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In the name of Al ah the Beneficent the Merciful
Col ected by
Youad Hashempour
Student of Shiraz technology Uni
In feb 2012
This pdf is in spanish language
but it is to easy
understand problems and that is a brief of book
Email: s872475@gmail.com
Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/
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Content
Excersice of ch 1
………………………6
Excersice 2of ch
……………………….22
Excersice of ch 3
……………………….50
Excersice of
4 ch
……………………….84
Excersice 5
ofof
chch
……………………….108
Excersice 6 of ch
……………………….no solve
Excersice 7 of ch
……………………….127
Excersice 8 of ch
……………………….no solve
Excersice 9
……………………….no solve
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Contents
ix
Introduction
1
1 OProblemaeospostulados 1.1
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Segunda Aula: 19/03/2008 . . . . . . .
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1.1.1 Quantitative Definition of Heat: . . .
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3.1.2 The Gibbs-Duhem relationship: . .
3.1.3 Summary of Formal Structure: . .
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CONTENTS
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4 Reversible Processes and Content. of Work Max.
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4.1.1 Possible processes and impossible processes: . .
4.1.2 Quasi-static and reversible processes: .
4.1.3 Relaxation and irreversibility times: .
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4.1.4 Heat flow: coupled systems and process reversal: 79
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4.2.2 Machine, cooler and heat pump coefficients: 81
4.2.3 The Carnot cycle: . . . .
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in
CONTENTS
147
9 First Order Phase Transitions: 9.1 Fifteenth Lecture (06/23/2008): .
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9.1.2 The discontinuity in entropy: .
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9.2 D9.é2c.1imaISsoetxetramAus lians (t2á5v/e0i6s/e20tr0a8n)s :i .çõ.es .de. pr .im. e.ir .aord.em: .
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1. 51353
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we
CONTENTS
Machine Translated by Google
Preface
vii
Machine Translated by Google
viii
PREFACE
Machine Translated by Google
Introduction
ix
Machine Translated by Google
x
INTRODUCTION
Machine Translated by Google
Chapter 1
The Problem and the Postulates
1.1 Second Class: 03/19/2008
1.1.1 Quantitative Definition of Heat:
We now wish to introduce a quantitative definition of heat, as well as
establish your units. Therefore, intuitively, we can say that the
heat (produced
or every
absorbed)
byenergy
a thermodynamically
number
of moles,
type of
that cannot be simple
writtensystem is, without changing the
in terms of the work done by or
mechanic
on the system.
So we have
Q
=
of
ÿ
WM.
(1.1)
However, two points are worth noting:
• This expression holds for non-exact differentials for heat and temperature.
work (). However,
energy is an exact differential. This
Q WM
implies that the variation values of are dependent
on the particular process being
Q WM
considered, whereas the variation value of
IN is independent of the particulars of the process. internal
energy
It is
It is
• Note that this expression should be valid only for situations where there is no change in
the number of moles. For cases where there is such a variation,
we must include another term in the equation.
t = fdx
Mechanical work can be written, as usual, in the form where is force and is
displacement.
However,x as we have already stated,
f
and displacement thermodynamic
they are not
variables. We can pass to the thermodynamic variables by
writing
t
=ÿ
f Adx
),
A (
1
,
Machine Translated by Google
CHAPTER 1. THE PROBLEM AND THE POSTULATES
2
where is
A the cross-sectional area that applies to the particular problem. In this case, taking the pressure
P as
f /A
and the volume as
t
=
dv
Adx
,
We have to
VAT.
=ÿ
(1.2)
it is
the bre
Ofeitsoinso
goatsiivsotedmeaveirsáeraucomloecnatdaor spuoarqeuneeregsitaam(soestausdsoumacionndtoecqeuresuemm ptrearbdaalhdoe
heat). In fact, by expression (1.1), we have
of
=
dwm
(In the absence of change in heat both internal energy and work are exact differentials). Now, if we
decrease the volume of a certain thermodynamic system ( 0), then we are acting on the system and, thus,
increasing
dV <
its energy, so that we must have, in this case,
of
VAT
=ÿ
dv
so that the negative value of implies an increase
in internal energy. It is important to point out that
expression (1.2) for mechanical work is only valid for variations in the system.
quasi-static
From expression (1.1) above it is clear that heat and energy have the same in the MKS system ( 1
terribly
in the cgs system, or ).
Joulecal unit, which can be 7 10
,
= 41858
.
bad The calorie is also used and we have 1
Example 1
1.
(a)
J
.
The example presents a curve that holds when there is no
adiabat
heat involved (called ), given by
3
5
P IN = const.
O system traversed this curve,
Through this expression, considering that we have
3 5
3 5
P OF A = P IN ,
at any point on the curve. So we can write
P = Well
AND
5/
3
.
IN
ÿ
As in the adiabat curve there is no changein heat,ÿ we have
that the energy is
simply
UB
ÿ
VAT
UA
=ÿÿ
as shown in the book.
AB
= ÿ1125 . J,
a variation
J =
Machine Translated by Google
3
1.1. SECOND CLASS: 03/19/2008
2.
With
a introduction of the mechanical system that will express tarnsform mechanical energy
a
in heat second
dP
=
dt
where t
and the
2
oh
3
IN
t,
torque, we have
2
dP =
2
right
3
dt dtÿV
dt =
3 o'clock
2
=
1
date
3
IN
=
of
2
IN
3
,
so that
of
3
=
2
V dP.
,
Now, we know that dU since V dPÿis0always
positive, t
as
we have
raise to give in the pressure direction. for C
he can
It is
dÿ have
ÿ0
,
O same sign. Thus, indicating
that
O process
only
a particular situation (constant
to
volume!),
with
on what
is performed between
O process
We have to
you points
It is
(by direct integration)
UA
UC =
ÿ
3
2
AND ( Well
ÿ
PC ) = 1453 . J
the of
same
form, OUT
3.
UB
ÿ
3
=
2
VB ( PD
ÿ
PB ) = 11625 . J
It is clear, therefore, that we can connect any two points in the PV plane shown in the book by simply
choosing an isochoric curve (same volume) adiabat ( Q ). at some point provide a thermodynamic path
is another curve
Such curves will
with meet
for the to go from point A=to0 D
to tados It(initial
final). For example,
is
we can use, through theIt adiabatic
process, UA isochoric process, UD UB J, yes, we you must be
have UD UA J, etc.
UB point
It is
by
= 11625 .
ÿ
A
= ÿ1125 . Yes ,
As-
as previously calculated.
= 1050
ÿ
4. In the process
ÿ
ÿ
D terwe must
heat production(not adiabat).But we
It is
a curve
O value
have UD UA, given above, as well
as weofhave (work value WAD WADB, given constant volume). Like this,
ÿ
=
O
due
OUT
It is,
a DB part does not contribute
to the
therefore, QAD
ÿ
UA =
WAD
+
HEIGHT
= 1750 J, like in the book.
1.1.2 Energy measurement:
The previous problem enables us to say that we can always control and measure the internal energy of a
thermodynamic system.
Machine Translated by Google
4
CHAPTER 1. THE PROBLEM AND THE POSTULATES
In fact, we can control to the extent that we know that there are walls that are
impermeable to heat exchange (walls ), as well as thoseadiabats
that are not impermeable (walls ).
diathermic
We also know there are
it seems that they do not allow heat exchange or work (called energy-restricting walls). Thus,
controlling the use of these walls, we can always
controllingsystem
the flow
of energy
intoto
orrestrictive
out of a thermodynamic
system.
that,
in addition
walls with respect
to thermodynamic
energy, it is also composed of restrictive walls to the exchange of number of moles is said
thermodynamically closed
measure
However, we also want to know the energy. Now,
the example made in the previous
section indicates how this can be done: we just need to link our
system to a particular mechanical system whose energy we are able to measure.
Thus, the energy variation of the thermodynamic system will imply the energy variation of the
mechanical system, which we will measure. Interestingly, as seen in the previous example
where changes could only be made in 'direction' ÿ 0, that in general thermodynamic changes
dP made by external elements
can only be given in one direction (showing the character of many irreversible
.
thermodynamic phenomena).
1.1.3 The basic problem of thermodynamics:
In thermodynamics, we are interested in knowing what equilibrium state eventually results
from the withdrawal of certain internal constraints from a system.
closed compound.
two thermodynamic systems 'in conTo clarify: consider that we have
touch' (but not necessarily exchanging energy, heat, work, etc... this will
depending on the links – walls – that make up the system). Such systems are in equilibrium
in the initial situation. However, if we remove certain ties
(eliminating a wall, making a wall permissive to heat exchange, etc.) the system will look for
another configuration in which it will be again
in balance. This new state (situation) is what we want to know.
