.628 +dt}{1+\left(C_P^\prime \cdot \mu^\prime / V\right)}\right\}
   1.629 +\end{equation}
   1.630 +
   1.631 +where
   1.632 +\begin{equation}
   1.633 +\alpha^\prime \equiv \frac{1}{V}\left(\frac{\partial V}{\partial t}\right)_P
   1.634 +\end{equation}
   1.635 +is the thermal expansion coefficient,
   1.636 +\begin{equation}
   1.637 +C_P^\prime \equiv \left(\frac{\partial H}{\partial t}\right)_P
   1.638 +\end{equation}
   1.639 +is the heat capacity at constant pressure, and
   1.640 +\begin{equation}
   1.641 +\mu^\prime \equiv \left(\frac{dt}{dP}\right)_H
   1.642 +\end{equation}
   1.643 +
   1.644 +is the coefficient measured in the Joule-Thompson porous plug
   1.645 +experiment, the primes denoting again that all are to be measured
   1.646 +in terms of the empirical temperature scale $t$.
   1.647 +Since $\alpha^\prime$, $C_P^\prime$, $\mu^\prime$ are all 
   1.648 +easily measured in the laboratory, Eq. (1-38) provides a 
   1.649 +feasible way of realizing the Kelvin temperature scale experimentally, 
   1.650 +taking account of the imperfections of real gases. 
   1.651 +For an account of the work of Roebuck and others based on this
   1.652 +relation, see [[http://books.google.com/books?id=KKJKAAAAMAAJ][Zemansky (1943)]]; pp. 252-255.
   1.653 +
   1.654 +Note that if $\mu^\prime = O$ and we heat the gas at constant
   1.655 +pressure, (1-38) reduces to
   1.656 +
   1.657 +\begin{equation}
   1.658 +\frac{T_2}{T_1} = \exp\left\{ \int_{t_1}^{t_2}
   1.659 +\frac{1}{V}\left(\frac{\partial V}{\partial t}\right)_P dt \right\} = \frac{V_2}{V_1}
   1.660 +\end{equation}
   1.661 +
   1.662 +so that, with a constant-pressure gas thermometer using a gas for
   1.663 +which the Joule-Thomson coefficient is zero, the Kelvin temperature is
   1.664 +proportional to the measured volume.
   1.665 +
   1.666 +Now consider another empirical fact, [[http://en.wikipedia.org/wiki/Boyle%27s_law][Boyle's law]]. For gases
   1.667 +sufficiently far from their condensation points---which is also
   1.668 +the condition under which (1-32) is satisfied---Boyle found that
   1.669 +the product $PV$ is a constant at any fixed temperature. This
   1.670 +product is, of course proportional to the number of moles $n$
   1.671 +present, and so Boyle's equation of state takes the form
   1.672 +
   1.673 +\begin{equation}PV = n \cdot f(t)\end{equation}
   1.674 +
   1.675 +where f(t) is a function that depends on the particular empirical
   1.676 +temperature scale used. But from (1-33) we must then have 
   1.677 +$f(t) = RT$, where $R$ is a constant, the universal gas constant whose
   1.678 +numerical value (1.986 calories per mole per degree K) , depends
   1.679 +on the size of the units in which we choose to measure the Kelvin
   1.680 +temperature $T$. In terms of the Kelvin temperature, the ideal gas
   1.681 +equation of state is therefore simply
   1.682 +
   1.683 +\begin{equation}
   1.684 +PV = nRT
   1.685 +\end{equation}
   1.686 +
   1.687 +
   1.688 +The relations (1-32) and (1-44) were found empirically, but
   1.689 +with the development of thermodynamics one could show that they
   1.690 +are not logically independent. In fact, all the material needed
   1.691 +for this demonstration is now at hand, and we leave it as an
   1.692 +exercise for the reader to prove that Joule‘s relation (1-32) is
   1.693 +a logical consequence of Boyle's equation of state (1-44) and the
   1.694 +first law.
