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πiXeL WallRiDeR Karlengs
SOLUTIONS MANUAL
THERMODYNAMICS
AND AN INTRODUCTION
THERMOSTATISTICS
SECOND EDITION
TO
HERBERT B. CALLEN

Manual de soluções para o livro Thermodyna-
mics and an introduction to thermostatistics -
second edition. [SUMÁRIO]
Este manual de soluções é essencialmente uma tradução do solucionário original, feito pelo
próprio Herbert B. Callen e publicado pela editora JOHN WILEY & SONS no ano de 1986
(ISBN 0 471 81993 X). O solucionário original foi feito e publicado na forma manuscrita, sendo
essa a principal razão pela qual decidimos editar esse documento. A praticidade e clareza deste
manual de soluções o torna uma excelente ferramenta auxiliar para as pessoas que usam o
livro-texto, porém deve-se tomar o cuidado de não fazer dessa ferramenta consultiva a principal
forma de resolver os problemas, sendo essa nossa principal preocupação em tornar esse material
público.
O livro thermodynamics and an introduction to thermostatistics apresenta 391 problemas ao
todo, e esse manual traz a solução de 346 desses problemas e ainda 7 problemas suplementares
e suas respectivas soluções. Tomamos o cuidado de deixar esse documento tão mais fiel ao
original quanto fosse possível, tanto nas explicações quanto na disposição das figuras, fazendo
pequenas correções e alterações quando necessário e conveniente.
Esse manual é um bem público feito por nós estudantes para vocês estudantes. Por favor
não o comercialize. Este material, bem como a ciência e não obstante o conhecimento, deve ser
acessado por todos e para todos, sendo portanto livre o seu compartilhamento.
Comentários, correções, ou mesmo sugestões de solução para qualquer problema do livro
ou deste manual, favor enviar para o e-mail solutionscallen@gmail.com, dessa forma podemos
mantê-lo sempre atualizado.
Goiânia, fevereiro de 2022
πiXeL
Karlengs
WallRiDeR
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Sumário [SUMÁRIO]
1 THE PROBLEM AND THE POSTULATES 20
1.3 THE COMPOSITION OF THERMODYNAMIC SYSTEMS . . . . . . 20
1.3.1 One tenth of a kilogram of NaCl and 0.15 kg of sugar (C12H22O11)... . . . 20
1.3.2 Naturally occurring boron has an atomic mass of 10.811 g... . . . . . . . 20
1.3.3 Twenty cubic centimeters each of ethyl alcohol... . . . . . . . . . . . . . . 20
1.3.4 A 0.01 kg sample is composed of 50 molecular percent H2... . . . . . . . 21
1.3.5 A solution of sugar (C12H22O11) in water is 20% sugar by weight... . . . . 21
1.3.6 An aqueous solution of an unidentified solute has a total mass... . . . . . 22
1.3.7 One tenth of a kg of an aqueous solution of HCl is poured... . . . . . . . 22
1.8 QUANTITATIVE DEFINITION OF HEAT-UNITS . . . . . . . . . . . . 23
1.8.1 For the system considered in Example 1, calculate... . . . . . . . . . . . . 23
1.8.2 Calculate the heat transferred to the system considered in... . . . . . . . 23
1.8.3 For a particular gaseous system it has been determined that... . . . . . . 24
1.8.4 For the system of Problem 1.8-3 find the equation of the adiabats... . . . 25
1.8.5 The energy of a particular system, of one mole, is given by ... . . . . . . . 25
1.8.6 For a particular system it is found that if the volume is kept ... . . . . . . 26
1.8.7 Two moles of a particular single-component system are found... . . . . . 26
1.8.8 (Supplementary Problem) A particular single-component system of
one mole has adiabats of the form P5/3
= constant . . . . . . . . . . . . 27
1.10 THE ENTROPY MAXIMUM POSTULATES . . . . . . . . . . . . . . . 28
1.10.1 The following ten equations are purported to be fundamental... . . . . . 28
1.10.2 For each of the five physically acceptable fundamental equations... . . . . 29
1.10.3 The fundamental equation of a system A is S =

R2
v0θ
1/3
(NV U)1/3
and
similarly... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2 THE CONDITIONS OF EQUILIBRIUM 31
2.2 EQUATIONS OF STATE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.2.1 Find the three equations of state for a system with the fundamental... . . 31
2.2.2 For the system of problem 2.2-1 find µ as a function of T, V e N . . . . . 31
2.2.3 Show by a diagram (drawn to arbitrary scale) the dependence of... . . . . 31
2.2.4 Find the three equations of state for a system with the fundamental
equation u =

θ
R

s2
−

Rθ
v2
0

v2
. . . . . . . . . . . . . . . . . . . . . . 32
2.2.5 Express µ as a function of T and P for the system of problem 2.2-4 . . . 32
equation u =

v0θ
R

s2
v
es/R
. . . . . . . . . . . . . . . . . . . . . . . . . 33
2.2.7 A particular system obeys the relation u = Av−2
exp(s/R)... . . . . . . . 33
2.2.8 Show that, in analogy with equation 2.25, for a system with r components... 33
2.2.9 Show that if a single-component system is such that PV k
is constant... . 34
2.3 ENTROPIC INTENSIVE PARAMETERS . . . . . . . . . . . . . . . . . 35
2.3.1 Find the three equations of state in the entropy representation for... . . . 35
2.3.2 Show by a diagram (drawn to arbitrary scale) the dependence of tempe-
rature on volume ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.3.3 Find the three equations of state in the entropy representation for ... . . 37
2.3.4 Consider the fundamental equation S = AUn
V m
Nr
where A is a positive
constant... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
relation... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.6 TEMPERATURE UNITS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
6

2.6.1 The temperature of a system composed of ice, water, and water vapor... . 38
2.6.2 The gas constantR is defined as the product of Avogrado’s number... . 38
2.6.3 Two particular systems have the following equations of state... . . . . . . 38
2.6.4 Two systems with the equations of state given in Problem 2.6-3 ... . . . . 39
2.7 MECHANICAL EQUILIBRIUM . . . . . . . . . . . . . . . . . . . . . . . . 39
2.7.1 Three cylinders are fitted with four pistons, as shown in Fig. 2.2 ... . . . 39
2.7.2 Two particular systems have the following equations of state : ... . . . . . 40
2.7.3 The hypothetical problem of equilibrium in a closed composite system ... 40
2.8 EQUILIBRIUM WITH RESPECT TO MATTER FLOW . . . . . . . . 41
2.8.1 The fundamental equation of a particular type of two-component system
is... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.8.2 A two-component system gaseous system has a fundamental equation of
the form... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.9 CHEMICAL EQUILIBRIUM . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.9.1 The hydrogenation of propane (C3H8) to form methane (CH4) proceeds
by the reaction... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3 SOME FORMAL RELATIONSHIPS, AND SAMPLE SYSTEMS 43
3.1 THE EULER EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.1.1 Write each of the five physically acceptable fundamental equations... . . . 43
3.2 THE GIBBS-DUHEM RELATION . . . . . . . . . . . . . . . . . . . . . . 44
3.2.1 Find the relation among T, P, and µ for the system with the fundamental
equation... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.3 SUMMARY OF FORMAL STRUCTURE . . . . . . . . . . . . . . . . . . 45
3.3.1 A particular system obeys the two equations of state... . . . . . . . . . . 45
3.3.2 It is found that a particular system obeys the relations U = PV and
P = BT2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.3.3 A system obeys the equations... . . . . . . . . . . . . . . . . . . . . . . . 46
3.3.4 A system obeys the two equations u =
3
2
Pv and u1/2
= BTv1/3
... . . . . 47
3.4 THE SIMPLE IDEAL GAS AND MULTICOMPONENT SIMPLE
IDEAL GASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.4.1 A constant volume ideal gas thermometeris contructed as shown (sche-
matically) in Fig. 3.3. ... . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.4.2 Show that the relation between the volume and the pressure of a mono-
atomic ideal gas ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.4.3 Two moles of a monoatomic ideal gas are at a temperature of 0℃... . . . 48
3.4.4 By carring out the integral
R
PdV , compute the work done by the gas in
Problem 3.4-3... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.4.5 In a particular engine a gas is compressed in the initial stroke of the piston... 49
3.4.6 Find the three equations of state of the simple ideal gas(equation 3.34... 50
3.4.7 Find the four equations of state of a two-component mixture if simple
ideal gases... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.4.8 If a monoatomic ideal gas is permitted to expand into an evacuated region,
thereby increasing its volume... . . . . . . . . . . . . . . . . . . . . . . . 51
3.4.9 A tank has a volume of 0.1m3
and is filled with He gas at a pressure of... 52
3.4.10 ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.4.11 Show that the pressure of a multicomponent simple ideal gas can be
written as the sum... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.4.12 Show that µj, the electrochemical potential of the jth component in a ... 53
3.4.13 An impermeable, diathermal, and rigid partition divides a container into
two subvolumes... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
7

3.4.14 Use the results of problem 3.4-11 and 3.4-12 to stablish the results Pα =
Pγ e... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.4.15 An impermeable, diathermal and rigid partition divides a container into
two subvolumes... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.5 THE ˝IDEAL VAN DER WAALS FLUID . . . . . . . . . . . . . . . . . 54
3.5.1 Are each of the listed pairs of equations of state compatible (recall equa-
tions 3.46)? ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.5.2 Find the relationship between the volume and the temperature of an ideal
Van Der Waals fluid... . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.5.3 Repeat the problem 3.4-3 for CO2, rather than for a monoatomic ideal
gas... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.5.4 Repeat parts (a), (b), e (c) do problema 3.4-5, assuming that η = −1/2
and that gas... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.5.5 Consider a Van Der Waals gas contained in the apparatus described in
Problem 3.4-1... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.5.6 One mole of a monoatomic ideal gas and one mole of Cl2 are contained
in a rigid cylinder and... . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.5.7 (Supplementary problem) Suppose we wish to account (crudely) for
a long-range... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.6 ELECTROMAGNETIC RADIATION . . . . . . . . . . . . . . . . . . . . 58
3.6.1 The universe is considered by cosmologists to be an expanding electro-
magnetic cavity containing radiation... . . . . . . . . . . . . . . . . . . . 58
3.6.2 Assuming the electromagnetic radiation filling the universe to be in equi-
librium at T = 2.7 K... . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.6.3 The density of matter (primarily hydrogen atoms) in intergalactic space
is such that... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.7 THE ˝RUBBER BAND . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
3.7.1 For the rubber band model, calculate the fractional change in (L − L0)
that results from an increase δT... . . . . . . . . . . . . . . . . . . . . . . 59
3.7.2 A rubber band is stretched by an amount dL, at constant T. Calculate
the heat transfer đQ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
3.7.3 If the energy of the unstretched rubber band were found to increase qua-
dratically with T... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
3.8 UNCONSTRAINABLE VARIABLES; MAGNETIC SYSTEMS . . . . 60
3.8.1 Calculate the three equations of state of the paramagnetic model of equa-
tion 3.66. That is, calculate... . . . . . . . . . . . . . . . . . . . . . . . . 60
3.8.2 Repeat problem 3.8-1 for a system with the fundamental equation U =
µ0I2
/2Nχ+... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
3.9 MOLAR HEAT CAPACITY AND OTHER DERIVATIVES . . . . . . 61
3.9.1 ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3.9.2 Corroborate equation 3.70 for a multicomponent simple ideal gas, showing
that both the right- and left-hand... . . . . . . . . . . . . . . . . . . . . . 62
3.9.3 Compute the coefficient of expansion α and the isothermal compressibility
κT in terms... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
3.9.4 Compute CP , CV , κS, e κT for the system in Problem 1.10-(a). With
these values... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.9.5 From equations 3.75 and 3.76 show that... . . . . . . . . . . . . . . . . . 63
3.9.6 A simple fundamental equation that exhibits some of the qualitative pro-
perties of typical crystaline solids... . . . . . . . . . . . . . . . . . . . . . 64
3.9.7 The density of mercury at various temperatures is given here in grams/cm3
... 65
3.9.8 For a particular material CP , α, e κT can be represented empirically by
power series... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
8

3.9.9 Calculate the molar entropy s(T, P0) for fixed pressure P0 and for tem-
perature T in the vicinity of ... . . . . . . . . . . . . . . . . . . . . . . . 66
3.9.10 By analogy with equations 3.70 and 3.71 show that for a paramagnetic
system... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.9.11 By analogy with equations 3.70 and 3.71 show that for a paramagnetic
system... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
3.9.12 The magnetic analogues of the molar heat capacities CP and Cv are CB
and CI. Calculate... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
3.9.13 The (isothermal) molar magnetic susceptibility is defined by ... . . . . . . 67
3.9.14 Calculate the adiabatic molar susceptibility... . . . . . . . . . . . . . . . 68
3.9.15 Calculate the isothermal and adiabatic molar susceptibilities (defined in
Problem 3.9-13 and 3.9-14)... . . . . . . . . . . . . . . . . . . . . . . . . 68
3.9.16 Show that for the system of Problem 3.8-2... . . . . . . . . . . . . . . . . 68
3.9.17 Calculate the heat transfer to a particular system if 1 mole is taken from
(T0, P0) to (2T0, 2P0)... . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.9.18 (Supplementary Problem) A particular substance is known to have
α = 5/(4T), κT = 2/(P + P0)... . . . . . . . . . . . . . . . . . . . . . . . 69
4 REVERSIBLE PROCESSES AND THE MAXIMUM WORK THEOREM 70
4.1 POSSIBLE AND IMPOSSIBLE PROCESSES . . . . . . . . . . . . . . . 70
4.1.1 One mole of a monatomic ideal gas and one mole of an idela Van der
Waals fluid (Section 3.5)... . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.1.2 A rubber band (Section 3.7) is initially at temperature TB and length
LB. One mole of a monatomic ideal gas... . . . . . . . . . . . . . . . . . 71
4.1.3 Supose the two systems in Example 1 were to have heat capacities of the
form C(T) = DTn
, with n 0 ... . . . . . . . . . . . . . . . . . . . . . . 71
4.2 QUASI-STATIC AND REVERSIBLE PROCESSES . . . . . . . . . . . 72
4.2.1 Does every reversible process coincide with a quasi-static locus? Does
every quasi-static locus coincide with a reversible process ?... . . . . . . . 72
4.2.2 Consider a monatomic ideal gas in a cylinder fitted with a piston. The
walls of the cylinder and the piston are adiabatic. ... . . . . . . . . . . . 72
4.2.3 A monatomic ideal gas is permitted to expand by a free expansion from
V to V + dV ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.2.4 In the temperature range of interest a system obeys the equations T =
Av2
/s and P = −2Av
ln(s/s0)... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3 RELAXATION TIMES AND IRREVERSIBILITY . . . . . . . . . . . . 73
4.3.1 A cylinder of length L and cross-sectional area A is divided into two
equal-volume chambers by a piston, held at... . . . . . . . . . . . . . . . 73
4.4 HEAT FLOW: COUPLED SYSTEMS AND REVERSAL OF PRO-
CESSES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
4.4.1 Each of two bodies has a heat capacity given, in the temperature range
of interest, by C = A + BT ... . . . . . . . . . . . . . . . . . . . . . . . . 74
4.4.2 Consider again the system of Problem 4.4-1. LEt a third body be avai-
lable, with heat capacity C3 = BT... . . . . . . . . . . . . . . . . . . . . 74
4.4.3 Prove that the entropy change in a heat flow process, as given in equation
4.5, is intrinsically positive. . . . . . . . . . . . . . . . . . . . . . . . . . 75
4.4.4 Show that if two bodies have equal heat capacities, eqach of which is
constant (independent of temperature), the equilibrium temperature... . . 75
4.4.5 Over a limited temperature range the heat capacity at constant volume
of a particular type of system is inversely proportional... . . . . . . . . . 76
9