1.1.4 Entropy and maximization postulates:
Axiom 2
From what was said earlier, we can see that a suppose a existence of a
It is
entropy
function, which we will call that depends
only on
the terextensive
S ( U,V,N whose
maximum gives
the problem
(S ) =modynamics
of)
equilibrium configuration of the thermodynamic system under analysis.
a
This assumption closely follows the idea dear to mechanics (to physics in general), that
principles of maximizing certain functions (generally energy - minimization), imply complete
knowledge of the physical system in question.
So, we must have
S
Thus, in a given situation
so active, we have an entropy
=
SU,V,N
,...,N
(
1
r
,
(1.3)
)
X in which certain constraints
vA, vB,...
SX which is maximal in terms of the variables
is-
Machine Translated by Google
5
1.1. SECOND CLASS: 03/19/2008
UX, VX,NX (that is, if we change the value of one of these variables (or several), the
ÿ
in A,
in ÿ ,..
smaller
entropy takes on a value ). If we modify
the bounds for the variables they have to
B
S
change their value
order
NXto keep the function for these bounds as still being a maximum.
UX, inVX,
Thus, it is worth noting that the
function isSa maximum in terms of the extensive thermodynamic parameters
HereO axthe
configuration
of seem
the links
anterior
ioma may
quite obvious, but it is far from being so.
In fact, in mechanics, it is the energy function whose minimization implies obtaining the relevant
results for the physical situation under consideration. Why, then, do we not start here by
.
considering that this energy is the sought-after function whose minimization will give the
relevant results? Simply because it wouldn't work! If we do that, we'll be writing something like
IN = IN( IN,{ In }) ,
S and
and we will never have introduced, at any time, a new function that we are calling entropy
which, as we will see, is directly linked to
,
with the notion of heat. We would not leave, in other words, the field of mechanics
traditional1 .
Axiom 3
a entropy must be additive over
you support
We also impose that
constituent themes of the composite system in addition to being a monotonic-differentiable
growing mind of the energy, continually, that
we must have
S (l IN,l IN,l N 1,...,l
or that is,aentropy is
extensive parameters.
It is
N r) =
a harmonic function
function. This immediately implies
ÿS U,V,N
,...,N
(
1
r
),
first order with respect to
Proof. In fact, to see this it suffices to consider the situation in which
l separate identical systems. Each has an entropy })
S (U,V, { In
have identical to the others. When we put these systems in contact, the com- ( }) system , due to
ÿS the
U,V,
{ In property, will have an entropy. However, we
must remember that the }
additive
we
U,V,an{ In
parameters are also extensive. Thus, the composite system should also have
entropy (ÿ
S
S IN,
IN,l V,ÿNi
IN, { ÿNi
{
}) = inÿS
}). Thus, we have that (ÿ( Instead
ofl writing
entropy
terms of energy, volume and number
U,V, { In }).
of moles,ofwe
can invert
the then
expression and write the energy in terms of entropy, volume, and
number
moles.
we have
IN = U (S,V,N ,...,N
1
r
)
(1.4)
1It should be remembered that the placement of a physical theory in the form of postulates (axioms) to the
posterity orally
is something that
elaboration of the theory by much more tortuous means (or
is done unsystematically). Thus, the postulates are already elaborated knowing that they will be sufficient to
the development of all the desired results. It is a theory-organizing process, much more than a discovery
process.
Machine Translated by Google
6
CHAPTER 1. THE PROBLEM AND THE POSTULATES
and, if we impose that entropy must be a monotonically increasing function of energy, in addition
to being continuous and differentiable, we must have
ÿ
ÿS
> 0
ÿ
ÿU
V,N,..,N
1
(1.5)
.
r
The inversion of expression (1.3) to obtain expression (1.4) is particularly interesting, as
it implies that entropy maximization corresponds to energy minimization, which is usually how
the minimization principles are introduced. in physics.
Axiom 4
with
The last postulate says only that it nullifies it entropy of any system
for a state in which we have
ÿ
ÿU
or that is, in the state for
O
=0
ÿ
ÿS
V,N,..,N
1
,
r
which a temperature is zero.
S ( U,V,
{ In
It is very important to point out that the function
}) or, equally,
the
function (IN{ }),S,provide
the relevant thermodynamic information
V, In for the system all
under consideration.
1.2 Chapter Exercises:
Exercise 5
(1.8-1) Consider from
PV new
thepoint
example
for system.
with O
a picture below. In it we present
O
The point is on an isochoric (same volume) that starts from the point
We know how to calculate energy over an isochoric using relation to
of
=
3
2
O
whichwe want to calculate
V dP,
diagram of
a energy
D.
a
Machine Translated by Google
1.2. CHAPTER EXERCISES:
7
obtained from the mechanical expression
dP
=
dt
2
oh
3
IN
t.
So, we have to
OUT
It is
= 3 IN ( PD
2
UE
ÿ
O value we already have for
like this, using
ON )
ÿ
OUT
= 1050 J (see example), we are left
with
3
5
UE = 1050 ÿ
8 × 10ÿ3 ÿ10
2
ÿ 0 .5 × 10 5ÿ = 450 J.
Exercise 6
to the system in process state
(1.8-2) To calculate which one it O heat transferred from
the action
no will (by a straight line) cially identify what kind state
And, ini- we have to
a said straight line implies between
a
of pressure relationship A point on this straight line
there
volume.
is that
such
P
((IN
ÿ
Well)
5ÿ 10 / 32
5 10
=
. ,
= ÿ138392857110ÿ3
ÿ 8 × 10ÿ3
ÿ
AND)
or it is
5 7 = 10
P
ÿ 13839285
.
× 10
with going
Therefore, the work done for
IN
IN
.
from ÿstateÿ 10ÿ3ÿto the state
and stay
AND
5 7 ÿ 13839285 × 10
ANYWAY
= ÿ ÿ INA
.
ÿ10
A devariation
foi (we are
energy
guessing
ÿ
IN ÿ 10ÿ3ÿÿ
=0
A
UE
ÿ
UA
,
.
dv = 3609375
as in the example)
= 450 J
Now
UE
UA =
ÿ
ANYWAY +
QAE
and,therefore,
. +
450 = ÿ3609375
QAE
or
QAE
Exercise 7
a has to
(1.8-3)
.
= 890625
.
The process now refers to which one
has
figure shown not text, for
IN = 25. PV + const.
with
values of
3
.
It is
. MPa
. m there.
= 02ofQ
= presented
001
pointA data byPA V Desired values
W for each
one
(The)
path (process)A
It is
let's take the
B; The energy at A is given by
UA
= 25.× 0
ÿ
.2 ÿ10 6ÿ × 0 .01 = 5000 J + const.
you
Initially, processes
Machine Translated by Google
CHAPTER 1. THE PROBLEM AND THE POSTULATES
8
of
B
energy in is
so that
given by
UB = 25. × 0
.2 ÿ10 6ÿ × 0 .03 = 15000 J + const.
a energy difference
(a constant disappears)
=
UAB
UB
UA = 10000 J.
ÿ
The work done along this path is
WAB
so that
O
Well ( VB
=ÿ
. 6ÿ × 0
= ÿ02 ÿ10
AND)
ÿ
heat associated with this process
QAB
=
UAB
WAB
ÿ
.
O pathoB
C, we have UC = 25×0
(b)For
B
const..
with
and
The work done for
line equation
PV)
ÿ
ÿ
.02 = ÿ4000
J
stay
= 14000 J.
.5 ÿ10 6ÿ×0 .01+ const. = 12500 J +
O calculationof
C is given by (after
INC
WBC
= ÿ ÿ INB
. × 10
6 ÿ05
ÿ 15 .× 10
7
( IN ÿ 0 .01)ÿ dv = 7000 J.
Like this, a The amount of heat involved in the process is given by
QBC
= ( UC
ÿ
UB ) ÿ WBC
= (12500 ÿ 15000) ÿ 7000 = ÿ9500
[ the other possibilities must be made by the readers
Exercise 8
a relationship
(1.8-4) We have worth
5
IN =
O
].
2
PV + const.
which means we should
5
ter U=
2
( VAT + V dP );
a relationship
now, we must also, on an adiabatic curve, have,
of
so that
ÿ
=
VAT
VAT,
=ÿ
5
( VAT + V dP
2
or yet
dv
ÿ7
IN
=5
dP
P
so that
ÿ7 ln V/V
0
= 5ln
P/P
0
)
J.
Machine Translated by Google
1.2. CHAPTER EXERCISES:
9
or
ÿ
IN
P
IN0 ÿÿ7 = ÿ
P 0 ÿ5
or yet
P 5 IN7
Exercise 9
P 05 IN07 =
=
const.