   1.695 +
   1.696 +
   1.697 +Historically, the advantages of the gas thermometer were
   1.698 +discovered empirically before the Kelvin temperature scale was
   1.699 +defined; and the temperature scale \theta defined by
   1.700 +
   1.701 +\begin{equation}
   1.702 +\theta = \lim_{P\rightarrow 0}\left(\frac{PV}{nR}\right)
   1.703 +\end{equation}
   1.704 +
   1.705 +was found to be convenient, easily reproducible, and independent
   1.706 +of the properties of any particular gas. It was called the
   1.707 +/absolute/ temperature scale; and from the foregoing it is clear
   1.708 +that with the same choice of the numerical constant $R$, the 
   1.709 +absolute and Kelvin scales are identical.
   1.710 +
   1.711 +
   1.712 +For many years the unit of our temperature scale was the
   1.713 +Centigrade degree, so defined that the difference $T_b - T_f$ of
   1.714 +boiling and freezing points of water was exactly 100 degrees.
   1.715 +However, improvements in experimental techniques have made another
   1.716 +method more reproducible; and the degree was redefined by the
   1.717 +Tenth General Conference of Weights and Measures in 1954, by
   1.718 +the condition that the triple point of water is at 273.l6^\circ K,
   1.719 +this number being exact by definition. The freezing point, 0^\circ C,
   1.720 +is then 273.15^\circ K. This new degree is called the Celsius degree.
   1.721 +For further details, see the U.S. National Bureau of Standards
   1.722 +Technical News Bulletin, October l963.
   1.723 +
   1.724 +
   1.725 +The appearance of such a strange and arbitrary-looking
   1.726 +number as 273.16 in the /definition/ of a unit is the result of
   1.727 +the historical development, and is the means by which much
   1.728 +greater confusion is avoided. Whenever improved techniques make
   1.729 +possible a new and more precise (i.e., more reproducible) 
   1.730 +definition of a physical unit, its numerical value is of course chosen
   1.731 +so as to be well inside the limits of error with which the old
   1.732 +unit could be defined. Thus the old Centigrade and new Celsius
   1.733 +scales are the same, within the accuracy with which the 
   1.734 +Centigrade scale could be realized; so the same notation, ^\circ C, is used
   1.735 +for both . Only in this way can old measurements retain their
   1.736 +value and accuracy, without need of corrections every time a
   1.737 +unit is redefined.
   1.738 +
   1.739 +#capitalize Joules?
   1.740 +Exactly the same thing has happened in the definition of
   1.741 +the calorie; for a century, beginning with the work of Joule,
   1.742 +more and more precise experiments were performed to determine
   1.743 +the mechanical equivalent of heat more and more accurately . But
   1.744 +eventually mechanical and electrical measurements of energy be
   1.745 +came far more reproducible than calorimetric measurements; so
   1.746 +recently the calorie was redefined to be 4.1840 Joules, this
   1.747 +number now being exact by definition. Further details are given
   1.748 +in the aforementioned Bureau of Standards Bulletin.
   1.749 +
   1.750 +
   1.751 +The derivations of this section have shown that, for any
   1.752 +particular substance, there is (except for choice of units) only
   1.753 +one temperature scale $T$ with the property that $dQ = TdS$ where
   1.754 +$dS$ is the exact differential of some state function $S$. But this
   1.755 +in itself provides no reason to suppose that the /same/ Kelvin
   1.756 +scale will result for all substances; i.e., if we determine a
   1.757 +\ldquo{}helium Kelvin temperature\rdquo{} and a 
   1.758 +\ldquo{}carbon dioxide Kelvin temperature\rdquo{} by the measurements 
   1.759 +indicated in (1-38), and choose the units so that they agree numerically at one point, will they then
   1.760 +agree at other points? Thus far we have given no reason to 
   1.761 +expect that the Kelvin scale is /universal/, other than the empirical
   1.762 +fact that the limit (1-45) is found to be the same for all gases.