4.4.6 A series of N + 1 large vats of water have temperatures T0, T1, T2, ..., TN
(with Tn Tn−1). A small body with heat capacity C... . . . . . . . . . . 76
4.5 THE MAXIMUM WORK THEOREM . . . . . . . . . . . . . . . . . . . . 77
4.5.1 One mole of a monatomic ideal gas is contained in a cylinder of volume
10−3
m3
at temperature of 400 K... . . . . . . . . . . . . . . . . . . . . . 77
4.5.2 Consider the following process for the system of Problem 4.5-1. The ideal
gas is first expanded adiabatically ... . . . . . . . . . . . . . . . . . . . . 77
4.5.3 Describe how the gas of the preceding two problems could be brought to
the desired final state by a free expansion. ... . . . . . . . . . . . . . . . . 78
4.5.4 The gasous system of Problem 4.5-1 is to be restored to its initial state.
Both states have temperature 400 K... . . . . . . . . . . . . . . . . . . . 78
4.5.5 If the thermal reservoir of Problem 4.5-1 were to be replaced by a rever-
sible heat source having a heat capacity of ... . . . . . . . . . . . . . . . . 78
4.5.6 A system can be taken from state A to state B (where SB = SA) either
(a) directly along the adiabat S = constant, or (b) along the isochore AC... 79
4.5.7 Consider the maximum work theorem in the case in which the specified
final state of the subsystem has lower... . . . . . . . . . . . . . . . . . . . 79
4.5.8 If SB SA and UB UA does this imply that the delivered work is
negative ? Prove your assertion... . . . . . . . . . . . . . . . . . . . . . . 79
4.5.9 Two identical bodies each have constant and equal heat capacities (C1 =
C2 = C, a constant). In addition a reversible work source is available... . 80
4.5.10 Two identical bodies eqch have heat capacities (at constant volume) of
C(T) = a/T... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
4.5.11 Two bodies have heat capacities (at constant volume) of C1 = aT and
C2 = 2bT... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
4.5.12 One mole of an ideal Van der Waals fluid is contained in a cylinder fitted
with a piston. The initial temperature of the gas... . . . . . . . . . . . . 81
4.5.13 A system has a temperature-independent heat capacity C. The system
is initially at temperature Ti and a heat reservoir... . . . . . . . . . . . . 81
4.5.14 If the temperature of the atmosphere is 5℃on a winter day and if 1 Kg
of water at 90 ℃is... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
4.5.15 A rigid cylinder contains an internal adiabatic piston separating it into
two chambers, of volume V10 and V20. ... . . . . . . . . . . . . . . . . . . 82
4.5.16 Each of three identical bodies has a temperature-independent heat capa-
city C. The three bodies... . . . . . . . . . . . . . . . . . . . . . . . . . . 82
4.5.17 Each of two bodies has a heat capacity given by C = A + 2BT... . . . . . 83
4.5.18 A particular system has the equations of state T = As/v1/2
and P =
T2
/4Av1/2
... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
4.5.19 A particular type of system obeys the equations T = u
b
and P = avT... . 84
4.5.20 Suppose that we have a system in some initial state (we may think of a
tank of hot, compressed gas as an example) and... . . . . . . . . . . . . . 84
4.5.21 An antartic meteorological station suddenly loses all of its fuel. It has N
moles of an inert ideal Van der Waals fluidat... . . . . . . . . . . . . . 85
4.5.22 A geothermalpower source is available to drive an oxygen production
plant. The geothermal source is simply a ... . . . . . . . . . . . . . . . . 85
4.6 COEFFICIENTS OF ENGINE, REFRIGERATOR, AND HEAT PUMP
PERFORMANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
4.6.1 A temperature of 0.001 K is accessible in low temperature laboratories
with moderate effort. If the price... . . . . . . . . . . . . . . . . . . . . . 86
4.6.2 A home is to be maintained at 70F, and the external temperature is 50F.
One method of heating... . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
10

4.6.3 A household refrigerator is maintained at a temperature of 35°F. Every
time the door is opened, warm material... . . . . . . . . . . . . . . . . . . 86
4.6.4 Heat is extracted from a bath of liquid helium at a temperature of 4.2 K.
The high-temperature reservoir is a ... . . . . . . . . . . . . . . . . . . . 86
4.6.5 Assume that a particular body has the equation of state U = NCT with
NC = 10 J/K and assume that this equation... . . . . . . . . . . . . . . 87
4.6.6 One mole of a monatomic ideal gas is allowed to expand isothermally
from an initial volume of 10 liter to a final... . . . . . . . . . . . . . . . . 87
4.6.7 Give a constructive solutionof Example 2 of Section 4.1. Your solution
may be based on the following procedure... . . . . . . . . . . . . . . . . . 87
4.6.8 Assume that 1 mole of an ideal Van der Waals fluid is expanded isother-
mally, at temperature Th, from an initial... . . . . . . . . . . . . . . . . . 88
4.6.9 Two moles of a monatomic ideal gas are to be taken from an initial state
(Pi, Vi) to a final state... . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4.6.10 Assume the process in Problem 4.6-9 to occur along the locus P = B/V 2
,
where B = PiV 2
i . Apply the... . . . . . . . . . . . . . . . . . . . . . . . . 90
4.6.11 Assume the process in Problem 4.6-9 to occur along a straight-line locus
in the T − V plane. Integrate along... . . . . . . . . . . . . . . . . . . . . 91
4.7 THE CARNOT CYCLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.7.1 Repeat the calculation of Example 5 assuming the working substanceof
the auxiliary system to be 1 mole... . . . . . . . . . . . . . . . . . . . . . 91
4.7.2 Calculate the work and the heat transfers in each stage of the carnot
cycle for the auxiliary system being an... . . . . . . . . . . . . . . . . . . 93
4.7.3 A primary subsystemin the initial state A is to be brought reversibly
to a specified final state B. A reversible... . . . . . . . . . . . . . . . . . . 93
4.7.4 The fundamental equation of a particular fluid is UN1/2
V 1/2
= A(S−R)3
where A... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
4.7.5 One mole of the simple paramagnetic model systemof equation 3.66 is
to be used as the auxiliary system of... . . . . . . . . . . . . . . . . . . . 94
4.7.6 Repeat Problem 4.7-4 using the rubber bandmodel of section 3.7 as the
auxiliary system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
4.7.7 (Supplementary Problem) Repeat Problem 4.7-4 if the working subs-
tance is one mole of.... . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
4.8 MEASURABILITY OF THE TEMPERATURE AND OF THE EN-
TROPY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
4.8.1 To corroborate equation 4.22 show that ∂P
∂s

T
= − ∂T
∂v

P
... . . . . . . . 96
4.9 OTHER CRITERIA OF ENGINE PERFORMANCE; POWER OUT-
PUT AND ˝ENDOREVERSIBLE ENGINES  . . . . . . . . . . . . . . 97
4.9.1 Show that the efficiency of an endoreversible engine, maximized for power
output, is always less than εCarnot... . . . . . . . . . . . . . . . . . . . . . 97
4.9.2 Suppose the conductance σh (= σc) to be such that 1 KW is transfered
to the system (as heat flux) if its... . . . . . . . . . . . . . . . . . . . . . 97
4.9.3 Consider an endoreversible engine for which the high temperature reser-
voir is boiling water (100℃) and... . . . . . . . . . . . . . . . . . . . . . . 97
4.9.4 Assume that one cycle of the engine of Problem 4.9-3 takes 20 s and that
the condutance σh = σc = 100 W/K... . . . . . . . . . . . . . . . . . . . 98
4.10 OTHER CYCLIC PROCESSES . . . . . . . . . . . . . . . . . . . . . . . . 98
4.10.1 Assuming that the working gas is a monatomic ideal gas, plot a T − S
diagram for the Otto cycle. . . . . . . . . . . . . . . . . . . . . . . . . . 98
4.10.2 Assuming that the working gas is a simple ideal gas (with temperature-
independent heat capacities), show that the engine efficiency of the Otto
cycle... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
11

4.10.3 Assuming that the working gas is a simples ideal gas(with temperature-
independent heat capacities), show that the engine efficiency of the Bray-
ton cycle ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
diagram of the Brayton cycle. . . . . . . . . . . . . . . . . . . . . . . . . 99
diagram of the air-standart diesel cycle. . . . . . . . . . . . . . . . . . . . 99
5 ALTERNATIVE FORMULATIONS AND LEGENDRE TRANSFORMATI-
ONS 99
5.1 THE ENERGY MINIMUM PRINCIPLE . . . . . . . . . . . . . . . . . . 99
5.1.1 Formulate a proof that the energy minimum principle implies the entropy
maximum ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
5.1.2 An adiabatic, impermeable and fixed piston separates a cylinder into two
chambers... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
5.2 LEGENDRE TRANSFORMATIONS . . . . . . . . . . . . . . . . . . . . . 100
5.2.1 The equation y = x2
/10 describes a parabola. ... . . . . . . . . . . . . . . 100
5.2.2 Let y = AeBx
... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.3 THERMODINAMIC POTENTIALS . . . . . . . . . . . . . . . . . . . . . 101
5.3.1 Find the fundamental equation of a monatomic ideal gas in the Helmholtz
representation, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.3.2 Find the fundamental equation of the ideal van der Waals fluid (Section
3.5) in the Helmholtz representation... . . . . . . . . . . . . . . . . . . . 102
5.3.3 Find the fundamental equation of electromagnetic radiation in the Helmholtz
representation... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
5.3.4 Justify the following recipe for obtaining a plot of F(V ) from a plot of
G(P)... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.3.5 From the first acceptable fundamental equation in Problem 1.10-1 calcu-
late the fundamental equation... . . . . . . . . . . . . . . . . . . . . . . . 103
5.3.6 From the second acceptable fundamental equation in Problem 1.10-1 cal-
culate the... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
5.3.7 The enthalpy of a particular system is H = AS2
N−1
ln(P/P0)... . . . . . 104
5.3.8 In Chapter 15 it is shown by a statistical mechanical calculation that the
fundamental equation... . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
5.3.9 Show, for the two-level system of Problem 5.3-8, that as the temperature
increases from zero... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.3.10 Show that the Helmholtz potential of a mixture of simple ideal gases is
the... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.3.11 A mixture of two monatomic ideal gases is contained in a volume V at
temperature T... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
5.3.12 A system obeys the fundamental relation (s − s0)4
= avu2
... . . . . . . . 106
5.3.13 For a particular system it is found that u = (3
2
)Pv and P = AvT4
... . . . 106
5.3.14 For a particular system (of 1 mole) the quantity (v + a)f is known to be
a function of... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.4 GENERALIZED MASSIEU FUNCTIONS . . . . . . . . . . . . . . . . . 107
5.4.1 Find the fundamental equation of a monatomic ideal gas in the represen-
tation... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.4.2 Find the fundamental equation of electromagnetic radiation (Section 3.6)...108
5.4.3 Find the fundamental equation of the ideal van der Waals fluid in the
representation ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
12