(1.8-5) We have to
2
IN = AP
Of similar mode
to the
IN.
problem above, we know that
of
2
= 2 OF PdP + AP dv
VAT
=ÿ
so that
2OF dP = ÿ(1 +
APdV
)
implying that
2AdP
dv
=
1 + AP
It is
IN
ÿ
like this
AP
IN
ÿ 1 +1 + AP 0 ÿ2 = ÿln ln IN0
or
(1 + AP
Exercise 10
AP 0)2 IN0
IN = (1 +
)2
=
a constant volume,
(1.8-6) We have to,
const.
a heat transfer
is given by
ÿ
Q
we also know that
=
A (P
ÿ
P 0)
ÿ
.
a adiabatic curve of the system is
PV
c
=
given by
const.
IN ( P, V we
To find
must remember that any point in space
)
It is an isochoric curve.
PV can be reached through an adiabaticcurve
a
in the first
Consider the figure
below, in which we indicate the system process involved.
O performs
part,
see the process
( 0 , P 0) ÿ ( V, P there
) (where
is realization
not
of
work –isochoric character– and
a
of energy can be written as change
ÿ
IN
ÿ
IN0
=
A (P
ÿ
ÿ
P 0)
.
(1.6)
Machine Translated by Google
CHAPTER 1. THE PROBLEM AND THE POSTULATES
10
In the process taken on
a adiabatic curve, we have
c IN
1ÿ
IN
IN
VAT
ÿ
ÿ
=ÿÿ
P0 INc
=ÿ
c
IN
ÿ
dv
P0 IN0c
=ÿ
ÿ
IN
1ÿ
0ÿ
c
IN0
Come
on O value
0
(note that a integration is done of the initial state volume value
the final volume is another
O volume cannot
with takes for an isochoric – and,
IN, Since
process so
O other way.
change in that
IN,
The final energy is
mass
ÿ
initial u is
IN
IN
ÿ
O
,
ÿ
=
que he is clear
P0 IN0c
1ÿ
IN
ÿ
c
ÿ1
IN
ÿ
IN
c
IN
ÿ
ÿ
=
PV
=
IN
c
1ÿ
c
IN
ÿ
,
0ÿ
ÿr
ÿ1 ÿ
ÿ1
ÿ1ÿ
,
a relationship
V/V 0 Also note that in expression (1.6) we have that it is worth
.
PV c
once the
result is, adding the
IN
ÿ
IN0
=
c
ÿ
P IN0
a adiabatic curve. Like this,
dots are on
P
there
1ÿ
ÿ
ÿ1
for obvious reasons. We are then left with
c
where r
PV c
ÿ=
c
P 0 IN0c by PV
where we replace
previous figure). We therefore have to make it
1ÿ0 c
c
ÿ
=
rc
P
two terms,
=
A(
rc
P
ÿ
P0) + [ PV/ (c ÿ 1)] ÿ1 ÿ
Exercise 11
ÿr
ÿ1ÿ
.
(1.8-7) We have two moles of a single-component system in the volume
It is
that has a dependence of the internal energy on the pressure given by
IN = APV
2
Machine Translated by Google
1.2. CHAPTER EXERCISES:
11
It is
we want to know how it should be from a dependency with2 moles)
the moles.
numberShall we
energy in terms of volume and energy as
O with that volume number of moles are variables 'we group'
remember that both are a
(which applies to full
extensive of the thermodynamic problem. So we know that systems, so we go from
U,V,NU,V,N We have to
(
) ÿ (l
2
IN = APV
where f
so that
and the
=
IN
demanding that
ÿ
IN
=
l
)
a ticket
2
APÿ
with
ÿ
.
(N ) ,
f
a be determined. let's do
function
ÿ
It is
l
(U,V,N
) ÿ (l
IN,l IN,l N
2
IN f ÿN
(
)
ÿU, we have
1
what
(ÿN
where is c
)=
,
ÿcN
any constant, such that
2
ÿ
IN
c
=2
=
2
APV
,
and,
therefore,
IN =
Exercise 12
ÿU.
cN
IN
so we are left with
=
ÿAPV
you can do it easily remembering that, so that
O
It remains only to determine c, ÿ we
must
= 2have
=
1
2
2
APV
/N.
(1.10-1) We must simply apply various properties relatedas
to the postulates to see which
It is
The
are satisfied and which are not.
we must
a
we must have that it must (1)
be entropy
differ-extensive
as a homogeneous
function properties are: to have first order of (2) of
parameters, (3) ÿS /ÿU (4) enciable we must have we must
and
ÿU/ÿS only when S do exercise: / ÿ, as
bethat
continuous,
to have
(2) =is0also
(required. so
) IN,that
{ N i}
(1)
It is
satisfied due to ÿ
3ÿ1ÿ/ 3
=0
to the
degree
withwe
1 3, then
=
) IN,{ N i} >
(
do
The satisfied
thermosensitive,
O also
Let's see
.
0
It is
(a)
(U,V,N ) ÿ (l IN,l IN,l N )
clearly continuous is
It is
different
function. Yet
ÿS
ÿ
ÿU
=
ÿ
A
IN,{ N i}
1
NV
3
/
(NUV )2
3 ,
where A is the multiplicative constant presented there, which is, evidently, greater (3) so that it is also valid.
than zero,
Finally, we can write
IN =
A ÿ3
S3
NV
)
Machine Translated by Google
CHAPTER 1. THE PROBLEM AND THE POSTULATES
12
so that
ÿU
ÿ
admissible. For
ÿS
ÿ
where B
=
3 A ÿ3 S 2
=
ÿ
NV
IN,{ N i}
= 0 , the satisfaction of (a) is therefore (4).
S implying
O item (h) we fear that
cancel if
only if
ÿS
A
B ÿ2
UV
NB
ÿ
=ÿ
ÿ
ÿU
1
2
The function of the item
IN,{ N i}
NOT
ÿ
UV NB
ÿ expÿÿ
ÿ
,
Rÿv 0 However, we cannot guarantee that this function is always such that this
is
a entropy.
expression (h) is admissible for greaterno
than Itzero,
For item (i) we have that the
.
ÿU
ÿ
only if
AS
IN,{ N i}
S =0
cancel if
=
ÿ
ÿS
S
No. + S
expÿ No. ÿÿ 2
To calculate
(prop. 4).
ÿ
NV R
( ÿS/ÿU
just using different
)
implicit ciation. So we have
1=
AS
IN
ÿS
ÿ2ÿ
ÿU
+
ÿ
IN,{ N i}
S
S
ÿ exp ÿ
No.
No.
ÿ
,
so that
ÿS
ÿ
ÿU
=
ÿ
A ÿ1 IN
S
2S
No.
IN,{ N i}
expÿÿ
S
ÿÿ
2 No.
a expression
which, for large values of of part (i) S will be a negative number. So entropy (or
A
is also not admissible for is (f) expression in is such a
a internal energy).
that
IN =
N 2B
exp( S/NO )
IN
,
where B is a constant. Like this,
ÿU
ÿ
that does not satisfy a prop.
same ideas.
follow the
ÿS
=
ÿ
IN,{ N i}
NB
INR
exp( S/NO
4, is also not admissible.
Exercise 13
It
is (1.10-2) a simple
inversion problem.
Exercise 14
(1.10-3) We have to
S
=
A ( NV IN)1/
)
the other items
functions.
3
Machine Translated by Google
1.2. CHAPTER EXERCISES:
13
3
m3,
=3
such NA VA
system
VB
that
UT total
J. ) composite
= 9(10ÿ6
m , NB = 2 . The energy
= 4 ÿ10ÿ6ÿ
We have that a total entropy of the system
It is
= 80
is
when it's made composite it's additive, so that
ST
=
on + SB
=
A
( navua
)1
/3
+ ( NBVBUB
)1
/
3ÿ
ÿ
It is
we also know that
VT
=
AND +
VB = 13 ÿ10ÿ6ÿ
OUT =
,
UA + UB = 80
NT
,
=
THAT +
NB = 5
.
Let's put,
× 3 ÿ10ÿ2ÿÿ
Note that to do
entropy by de
chart in terms
UA )1
(80 ÿ
3
3ÿ ;
AU/ 80 , since the
AU (and not UA/ ),80
UBenergia total
with
80 we have O result shown in the figure below
The balance must
with
will give for
ÿS
=0
,
ÿUA
so that
1
100
ÿ2
INA
/3
1
ÿ
150
(80 ÿ
/
UA )ÿ2
or yet
/2
=
ÿ 15 10ÿ3
there
UA
80 ÿ
result is
.
UA = 5180236437
de
what is a maximum point.
/
UA/ ( UA +
O graph of entropy in terms of da
O graphic
Doing the
2
1
INA/ 3 +
agoraST = A
UA
3
=0
)
it's the same
it is conserved.