   1.763 +In section 2.0 we will see that this universality is a conse
   1.764 +quence of the second law of thermodynamics (i.e., if we ever
   1.765 +find two substances for which the Kelvin scale as defined above
   1.766 +is different, then we can take advantage of this to make a 
   1.767 +perpetual motion machine of the second kind).
   1.768 +
   1.769 +
   1.770 +Usually, the second law is introduced before discussing
   1.771 +entropy or the Kelvin temperature scale. We have chosen this
   1.772 +unusual order so as to demonstrate that the concepts of entropy
   1.773 +and Kelvin temperature are logically independent of the second
   1.774 +law; they can be defined theoretically, and the experimental
   1.775 +procedures for their measurement can be developed, without any
   1.776 +appeal to the second law. From the standpoint of logic, there
   1.777 +fore, the second law serves /only/ to establish that the Kelvin
   1.778 +temperature scale is the same for all substances.
   1.779 +
   1.780 +
   1.781 +** COMMENT Entropy of an Ideal Boltzmann Gas
   1.782 +
   1.783 +At the present stage we are far from understanding the physical 
   1.784 +meaning of the function $S$ defined by (1-19); but we can investigate its mathematical
   1.785 +form and numerical values. Let us do this for a system con
   1.786 +sisting cf n moles of a substance which obeys the ideal gas
   1.787 +equation of state
   1.788 +and for which the heat capacity at constant volume CV is a
   1.789 +constant. The difference in entropy between any two states (1)
   1.790 +and (2) is from (1-19),
   1.791 +
   1.792 +
   1.793 +where we integrate over any reversible path connecting the two
   1.794 +states. From the manner in which S was defined, this integral
   1.795 +must be the same whatever path we choose. Consider, then, a
   1.796 +path consisting of a reversible expansion at constant tempera
   1.797 +ture to a state 3 which has the initial temperature T, and the
   1.798 +.L ' "'1 final volume V2; followed by heating at constant volume to the final temperature T2. Then (1-47) becomes
   1.799 +3 2 I If r85 - on - db — = d — -4 S2 51 J V [aT]v M (1 8)
   1.800 +1 3
   1.801 +To evaluate the integral over (1 +3) , note that since
   1.802 +dU = T :15 — P dV, the Helmholtz free energy function F E U — TS
   1.803 +has the property dF = --S - P 61V; and of course is an exact
   1.804 +differential since F is a definite state function. The condition
   1.805 +that dF be exact is, analogous to (1-22),
   1.806 +which is one of the Maxwell relations, discussed further in
   1.807 +where CV is the molar heat capacity at constant volume. Collec
   1.808 +ting these results, we have
   1.809 +3
   1.810 +l 3
   1.811 +1 nR log(V2/V1) + nCV log(T2/Tl) (1-52)
   1.812 +since CV was assumed independent of T. Thus the entropy function
   1.813 +must have the form
   1.814 +S(n,V,T) = nR log V + n CV log T + (const.) (l~53)
   1.815 +
   1.816 +
   1.817 +From the derivation, the additive constant must be independent
   1.818 +of V and T; but it can still depend on n. We indicate this by
   1.819 +writing
   1.820 +where f (n) is a function not determined by the definition (1-47).