6 THE EXTREMUM PRINCIPLE IN THE LEGENDRE TRANSFORMED
REPRESENTATIONS 109
6.2 THE HELMHOLTZ POTENTIAL . . . . . . . . . . . . . . . . . . . . . . . 109
6.2.1 Calculate the pressure on each side of the internal piston in Example 1,
for ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
6.2.2 Two ideal van der Waals fluids are contained in a cylinder, separated by
an internal ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
6.2.3 Two subsystems are contained within a cylinder and are separated by an
internal piston... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
6.3 THE ENTHALPY: THE JOULE-THOMSOM OR ˝THROTTLING
PROCESS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
6.3.1 A hole is opened in the wall separating two chemically identical single-
component subsystems ... . . . . . . . . . . . . . . . . . . . . . . . . . . 111
6.3.2 A gas has the following equations of state ... . . . . . . . . . . . . . . . . 111
6.3.3 Show that for an ideal van der Waals fluid ... . . . . . . . . . . . . . . . . 111
6.3.4 One mole of a monatomic ideal gas is in a cylinder with a movable piston
on the other ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
6.3.5 Assume that the gas of Problem 6.3-4 is an ideal van der Waals fluid with
the van der Waals... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
6.4 THE GIBBS POTENTIAL; CHEMICAL REACTIONS . . . . . . . . . 112
6.4.1 One half mole of H2S, 3
4
mole of H2O, 2 moles of H2, and... . . . . . . . 113
7 MAXWELL RELATIONS 113
7.2 A THERMODYNAMIC MNEMONIC DIAGRAM . . . . . . . . . . . . 113
7.2.1 In the immediate vicinity of the state T0, v0 the volume of a particular ... 113
7.2.2 For a particular system of 1 mole, in the vicinity of a particular state ... . 114
7.2.3 Show that the relation α = 1
T
implies ... . . . . . . . . . . . . . . . . . . 114
7.3 A PROCEDURE FOR THE REDUCTION OF DERIVATIVES IN
SINGLE-COMPONENT SYSTEMS . . . . . . . . . . . . . . . . . . . . . 114
7.3.1 Thermodynamicists sometimes refer to the first TdS equationand ... . 114
7.3.2 Show that the second equation in the preceding problem leads directly ... 115
7.3.3 Calculate (∂H/∂V )T,N in terms of the standart quantities... . . . . . . . 115
7.3.4 Reduce the derivative (∂v/∂s)P ... . . . . . . . . . . . . . . . . . . . . . . 115
7.3.5 Reduce the derivative (∂s/∂f)v ... . . . . . . . . . . . . . . . . . . . . . . 115
7.3.6 Reduce the derivative (∂s/∂f)P ... . . . . . . . . . . . . . . . . . . . . . 116
7.3.7 Reduce the derivative (∂s/∂v)h ... . . . . . . . . . . . . . . . . . . . . . . 116
7.4 SOME SIMPLE APPLICATIONS . . . . . . . . . . . . . . . . . . . . . . . 116
7.4.1 In the analysis of a Joule-Thomson experiment we may be given the initial
... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
7.4.2 The adiabatic bulk modulus is defined by ... . . . . . . . . . . . . . . . . 117
7.4.3 Evaluate the change in temperature in an infinitesimal free expansion of ...117
7.4.4 Show that equation 7.46 can be written as ... . . . . . . . . . . . . . . . . 117
7.4.5 A 1% decrease in volume of a system is carried out adiabatically. Find ... 117
7.4.6 Two moles of an imperfect gas occupy a volume of 1 liter and are at ... . 118
7.4.7 Show that (∂cv/∂v)T = T(∂2
P/∂T2
)v and evaluate this ... . . . . . . . . 118
7.4.8 Show that ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
7.4.9 One mole of the system of Problem 7.4-8 is expanded isothermally ... . . 119
7.4.10 A system obeys the van der Waals equation of state. One mole ... . . . . 119
7.4.11 Two moles of O2 are initially at a pressure of 105
Pa and a temperature ...119
7.4.12 A ball bearing of mass 10g just fits in a vertical glass tube of ... . . . . . 119
7.4.13 Calculate the change in the molar internal energy in a throttling ... . . . 120
7.4.14 Assuming that a gas undergoes a free expansion and that the ... . . . . . 120
13

7.4.15 One mole of an ideal van der Waals fluid is contained in a vessel ... . . . 120
7.4.16 Assuming the expansion of the ideal van der Waals fluid of ... . . . . . . 121
7.4.17 It is observed that an adiabatic decrease in molar volume of 1% ... . . . . 121
7.4.18 A cylinder is fitted with a piston, and the cylinder contains helium ... . . 121
7.4.19 The cylinder in Problem 7.4-18 is initially filled with 1
10
mole of ... . . . . 122
7.4.20 A system is composed of I mole of a particular substance. In the ... . . . 122
7.4.21 A system is composed of I mole of a particular substance. Two ... . . . . 123
7.4.22 The constant-volume heat capacity of a particular simple system is ... . . 124
7.4.23 A system is expanded along a straight line in the P − v plane, from ... . 124
7.4.24 A nonideal gas undergoes a throttling process (i.e., a Joule-Thomson
expansion) ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
7.5 GENERALIZATIONS: MAGNETIC SYSTEMS . . . . . . . . . . . . . . 126
7.5.1 Calculate the magnetic Gibbs potentialU(T, Be) for the paramagnetic ...126
7.5.2 Repeat Problem 7.5-1 for the system with the fundamental equation given
... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
7.5.3 Calculate (∂I/∂T)s for the paramagnetic model of equation 3.66. ... . . . 126
7.5.4 Show that ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
8 STABILITY OF THERMODYNAMIC SYSTEMS 127
8.1 INTRINSIC STABILITY OF THERMODYNAMIC SYSTEMS . . . . 127
8.1.1 To establish the inequality 8.6 expand the left-hand side of 8.5 in a Taylor...127
8.1.2 Consider the fundamental equation of a monatomic ideal gas and show... 128
8.2 STABILITY CONDITIONS FOR THERMODYNAMIC POTENTIALS128
8.2.1 a) Show that in the region X 0 the function Y = Xn
is concave... . . . 128
8.2.2 Prove that... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
8.2.3 Show that stability requires equations 8.15 and... . . . . . . . . . . . . . 129
8.2.4 (Supplementary Problem) Of the five nominally-acceptable funda-
mental.... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
8.3 PHYSICAL CONSEQUENCES OF STABILITY . . . . . . . . . . . . . 130
8.3.1 Explain on intuitive grounds why cP ≥ cv and κT ≥ κS ... . . . . . . . . . 130
8.3.2 Show that the fundamental equation of a monatomic ideal gas satisfies
the... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
8.3.3 Show that the van der Waals equation of state does not satisfy the criteria...131
8.3.4 (Supplementary Problem)Show that the stability conditions 8.20-8.21
imply... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
8.3.5 (Supplementary Problem) An isochore and an adiabat intersect at... . 132
8.5 THE LE CHATELIER-BRAUN PRINCIPLE . . . . . . . . . . . . . . . . 132
8.5.1 A system is in equilibrium with its environment at a common temperature...132
9 FIRST-ORDER PHASE TRANSITIONS 133
9.1 FIRST-ORDER PHASE TRANSITIONS IN SINGLE COMPONENT
SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
9.1.1 The slopes os all three curves in Fig. 9.5 are... . . . . . . . . . . . . . . . 133
9.2 THE DISCONTINUITY IN THE ENTROPY - LATENT HEAT . . . 133
9.2.1 In a particular solid-liquid phase transition the... . . . . . . . . . . . . . 133
9.2.2 Discuss the equilibrium that eventually results if a solid is... . . . . . . . 134
9.3 THE SLOPE OF COEXISTENCE CURVES; THE CLAYPEYRON
EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
9.3.1 A particular liquid boils at 127 °C at a pressure of 800 mmHg ... . . . . . 134
9.3.2 A long vertical column is closed at the bottom and open the top ... . . . 134
9.3.3 It is found that a certain boils at a temperature of 95 °C at the top ... . . 135
9.3.4 Two weights are hung on the ends of a wire, which passes ... . . . . . . . 136
14

9.3.5 In the vicinity of the triple point the vapor presure of liquid ammonia... . 136
9.3.6 Let x be the mole fraction of solid phase in a solid-liquid two-phase system...137
9.3.7 A particular material has a latent heat of vaporization of... . . . . . . . . 138
9.3.8 Draw the phase diagram, in the Be − T plane, fora simple ... . . . . . . . 138
9.3.9 A system has cohexistence curves similar to those shown ... . . . . . . . . 139
9.4 UNSTABLE ISOTHERMS AND FIRST-ORDER PHASE TRANSI-
TIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
9.4.1 Show that the difference in molar volumes across a coexistente curve is ... 139
9.4.2 Derive the expression for vc, Pc and Tc given in Example 1 ... . . . . . . . 140
9.4.3 Using the van der Waals constants for H2O as given in table 3.1... . . . . 140
9.4.4 Show that for sufficiently low temperature the van der Waals isotherm... 141
9.4.5 Is the fundamental equation of an ideal van der Waals fluid... . . . . . . . 141
9.4.6 Explicitly derive the relationship among ṽg, ṽl, as given ... . . . . . . . . . 141
9.4.7 A particular substance satisfies yhe van der Waals equation of state... . . 142
9.4.8 Using the two points as T̃ = 0, 95 and T̃ = 1 on... . . . . . . . . . . . . . 142
9.4.9 Plot the van der Waals isotherm, in reduced variables... . . . . . . . . . . 143
9.4.10 Repeat problem 9.4-8 in the range... . . . . . . . . . . . . . . . . . . . . 143
9.4.11 Two mols of a van der Waals fluid are maintained at ... . . . . . . . . . . 143
9.6 FIRST-OERDER PHASE TRANSITIONS IN MULTICOMPONENT
SYSTEMS - GIBBS PHASE RULE . . . . . . . . . . . . . . . . . . . . . . 144
9.6.1 In a particular system, solute A and solute B are each dissolved... . . . . 144
9.6.2 If g, the molar Gibbs function, is a convex of x1, x2 ... . . . . . . . . . . . 145
9.6.3 Show that the conditions of stability in a multicomponent system ... . . . 145
9.7 PHASE DIAGRAMS FOR BINARY SYSTEMS . . . . . . . . . . . . . . 146
9.7.1 In a particular system, solute A and solute B are each dissolved... . . . . 146
9.7.2 Show that if a small fraction (−dN/N) of the material is boiled off ... . . 146
9.7.3 The phase diagram of a solution of A in B, at a pressure of 1 atm ... . . . 147
10 CRITICAL PHENOMENA 148
10.6 SCALING AND UNIVERSALITY . . . . . . . . . . . . . . . . . . . . . . 148
10.6.1 Show that the following identities hold among ... . . . . . . . . . . . . . . 148
10.6.2 Are classical values of the critical exponents consistent... . . . . . . . . . 148
11 THE NERNST POSTULATE 149
11.1 NERNST’S POSTULATE, AND THE PRINCIPLE OF THOMSEN
AND BERTHELOT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
11.1.1 Does the two-level system of problem 5.3-8 satisfy ... . . . . . . . . . . . 149
13 PROPERTIES OF MATERIALS 149
13.2 CHEMICAL REACTIONS IN IDEAL GASES . . . . . . . . . . . . . . 149
13.2.1 How is the equilibrium constant of the reaction in the Example ... . . . . 149
13.2.2 What are the mole fractions of the reaction in the Example ... . . . . . . 149
13.2.3 In the Example, what would the final mole fractions ... . . . . . . . . . . 150
13.2.4 In an ideal gas reaction an increase in presure at constant temperature ... 150
13.2.5 In an ideal gas reaction an increase in presure at constant temperature ... 151
13.2.6 At temperatures above ∼ 500 K phosphorus pentachloride dissociates ... 151
13.2.7 A system containing 0.02 Kg of CO and 0.02 Kg of O2... . . . . . . . . . 152
13.2.8 Apply equation 13.8 to a single-component general ideal gas ... . . . . . . 153
13.2.9 An experimenter finds that water vapor is 0.53% dissociated... . . . . . . 153
13.3 SMALL DEVIATIONS FROM ˝IDEALITY -THE VIRIAL EXPAN-
SION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
13.3.1 In a thermostatistical model in which each atom is treated as ... . . . . . 153
13.3.2 Expand the mechanical equation of state of Van de Waals gas ... . . . . . 154
15

13.3.3 Show that the second virial coefficient of gaseous nitrogen (Fig. 13.2) ... . 154
13.3.4 The simplest analytic form suggested by the qualitative shape of B(T) ... 154
13.3.5 A porous plugexperiment is carried out by installing a porous plug in ...155
13.5 DILUTE SOLUTIONS: OSMOTIC PRESSURE AND VAPOR PRES-
SURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
13.5.1 Assuming the latent heat of vaporization of a fluid to be constant ... . . . 156
13.5.2 One hundred grams of a particular solute are dissolved in one ... . . . . . 156
13.5.3 If 20 grams of sugar (C12H22O11), are dissolved in 250 cm3
of water ... . 156
15 STATISTICAL MECHANICS IN THE ENTROPY REPRESENTATION:
THE MICROCANONICAL FORMALISM 157
15.1 PHYSICAL SIGNIFICANCE OF THE ENTROPY FOR CLOSED
SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
15.1.1 A system is composed of two harmonic oscillators each of natural fre-
quency ω0 ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
15.1.2 A system is composed of two harmonic oscillators of natural frequencies
ω0 e 2ω0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
15.2 THE EINSTEIN MODEL OF A CRYSTALLINE SOLID . . . . . . . . 158
15.2.1 Calculate the molar heat capacity of the Einstein model by equation 15.7. 158
15.2.2 Obtain an equation for the mean quantum number n̄ of an Einstein ... . . 158
15.2.3 Assume that the Einstein frequency ω0 for a particular crystal depends
upon the molar volume... . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
15.3 THE TWO-STATE SYSTEM . . . . . . . . . . . . . . . . . . . . . . . . . . 160
15.3.1 In the two-state model system of this section suppose the excited state
energy ε . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
15.4 A POLYMER MODEL-THE RUBBER BAND REVISITED . . . . . . 160
15.4.1 Is the sign correct in equation 15.19 ? Explain . . . . . . . . . . . . . . 160
15.4.2 Eliminate U/ε between equations 15.23 and 15.24 and show that the
formal solution is equation 15.25... . . . . . . . . . . . . . . . . . . . . . 160
15.4.3 A rubber band consisting of n polymer chains is stretched from zero ... . 161
15.4.4 Calculate the heat capacity at constant length for a rubber bandconsisting
of n polymer... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
15.4.5 Calculate the heat capacity at constant length for a rubber bandconsisting
of n polymer... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
15.5 COUNTING TECHNIQUES AND THEIR CIRCUMVENTION; HIGH
DIMENSIONALITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
15.5.1 To stablish equation 15.29 le Ωn be the hyperplane in n dimensions. ... . 162
15.5.2 Recalling that lim
x→0
(1 + x)1/x
= e... . . . . . . . . . . . . . . . . . . . . . 162
15.5.3 Calculate the fraction of the hypervolume between the radii 0.9r and r ... 162
16 THE CANONICAL FORMALISM: STATISTICAL MECHANICS IN HELMHOLTZ
REPRESENTATION 162
16.1 THE PROBABILITY DISTRIBUTION . . . . . . . . . . . . . . . . . . . 162
16.1.1 Show the equation 16.13 is equivalent to U = F + TS... . . . . . . . . . . 162
16.1.2 From the canonical algorithm expressed by equations... . . . . . . . . . . 162
16.1.3 Show that S/kB = β2
∂F/∂β and thereby express S... . . . . . . . . . . . 163
16.1.4 Show that cv = −β(∂s/∂β)v and thereby express cv in ... . . . . . . . . . 163
16.2 ADDITIVE ENERGIES AND FACTORIZABILITY OF THE PAR-
TITION SUM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
16.2.1 Consider a system of three particles, each different. The first ... . . . . . 163
16.2.2 Show that for the two level system the Helmholtz potential calculated in...163
16.2.3 Is the energy additive over the particles of a gas if the particles are ... . . 164
16