AU ranging from
0
until
Machine Translated by Google
14
CHAPTER 1. THE PROBLEM AND THE POSTULATES
Machine Translated by Google
Chapter 2
The Equilibrium Conditions
2.1 Third Class: 03/24/2008
2.1.1 Intensive Parameters:
As we said at the end of the previous chapter, the main problem in thermodynamics is to find
the values of the thermodynamic parameters when a system moves from an equilibrium
situation A to an equilibrium situation B, respecting the constraints in both situations. Thus,
we are interested in
processes by which systems change extensive parameters. It is clear, therefore, that we are
interested, finally, not exactly in the fundamental equation, for example:
IN = IN( S, V, { In
})
,
but the differential form
of
ÿU
= ÿ ÿS
dS
ÿ
IN,{ N i}
r
ÿU
+ ÿ ÿV
ÿU
dv + ÿ
ÿ
S, { N i}
ÿ
j =1
ÿNj
dnj
ÿ
where we use the chain rule. The partial derivatives are called and are defined as
intensive rivers
parameter-
T
ÿU
=
ÿS
IN,{ N i}
ÿU ÿÿ
=ÿ
ÿV
S, { N i}
ÿ
ÿ
(2.1)
S, V,{ N i}iÿ=j
P
,
(2.2)
=
ÿU ÿN
j
ÿ
µj
S, V, N
i}iÿ=j
ÿ
T pressure
component calls
If we assume
that there isPno change
in potential
the number
temperature
chemical
µj of moles, we can write j
{ and
,
the
equation (2.1) as
TdS
=
of
ÿ
WM,
so we can identify
TdS
=
15
Q,
.
Machine Translated by Google
16
CHAPTER 2. THE CONDITIONS OF BALANCE
the system is associated with a
for
that is, a quasi-static heat flow of entropy in
increase this system.
The term is often called
r
=
Wc
µjdNj
j =1
ÿ
of quasi-static chemical work, so that equation (2.1) can be written
as
of
=
WM +
Q+
Wc.
2.1.2 Equations of State:
From equations (2.2) it is clear that the intensive parameters of the system are also functions
of the extensive parameters in the form
T
P
=
=
=
µj
T S, V, In })
P S, V, In })
µj (((S, V,{{{In })
. If we know all that are
called the equations ofequations of state
state, we will know everything about the thermodynamic behavior of the system. are hoIt is also clear from expressions (2.2) that the zero-order
mogeneous functions, that is, are such that
T, P, m j
T (l S,l IN, { ÿNi }) = T ( S, V, { In }) ,
for the simple reason that the equation of state, from which they were taken, is first order
homogeneous and already appears in it (in each term) a differential involving an extensive
parameter. So, for example, as in the definition of
dS with parameter,
S
temperature we have the appearance of the term an extensive
we can only have
intensive multiplicative factors (temperature, in this case), the same being true for the other
variables.
,
1
V, N,...,N
It is often adopted the convention of writing {
so we are left with
IN = IN ( S,
Xi
{
)}
and the intensive parameters become
T =ÿ
Pj
being that one chooses X
1
=ÿ
P
=ÿ
.
ÿU
ÿS
ÿU
ÿX j ÿ
ÿ
{ X i}
,
S, { N i}iÿ=j
r
1
} = { X,...,X
t}
Machine Translated by Google
17
2.1. THIRD CLASS: 03/24/2008
2.1.3 Intensive Entropic Parameters:
In the same way that we can write the fundamental equation having the energy
internal as a variable we can write
dependent
the fundamental equation
,
taking a as the entropy
dependent variable. A representation we call
representation of energyentropy while the other we call
representation of
,
. From a formal point of view, both are equivalent in terms of what information can be
extracted from them, but often a problem is much more
easily solved using one or the other of them. In entropy representation, we have
S
where X
0
=
S ( X 0 , X,...,X
1
=
t)
,
IN. Note that equation (2.1) implies that
r
TdS
ÿÿ
of + VAT
=
µjdNj,
j =1
so that
dS
1
=
T
of +
r
P
dv
T
µj
ÿÿ
T dNj
j =1
or, according to convention, with the entropic intensive parameters given by
F0 = 1
T
=ÿ
where P 1
P
=ÿ
Fk
,
ÿSX
=ÿ
PkT ,
X i}iÿ=0
k = 1 , 2,
..
0ÿ{
.
2.1.4 Thermal Equilibrium and Intuitive Concept of Temperature
stop:
Once done the temperature,
from the equation of state in the
definition
energy, it remains to be shown that this state function (thermodynamic variable
intensive) behaves as our intuition perceives the temperature. This can
be seen easily if we take a system in which there is no change in the number of moles, nor in the
IN
volume ('restrictive variable walls') and such that the
It is
N
complete system consists of two subsystems, whose internal energies are
IN(1)
It is
IN(2) . The complete system is closed, so that (1) +
IN
IN(2)
const.
=
(2.3)
If we remove the wall that prevented the exchange of heat between the two systems, their free
energies will adjust in such a way that the entropy is a
maximum, that is, so that
dS = 0
(2.4)
.
We know, however, that
S
=
S
(1) ÿ
IN(1) , IN(1) , ÿ N i(1) ÿÿ + (2)Sÿ
IN(2) , IN(2) , ÿ N i(2) ÿÿ
Machine Translated by Google
18
CHAPTER 2. THE CONDITIONS OF BALANCE
and so
dS
ÿS
(1)
ÿS
of
(1)
of
(1) +
= ÿ ÿU (1) ÿ IN(1) , ÿjNÿ(1)
(2)
+ ÿ ÿU (2) ÿ IN(2) , ÿjNÿ(2)
of
(2)
and therefore,
dS
But (2.3) implies that
of
1
=
(1)
T (1)
=ÿ
of
(2)
1
1
T (2)
of
(2)
.
(2.5)
and (2.4) implies that
=
T (1)
1
T (2)
(2.6)
,
so the systems will seek equilibrium in order to equalize the temperatures. It is interesting to
note that,
although this is not common in the field of thermodynamics, the relations (2.3) and (2.6),
when seen as equations relating
functions of the extensive variables imply the possibility of obtaining numerical values for the
energies of each system in the final situation of equilibrium.
If in the above situation we have
T (1) ÿ T (2) ,
then, when the wall separating the two systems is removed, the total entropy increases, i.e. ÿ0
(until an equilibrium pointSis>found for
the new situation of the removed link, in which ÿ= 0, as already
S seen). But is
It is easy to see from equation (2.5) that
ÿ S
1
ÿ
ÿ ÿ 1T (1)
T (2) ÿ ÿ
IN(1)
(in first order), so
queÿ IN(1) < 0,
that is, heat will flow from the higher temperature system to the lower temperature system
(which is, of course, necessary for them to equalize
at some point in the process of going to equilibrium.)
Both marked properties represent the behaviors that we intuitively expect from the
temperature function, indicating that our definition of temperature is adequate.
2.1.5 Temperature units:
Since we have not yet chosen the dimensional unit of entropy, the dimensional unit of
temperature also remains to be determined, since temperature, by definition, has, evidently, a
dimension of energy by entropy.
Results of statistical physics imply the choice of
entropy as being
Machine Translated by Google
19
2.2. FOURTH CLASS: 03/26/2008
so that the temperature takes on the dimension
a dimensionless greatness
power.
unit is not commonly chosen like the
Even having an energy dimension, temperature
Joule, for example. From the fact that there is a zero
absolute temperature, one is left with the arbitrary choice of the unit of
temperature,
whichthe
canunit,
be chosen
byitassigning
any value to a certain state of a standard
system. Whatever
however,
is
T = 0.
It is evident that all temperature scales must coincide in
Thus we have the Kelvin scale of temperature, defined by assigning the number
. to the temperature of a mixture of pure ice, water and water vapor in
27316
mutual balance (triple point of water). The unit on this scale is called a and the symbol is
K in to a Joule is a number
Kelvin The ratio of a Kelv
.3806
× 10ÿ23
constant of
dimensionless given by 1 and
is known
asJ/K
and represented by
Boltzmann simply 9
kB . There is also the Rankine scale, which is sim.
,
/ 5 Kelvin temperature (it is usually represented by the symbol
oR
).However, these scalars end up implying large numbers when
We consider common everyday situations. Thus, usual temperatures are in the
oR ); we then have two other scales: the Celsius scale, given
region of 300 (540K
by
T o( C ) = T ( K ) ÿ 27315
.
O
,
C . On this scale the zero term
whose unit is the degree Celsius, represented by
inamic is shifted, so that, strictly speaking, the Celsius scale does not
is a thermodynamic temperature scale. However, the differences between
temperatures represented on the Celsius scale are correct. ) is also
The Fahrenheit scale ( oF a practical scale, defined by
9
T oF
( ) = 5 T o( C ) + 32 ,
the same considerations as we made regarding the Celsius temperature are valid.