   1.821 +The form of f (n) is , however, restricted by the condition that
   1.822 +the entropy be an extensive quantity; i .e . , two identical systems
   1.823 +placed together should have twice the entropy of a single system;
   1.824 +Substituting (l—-54) into (1-55), we find that f(n) must satisfy
   1.825 +To solve this, one can differentiate with respect to q and set
   1.826 +q = 1; we then obtain the differential equation
   1.827 +n f ' (n) — f (n) + Rn = 0 (1-57)
   1.828 +which is readily solved; alternatively, just set n = 1 in (1-56)
   1.829 +and replace q by n . By either procedure we find
   1.830 +f (n) = n f (1) — Rn log n . (1-58)
   1.831 +As a check, it is easily verified that this is the solution of
   1.832 +where A E f (l) is still an arbitrary constant, not determined
   1.833 +by the definition (l—l9) , or by the condition (l-55) that S be
   1.834 +extensive. However, A is not without physical meaning; we will
   1.835 +see in the next Section that the vapor pressure of this sub
   1.836 +stance (and more generally, its chemical potential) depends on
   1.837 +A. Later, it will appear that the numerical value of A involves
   1.838 +Planck's constant, and its theoretical determination therefore
   1.839 +requires quantum statistics .
   1.840 +We conclude from this that, in any region where experi
   1.841 +mentally CV const. , and the ideal gas equation of state is
   1.842 +
   1.843 +
   1.844 +obeyed, the entropy must have the form (1-59) . The fact that
   1.845 +classical statistical mechanics does not lead to this result,
   1.846 +the term nR log (l/n) being missing (Gibbs paradox) , was his
   1.847 +torically one of the earliest clues indicating the need for the
   1.848 +quantum theory.
   1.849 +In the case of a liquid, the volume does not change appre
   1.850 +ciably on heating, and so d5 = n CV dT/T, and if CV is indepen
   1.851 +dent of temperature, we would have in place of (1-59) ,
   1.852 +where Ag is an integration constant, which also has physical
   1.853 +meaning in connection with conditions of equilibrium between
   1.854 +two different phases.
   1.855 +1.1.0 The Second Law: Definition. Probably no proposition in
   1.856 +physics has been the subject of more deep and sus tained confusion
   1.857 +than the second law of thermodynamics . It is not in the province
   1.858 +of macroscopic thermodynamics to explain the underlying reason
   1.859 +for the second law; but at this stage we should at least be able
   1.860 +to state this law in clear and experimentally meaningful terms.
   1.861 +However, examination of some current textbooks reveals that,
   1.862 +after more than a century, different authors still disagree as
   1.863 +to the proper statement of the second law, its physical meaning,
   1.864 +and its exact range of validity.
   1.865 +Later on in this book it will be one of our major objectives
   1.866 +to show, from several different viewpoints , how much clearer and
   1.867 +simpler these problems now appear in the light of recent develop
   1.868 +ments in statistical mechanics . For the present, however, our
   1.869 +aim is only to prepare the way for this by pointing out exactly
   1.870 +what it is that is to be proved later. As a start on this at
   1.871 +tempt, we note that the second law conveys a certain piece of
   1.872 +informations about the direction in which processes take place.
   1.873 +In application it enables us to predict such things as the final
   1.874 +equilibrium state of a system, in situations where the first law
   1.875 +alone is insufficient to do this.
   1.876 +A concrete example will be helpful. We have a vessel
   1.877 +equipped with a piston, containing N moles of carbon dioxide.
   1.878 +
   1.879 +
   1.880 +The system is initially at thermal equilibrium at temperature To, volume V0 and pressure PO; and under these conditions it contains
   1.881 +n moles of CO2 in the vapor phase and moles in the liquid
   1.882 +phase . The system is now thermally insulated from its surround
   1.883 +ings, and the piston is moved rapidly (i.e. , so that n does not
   1.884 +change appreciably during the motion) so that the system has a
   1.885 +new volume Vf; and immediately after the motion, a new pressure
   1.886 +PI . The piston is now held fixed in its new position , and the
   1.887 +system allowed to come once more to equilibrium. During this
   1.888 +process, will the CO2 tend to evaporate further, or condense further? What will be the final equilibrium temperature Teq, the final pressure PeCE , and final value of n eq?