16.2.4 Calculate the heat capacity per mode from the fundamental equation ... . 164
16.2.5 Calculate the energy per mode from equation 16.24. What is the leading...164
16.2.6 A binary alloy is composed of ÑA atoms of type A and ÑB... . . . . . . . 164
16.2.7 A paramagnetic salt is composed of 1 mole of noninteracting ions, each ... 165
16.3 INTERNAL MODES IS A GAS . . . . . . . . . . . . . . . . . . . . . . . . 165
16.3.1 Calculate the average rotational energy per molecule and the rotational ...165
16.3.2 Calculate the rotational contribution of the Helmholtz potential per ... . 166
16.3.3 A particular heteronuclear diatomic gas has one vibrational mode, of ... . 166
16.4 PROBABILITIES IN FACTORIZABLE SYSTEMS . . . . . . . . . . . . 166
16.4.1 The probability that the ith element is in its jth orbital... . . . . . . . . 166
16.4.2 Demonstrate the equivalence of the fundamental equations found in... . . 166
16.5 STATISTICAL MECHANICS OF SMALL SYSTEMS: ENSEMBLES 167
16.5.1 The energies of the orbital states of a given molecule ar such that ... . . . 167
16.5.2 A hydrogen atim in equilibrium with a radiation field at temperature ... . 167
16.5.3 A small system has two normal modes of vibration, with natural... . . . . 167
16.5.4 DNA, the genetic molecule deoxyribonucleic acid, exists as a ... . . . . . 168
16.5.5 Calculate the probability that a harmonic oscillator of natural ... . . . . . 168
16.5.6 A small system has two energy levels, of energies 0 and ε... . . . . . . . . 169
16.5.7 Two simple harmonic oscillators, each of natural frequency ω... . . . . . . 169
16.6 DENSITY OF STATES AND DENSITY OF ORBITAL STATES . . . 170
16.6.1 Show that the number of orbital states in the energy interval... . . . . . . 170
16.6.2 For the particles of a gas ε = p2
/2m = (~2
/2m)k2
... . . . . . . . . . . . . 170
16.6.3 For excitations obeying the spectral relation ω = Akn
, n 0... . . . . . . 170
16.7 THE DEBYE MODEL OF NONMETALLIC CRYSTALS . . . . . . . . 170
16.7.1 Calculate the energy of a cristal in the Debye approximation. Show ... . . 170
16.7.2 Calculate the entropy of a cristal in the Debye approximation, and... . . 171
16.7.3 The frequency ω(λ) of the vibrational mode of wave ... . . . . . . . . . . 171
16.8 ELECTROMAGNETIC RADIATION . . . . . . . . . . . . . . . . . . . . 171
16.8.1 Show that including the zero-point energies˝of the electromagnetic... . . 171
16.8.2 Show that the energy per unit volume of electromagnetic radiation... . . 172
16.8.3 Evaluating the number of photons per unit volume in the frequency... . . 172
16.8.4 Since radiation within a cavity propagates isotropically with velocity c... 172
16.10THE CLASSICAL IDEAL GAS . . . . . . . . . . . . . . . . . . . . . . . . 172
16.10.1Show that the calculation of Z = zÑ
, with z given by equation... . . . . . 172
16.10.2Show that the fundamental equations of a “multicomponent simple gas”... 172
16.10.3The factors (1/Ñ1!)(1/Ñ2!) in equation 16.74 give and additive ... . . . . 173
16.10.4Consider a particle of mass m in a cubic container of volume V... . . . . 173
16.10.5A single particle is contained in a vessel of volume 2V which... . . . . . . 173
17 ENTROPY AND DISORDER: GENERALIZED CANONICAL FORMU-
LATIONS 174
17.1 ENTROPY AS A MEASURE OF DISORDER . . . . . . . . . . . . . . . 174
17.1.1 Consider the quantity x ln x in the limit x → 0.Show... . . . . . . . . . . 174
17.1.2 Prove that the disorder, defined in equation 17.6, is nonnegative... . . . . 174
17.1.3 Prove that the quantity −k
P
j fj is maximum if all the... . . . . . . . . . 174
17.2 DISTRIBUTIONS OF MAXIMAL DISORDER . . . . . . . . . . . . . . 175
17.2.1 Show that the maximum value of disorder, as calculate in this section... . 175
17.2.2 Given the identification of the disorder as the entropy, and of ... . . . . . 175
17.3 THE GRAND CANONICAL FORMALISM . . . . . . . . . . . . . . . . 175
17.3.1 Calculate (∂ log Z/∂β)βµ directly from equation 17.27... . . . . . . . . . . 175
17.3.2 A system is contained in a cylinder with diathermal impermeable walls... 175
17.3.3 For the surface adsorption model of the preceding example, investigate... 175
17

17.3.4 Suppose the adsorption model to be augmented by assuming that two... . 176
17.3.5 Calculate the fundamental equation of the polymer model of section ... . 177
17.3.6 A system contains Ñ sites and Ñ electrons. At a given site... . . . . . . . 177
17.3.7 Carbon monoxide molecules (CO) can be adsorbed at specific sites on a
solid... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
18 QUANTUM FLUIDS 178
18.1 QUANTUM PARTICLES: A “FERMION PRE-GAS MODEL” . . . . 178
18.1.1 Obtain the mean number of particles in the fermion pre-gras model... . . 178
18.1.2 The entropy of a system is given by S = −kB
P
j fj, where fj... . . . . . 178
18.1.3 Apply equation 17.27 for U to the fundamental equation of the... . . . . 179
18.1.4 Show that df/dε = −β/4 at ε = µ. With this result ... . . . . . . . . . . 179
18.1.5 Show that Fig. 17.2 (of f(ε, T) as a function of ε) is ... . . . . . . . . . . 179
18.1.6 Suppose f(ε, T) is to be approximated as a function of... . . . . . . . . . 179
18.2 THE IDEAL FERMI FLUID . . . . . . . . . . . . . . . . . . . . . . . . . . 180
18.2.1 Prove equations c, g, h, i and j of table 18.1 (for fermions only)... . . . . 180
18.3 THE CLASSICAL LIMIT AND THE QUANTUM CRITERION . . . 180
18.3.1 Calculate the definite integrals appearing in equations 18.25 and... . . . . 180
18.3.2 Validate the interpretation of λT as the “thermall wavelength”by... . . . . 180
18.4 THE STRONG QUANTUM REGIME: ELECTRONS IN A METAL . 180
18.4.1 Show that equation 18.32 can be interpreted as µ0 = ~2
k2
f where... . . . . 180
18.4.2 Derive equation 18.35 by the following sequence of operations: ... . . . . 180
18.6 NONCONSERVED IDEAL BOSON FLUIDS: ELECTROMAGNE-
TIC RADIATION REVISITED . . . . . . . . . . . . . . . . . . . . . . . . 181
18.6.1 Calculate the number of photons in the lowest orbital state... . . . . . . . 181
18.6.2 (a) In applying the grand canonical formalismo...(b) Denoting the ... . . 181
18.7 BOSE CONDENSATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
18.7.1 Show that equations 18.56 and 18.58, for Ñe and U, respectively... . . . . 181
18.7.2 Show that F3/2(1), F5/2(1) and F0
5/2(1) are all finite... . . . . . . . . . . . 181
18.7.3 Show that the explicit inclusion of the orbital ground state... . . . . . . . 182
19 FLUCTUATIONS 182
19.2 MOMENTS OF THE ENERGY FLUCTUATIONS . . . . . . . . . . . . 182
19.2.1 A molecule has a vibrational mode of natural frequency ω. The ... . . . . 182
19.2.2 Calculate the third central moment for the molecule in the preceding
problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
19.2.3 Calculate the mean square deviation of the energy contained within a... . 182
19.3 GENERAL MOMENTS AND CORRELATION MOMENTS . . . . . . 183
19.3.1 An ideal gas is in contact with a thermal and a pressure reservoir... . . . 183
19.3.2 Repeat Problem 19.3-1 for a van der Waals gas (recall Problem 3.8-3). . 183
19.3.3 A conceptual subsystem of N moles in a single-component simple... . . . 183
19.3.4 What is the order of magnitude of the mean square deviation of the... . . 184
19.3.5 Consider a small volume V within a two-component simple system... . . . 184
19.3.6 Consider a small quantity of matter consisting of a fixed number N... . . 184
19.3.7 Show that the density fluctuations of an ideal gas are given by... . . . . . 185
19.3.8 Show that the relative root mean square deviation in density of ... . . . . 185
19.3.9 The dielectric constant ε of a fluid varies with... . . . . . . . . . . . . . . 185
19.3.10If light of intensity I0 is incident on a region of volume V, which... . . . . 185
19.3.11The classical theory of fluctuations, due to Einstein, proceeds from... . . 186
18

20 VARIATIONAL PROPERTIES, PERTURBATION EXPANSIONS, AND
MEAN FIELD THEORY 186
20.1 THE BOGOLIUBOV VARIATIONAL THEOREM . . . . . . . . . . . . 186
20.1.1 Derive equation (h) of Example 1, fisrt showing that for a harmonic... . . 186
20.1.2 Solve the quartic potential problem of Example 2 assuming the... . . . . 186
20.1.3 Complete Example 2 by writing the Helmholtz potential F(T, α)... . . . . 187
20.1.4 Consider a particle in a quadratic potential V (x) = ax2
/2a2
. Despite... . 187
20.2 MEAN FIELD THEORY . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
20.2.1 Formulate the exact solution of the two-particle Ising model... . . . . . . 188
20.2.2 Formulate mean field theory for the three state Ising model... . . . . . . . 188
20.2.3 For the Heisenberg ferromagntic model the Hamiltonian is... . . . . . . . 189
20.2.4 A metallic surface is covered by a monomolecular layer of Ñ ... . . . . . . 189
20.2.5 Solve the preceding problem if the molecules can exist in three steric... . 190
20.2.6 In the classic Heisenberg model each spin can take any orientation... . . . 191
20.2.7 2Ñ two-valued Ising spins are arranged sequentially on a circle... . . . . . 191
20.2.8 Consider a sequence of 2Ñ alternating A sites and B sites... . . . . . . . 192
21 POSTLUDE: SYMMETRY AND THE CONCEPTUAL FOUNDATIONS
OF THERMOSTATISTICS 192
21.1 STATISTICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
21.1.1 The probability of throwing a “seven ”on two dice can be viewed as ... . . 192
21.1.2 Associate the value +1 with one side of a coin (head) and... . . . . . . . 192
21.5 BROKEN SYMMETRY AND GOLDSTONE’S THEOREM . . . . . . 192
21.5.1 Draw a longitudinal vibrational mode in a one-dimensional system... . . . 192
19

1 THE PROBLEM AND THE POSTULATES [SUMÁRIO]
1.3 THE COMPOSITION OF THERMODYNAMIC SYSTEMS
[SUMÁRIO]
1.3.1 One tenth of a kilogram of NaCl and 0.15 kg of sugar (C12H22O11)... [SUMÁRIO]
Solução:
Os pesos moleculares são :
Na : 0, 023 kg; Cl : 0, 035 kg;
C : 0, 012 kg; O : 0, 016 kg;
Então temos que
NaCl : 0, 058 kg
C12H12O11 : 0, 34 kg
Temos que o número de mols dos três componentes do sistema é :
0, 1 kg NaCl = 1, 7 mols
0, 15 kg C12H12O11 = 0, 44 mol
0, 5 kg H2O = 28 mol
∴ Total de mol = 30
As frações molares são :
Fração molar de NaCl = 0, 057
Fração molar de C12H12O11 = 0, 015
Fração molar de H2O = 0, 93
∴ V olume molar =
0, 55 × 10−3
m3
30 mol
= 18 × 10−6
m3
/mol
1.3.2 Naturally occurring boron has an atomic mass of 10.811 g... [SUMÁRIO]
Solução:
X × 10, 0129 + (1 − X) × 11, 0093 = 10, 811
Assim a fração molar de 10
B é
X = 0, 199
1.3.3 Twenty cubic centimeters each of ethyl alcohol... [SUMÁRIO]
Solução:
Álcool etílico, C2H5OH : densidade = 0, 79 g/cm3
Álcool metílico CH3OH : densidade = 0, 81 g/cm3
Água H2O : densidade = 1 g/cm3
NC2H5OH = (0, 79 g/cm3
) × (20 cm3
)/(46 g/mol) = 0, 34 mol
NCH3OH = (0, 81 g/cm3
) × (20 cm3
)/(32 g/mol) = 0, 51 mol
NH2O = (1 g/cm3
) × (20 cm3
)/(18 g/mol) = 1, 11 mol
O número de mols é
∴ N = NC2H5OH + NCH3OH + NH2O = 1, 96 mol
20

e as frações molares dos componentes do sistema são
XC2H5OH =
0, 34 mol
1, 96 mol
= 0, 17
XCH3OH =
0, 51 mol
1, 96 mol
= 0, 26
XH2O =
1, 11 mol
1, 96 mol
= 0, 57
1.3.4 A 0.01 kg sample is composed of 50 molecular percent H2... [SUMÁRIO]
Solução:
Seja N o número de mols na amostra inicialmente. Portanto
massa do H2 = 0, 5N × 2 × 10−3
kg = 10−3
N kg
massa do HD = 0, 3N × 2 × 10−3
kg = 0, 9 × 10−3
N kg
massa do D2 = 0, 2N × 4 × 10−3
kg = 0, 8 × 10−3
N kg
Massa total = 0, 01 kg = (1 + 0, 9 + 0, 8) × 10−3
N kg
∴ N =
10−2
2, 7 × 10−3
= 3, 7 mol
Número inicial de mols de D2 = 0, 2 × 3, 7 = 0, 74 mol
Seja N0
o número de mols de D2 a ser adicionado.
Então
0, 3 =
0, 74 mol + N0
3, 7 mol + N0
N0
= 0, 53mol
∴ Massa de D2 = 0, 53 × 4 × 10−3
kg = 2, 1 × 10−3
kg
1.3.5 A solution of sugar (C12H22O11) in water is 20% sugar by weight... [SUMÁRIO]
Solução:
Massa da Água(water) = M(water)
= 0, 8M(solution)
Massa de açúcar(sugar) = M(sugar)
= 0, 2M(Solution)
N(water)
=
0, 8M(solution)
0, 018
= 44, 44M(solution)
N(sugar)
=
0, 2M(solution)
0, 34
= 0, 588M(solution)
∴ Xsugar
=
N(sugar)
N(sugar) + N(water)
=
0, 588M(solution)
0, 588M(solution) + 44, 44M(solution)
= 0, 013
Ou
∴ Xsugar
= 1, 3%
21