2.2 Fourth class: 03/26/2008
2.2.1 Mechanical Balance:
The principle of maximum entropy can also be used to analyze the conditions for a
mechanical equilibrium (involving the variables related to the
mechanical work.) To do so, it suffices to consider a thermodynamic system
with restrictive constraints for changing the number of moles, but which has
movable diathermy walls (being non-restrictive to temperature variations
and the volume). In this case, we have to
IN = IN(1) + IN(2)
IN = IN(1) + IN(2)
Machine Translated by Google
20
CHAPTER 2. THE CONDITIONS OF BALANCE
are constant, even though each of the variables relating to one of the systems may vary. In
any case, it must
dS = 0 ,
so that we get, using
of
dv
(1)
(1)
of
dv
=ÿ
=ÿ
(2)
(2)
the relationship
dS
1
ÿ
(1)
= ÿ 1T
T (2) ÿ
of
P (1)
(1)
(1) + ÿ T
P (2)
dv (1) = 0
T (2) ÿ
ÿ
,
or yet
=
T1
P (1)
=
T (1) (1)
T1
P (2)
T (2) (2)
,
which are the conditions for equilibrium. Evidently, equal pressures
are precisely the result one would expect from mechanical notions.
It should be noted that the case of a movable wall in an adiabatic system (and restrictive
for the variation in the number of moles) is a problem with special characteristics, in that it
does not have a unique well-defined solution (see exercise ( 2.7- 3)).
Note that, in the case of a system where we have the same temperature, the change
in entropy in a process where a diathermal wall is mobile
is given by
dS
=
P (1) ÿ
P (2)
T
dv
T
,
(1)
,
so that, knowing that in the process of modifying the thermodynamic system
(to adjust to the new links and the contact situation between the systems)
P (1) > P (2) then the wall will tend to move in
dS > 0, we set a pressure in
such a direction that implies
dv (1) > 0 (which we expected, of course).
,
2.2.2 Equilibrium with respect to the flow of matter:
The flow of matter is linked to the concept of chemical potential. If we consider an equilibrium
situation related to two systems connected by
a
rigid wall (restrictive
to volume variations), but diathermal and permeable (being impermeable
a only one
of the types
of matter
to all the others – this hypothesis serves for the
isolate
alteration due only to the flow of one of the types of matter), then, by considerations absolutely
analogous to those made for
the mechanical problem, we have
1
(1)
=
1
(2)
BILLION
BILLION
T
T (2) j (2)
=
(1) j (1)
,
Machine Translated by Google
21
2.2. FOURTH CLASS: 03/26/2008
where we choose the type
aveis.
j as being the one for whom the walls are permeable.
T ),
If we think of a situation in which the temperatures are equal (a then the entropy change
is given by
(2)
dS
µj
=
(1)
j
T µ
dN j (1)
ÿ
,
(1)
so that, as in the process of alteration entropy grows, if (1)
than will
(1)
have
mj
mj
is bigger
dN j showing that there will be a flow of particles from regions of higher
tothen
be negative,
,
chemical potential to regions of lower chemical potential.
Thus, in the same way that temperature can be seen analogously
as a potential for heat flow, and pressure can be seen as a
type of potential for volume change, chemical potential can be seen
as a potential for the flow of matter. Differences in temperature generate
a heat flow from the hottest region to the less hot one, pressure differences generate a wall
movement from the region with higher pressure to the
region with lower pressure and differences in chemical potential generate a force
generalized that sets matter in motion from the region with the greatest potential
chemical to the region with lower chemical potential (precisely in the sense of
tqourn
chemical
potential in the composite system, since this potential is
ymicthe
anunique
intensive
variable.)
Chemical Equilibrium:
Linked to the question of equilibrium related to the numbers of moles of a
thermodynamic system, there is the question of the chemical equilibrium that is established
in chemical reactions (which can be seen, of course, as the passage of a thermodynamic system from
a situation of equilibrium (one side of the equation) to another situation of equilibrium (the other side
of the equation)). So, in
chemical reaction
2H 2 + O 2 ÿ 2 H 2 O,
we can write
ÿjAj,
0ÿÿ
j
where
ÿj
is the stoichiometric coefficient related to the chemical type = ÿ2,
previous chemical equation, we would have, for example, = 2,
A 1 = O 2, n 3
A 3 = H 2O
n
1
A
Aye
1
=
H
2,
n
2
. Already
= ÿ1,
.
The relationship between these chemical reactions and our thermodynamic systems is because
the stoichiometric coefficients are such that their changes must be proportional to the change in the
number of moles (evidently, since it is a question of the change in the number of molecules per mole,
etc.). for an equation
fundamental
S
=
S ( U,V,
{
In })
Machine Translated by Google
22
CHAPTER 2. THE CONDITIONS OF BALANCE
IN
of a chemical system in which both the total energy and the volume remain fixed, the change
in entropy in a chemical process is given
by
r
dS
µj
=ÿ
T dNj
j
ÿ
and, with the proportionality between the stoichiometric coefficients and the variations in the
number of moles, we have
ÿ
njdN
=
dNj
,
so that, in equilibrium,
dS
dN
r
ÿ
ÿ
=ÿ
T
=0
µjÿj
j
and so,
r
ÿ
=0
µjÿj
.
j
If we know the equations of state of a mixture, then these last conditions allow a complete
solution for the final number of number of the
moles.
2.3 Chapter Exercises:
Exercise 15
(2.2-1)
The basic equation is given by
IN = AS 3/ ( NV ) ,
Like this,
T
P
=
P S,N,V
=
m
T S,N,V
=
ÿU
=
) = ÿ ÿS
ÿU
ÿ) = ÿ ÿ ÿV
ÿ) =
m(((S,N,V
ÿU
ÿ ÿN
ÿ
ÿ
=
3
Al 2 S 2
N l IN
there
even for the
Exercise 16
too much. Like this,
T,P
AS
NV
3
AS
N 2 IN
=ÿ
=ÿ
S.V
3
2
.
Indeed,
l
3AS
NV
=
min
2
NV
S,N
Each of these equations is zero-order homogeneous.
l
a ticket
(
) ÿ (l
hold on, let's do S,N,VS,N,V
to get
T
3 AS
V,N
2
to the
=
T,
are intensive parameters. µ
It is
(2.2-2) We must find µ as a function of T, V and We know that
AS
m
temper-
)
3
=ÿ
N 2 IN
N only.
Machine Translated by Google
2.3. CHAPTER EXERCISES:
It is
23
a the temperature
of
equation
a entropy. For this we use
we need to eliminate
to write
S
Volunteers
= ÿ 1 3A
so that
m
A
N 2 IN
=ÿ
3A
T 3/ 2
3ÿ 3 A
2
/
Volunteers
ÿ
=ÿ
ÿ3
IN
,
N
ÿ
is that O desired result.
Exercise 17
a pressure dependency
with
(2.2-3) We want
a temperature.
We have to
S
IN = A
It is
3
NV
we already know that
T
Differentiating with respect
to the
.
NV
volume, we have to
ÿU
P
So we have
A
=
NV
NV
S,N
2
/
Volunteers
2ÿ
AS
=
ÿ
ÿV
=ÿÿ
P
2
3 AS
=
3A
T
=
ÿ3
3/
3
2
.
2
N
ÿ
IN
3ÿ 3 A
or yet
PV
1/
2
=
const.
the greater the greater the constant a
for an isotherm curve, where temperature.
is the
a picture below.
check out
Exercise 18
a fundamental equation
(2.2-4) We have
in =
where s = Status
and v
=
V/N
It is
Y/N are
T =
P
=ÿ
As
2
A andB are
ÿU ÿ
ÿ ÿS V,N
ÿU
ÿV
S,N
ÿ
2
Bv
ÿ
,
The equations
any two constants.
=ÿ
=ÿ
ÿu ÿ
= 2 As
ÿs
In
ÿu
= ÿ2
ÿv
S
ÿ
Bv
;
ÿ
ÿ
I don't case chemical potential, we must be a little more careful. We have
that
ÿU
m
So we make the
= ÿ ÿN
ÿ
.
S.V
a order to obtain
dathe
(per mole) return internal energy expression in the expression
internal energy,
IN = ÿ AS
2
ÿ
BV
2ÿ
/N
Machine Translated by Google
24
CHAPTER 2. THE CONDITIONS OF BALANCE
Figure 2.1:
It is
ÿ
we perform to
derivationµ = ÿ
ÿU ÿ
ÿN
S.V
2
= ÿ ÿ AS
BV
ÿ
2ÿ
/N
2
= ÿ ÿ As
2
ÿ
Bv 2ÿ =
in.
Exercise 19
(2.2-5) We have to
m=
in =
2
As
2
+ Bv
T
=
2
+
P2
.
4A
4B
a
Thus, although
it can be written in terms of extensive parameters,
µ since it can be written in
terms of only intensive parameters, the exercise shows is that, in general,
it must be, necessarily intensive. of a
What O
characterthermodynamic parameter just by looking
we can't judge
for the the way it can be written in terms of extensive variables, being
a from his writing in terms of his writing in
most appropriate judge it
maximum number of intensive parameters.
function of the
ÿ
Exercise 20
ÿ
ÿ
(2.2-6) We have, by the statement, that
A
in =
s2
s/R
It is
,
in
so that
T
ÿu
ÿs
=
P
It is like
=
In
ÿÿ ÿ
ÿ
IN = A
s/R
2ÿ
United States
ÿu
=
ÿv ÿ S
As
=
2
2
ÿ
S2
in
It is
sR
s/R
It is
ÿ
S
IN expÿ
No.