   1.889 +It is clear that the firs t law alone is incapable of answering
   1.890 +these questions; for if the only requirement is conservation of
   1.891 +energy, then the CO2 might condense , giving up i ts heat of vapor
   1.892 +ization and raising the temperature of the system; or it might
   1.893 +evaporate further, lowering the temperature. Indeed, all values
   1.894 +of neq in O i neq i N would be possible without any violation of the first law. In practice, however, this process will be found
   1.895 +to go in only one direction and the sys term will reach a definite
   1.896 +final equilibrium state with a temperature, pressure, and vapor
   1.897 +density predictable from the second law.
   1.898 +Now there are dozens of possible verbal statements of the
   1.899 +second law; and from one standpoint, any statement which conveys
   1.900 +the same information has equal right to be called "the second
   1.901 +law." However, not all of them are equally direct statements of
   1.902 +experimental fact, or equally convenient for applications, or
   1.903 +equally general; and it is on these grounds that we ought to
   1.904 +choose among them .
   1.905 +Some of the mos t popular statements of the s econd law be
   1.906 +long to the class of the well-—known "impossibility" assertions ;
   1.907 +i.e. , it is impossible to transfer heat from a lower to a higher
   1.908 +temperature without leaving compensating changes in the rest of
   1.909 +the universe , it is imposs ible to convert heat into useful work
   1.910 +without leaving compensating changes, it is impossible to make
   1.911 +a perpetual motion machine of the second kind, etc.
   1.912 +
   1.913 +
   1.914 +Suoh formulations have one clear logical merit; they are
   1.915 +stated in such a way that, if the assertion should be false, a
   1.916 +single experiment would suffice to demonstrate that fact conclu
   1.917 +sively. It is good to have our principles stated in such a
   1.918 +clear, unequivocal way.
   1.919 +However, impossibility statements also have some disadvan
   1.920 +tages . In the first place, their_ are not, and their very
   1.921 +nature cannot be, statements of eiperimental fact. Indeed, we
   1.922 +can put it more strongly; we have no record of anyone having
   1.923 +seriously tried to do any of the various things which have been
   1.924 +asserted to be impossible, except for one case which actually
   1.925 +succeeded‘. In the experimental realization of negative spin
   1.926 +temperatures , one can transfer heat from a lower to a higher
   1.927 +temperature without external changes; and so one of the common
   1.928 +impossibility statements is now known to be false [for a clear
   1.929 +discussion of this, see the article of N. F . Ramsey (1956) ;
   1.930 +experimental details of calorimetry with negative temperature
   1.931 +spin systems are given by Abragam and Proctor (1958) ] .
   1.932 +Finally, impossibility statements are of very little use in
   1.933 +applications of thermodynamics; the assertion that a certain kind
   1.934 +of machine cannot be built, or that a -certain laboratory feat
   1.935 +cannot be performed, does not tell me very directly whether my
   1.936 +carbon dioxide will condense or evaporate. For applications,
   1.937 +such assertions must first be converted into a more explicit
   1.938 +mathematical form.
   1.939 +For these reasons, it appears that a different kind of
   1.940 +statement of the second law will be, not necessarily more
   1.941 +"correct,” but more useful in practice. Now both Clausius (3.875)
   1.942 +and Planck (1897) have laid great stress on their conclusion
   1.943 +that the most general statement, and also the most immediately
   1.944 +useful in applications, is simply the existence of a state
   1.945 +function, called the entropy, which tends to increase. More
   1.946 +precisely: in an adiabatic change of state, the entropy of
   1.947 +a system may increase or may remain constant, but does not
   1.948 +decrease. In a process involving heat flow to or from the
   1.949 +system, the total entropy of all bodies involved may increase
   1.950 +
   1.951 +
   1.952 +or may remain constant; but does not decrease; let us call this
   1.953 +the “weak form" of the second law.
   1.954 +The weak form of the second law is capable of answering the
   1.955 +first question posed above; thus the carbon dioxide will evapo
   1.956 +rate further if , and only if , this leads to an increase in the
   1.957 +total entropy of the system . This alone , however , is not enough
   1.958 +to answer the second question; to predict the exact final equili
   1.959 +brium state, we need one more fact.