1.3.6 An aqueous solution of an unidentified solute has a total mass... [SUMÁRIO]
Solução:
Seja Ms a massa molecular (peso molecular) do soluto e N o número total de mols na amostra
original
∴ (0, 1N) × Ms + (0, 9N) × 18 × 10−3
= 0, 1029 × 10−3
kg
Depois de adicionar 2 mols (= 36 × 10−3
kg) de H2O :
0, 1N
N + 2
= 0, 07 ou N = 4, 67
Onde Ms = 0, 0584 kg
Consequentemente NaCl é um palpite razoável.
1.3.7 One tenth of a kg of an aqueous solution of HCl is poured... [SUMÁRIO]
Solução:
Massa molar de HCl = 1, 01 + 35, 45 = 36, 46
Massa molar de NaOH = 22, 99 + 16 + 1, 01 = 40, 00
Massa molar de H2O = 2, 02 + 16 = 18, 02
Massa molar de NaCl = 22, 99 + 35, 45 = 58, 44
Solução ácida: Massa= 0, 9N×MH2O+0, 1N×MHCl = (0, 9×18, 02+0, 1×36, 46)N×10−3
=
0, 1 kg
N = 5, 034 mol
∴ N0
H2O = 0, 9N = 4, 53 mol N0
HCl = 0, 1N = 0, 50 mol
Solução básica: 0, 2 kg = (0, 75 × 18, 02 + 0, 25 × 40) × 10−3
= 23, 52N0
N0
= 8, 503 mol
∴ N0
H2O =
3
4
× 8, 503 = 6, 38 mol N0
NaOH = 0, 25N0
= 2, 13 mol
Então
NaOH + HCl −→ NaCl + H2O
O HCl é a quantidade limite, e apenas 0, 5 mol de NaOH podem reagir.
A composição final é
NaCl : 0, 5 mol NaOH : 1, 63 mol HCl : 0 mol
H2O : 4, 53 + 6, 38 + 0, 5 = 11, 41 mol
O número total de mols é = 13, 54, que está de acordo com o número original (Note que a
reação química preserva o número total de mols). Então
XNaCl = 0, 04 ; XNaOH = 0, 12 ; XH2O = 0, 84
22

1.8 QUANTITATIVE DEFINITION OF HEAT-UNITS [SUMÁRIO]
1.8.1 For the system considered in Example 1, calculate... [SUMÁRIO]
Solução:
Denote este estado por E; Note que é a meio caminho entre os estados B e D. Note também
que ao longo da linha de V constante, a energia é diretamente proporcional a P. Isto é
dU =
3
2
V dP (V constante)
Consequentemente
UE =
1
2
(UD + UB) =
1
2
(967, 5 − 112, 5)
Tomando UA = 0 temos
UE = 427, 5J
1.8.2 Calculate the heat transferred to the system considered in... [SUMÁRIO]
Solução:
10⁵
5x10⁴
10⁴
V
P
10⁻³ 8x10⁻³
A
E
W = −
R
Pdv = −(Área sombreada)
A área sombreada é um trapézio.
Áreatrapézio =
B + b
2
× h
W=−7, 5 × 104
× 7 × 10−3
= −525J
UE = 427, 5J (Do problema 1.8-1)
UE − UA = WA→E + QA→E
427, 5 = −525 + QA→E
∴ QA→E = 952, 5J
23

1.8.3 For a particular gaseous system it has been determined that... [SUMÁRIO]
Solução:
P(MPa)
V(m³)
0,01 0,03
0,5
0,2
A B
C
Faça UA ≡ 0,
Então U = 2, 5PV − 5 × 103
J
∴ UB = 104
J, UC = 7, 5 × 103
J
WAB = −
R B
A
PdV = −P∆V = −4 × 103
Joules
e UB − UA = WAB + QAB
∴ QAB = 14 × 103
J (Um fluxo de energia na
forma de calor entra no sistema)
de forma similar :
WBC = −
R C
B
PdV = área abaixo da linha BC = 7 × 103
J
QBC = UC − UB − WBC = −2, 5 × 103
− 7 × 103
= −9, 5 × 103
Joules
Finalmente WCA = 0
UA−UC = QCA+WCA, QCA = −7, 5×103
J (Fluxo de energia na forma de calor sai do sistema)
Para checar, note que o trabalho realizado no ciclo é WAB +WBC +WCA = 3×103
J e a energia
na forma de calor que entrou no ciclo é 14 × 103
J − 9, 5 × 103
J − 7, 5 × 103
J = −3 × 103
J. A
Soma do trabalho e da energia na forma de calor transferidos é zero, exatamente como deveria
ser.
Agora nos resta calcular Q e W para o processo de A para B ao longo da parábola
P = 105
+ 109
(V − 0, 02)2
P(MPa)
V(m³)
0,01 0,03
0,2
A B
0,1
WAB = −
R 0,03
0,01
PdV
= −
R 0,03
0,01
[105
+ 109
(V − 0, 02)2
]dV
WAB = −

105
(VB − VA) +
109
(V − 0, 02)3
3
0,03
0,01
Então WAB = −2, 7 × 103
J
Q = ∆U − W = 104
+ 2, 7 × 103
∴ Q = 12, 7 × 103
Joules
24

1.8.4 For the system of Problem 1.8-3 find the equation of the adiabats... [SUMÁRIO]
Solução:
U = 2, 5PV + const.
Então dU = 2, 5PdV + 2, 5V dP
Mas ao longo de uma adiabata, temos que dU = −PdV
∴ 2, 5PdV + 2, 5V dP = −PdV , assim
7
dV
V
+ 5
dP
P
= 0
Integrando, ficamos com 7 ln

V
V0

+ 5 ln

P
P0

= 0
∴ V 7
P5
= V 7
0 P5
0
1.8.5 The energy of a particular system, of one mole, is given by ... [SUMÁRIO]
Solução:
dU = 2APV dP + AP2
dV = −PdV
∴ A
dP
1 + AP
= −
1
2
dV
V
ln(1 + AP) = −
1
2
ln V + const.
(1 + AP)V 1/2
= const
Obs : Para onde foi o A que multiplica
dP
1 + AP
na segunda linha?
Figura 1: reprodução da solução original, feita pelo próprio Callen, onde ele questiona sobre o
“A”
25

1.8.6 For a particular system it is found that if the volume is kept ... [SUMÁRIO]
Solução:
P
V
P'
P0
V0
(P',V0)
(P0,V0)
(P,V)
Primeiro, encontre a energia ao longo da isocórica V = V0.
Ao longo da isocórica nenhum trabalho é realizado, portanto :
U(P0
, V0) = U0 + Q0
= U0 + A(P0
− P0)
Agora considere o ponto (P, V ). A adiabata através desse
ponto intercepta a isocórica V0 em P0
= P(V/V0)γ
.
Consequentemente
U(P, V ) − U(P0
, V0) = −
R V
V0
P00
dV 00
= −P
R V
V0
(V/V 00
)γ
dV 00
= −PV γ [V 1−γ
− V 1−γ
0 ]
1 − γ
= −
P
1 − γ

V − V0

V
V0
γ
ou U(P, V ) = U0 + A

P

V
V0
γ
− P0

+
PV
γ − 1

1 −

V
V0
γ−1
#
1.8.7 Two moles of a particular single-component system are found... [SUMÁRIO]
Solução:
Resposta U =
2
N
APV 2
26

1.8.8 (Supplementary Problem) A particular single-component system of one
mole has adiabats of the form P5/3
= constant (that is, the pressure and the
volume obey the given relationship when when the system is enclosed by an
adiabatic, impermeable wall). The system is fitter with a stirrer, as in the
example of this section. If the system is adiabatically insulated and main-
tained at constant volume, and if an amount of work dW (= torque × dθ)
is transmitted (by an external torque turning the stirrer through an angle
dθ), the pressure of the system is observed to increase by an amount dP,
where
(AV + BV5/3
)dP = d̄W (torque × dθ)
Find the internal energy as a function of P, V and N. [SUMÁRIO]
Solução:
P
v
P'
P0
v0
(P',v0)
(P0,v0)
(P,v)
Primeiro encontramos a escala de energia ao
longo da isocórica v = v0:
u = u0 + (Av0 + Bv
5/3
0 ) × (P − P0)
Agora escolhemos um ponto arbitrário P1, v1.
A adiabata através desse ponto intercepta a
isocórica v = v0 no ponto P0
= P1

v1
v0
5/3
.
O trabalho realizado no processo de levar
o sistema de (P0
, v0) para (P1, v1) ao longo
da adiabata é :
W = −
Z v1
v0
Pdv = −
Z v1
v0
P1
v1
v
5/3
dv = −
3
2
P1v1

v1
v0
2/3
− 1
#
então
u(P1, v1) = u(P0
, v0) −
3
2
P1v1

v1
v0
2/3
− 1
#
E portanto
u(P1, v1) = u0 + (Av0 + Bv
5/3
0 )

P1

v1
v0
5/3
− P0
#
−
3
2
P1v1

v1
v0
2/3
− 1
#
Então, para um sistema de N moles (e também retirando os subscritos em P1 e v1):
U(P, V, N) = Nu0 + N(Av0 + Bv
5/3
0 )

P

v
v0
5/3
− P0
#
−
3
2
NPv

v
v0
2/3
− 1
#
27

1.10 THE ENTROPY MAXIMUM POSTULATES [SUMÁRIO]
1.10.1 The following ten equations are purported to be fundamental... [SUMÁRIO]
Solução:
Precisamos encontrar as 5 equações supostamente fundamentais que são fisicamente inadimis-
síveis e indicar quais postulados são violados por cada uma.
Fisicamente aceitável
(a) S =

R2
v0θ
1/3
(NV U)1/3
Não viola nenhum postulado.
Viola algum postulado
(b) S =

R
θ2
1/3
NU
V
2/3
Não é homogênea de primeira ordem.
(c) S =

R
θ
1/2
NU +
RθV 2
v2
0
1/2
(d) S =

R2
θ
v3
0

V 3
/NU Não é uma função monotonicamente crescente e
também viola o postulado de Nernst.
(e) S =

R3
v0θ2
1/5
[N2
V U2
]1/5
(f) S = NRln(UV/N2
Rθv0) Viola o postulado de Nernst.
(g) S =

R
θ
1/2
[NU]1/2
exp(−V 2
/2N2
v2
0) Não viola nenhum postulado
(h) S =

R
θ
1/2
(NU)1/2
exp

−
UV
NRθv0

Não é homogênea de primeira ordem nem
uma função monotonicamente crescente
de U
28

(i) U =

v0θ
R

S2
V
exp(S/NR) Não viola nenhum postulado.
(j) U =

Rθ
v0

NV

1 +
S
NR

exp(−S/NR) Não é homogênea de primeira ordem nem
uma função monotonicamente crescente
de U
Obs: O postulado de Nernst ao qual o autor se refere é o postulado reformulado por Planck,
e que o Callen chama de Postulado IV. Discussões mais detalhadas a respeito desse postulado
podem ser vistas na página 30 do capítulo 1 e no capítulo 11 do livro.
Postulado IV.A entropia de qualquer sistema se anula no estado para o qual
(∂U/∂S)V,N1,...,Nr = 0 (isto é, no zero de temperatura)
1.10.2 For each of the five physically acceptable fundamental equations... [SUMÁRIO]
Solução:
(a) U =
1
NV

v0θ
R2

S3
(c) U =
θS2
NR
−
RθV 2
Nv2
0
(e) U = N−1
V −1/2

v0θ
R3
1/2
S5/2
(g) U =

θS2
NR

exp

V 2
N2v2
0

(i) U =

v0θ
R

S2
V
exp

S
NR

29

1.10.3 The fundamental equation of a system A is S =

R2
v0θ
1/3
(NVU)1/3
and
similarly... [SUMÁRIO]
Solução:
SA =

R2
v0θ
1/3
(NAVAUA)1/3
VA = 9 × 10−6
m3
, NA = 3
SB =

R2
v0θ
1/3
(NBVBUB)1/3
VB = 4 × 10−6
m3
, NB = 2
UA + UB = 80J
Faça S0
=
S
R2v0θ
∴ S0
A = 3 × 10−2
U
1/3
A e S0
B = 2 × 10−2
U
1/3
B
S0
= S0
A + S0
B = 3 × 801/3
× 10−2
×

ξ1/3
+
2
3
(1 − ξ)1/3

Onde ξ = UA/80
1 − ξ = UB/80
Para maximizar fazemos
dS0
dξ
= 0
dS0
dξ
= 3 × 801/3
× 10−2
×

1
3
ξ−2/3
−
2
9
(1 − ξ)−2/3

= 0
Então
ξ =

2
3
−3/2
(1 − ξ) ⇒ ξ = 0, 6475295549 ou arredondando ξ = 0, 65
∴ UA = 80ξ = 52J e UB = 80(1 − ξ) = 28J
________
3x(80)1/3
x10-2
S'
Figura 2: Entropia como função de UA/(UA + UB) com o máximo em ξ = 0, 65
30