ÿ
,
we have
ÿU
whatµ
= ÿ ÿN
=ÿ
ÿ
S.V
A
S3
S
N 2 INR exp ÿ No. ÿ = ÿ
A
R
sv
exp ÿ
sR ÿ
=ÿ
uR .
Machine Translated by Google
2.3. CHAPTER EXERCISES:
Exercise 21
25
(2.2-7) We have
queue
= Of
It is
ÿ2 expÿ ÿ sR
N moles of the substance, initially dispersed
isentropically
=
that
Pf
is, we bring the
T = ÿ ÿu
ÿs
In
ÿ
P ÿ= ÿ ÿ ÿu
ÿv
s.
=
in
A
R exp ÿ
It is
P halved
pressure are
0
ÿ
sR
= 22A
S
in
.
ÿ
3 exp ÿ
sR
so that we always have
a isentropic expansion,
was
We know that
entropy
as
T0
to temperature (s
to have
=
or
const.)
to a / We want expressions,
.
P 0to2know
the final temperature.a Initially, find equations of state, making
Pe
Thus, we obtain a relationship between
the same
T, that can be written
Pv = 2 RT.
We need to know what happened to the
volume in the process. But this information
a was pressure
diminished to
half. Like this
it
is given by the condition that
P0
=
2A
expÿ
in03
sR
ÿ
2A
=
Pf
,
P
expÿ
in 3
f
sR ÿ
=
02
so that
/
A
vf
= ÿ 4P 0 expÿ
3
sR ÿÿ1
A Pv0
/
3
0
3
= ÿ3 2 in0
2A ÿ1
= ÿ 4P 0
.
Thus, we are left with
Tf
As
P 0 in20=
=
P0
4 Rvf
P 0 in0
=
4R
ÿ3 2 .
RT 0 , We have to
Tf
Exercise 22
=
ÿ3 2
T 0 = 06299605250
.
2
T0
.
r components. So we know that we can
(2.2-8) We have a system of
write its fundamental equation in energy as
r
of
=
T dS
ÿ
VAT +
µidNi
.
i =1
However, we know that
ÿ
r
N
In
=ÿ
i =1
and whatdN
=0
.
But then there is a linear dependence between
r
days
.
0=ÿ
i =1
you dNi, given that
Machine Translated by Google
26
CHAPTER 2. THE CONDITIONS OF BALANCE
Let's choose
dNr of others; we have to be written in terms
O
r ÿ1
dNr = ÿÿ
days
i =1
It is
r ÿ1
like this,
µrdNr
µrdNi
= ÿÿ
.
i =1
Substituting this expression into the original expression for internal energy, we have
dU
=
T dS
=
T dS
ÿ
r
i =1
VAT + ÿ
r ÿ1
i =1
VAT + ÿ
ÿ
now dividing by
=
µidNi
µidNi
ÿÿ
T dS
=
of
r ÿ1
i =1
µrdNi
ri ÿ1
PdV + ÿ+ =1 µidNi + µrdNo .
r ÿ1
VAT ÿ
µr ) days
i =1 ( i
ÿ
T dS
=
ÿ
ÿ
No, we fear
r ÿ1
of
T ds
=
VAT
ÿ
i
+ ÿ(
µr
ÿ
dxi ,
)
i =1
as wished.
k =
Exercise 23
(2.2-9) We know that PV
a is given by
We have to show that
1
IN =
using to
const. in an adiabatic process. energy
PV + Nf
k ÿ1
suggestion, we have to PV
ÿ PV
k =
Sg()
ÿV
=ÿÿ
.
or it is,
,
ÿU
P
kÿ
k /N
=
ÿ
S,N
Sg()
In k
.
But then, integrating, we have
IN
dv
=ÿ
Sg(
=
In k
)ÿ
Sg()
IN1ÿ k + Nf
k ÿ1
S
Note that it is
1
k ÿ1
PV k
a function of ,
IN =
PV + Nf
( S,N )
.
so that we can write
1
k ÿ1
PV + Nf
ÿ PV
But note that f , due to the siva multiplicative term
of extensive variables,
so we must have
f
,
k
( ) = P IN We have to
wheref is an arbitrary function. as sg
IN =
( S,N )
ÿ PV
k N
,
ÿ= F
ÿ PV
N
k , Nÿ
,
k /N
.
must be an intentional function
kÿ
;
Machine Translated by Google
2.3. CHAPTER EXERCISES:
27
and,
therefore,
1
IN =
k ÿ1
+ NF
PV
ÿ
PV k /N
kÿ
,
as wished.
Exercise 24
The equation is given by
(2.3-1)
in =
A
5
s /
2
in 1/
2
so that we have
1 5 2
s = Bv / in /
5
and,
therefore,
1
1
in /
2
=
B
T
5
.
5
in
35/
And in the same way, we have to
P
=
B
T
2
in /
5
in4/
5
and for the chemical Potential,
1 5 2 5 2
N
S = BV /// IN
It is
5
like this
m
2
=ÿ
T
5
BV
1// 5 2 5
N
IN
ÿ3
/
2
5
Bv
=ÿ
1/ 5 2/
in
5
.
5
x volume (fixed pressure) can be obtained
The temperature
(2.3-2)
a from the state equations presented above. We obviously have to
Exercise 25
variable
u
From expression
to inverse
the temperature, we have
O
(internal energy per mole). delete
in =
Cv
1// 3 5
T
3
,
where isCa constant, therefore, and,
P
T
where
AND =
BC
25/
and is
=
B
2 5 2/ 15 2/
T
Cv /
in 12/
3
ÿ
2
in /
or
T
graph stays
=
a constant. Like this,
5
T /
there
3
15
ÿ
1
in /
5
3
T 2/
AND
in 2/
3
3
,
Machine Translated by Google
28
CHAPTER 2. THE CONDITIONS OF BALANCE
Exercise 26
(2.3-3) We have
=
queue
2
As
in
ÿ
2
It is
/in
0
,
so we can invert it to get
1
s =
It is
so get
ÿ ue in
ÿA
2
/ 2 in 0
as equations of state as being
1
ÿsÿu
1
=
ÿ
T =ÿ
in
2ÿ At
{}in
It is
2
/ 2 in 0
,
etc.
Exercise 27
(2.3-4) We
is the
given
S
fundamental equation
=
AT
n
Vm
N
r
It is
you thermodynamic postulates, in addition to
we want this equation to satisfy
P that increases with of U/V with N constant (taking
O
supply a pressure
zero energy as being the one related to
the zero temperature).
Not,
we know we must have
n + m + r =1 ,
(2.7)
so that
a function S is first order harmonic. We still owe
what
ÿ
ÿS
> 0,
ÿ
ÿU
V,N
so that
nAU
n
to have
ÿ1
Vm
N r>
0
and,
therefore,
n> 0
.
(2.8)
Machine Translated by Google
2.3. CHAPTER EXERCISES:
29
The fourth postulate tells us that
ÿU
ÿ
s =0
just in case
.
T =0
=
ÿ
ÿS
V,N
Then we write in the energy representation
S 1/n
A 1/n
=
IN
INm/nNr/n
to get
T
1
=
/n
A
ÿ
n
INm No.
S (1ÿ)
ÿ1
n /n
and,
therefore,
1ÿ n
n
>
0
It is
n < 1
(2.9)
.
we must also have
P
It is
ÿS
=
ÿ
= ÿ ÿV
P increase with how
we want
Vm
ÿ1
N
r
we should have
U/V ,
m =1ÿ
given that
n
me
U,N
n,
(2.10)
n > 0 Thus, we are left with
.
=
P
that satisfies
n
IN
mA
ÿ
N
ÿ
IN
O required.
The (2.7), (2.8), (2.9) relations
(2.3-5)
The equation of state is given by
r
,
It is
(2.10)
they are
as relations
demand.
Exercise 28
=
S
so that
R
N
UV
ÿ
3
ÿ
ÿ
N
,
UV
as state equations are
1 =
R
T
P =
R
T
m
T
=ÿ
R
N
ÿ
3
VN
+ IN IN
AND
+ UV
2
N
ÿ
ÿ
2
UVN
+
3
ÿ2ÿ
3N
UV
.
2
ÿ
(a) Evidently, such intensive parameters are zero-order homogeneous.