   1.960 +The strong form of the second law is obtained by adding the
   1.961 +further assertion that the entropy not only “tends" to increase;
   1.962 +in fact it will increase, to the maximum value permitted E2 the
   1.963 +constraints imposed.* In the case of the carbon dioxide, these
   1.964 +constraints are: fixed total energy (first law) , fixed total
   1.965 +amount of carbon dioxide , and fixed position of the piston . The
   1.966 +final equilibrium state is the one which has the maximum entropy
   1.967 +compatible with these constraints , and it can be predicted quan
   1.968 +titatively from the strong form of the second law if we know,
   1.969 +from experiment or theory, the thermodynamic properties of carbon
   1.970 +dioxide (i .e . , heat capacity , equation of state , heat of vapor
   1.971 +ization) .
   1.972 +To illus trate this , we set up the problem in a crude ap
   1.973 +proximation which supposes that (l) in the range of conditions
   1.974 +of interest, the molar heat capacity CV of the vapor, and C2 of
   1.975 +the liquid, and the molar heat of vaporization L, are all con
   1.976 +stants, and the heat capacities of cylinder and piston are neg
   1.977 +ligible; (2) the liquid volume is always a small fraction of the
   1.978 +total V, so that changes in vapor volume may be neglected; (3) the
   1.979 +vapor obeys the ideal gas equation of state PV = nRT. The in
   1.980 +ternal energy functions of liquid and vapor then have the form
   1.981 +UPb = + A} (1-61)
   1.982 +T T U = n‘ C '1‘ A + L] (1-62)
   1.983 +v , v
   1.984 +where A is a constant which plays no role in the problem. The
   1.985 +appearance of L in (1-62) recognizes that the zero from which we
   1.986 +*Note , however , that the second law has nothing to say about how rapidly this approach to equilibrium takes place.
   1.987 +
   1.988 +
   1.989 +measure energy of the vapor is higher than that of the liquid by
   1.990 +the energy L necessary to form the vapor. On evaporation of dn
   1.991 +moles of liquid, the total energy increment is (ill = + dUV= O,
   1.992 +or
   1.993 +[n CV [(CV — CQ)T + = O (l—63)
   1.994 +which is the constraint imposed by the first law. As we found
   1.995 +previously (l~59) , (1-60) the entropies of vapor and liquid are
   1.996 +given by
   1.997 +S = n [C 1n T + R ln (V/n) + A ] (1-64)
   1.998 +v v v
   1.999 +where AV, ASL are the constants of integration discussed in the
  1.1000 +Si
  1.1001 +last Section.
  1.1002 +We leave it as an exercise for the reader to complete the
  1.1003 +derivation from this point , and show that the total entropy
  1.1004 +S = 82 + SV is maximized subject to the constraint (1-6 3) , when
  1.1005 +R
  1.1006 +the values 11 , T are related by
  1.1007 +eq eq
  1.1008 +Equation (1-66) is recognized as an approximate form of the Vapor
  1.1009 +pressure formula .
  1.1010 +We note that AQ, AV, which appeared first as integration
  1.1011 +constants for the entropy with no parti cular physical meaning ,
  1.1012 +now play a role in determining the vapor pressure.
  1.1013 +l.ll The Second Law: Discussion. We have emphasized the dis
  1.1014 +tinction between the weak and strong forms of the second law
  1.1015 +because (with the exception of Boltzmann ' s original unsuccessful
  1.1016 +argument based on the H—theorem) , most attempts to deduce the
  1.1017 +second law from statis tical mechanics have considered only the
  1.1018 +weak form; whereas it is evidently the strong form that leads
  1.1019 +to definite quantitative predictions, and is therefore needed
  1.1020  
  1.1021  
  1.1022