2 THE CONDITIONS OF EQUILIBRIUM [SUMÁRIO]
2.2 EQUATIONS OF STATE [SUMÁRIO]
2.2.1 Find the three equations of state for a system with the fundamental equation
U =

v0θ
R2

S3
NV
... [SUMÁRIO]
Solução:
T =

∂U
∂S

= 3

v0θ
R2

S2
NV
−P =

∂U
∂V

= −

v0θ
R2

S3
NV 2
µ =

∂U
∂N

= −

v0θ
R2

S3
N2V
Para corroborar com o fato de que as três equações de estado são homogêneas de ordem zero
(Isto é, T, P e µ são parâmetros intensivos) fazemos :
T(λS, λV, λN) = 3

v0θ
R2

(λS)2
(λN)(λV )
= 3

v0θ
R2

S2
NV
= T(S, V, N)
−P(λS, λV, λN) = −

v0θ
R2

(λS)3
(λN)(λV )2
= −

v0θ
R2

S3
NV 2
= −P(S, V, N)
µ(λS, λV, λN) = −

v0θ
R2

(λS)3
(λN)2(λV )
= −

v0θ
R2

S3
N2V
= µ(S, V, N)
2.2.2 For the system of problem 2.2-1 find µ as a function of T, V e N [SUMÁRIO]
Solução:
Eliminando S das equações de estado T(S, V, N) e µ(S, V, N) do problema 2.2-1,
µ = −3−3/2

v0θ
R2
−1/2
T3/2
v1/2
2.2.3 Show by a diagram (drawn to arbitrary scale) the dependence of... [SUMÁRIO]
Solução:
Eliminando S das equações de estado T(S, V, N) e −P(S, V, N) do problema 2.2-1
P =
1
3

R2
3v0θ
1/2
NT3
V
1/2
31

P
V
T1
T2
T2T1
Figura 3: Isotermas no diagrama PV
u =

θ
R

s2
−

Rθ
v2
0

v2
[SUMÁRIO]
Solução:
u =

θ
R

s2
−

Rθ
v2
0

v2
ou U =
θ
R
S2
N
−
Rθ
v2
0
V 2
N
T =

∂u
∂s

v
= 2
θ
R
s , P = −

∂u
∂v

s
= 2
Rθ
v2
0
v
µ =
∂U
∂N
= −
θ
R
S2
N2
+
Rθ
v2
0
V 2
N2
= −
θ
R
s2
+
Rθ
v2
0
v2
∴ µ = −u
2.2.5 Express µ as a function of T and P for the system of problem 2.2-4 [SUMÁRIO]
Solução:
Eliminando S e v das três equações de estado do problema anterior, temos que :
µ = −
1
4
R
θ
T2
+
v2
0
4Rθ
P2
32

u =

v0θ
R

s2
v
es/R
[SUMÁRIO]
Solução:
T =

∂u
∂s

v
= 2
v0θ
R
s
v
es/R
+
v0θ
R2
s2
v
es/R
−P =

∂u
∂v

s
= −
v0θ
R
s2
v2
es/R
Para encontrar µ, reescreva a equação fundamental como U =

v0θ
R

S2
V
eS/NR
∴ µ =

v0θ
R

S2
V
eS/NR
×

−
S
N2R

= −
v0θ
R2
S3
N2
eS/NR
2.2.7 A particular system obeys the relation u = Av−2
exp(s/R)... [SUMÁRIO]
Solução:
T =
∂u
∂s
=
A
R
v−2
es/R
P = −
∂u
∂v
= 2Av−3
es/R
Com s constante, então
Tf
T0
=
v2
0
v2
f
!
e
Pf
P0
=

v0
vf
3
∴
Tf
T0
=

Pf
P0
2/3
Ou Tf =
T0
22/3
= 0, 63T0 ⇒ Tf = 0, 63T0
2.2.8 Show that, in analogy with equation 2.25, for a system with r components...
[SUMÁRIO]
Solução:
Seja u = u(s, v, x1, ..., xr), xj ≡ Nj/N, N ≡
P
j=1
Nj
du =
∂u
∂s
ds +
∂u
∂v
dv +
r
P
j=1
∂u
∂xj
dxj
= Tds − Pdv +
r
P
j=1
µjdxj
33

mas
r
P
j=1
dxj = 0 ou dxr = −
r−1
P
j=1
dxj
∴ du = Tds − Pdv +
r−1
X
j=1
(µj − µr)dxj
2.2.9 Show that if a single-component system is such that PVk
is constant...
[SUMÁRIO]
Solução:
Nos é dado que PV k
= g(s), onde g é uma função arbitrária.
Agora seja u = u(s, v) e P = −
∂u
∂v
= −
g(s)
vk
Integrando,
u = −
g(s)
(k − 1)vk−1
+ h(s), h é uma função arbitrária
∴ u =
Pvk
(k − 1)vk−1
+ h(s) =
Pv
k − 1
+ h(s)
Mas de Pvk
= g(s) sabemos que s é uma função de Pvk
∴ u =
Pv
k − 1
+ f(Pvk
), f é uma função arbitrária
ou U =
1
k − 1
PV + Nf(PV k
/Nk
)
34

2.3 ENTROPIC INTENSIVE PARAMETERS [SUMÁRIO]
2.3.1 Find the three equations of state in the entropy representation for... [SUMÁRIO]
Solução:
s =
R3/2
v
1/2
0 θ
!2/5
u2/5
v1/5
1
T
=
∂s
∂u
=
2
5
R3/2
v
1/2
0 θ
!2/5
u−3/5
v1/5
Reescrevendo
−
−
−
−
−
−
−
−
−
→
1
T
=
2
5
v
1/2
0 θ
R3/2
!−2/5
v1/5
u3/5
P
T
=
∂s
∂v
=
1
5
R3/2
v
1/2
0 θ
!2/5
u2/5
v−4/5
Reescrevendo
−
−
−
−
−
−
−
−
−
→
P
T
=
1
5
v
1/2
0 θ
R3/2
!−2/5
u2/5
v4/5
Para a equação de estado
µ
T
= −
∂S
∂N
utilizamos as relações s =
S
N
= SN−1
, v =
V
N
= V N−1
e
u =
U
N
= UN−1
, que para esse problema, são, s = SN−1
, v1/5
= V 1/5
N−1/5
e u2/5
= U2/5
N−2/5
.
Assim, temos
SN−1
=
R3/2
v
1/2
0 θ
!2/5
U2/5
N−2/5
V 1/5
N−1/5
=⇒ SN−1
=
R3/2
v
1/2
0 θ
!2/5
U2/5
V 1/5
N−2/5
N−1/5
SN−1
=
R3/2
v
1/2
0 θ
!2/5
U2/5
V 1/5
N−3/5
=⇒
S =
R3/2
v
1/2
0 θ
!2/5
U2/5
V 1/5
N2/5
| {z }
Calcularemos
µ
T
a partir dessa equação
µ
T
= −
∂S
∂N
= −
2
5
R3/2
v
1/2
0 θ
!2/5
U2/5
V 1/5
N−3/5
, Note que (N−3/5 = N−2/5N−1/5)
=⇒
µ
T
= −
2
5
R3/2
v
1/2
0 θ
!2/5
U2/5
V 1/5
N−2/5
N−1/5
=⇒
µ
T
= −
2
5
R3/2
v
1/2
0 θ
!2/5
U2/5
N−2/5
V 1/5
N−1/5
=⇒
µ
T
= −
2
5
R3/2
v
1/2
0 θ
!2/5
u2/5
v1/5
Reescrevendo
−
−
−
−
−
−
−
−
−
→
µ
T
= −
2
5
v
1/2
0 θ
R3/2
!−2/5
u2/5
v1/5
Portanto, as três equações de estado na representação entrópica são :
1
T
=
2
5
v
1/2
0 θ
R3/2
!−2/5
v1/5
u3/5
,
P
T
=
1
5
v
1/2
0 θ
R3/2
!−2/5
u2/5
v4/5
,
µ
T
= −
2
5
v
1/2
0 θ
R3/2
!−2/5
u2/5
v1/5
35

2.3.2 Show by a diagram (drawn to arbitrary scale) the dependence of tempera-
ture on volume ... [SUMÁRIO]
Solução:
T
v
P=P2
P=P1
P2P1
Obtemos a relação de dependência entre T
e v eliminando u das equações de estado
1
T
e
P
T
do problema anterior, que são :
1
T
=
2
5
R3/2
v
1/2
0 θ
!2/5
u−3/5
v1/5
P
T
=
1
5
R3/2
v
1/2
0 θ
!2/5
u2/5
v−4/5
Isolando u na primeira equação
1
T
, temos :
u =

2
5
5/3
R3
v0θ2
1/3
v1/3
T5/3
Substituindo u na equação de estado
P
T
, encontramos :
P
T
=
1
5

R3
v0θ2
1/5

2
5
2/3
R3
v0θ2
2/15
v2/15
T2/3
#
v−4/5
=⇒ P =

2
5
2/3
1
5

R3
v0θ2
1/3
| {z }
= A (=Constante)
T5/3
v2/3
Assim
T5
v2
= AP3
ou ainda T = AP3/5
v2/5
, onde A = Cte =

2
5
2/3
1
5

R3
v0θ2
1/3
36

2.3.3 Find the three equations of state in the entropy representation for ... [SUMÁRIO]
Solução:
s =

R
θ
1/2
u1/2
e−v/v0
A partir dessa equação fundamental, as equações de estado calculadas são :
1
T
=

∂s
∂u

v
=
1
2

R
θ
1/2
u−1/2
e−v/v0
P
T
=

∂s
∂v

u
= −
1
v0

R
θ
1/2
u1/2
e−v/v0
µ
T
= −

∂S
∂N

V,U
= −
∂
∂N

R
θ
1/2
U1/2
N1/2
e−V/Nv0
#
=

R
θ
1/2
u1/2
e−v/v0

1
2
−
v
v0

2.3.4 Consider the fundamental equation S = AUn
Vm
Nr
where A is a positive
constant... [SUMÁRIO]
Solução:
a) Para que S seja homogênea de primeira ordem, devemos ter :
n + m + r = 1 , ou r = 1 − n − m daí
S = NA

U
N
n
V
N
m
b)
1
T
=
∂S
∂U
= nA

U
N
n−1
V
N
m
ou T =
1
nA
u1−n
vm
mas se u → 0 em T = 0 devemos ter 1 − n 0 ou
n 1
Também, multiplicando as equações acima por T e S
TS =
1
n
U , ou U = nTS
e eliminando U na equação de estado acima
Tn
= n−n
A−1
s1−n
v−m
e se T → 0 na medida em que s → 0 devemos ter
0 n 1
c)
Finalmente
P
T
=
∂S
∂V
= mA

U
N
n
V
N
m−1
e eliminando T da equação térmica de estado (primeira linha de (b))
P =
m
n

U
V

Se P cresce com U/V , m/n 0, ou m 0
Consequentemente devemos ter m 0 e 0 n 1
37

2.3.5 Find the three equations of state for a system with the fundamental rela-
tion... [SUMÁRIO]
Solução:
a,b)
1
T
= R

v +
1
u2v

0,
P
T
= R

u +
1
uv2

, −
µ
T
= R

−uv −
3
uv

c) s =
1
T
u +
P
T
v −
µ
T
= Ru

v +
1
u2v

+ R

uv +
1
uv

− R

uv +
3
uv

= uv − (uv)−1
d) Dividindo as duas primeiras equações de estado, P = u/v ou u = Pv e substituindo na
relação fundamental
s
R
= uv −
1
uv
= Pv2
−
1
Pv2
P
v
P=const/v2
Ou ainda Pv2
=Função de s.
Portanto as adiabatas são dadas por
Pv2
= const
2.6 TEMPERATURE UNITS [SUMÁRIO]
2.6.1 The temperature of a system composed of ice, water, and water vapor...
[SUMÁRIO]
Solução:
T(℃) = T(K) − 273, 15 = (373, 15 − 273, 15) ± 0, 01 = (100, 00 ± 0, 01)℃
T(°R) =
9
5
T(K) = (671, 67 ± 0, 02)°R
T(°F) = T(°R) − 459, 67 =
9
5
T(℃) − 32 = (212, 00 ± 0, 02)°F
2.6.2 The gas constantR is defined as the product of Avogrado’s number...
[SUMÁRIO]
Solução:
R = 8, 314 ×
5
9
J/mol °F = 4, 619J/mol °F
2.6.3 Two particular systems have the following equations of state... [SUMÁRIO]
Solução:
T(1)
= T(2)
=⇒
2
3
U(1)
RN(1)
=
2
5
U(2)
RN(2)
ou 5U(1)
= 2U(2)
.
Também U(1)
+ U(2)
= 2, 5 × 103
J =⇒ U(1)
= 714, 3J e U(2)
= 1785, 7J
38

2.6.4 Two systems with the equations of state given in Problem 2.6-3 ... [SUMÁRIO]
Solução:
U
(1)
0 =
3
2
RN(1)
T
(1)
0 =
3
2
× 8, 314 × 2 × 250 = 6235, 5J
U
(2)
0 =
5
2
RN(2)
T
(2)
0 =
5
2
× 8, 314 × 3 × 350 = 21824, 25J
∴ U(1)
+ U(2)
= 28060J
e novamente 5U(1)
= 2U(2)
=⇒ U(1)
= 8017J e U(2)
= 20043J
2.7 MECHANICAL EQUILIBRIUM [SUMÁRIO]
2.7.1 Three cylinders are fitted with four pistons, as shown in Fig. 2.2 ... [SUMÁRIO]
Solução:
1 3 2
Figura 4: Three volume-coupled system
δU(1)
+ δU(2)
+ δU(3)
= 0
e δV (3)
= −3δV (1)
−
3
2
δV (2)
Então
δS =
1
T(1)
δU(1)
+
1
T(2)
δU(2)
+
1
T(3)
δU(3)
−
P(1)
T(1)
δV (1)
−
P(2)
T(2)
δV (2)
−
P(3)
T(3)
δV (3)
Eliminando δU(3)
e δV (3)
das duas últimas equações,
δS =