To see it, just do, in each equation, substitution U,V,NU,V,N this,
a
(
) ÿ (l
l
l
)
Machine Translated by Google
30
It is
CHAPTER 2. THE CONDITIONS OF BALANCE
there
is to show
thathave
there is no variation in the
equation.
for we
1
T
etc. (b)
It is
=
ÿV
R
ÿ
ÿ
a temperature, for example,
l
+
ÿN
2
l
(c) To find the PT, just
P
3
N
1
ÿ=
2
UÿV
T
,
T is intrinsic-
R is a positive constant,
also evident that, being
caly positive.
write
( in ) ,
3
RU
=
T
mechanicalequation of state, given by
3
N
IN
ÿ
IN
+
N
IN
IN2 IN
ÿ=
INT
so that
=
IN
It is
P IN
like this
1
=
IN
R
T
N
3
N
+
2
P IN
3ÿ
where do we get
RV T
N
P =ÿ
IN
ÿ ÿ2 ÿ
N
RV T
ÿ
(d) We can write
PV
S= R
2
2
N
so that
ÿ
N
1
=
P
IN
2ÿ
3
N
ÿS
2R
PV
ÿ
+ÿS
ÿ
2
+ 4 R N 2ÿÿ
2
so that
PV
2
=
const.
represents the locus related to adiabatic curves.
Exercise 29
A ofAvogadro's
the R gas constant
number
(2.6-2)
ÿ
BY = 6
,
(
)
It is
´
O
product of the
constant of
defined as cells/
ÿ
230225 × 10 grind mole
of
=
.
= 8314
Boltzmann, i.e. NAkB. We therefore
have that R We also have thatR J/
J/mole.K.
.
= 8314
mole.o C.
To express in terms of the unit J/mole.oF, we
oR
can simply remember that there is a relationship between
9
It is K
oR = 5 K It is what
given by
(Rankine)
O
T ( F ) = T o( R ) ÿ 45967.
,
so
.
= 8314
J
.
= 8314
he wants
mole.K
J
mole.
5
9
=
O
R
9
5
.
× 8314
J
mole.oR
and,
therefore,
R
since the
Fahrenheit scale has the
=
9
. 5
× 8314
J
mole.oF
,
same 'greatness' of the Rankine scale.
Machine Translated by Google
2.3. CHAPTER EXERCISES:
Exercise 30
31
(2.6-3) We have two systems given by
1
T
with
N (1) = 2
It is
=
(1)
N (2) = 3
(1)
3
RN (1)
2
IN
1
,
T
(2)
5
=
RN (2)
2
IN
(2)
,
The systems are separated by energy diathermal walls
.
J.
25 .× internal
IN =what
103
a total system energy is We want to know
energy
each system has in
equilibrium.For this, the fact that, in
we use
O
It is
a
balance,
1
T
1
=
(1)
(2) ,
T
so that
2
3
2
R
3
5
=
IN (1)
R
2
IN (2)
It is like
IN (1) +
IN
(2) 3
= 25 .× 10
,
stay with
6
IN (1)
what
15
=
ÿ2 . 5 × 10 3
IN (1)ÿ
ÿ
from the
.
IN (1) = 7142857143
J.
The same two systems are separated by a wall
(2.6-4)
diaterma are and with the same numbers of
soft. Now we want as initial temperatures
Exercise 31
T
after
(1)
O
= 250
Ke
T
(2)
= 350
(1)
Ke
IN
to know balance has you values of
In equilibrium, we know that we must have
been established.
1
T
It is
1
=
(1)
T
(2)
(2)
=
;
we also know that we must in any case have
IN (1) + IN
as energies
So, we can calculate
We have
1
=
250
IN
(1)
2
3
.
8314
2
IN
It is
1
IN (1)
,
IN
=
350
(2)
initials to calculate U.
3
5
.
8314
2
IN (2)
so that
.5 ,
Ui (1)= 6235
Ui (2)= 21824 25 .
so that
IN = 2805975 .
Now we can use
you
J.
from the previous problem to write same steps
6
INf
(1)
15
=
ÿ 28059
. 75 ÿ
INf
(1)
ÿ
IN
(2)
Machine Translated by Google
32
CHAPTER 2. THE CONDITIONS OF BALANCE
to get
.
IN(1) = 8017 071429
J
f
It is
IN(2) = 2004267857.
J
a Final temperature can be calculated by any of the following formulas:
so you look like
1
Tf
1
Tf
Exercise 32
(1) =
.
8314
3
8 .314
5
.
= 3214285714
2
.
8017071429
3
2
=
.
.
= 3214285714
.
2004267857
2
The relationships between you volumes are given by
(2.7-1)
V
=
Aÿÿ
1
(2) =
V
1,
Aÿÿ
2
V
2,
(3) =
Aÿÿ
3
3,
where we know that
given that
ÿÿwith
of
1
=
ÿÿ 3
ÿ
ÿÿ 1 + ÿÿ 2 ,
or ÿÿ 2 increase, ÿÿ
must decrease. So, assuming how
3
habit, that
U
=
U
U
(1) +
U
(3) = 0
P (2)
(2)
(2) +
,
we have
(1)
=
S
T
=ÿ
of
1
+
(1)
P (1)
T (1)
1
ÿ
T
1 (2)
T (1) ÿ
(1)
(2)
dv
of
T
+
1
ÿ
(2) + ÿ
T
(2)
P
1
ÿ
It is,
of
1
T (1) ÿ
A
P (3) A
ÿ
T
2
ÿ
T
+ T (2) dv
of (3) + ÿ (3)
(3) (2)(2)
(3)
+ T
P (1)1A(1)
T
1 (3)
ÿ
of
+
P (3) A 3
T (3) ÿ
P (3)
(3)
T (3) dv
dÿ 1+
dÿ 2
3
then, in equilibrium:
1
1
1
=
T
=
T
(1)
T
(2)
(3)
It is
P
pressão×
that, once
(1)
A
´
area
=
=
It is
Exercise 33
(3)
=
f
1
a force exerted by cylinder i
1
=
T
3
2
(1)
=
P
(2)
A 2,
=
2
f
3,
on the wall(s) of the piston(s).
you two systems are given by the equations of
(2.7-2) We have to
state
A3
strength , we have a mechanical balance equation
f
where be
P
1
(1)
RN
IN(1)
,
P
(1)
T
(1)
(1)
=
RN
IN (1)
=
RN
IN (2)
It is
1
T
(2)
=
5
2
(2)
RN
IN(2)
,
P
(2)
T
(2)
(2)
(2) = 07520
T (1) = 200 K e
N (1) = 05 . It is
. , It is that initially
Nvolume total
= 300 K, being the
in exchange
ÿ.
The released how much system is
being that
T
(2)
Machine Translated by Google
2.3. CHAPTER EXERCISES:
33
of heat and movement of the walls (but impermeable). O system remains with energies as walls
as are
So we have to initially
IN (1) =
3 × 83 .× 05 × .200
= 1245
of
a total energy
(what
with
IN (2) =
J,
2
5 × 83 .× 075 ×. 300
= 4668
. 75 J
2
preserved)
IN = 591375.
you links
We have, therefore,
IN (1) + IN (2)
. =
591375
It is
IN (2) = 20
IN (1) +
It is
as energies, pressures
you volumes,
we want to know
as
e as temperatures
decade
subsystem in equilibrium. equations The equilibrium condition gives us two new
1
T
1
=
(1)
T
(2)
It is
P
(1)
T
(1)
=
P
(2)
T
(2)
.
From the first equilibrium equation we have
05.
3
so that
.
IN (1) = 1689643
.
075
=5
IN (1)
ÿ5913 . 75 ÿ IN (1)ÿ
J
,
.
IN (2) = 4224107
therefore,
J; We get, then,
as and final temperatures as
T
(1) =
.
2 × 1689642857
3 × 83 .× 05
.
= 271
T
.43 K,
(2) =
2 × 4224107143
.
5 × 83 .× 075 .
= 271 43.
K.
From the equilibrium of pressures, we have
83. × 05 .
IN (1)
=
83. × 0
. 75
20 ÿ IN (1)
giving
IN (1) = 8 ÿ,
so that
P
(1) =
IN (2) = 12 ÿ
as pressures stay
. × 83 ×. 05
27143
8
.
= 140
.80 ,
P
(2) =
. × 875 .3 ×
27143
12
.
0= 140
.80
.
Machine Translated by Google
34
CHAPTER 2. THE CONDITIONS OF BALANCE
Exercise 34
of
It is
No case of an adiabatic system, we have
(2.7-3)
(1)
P (1) dv
=ÿ
(1)
(2)
of
,
P (2) dv
=ÿ
(2)
(2.11)
like this
P (1) dv
As
P (2) dv
(1)
ÿ
(1)
dv
(1)
(2)
= P
.
have pressure.P
So we
With regard to temperature, we have
dS
no case
1
=
T
(1)
(2) =
of = 0
.