1
T(1)
−
1
T(3)

δU(1)
+

1
T(2)
−
1
T(1)

δU(2)
−

P(1)
T(1)
− 3
P(3)
T(3)

δV (1)
−

P(2)
T(2)
−
3
2
P(3)
T(3)

δV (2)
portanto δS = 0
Cada uma das quatro diferenciais δU(1)
, δU(2)
, δV (1)
, δV (2)
são arbitrárias, daí
T(1)
= T(2)
= T(3)
, P(1)
= 3P(3)
, 2P(2)
= 3P(3)
39

2.7.2 Two particular systems have the following equations of state : ... [SUMÁRIO]
Solução:
1
T(1)
=
3
2
R
N(1)
U(1)
e
P(1)
T(1)
= R
N(1)
V (1)
, N(1)
= 0, 5mol =
1
2
mol
1
T(2)
=
5
2
R
N(2)
U(2)
e
P(2)
T(2)
= R
N(2)
V (2)
, N(2)
= 0, 75mol =
3
4
mol
T
(1)
0 = 200K , T
(2)
0 = 300K , V
(1)
0 + V
(2)
0 = 20 litros = 20 × 10−3
m3
Se o pistão intermediário é móvel e diatérmico, encontrar os volumes, energias, pressões e
temperaturas finais.
1
T(1)
=
1
T(2)
e
P(1)
T(1)
=
P(2)
T(2)
. Logo, T(1)
= T(2)
e P(1)
= P(2)
Mas U(1)
+ U(2)
=
3
2
N(1)
RT(1)
+
5
2
N(2)
RT(2)
= (1245 + 4669)J = 5914J
∴

3
2
×
1
2
+
5
2
×
3
4

× R × Tf
= 5914J R = 8, 314472 Joule/mol − Kelvin
Tf
= 271 K
P(1)
T(1)
=
P(2)
T(2)
ou
N(1)
V (1)
=
N(2)
V (2)
∴ V (2)
=
3
2
V (1)
Também V (1)
+ V (2)
= 20 × 10−3
m3
Consequentemente V (1)
= 8 × 10−3
m3
= 8 litros , V (2)
= 12 × 10−3
m3
= 12 litros
P(1)
= P(2)
=
0, 5 × R × 271K
8 × 10−3m3
P(1)
= P(2)
= 1, 41 × 105
Pa
U(1)
=
3
2
N(1)
RT e U(2)
=
5
2
N(2)
RT
U(1)
= 1690J e U(2)
= 4220J
2.7.3 The hypothetical problem of equilibrium in a closed composite system ...
[SUMÁRIO]
Solução:
A equação fundamental é
S = S(1)
(U(1)
, V (1)
) + S(2)
(U(2)
, V (2)
)
note que suprimimos o número de mols na notação. A condição para o volume é (condição de
fechamento ou completeza)
V (1)
+ V (2)
= V = constante
40

E nós devemos formalizar o significado de restrição adiabática. Desde que nenhum fluxo de
energia na forma de calor seja permitido pela parede, a troca de energia de cada subsistema é
devida inteiramente pelo termo do trabalho, e a restrição adiabática é expressa pela condição
que (condição adiabática)
dU(1)
= −P(1)
dV (1)
; dU(2)
= −P(2)
dV (2)
A mudança na entropia total em um processo virtual é agora,
dS =
1
T(1)
dU(1)
+
P(1)
T(1)
dV (1)
+
1
T(2)
dU(2)
+
P(2)
T(2)
dV (2)
Inserindo a condição adiabática, a quantidade dS se anula identicamente. Era de se esperar
inserir a condição do volume (condição de fechamento ou completeza) e então fazer dS = 0 para
obter a condição de equilíbrio, mas o desaparecimento da quantidade dS no meio do caminho
durante os cálculos certamente inviabiliza a continuação da análise. Como esperado dos fun-
damentos físicos, uma solução matemática do problema não existe. Entretanto, uma solução
parcial pode ser obtida, para a segunda condição de fechamento que ainda não escrevemos,
que deve ser aplicada para todo problema na representação entrópica. Esta é a condição de
fechamento para a energia
U(1)
+ U(2)
= U = constante
Assim, temos
dU(1)
= −dU(2)
E, inserindo as condições adiabáticas
−P(1)
dV (1)
= P(2)
dV (2)
Finalmente, da equação da condição do volume encontramos dV (1)
= −dV (2)
, Consequente-
mente
P(1)
= P(2)
Novamente, esta caracterização necessária mas não suficiente do estado de equilíbrio está em
concordância com nossas expectativas nos fundamentos físicos.
2.8 EQUILIBRIUM WITH RESPECT TO MATTER FLOW [SUMÁRIO]
2.8.1 The fundamental equation of a particular type of two-component system
is... [SUMÁRIO]
Solução:
Para cada subsistema temos
1
T
=
∂S
∂U
=
3
2
NR
U
e
P
T
=
NR
V
,
e ainda −
µ1
T
= A + R ln

U3/2
V
N5/2

−
5
2
R − R ln

N1
(N1 + N2)

= R ln
U3/2
V
N1N3/2
+

A −
5
2
R

Mas
1
T(1)
=
1
T(2)
ou
N
(2)
1 + N
(2)
2
U(1)
=
N
(2)
1 + N
(2)
2
U(2)
e
µ
(1)
1
T(1)
=
µ
(2)
1
T(2)
ou
[U(1)
]3/2
[N
(1)
1 + N
(1)
2 ]3/2N
(1)
1
=
[U(2)
]3/2
[N
(2)
1 + N
(2)
2 ]3/2N
(2)
1
41

Elevando a primeira dessas equações a 3/2, e multiplicando as equações, obtemos
N
(1)
1 = N
(2)
1 , consequentemente N
(1)
1 = N
(2)
1 = 0, 75
Para encontrar as energias e temperaturas :
As energias iniciais eram :
U(1)
=
3
2
N(1)
RT(1)
= 562, 5R
U(2)
=
3
2
N(2)
RT(2)
= 562, 5R
As energias finais são : U(1)
=
3
2
× 1, 5 × RT , U(2)
=
3
2
× 1, 5 × RT Portanto,
U(1)
+ U(2)
= 4, 125RT
Então,
4, 125RT = 2 × 562, 5R =⇒ T = 272, 7K
Encontrando as pressões : (PV = NRT)
P(1)
V (1)
= N(1)
RT(1)
=⇒ P(1)
= 6, 8 × 105
Pa
P(2)
V (2)
= N(2)
RT(2)
=⇒ P(2)
= 5, 7 × 105
Pa
2.8.2 A two-component system gaseous system has a fundamental equation of the
form... [SUMÁRIO]
Solução:
Para cada subvolume
1
T
=
1
3
AU−2/3
V 1/3
N1/3
,
P
T
=
1
3
AU1/3
V −2/3
N1/3
Ou ainda T =
3
A

U2
NV
1/3
, P = U/V , −
µ1
T
=
1
3
AU1/3
V 1/3
N−2/3
+ B

N2
N
−
N1N2
N2

Então
U =

A
3
3
T3
(NV )1/2

E isto é suficiente para determinar Te :
2T3
e = T3
l + T3
r
Se Tr = 2Tl =⇒ 2T3
e = 9T3
l =⇒ Te = 1, 651Tl
Também, eliminando U da equação para µ/T :
−
µ1
T
=
1
9
A2
T

V
N
1/2
+ B

N2
N
−
N1N2
N2

Então, igualando µ1/T em cada subsistema,
1
9
A2
TeV 1/2
N
−1/2
1l =
1
9
A2
TeV 1/2

N1r +
1
2
−1/2
+ B

N2
N1r + N2
−
N1rN2
(N1r + N2)2

42

Mas
B =

A
3
2
TlV 1/2
Portanto,
TeN
−1/2
1l = Te

N1r +
1
2
−1/2
+ Tl

1
2N1r + 1
−
2N1r
(2N1r + 1)2

Assim temos que,

3
2
− N1r
−1/2
=

N1r +
1
2
−1/2
+
Tl
Te
(2N1r + 1)−2
Então,

3
2
− N1r
−1/2
−

N1r +
1
2
−1/2
− 0, 6057(2N1r + 1)2
= 0
Isto pode ser resolvido por tentativa e erro utilizando uma calculadora ou um computador.
Feito corretamente, você deve concluir que,
N1r = 0, 84
N1l = 1, 5 − 0, 84 = 0, 66
2.9 CHEMICAL EQUILIBRIUM [SUMÁRIO]
2.9.1 The hydrogenation of propane (C3H8) to form methane (CH4) proceeds by
the reaction... [SUMÁRIO]
Solução:
µC1H8
+ 2µH2
= 3µCH4
O que é formalmente indentico ao exemplo 1 da seção 2.7
3 SOME FORMAL RELATIONSHIPS, AND SAMPLE
SYSTEMS [SUMÁRIO]
3.1 THE EULER EQUATION [SUMÁRIO]
3.1.1 Write each of the five physically acceptable fundamental equations... [SUMÁRIO]
Solução:
(a)
S = C[UV N]1/3
, C ≡

R2
v0θ
1/3
43

Então,
S =

C
3
U−2/3
V 1/3
N1/3

| {z }
1/T
U +

C
3
U1/3
V −2/3
N1/3

| {z }
P/T
V +

C
3
U1/3
V 1/3
N−2/3

| {z }
−µ/T
N
(c)
S = C

NU +
Rθ
v2
0
V 2
1/2
, C ≡

R
θ
1/2
Então,
S =
CN
2

NU +
Rθ
v2
0
V 2
−1/2
| {z }
1/T
U +
CRθ
v2
0
V

NU +
Rθ
v2
0
V 2
−1/2
| {z }
P/T
V +
C
2
U

NU +
Rθ
v2
0
V 2
−1/2
| {z }
−µ/T
N
(e)
S = C[N2
V U]1/5
S =
2
5
CU[N2
V U]−4/5
| {z }
1/T
U +
1
5
C[N2
V U]−4/5
| {z }
P/T
V +
2
5
CN[N2
V U]−4/5
| {z }
−µ/T
N
(g) e (i) são similares
3.2 THE GIBBS-DUHEM RELATION [SUMÁRIO]
3.2.1 Find the relation among T, P, and µ for the system with the fundamental
equation... [SUMÁRIO]
Solução:
A equação fundamental é :
U =

v2
0θ
R3

S4
NV 2
Então,
T = 4

v2
0θ
R3

S3
NV 2
P = 2

v2
0θ
R3

S4
NV 3
µ = −

v2
0θ
R3

S4
N2V 2
Das duas primeiras equações =⇒ T/P = 2V/S
Da primeira e terceira equação =⇒ µ/T = −S/4N
Reescrevendo a primeira equação temos,
44

T = 4

v2
0θ
R3

S
N

S
V
2
= 4

v2
0θ
R3

−4
µ
T

2
P
T
2
∴ T = −(2)6

v2
0θ
R3

µP2
3.3 SUMMARY OF FORMAL STRUCTURE [SUMÁRIO]
3.3.1 A particular system obeys the two equations of state... [SUMÁRIO]
Solução:
Dadas as equações T = 3As2
/v e P = As3
/v, encontre µ(s, v) e a equação fundamental.
(a)
dµ = −sdT + vdP
= −sd[3As2
/v] + vd[As3
/v]
= −3A
s2
v
ds + A
s3
v2
dv
= −d[As3
/v]
⇒ dµ = −d[As3
/v]
Z
dµ =
Z
−d[As3
/v]
∴ µ = −As3
/v + cte
Podemos tomar a constante igual a zero (cte = 0)
Assim,
µ(s, v) = −As3
/v
Então a equação de Euler será:
U = TS − PV + µN
ou ainda
u = Ts − Pv + µ
∴ u =

3As2
v

s −

A
s3
v

v −

A
s3
v

Ou de forma mais sucinta,
U = A
S3
NV
(b)
45

Pela integração direta da equação de Euler na forma molar :
du = Tds − Pdv
=

3A
s2
v

ds −

A
s3
v

dv
= d[A
s3
v
]
⇒
Z
du =
Z
d

A
s3
v

Fazendo a constante de integração igual a zero, concluímos que :
U = A
S3
NV
3.3.2 It is found that a particular system obeys the relations U = PV and P = BT2
[SUMÁRIO]
Solução:
As equações envolvem U e estão portanto na representação entrópica. Escreva como :
1
T
= B1/2 v1/2
u1/2
e
P
T
= B1/2 u1/2
v1/2
Então,
ds =
1
T
du +
P
T
dv
= B1/2 v1/2
u1/2
du + B1/2 u1/2
v1/2
dv
= 2B1/2
d[u1/2
v1/2
]
∴ s = 2B1/2
u1/2
v1/2
+ const. ou S = 2B1/2
U1/2
V 1/2
+ Ns0
3.3.3 A system obeys the equations P = −
NU
NV − 2AVU
and T = 2C
U1/2
V1/2
N − 2AU
eAU/N
...
[SUMÁRIO]
Solução:
Escreva as equações na forma
P
T
= −
1
2
C−1 u1/2
v3/2
eAu
e
1
T
=
1
2
C−1
u−1/2
v−1/2
[1 − 2Au]e−Au
∴ ds =
1
2
C−1
e−Au

1 − 2Au
√
uv
du −
u1/2
v3/2
dv

46

Faça s = Dun
vm
e−Au
(tentativa e erro!)
∴ ds = D[(nun−1
vm
− Aun
vm
)]e−Au
du + mun
vm−1
e−Au
dv
Isto está de acordo se tomarmos n = 1/2, m = −1/2, D = C−1
∴ s = C−1
r
u
v
e−Au
3.3.4 A system obeys the two equations u =
3
2
Pv and u1/2
= BTv1/3
... [SUMÁRIO]
Solução:
As equações estão na representação entrópica, então escrevemos
1
T
= Bu−1/2
v1/3
e
P
T
=
2
3
Bu1/2
v−2/3
Então
ds =
1
T
du +
P
T
dv
= Bv1/3 du
u1/2
+
2
3
Bu1/2 dv
v2/3
= 2Bd(u1/2
v1/3
)
∴ s = 2Bu1/2
v1/3
+ s0 ou S = 2BU1/2
V 1/3
N1/6
+ Ns0
3.4 THE SIMPLE IDEAL GAS AND MULTICOMPONENT SIM-
PLE IDEAL GASES [SUMÁRIO]
3.4.1 A constant volume ideal gas thermometeris contructed as shown (sche-
matically) in Fig. 3.3. ... [SUMÁRIO]
Solução:
NENHUMA SOLUÇÃO PARA ESSE PROBLEMA NO MOMENTO
3.4.2 Show that the relation between the volume and the pressure of a monoato-
mic ideal gas ... [SUMÁRIO]
Solução:
Para um gás simples S = Ns0 + NR ln

U
U0
−c
V
V0

N0
N
#c+1
47

Se S = const, uc
v = uc
0v0e[(s−s0)/R]
∴ (cRT)c
v = (cRT0)c
v0e[(s−s0)/R]
(Pv)c
v = (P0v0)c
v0e[(s−s0)/R]
Pv
c+1
c = P0v
c+1
c
0 e(s−s0)/c
= const.
Então, para um gás ideal monoatômico, c = 3/2
Pv5/3
=
h
P0v
5/3
0 e−2s0/3R
i
e2s/3R
= const
3.4.3 Two moles of a monoatomic ideal gas are at a temperature of 0℃and a
volume... [SUMÁRIO]
Solução:
S =
N
N0
S0 + NR ln

U
U0
3/2
V
V0

N
N0
−5/2
#
Em S = const.,
U3/2
V = const =⇒

3
2
NRT
3/2
V = const. =⇒ T3/2
V = const.
Vf = vi

Ti
Tf
3/2
= 45

273
223
3/2
litros = 61 litros = 61 × 10−3
m3
Também,
PV = NRT
Portanto,
Pi =
NRTi
Vi
= 0, 101MPa e Pf =
NRTf
Vf
= 0, 061MPa
3.4.4 By carring out the integral
R
PdV, compute the work done by the gas in
Problem 3.4-3.. [SUMÁRIO]
Solução:
Do problema 3.4-2, PV 5/3
= const.
(−W) =
Z
PdV =
Z Vf
Vi
Pi

Vi
V
5/3
dV = −
3
2
PiV
5/3
i [V
−2/3
f −V
−2/3
i ] = −
3
2
PiVi

Vf
Vi
−2/3
− 1
#
Então
(−W) = −
3
2
× (1, 01 × 105
) × (45 × 10−3
)

61
45
−2/3
− 1
#
48

Logo
(−W) = 12, 5 × 102
Joules
Também
Ui =
3
2
NRTi = 3 × 8, 314 × 273 = 6810 Joules
Uf =
3
2
NRTf = 3 × 8, 314 × 223 = 5560 Joules
∆U = 12, 4 × 102
Joules
3.4.5 In a particular engine a gas is compressed in the initial stroke of the piston...
[SUMÁRIO]
Solução:
(a)
W = −
Z V1
V0
PdV = −NR
Z V1
V0
T
dV
V
= −
NRT0
V η
0
Z V1
V0
V η−1
dV = −
NRT0
η

V1
V0
η
− 1

W =
NR
η
(T0 − T1)
(b)
∆U =
3
2
NR(T1 − T0) Onde T1 =

V1
V0
η
T0
(c)
Q = ∆U − W =
3
2
NR(T1 − T0) +
NR
η
(T1 − T0)
∴ Q =

3
2
+
1
η

NR(T1 − T0)
(d)
S = Ns0 +
3
2
NR ln

U
U0

+ NR ln

V
V0

−
5
2
NR ln

N
N0

= Ns0 +
3
2
NR ln

T
T0

+ NR ln

V
V0

−
5
2
NR ln

N
N0

49

∴ dS =
3
2
NR
dT
T
+ NR
dV
V
=
3
2
NRη
dV
V
+ NR
dV
V
=

3
2
η + 1

NR
dV
V
∴ Tds =

3
2
η + 1

NRT0
V η−1
V η
0
dV
Então
Q =
Z
TdV =

3
2
η + 1

NRT0
1
η

V
V0
ηV1
V0
=

3
2
+
1
η

NRT0

V1
V0
η
− 1

∴

3
2
+
1
η

NR(T1 − T0) O que bate com a letra (c)
(e) Para checar η = −
2
3
se Q = 0
Então
PV = NRT = NR

V
V0
η
T0
ou ainda
PV 1−η
= const ou PV 5/3
= const.
E esta é a conhecida forma de uma adiabata.
3.4.6 Find the three equations of state of the simple ideal gas(equation 3.34...
[SUMÁRIO]
Solução:
S = Ns0 + NcR ln

U
Nu0

+ NR ln

V
Nv0

Então
1
T
=
NcR
U
e
P
T
=
NR
V
∴ −
µ
T
=
∂S
∂N
= s0 + cR ln

U
Nu0

+ R ln

V
Nv0

− (1 + c)R
3.4.7 Find the four equations of state of a two-component mixture if simple ideal
gases... [SUMÁRIO]
Solução:
S =
X
j
Njsj0 +
X
j
NjcjR ln

T
T0

+
X
j
NjR ln

V
NjV0

; U =
X
j
Njcj
!
RT
50

∴ S =
X
j
Njsj0 +
X
j
Njcj
!
R ln






U
P
j
Njcj
!
RT0






+
X
j
NjR ln

V
NjV0

Então
1
T
=
X
j
Njcj
R
U
P
T
=
X
j
Nj
R
V
−
µ
T
= sj0 + cjR ln



U
P
j
NjcjRT0


 − cjR + R ln

V
NjV0

− R
Para verificar a equação de Euler, calculamos
U
T
+
P
T
V −
X
j
µjNj
T
U
T
+P
T
V −
P
j
µjNj
T
=U
P
j
NjcjR
U
+V
P
j
NjR
V
+
P
j

sj0
+cjR ln

 U
P
j
NjcjRT0

−cjR+R ln

V
NjV0

−R

Nj
O lado direito desta equação é exatamente o S da segunda linha da solução.
3.4.8 If a monoatomic ideal gas is permitted to expand into an evacuated region,
thereby increasing its volume... [SUMÁRIO]
Solução:
Responda na ordem inversa, ou seja, (c), (b) e depois (a).
S = Ns0 + NR ln

U
U0
3/2
V
V0

N
N0
−5/2
#
(a) De PV = NRT
Pf
Pi
=
Vi
Vf
=
1
λ
=⇒ Pf =
Pi
λ
(b) de U =
3
2
NRT =⇒ Tf = Ti
(c) Se Uf = Ui e Vf /Vi = λ então
∆S = Sf − Si = NR ln(Vf ) − NR ln(Vi) = NR ln

Vf
Vi

∴ ∆S = NR ln(λ)
51

3.4.9 A tank has a volume of 0.1m3
and is filled with He gas at a pressure of...
[SUMÁRIO]
Solução:
U1 =
3
2
P1V1 , U2 =
3
2
P2V2, Uf =
3
2
Pf Vf =
3
2
Pf (V1 + V2) e Uf = U1 + U2
Então
3
2
Pf (V1 + V2) =
3
2
P1V1 +
3
2
P2V2
∴ Pf =
P1V1 + P2V2
V1 + V2
=
5 × 106
× 0.1 + 6 × 106
× 0, 15
0, 25
= 5, 6 × 106
Pa
3.4.10 ... [SUMÁRIO]
Solução:
(a)
P1V1 = N1RT1 ou 5 × 106
× 0, 1 = N1 × 8, 314 × 300 então N1 = 200, 5 mols
P2V2 = N2RT2 ou 6 × 106
× 0, 15 = N2 × 8, 314 × 350 então N2 = 309, 3 mols
Pf Vf = (N1 + N2)RTf ou 5, 6 × 106
× 0, 25 = 509, 8 × 8, 134 × Tf
∴ Tf = 330 K
(b)
Novamente N1 = 200, 5 e N2 = 309, 3 Então U1 + U2 = Uf
Logo
3
2
N1RT1 +
5
2
N2RT2 =
3
2
N1RTf +
5
2
N2RTf =⇒ Tf = 336 K
Note também que
3
2
P1V1 +
5
2
P2V2 =

3
2
N1 +
5
2
N2

Pf Vf
N1 + N2
=⇒ Pf = 5, 7 × 106
Pa
3.4.11 Show that the pressure of a multicomponent simple ideal gas can be written
as the sum... [SUMÁRIO]
Solução:
P =
NRT
V
e N =
X
i
Ni =⇒ P =
X
i

NiRT
V

≡
X
i
Pi
52

3.4.12 Show that µj, the electrochemical potential of the jth component in a ...
[SUMÁRIO]
Solução:
Nós temos S =
X
j
Njsj0 +
X
j
NjcjR ln

T
T0

+
X
j
NjR ln

V
Njv0

quando T =
U
R
P
j
Njcj
−
µj
T
=

∂S
∂Nj

U1,...
= sj0 + cjR ln

T
T0

+
1
T
X
j
Njcj
!
R

∂T
∂Nj

U
+ R ln

V
Njv0

− R
Mas

∂T
∂Nj

= −
U
R
cj
(
P
j
Njcj)2
= −
Tcj
P
j
Njcj
∴ −
µj
T
= sj0 + cjR ln

T
T0

− cjR + R ln

V
Njv0

− R
=⇒ µj = −RT ln

V
Njv0

+ RT

1 + cj − cj ln

T
T0

−
sj0
R

∴ µj = RT ln

Pj
RTv0

+ RT

1 + cj − cj ln

T
T0

−
sj0
R

3.4.13 An impermeable, diathermal, and rigid partition divides a container into
two subvolumes... [SUMÁRIO]
Solução:
Do problema anterior µj = −RT ln

V
Njv0

+ RT

1 + cj − cj ln

T
T0

−
sj0
R

e µ
(1)
H2
= µ
(2)
H2
∴ −RT ln
V0
N
(1)
H2
v0
!
= −RT ln
V0
N
(2)
H2
v0
!
Ou N
(1)
H2
= N
(2)
H2
=
1
2
mol
∴ P(1)
=
N
(1)
H2
RT
V0
=
1
2
RT
V0
e P(2)
=
[N
(2)
H2
+ N
(2)
Ne]RT
V0
=
7
2
RT
V0
3.4.14 Use the results of problem 3.4-11 and 3.4-12 to stablish the results Pα = Pγ
e... [SUMÁRIO]
Solução:
A pressão parcial de A em β, ou PA
β , é igual a Pα
53

A pressão parcial de B em β, ou PB
β , é igual a Pγ
∴ Pβ = PA
β + PB
β = Pα + Pγ
Também, por simetria,
Pα = Pγ =⇒ Pβ = 2Pα
3.4.15 An impermeable, diathermal and rigid partition divides a container into
two subvolumes... [SUMÁRIO]
Solução:
Note que U =
X
j
NjcjRT, e desde que a energia total seja constante, a temperatura é constante.
Inicialmente PH2 =
nRT
nV0
=
RT
V0
PHe =
mRT
mV0
=
RT
V0
Finalmente P =
(n + m)RT
(n + m)V0
=
RT
V0
=⇒ não há variação na pressão
Então a variação na entropia é igual a entropia de mistura. Veja o texto logo abaixo da
equação 3.40 no livro-texto (entropy of mixing-pg 69).
∴ ∆s = −R
X
j
Nj ln

Nj
N

= −R

n ln

n
n + m

+ m ln

m
n + m

3.5 THE IDEAL VAN DER WAALS FLUID[SUMÁRIO]
3.5.1 Are each of the listed pairs of equations of state compatible (recall equations
3.46)... [SUMÁRIO]
Solução:
(a)
U = aPv e Pv2
= bT
Escreva as equações como
P
T
=
b
v2
, e (dividindo por P = u/av), =⇒
1
T
=
ab
uv
Então
∂
∂u

P
T

?
=
∂
∂v

1
T

se torna
∂
∂u

b
v2

?
=
∂
∂v

ab
uv

, ou 0
?
= −
ab
uv2
=⇒ não é compatível
(b)
U = aPv2
e Pv2
= bT
54

∴
P
T
=
b
v2
e
1
T
=
ab
u
Então
∂
∂u

P
T

=
∂
∂v

1
T

, ou 0 = 0 =⇒ Compatível
(c)
P =
u
v
c + buv
a + buv
e T =
u
a + buv
∴
P
T
=
c
v
+ bu e
1
T
=
a
u
+ bv
∴
∂
∂u

P
T

=
∂
∂v

1
T

(= b) =⇒ compatível
Para encontrar as equações fundamentais para os casos (b) e (c) :
(b) ds =
1
T
du +
P
T
dv =
ab
u
du +
b
v2
dv ⇒ S − S0 = abN ln

U
N

− b
N2
V
(c) ds =
a
u
+ bv

du+
c
v
+ bu

dv ⇒ S = Ns0 + a ln

U
U0

+ b
UV
N
− bu0v0 + c ln

V
Nv0

3.5.2 Find the relationship between the volume and the temperature of an ideal
Van Der Waals fluid... [SUMÁRIO]
Solução:
S = NR ln[(v − b)(cRT)c
] + Ns0 =⇒ (v − b)Tc
= const.
3.5.3 Repeat the problem 3.4-3 for CO2, rather than for a monoatomic ideal gas...
[SUMÁRIO]
Solução:
Recordando S = NR ln[(v − b)(u + a/v)c
] + S0 Consequentemente, numa expansão adiabática,
(v − b)
h
u +
a
v
ic
= const.
também u +
a
u
= cRT
de onde (v − b)Tc
= const.
Então
(vf − b)Tc
f
(vi − b)Tc
i
= 1, e da tabela 3.1(pg.77 do livro), c = 3, 5 e b = 42, 7 × 10−6
m3
Tem-se também que vi = 45 × 10−3
m3
Neste caso b pode ser desprezado, então
vf
vi
'

Ti
Tf
c
=

273
223
3,5
= 2, 03
e ainda vf = 91, 4 litros = 0, 091m3
55

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