ÿ
(1)
of
,
O well-defined problem
and point
of sight
P (1)
dv
T (1)
+
(2)
dv
=ÿ
(1)
1
+
T
(2)
of
(2)
P (2)
dv
T (2)
+
(2) = 0
,
=0
balance. However, using (2.11), we have that dS
as temperatures.
identi-
dently, there being no equation for
Exercise 35
(2.8-1)
The basic equation is
3 2
S=
No. ln
THAT +
IN / IN
N 52/
ÿ
N 1 R ln
N
N
1
N2
N
N 2 R ln
ÿ
where
N
= 10 ÿ of
O volume total
is V
ment (only
Nÿ)1 he can
IN(2) = 5(1)
in each chamber and
.75
T = 300
It is
K.
N1+
=
N 2.
permeable membrane is only
first he(1)
(1)
1
the first are placed N
(2)
In the second chamber are placed N
0(2) = 250(2) K.
T
After balancing, we want to know ,T,P
In order not to vary
in any that
of this, it will
P
N 2go chambers.
remember
It is
to the
is divided in half
to alter). in O volume
with
= 05 (2)
.
= 11
,
N
(1)
1
,
you values of N
It is
=
(IN
(1)
N2
05 2 (2) .1
=
=
It is
(1)
N
.
We must have, in
(2)
(1)
balance,µ
=
T
i
It is
N
()i
1
,
N
() 2
, with
m
,
T 1 (2)
1
i =1, 2
.
,
1(1)
=
A + R
moles
1
=
32
(1)
RN IN(1)
IN
, 3/2
(1))
N
ln ÿ ( ( (1))
P (1)
T
IN (1)
3
2
× 83 ×.
IN()
,
(1)
RN IN (1)
=
5
2R
ÿ
R ln
N (1)
N 1 (1)
.
125
IN(1)
1
,
,
5/2 ÿ ÿ
IN(2) ,
O second system in terms of its variables
as initial energies with
as temperatures
We calculate
300
,
We have a that
(1)
O even valid for V
(2)
etc.
T (2)
equation of state, but for the fundamental equation values, so
T
m (1)
=
T 1 (1)
1
=
T (1)
as equations of state (in the same i of of separation, act
but first we need to find
each subsystem will have the i
IN()
1 (1)
250
=
3
2
× 83 ×.
It is
.
15
IN(2)
you numbers
Machine Translated by Google
2.3. CHAPTER EXERCISES:
35
giving
IN(1) = 466875
they are the same.
.
(1)
J
,
(2)
T
(2)
m1
1
=
(1)
so that we have
.
IN = 93375 . J. In equilibrium we must have
The total energy is that
1
IN(2) = 466875
J,
m1
=
(1)
T
(2)
T
T
as equations
N 1(1) + 075 .
N 1(2) + 05 .
=
IN(1)
IN(2)
It is
/
ÿ IN
ÿN
(1)
1
ÿ IN
=
/
.
2
N 1(1)
.
+0
/ 2
(2)ÿ35ÿ3
2
(1)ÿ375ÿ3
(2)
ÿN
1
/
2
N 1(2)
.
+0
Fromthe first, we have
IN(1)
It is
=
N 1(1) + 075
.
IN(2)
N 1(2) + 05 .
like this, a second equation gives
1
1
=
N 1(1)
(1)
=
or, N 1
on both sides.
N 1(2)
N 1(2)
so that, in the end, we will have as we have,
,
Not,
N1
a same amount of
initially
N1
=
N
(1)
N
+1
(2)
.
= 151
,
then we are left, in the final state, with
N 1(1)
=
N
IN(2)
,
(2)
1
.
= 075
.
The final internal energy can be calculated using
IN(1)
=
. + 075.
075
. + 05 .
075
IN(2) = 93375
IN(1) +
giving
IN(2) = 424432
Like this,
.
,IN
(1) = 509318
.
a final equilibrium temperature is
(1)
1
T = 32 RNIN(1) = 3
giving
T = 27273.
.
2
× 83. × 15
.
.
509318
The pressures can then be calculated easily, giving
P (1) = 2723 ×. 83 × .
15.
= 6791 .
5
It is
P (2) = 2723 ×. 83 × .
.
125
5
= 565 .9
.
.
,
Machine Translated by Google
36
CHAPTER 2. THE CONDITIONS OF BALANCE
Exercise 36
(2.8-2) Similar
Exercise 37
(2.9-1)
cos
to the
previous. Do it!
you stoichiometric
The equation related to
It is
nC H
3
It is like
= ÿ1
8
,
nH
proportionality factor we have dNdN
2
= ÿ2
n CH
,
4
=3
ÿ
ÿ
dS
so we reproduce
=ÿ
T
ÿÿ µC
3
H8
ÿ2 H
2
a equation
P (1) + 2 P (2) = 3 P
onlythat
chemical
potentials.
in terms
+ 3 µCH
(3)
,
4
ÿ=0
parameters
Machine Translated by Google
Chapter 3
Formal Relations and
Exemplary Systems
3.1 Fifth Class: (02/04/2008)
3.1.1 Euler's equation:
We are now interested in exploring the mathematical properties of the equations
fundamental. Basically, we are interested in finding ways in which we can obtain the
fundamental equations from information about the equations of state.
A first important tool in this sense is the Euler equation, which
uses the property of first-order homogeneity that the equations
fundamentals have. So, from
l V,ÿNi
{
}) =
we have, by differentiation with respect to
l
IN(l
ÿU
ÿ ÿS
(
ÿ ÿSÿÿ
( )
+
)
S,
ÿU
ÿ ÿV
(
,
r
ÿ ÿV
( ÿÿ)
+ÿ
)
i =1
(S, V, { In }) ,
ÿU
ÿ ÿNiÿÿ
(
)
ÿU
ÿ ÿNi
(
IN (S, V, { In })
=
)
so that
ÿU
ÿ ÿS
(
)
S+
r
ÿU
ÿ ÿV
(
)
V
+ÿ
i =1
ÿU
IN ( S, V, { In })
=
NiÿÿNi
(
)_
l
and since the equation holds for any value of
ÿU
,
r
ÿU
IN (S, V, { In }) =
S +
In
+ÿ
ÿS
ÿV
=1
i
37
l = 1 we have
putting
ÿU
ÿNiNi
Machine Translated by Google
38 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
or yet
r
IN(S, V, { In }) = T S
ÿ
PV + ÿ
µiNi
(3.1)
,
i =1
which is Euler's equation.
In the entropy representation, using the same ideas, we are left with
P
1
r
S (U,V, { In }) =
IN
+
TU
T ÿÿ
i =1
(3.2)
µiTNi ,
S in the representation of energy.
as can be seen also isolating itself
3.1.2 The Gibbs-Duhem relationship:
Another relation that helps us in the investigation process of the mathematical properties of the
fundamental equations (and, therefore, of thermodynamic systems in general), is the socalled Gibbs-Duhem relation, showing that the intensive variables are not all independent of
each other. .
The existence of dependence can be understood from a simple count of the variables
and equations associated with the thermodynamic problem and is, in fact, a direct consequence
of the property of first order homogeneity that these equations have. In the case of a system
with
only one component (particle type), the fundamental equation can be written in the form
in =
in
(s, v ) ,
where the number of moles wasN
'embedded' in the variables due to
in, s
It is
in exactly
the property of homogeneity (all can be divided without
by the extensive variable must also be N ). But then the three intensive parameters problem
s for
in the intensive parameters and two
functions of just . However, we have three equations
intensive parameters: this implies that there must be a relationship between the problem using
P mthe
the equations for the intensive T,
variables.
obtained
same argument
by eliminating the variables
s, in
It is
It is
,
applies to systems with several components, and in this case, we simply choose to 'normalize'
each of the extensive variables using the number of moles of one of the components. In fact,
as the intensive parameters are zero-order homogeneous, we have that
T ( U,V, { In }) = T (l IN,l IN,l N
1, ÿN 2 ,...,l
N
r
)
ÿ1
quickl = Nj
(and the same for the other equations of state) and choosing mos that
r
each intensive parameter depends on (+ 2) ÿ 1 extensive variables,
however there are ( + 2)
r
intensive parameters (or equations
of state).
,
Evidently, the type of relationship that exists between the intensive parameters will depend
on the system in question. There is, however, a relationship that holds for
Machine Translated by Google
39
3.1. FIFTH CLASS: (02/04/2008)
any system, since it is based on the equation of state itself in its mathematical
representation. If we take Euler's equation and take his differential, we have
r
of
=
T dS + SdT
VAT
r
V dP +
ÿ
+
µjdNj
ÿ
Judge
ÿ
j =1
ÿ
j =1
and, as we know that
r
of
T dS
=
VAT
ÿ
µjdNj
+ÿ
j =1
we conclude by the Gibbs-Duhem equation, given by
r
SdT
ÿ
V dP
+ÿ
Judge = 0
.
j =1
In the case of a system with only one component, we immediately have
dµ
sdT + vdP,
=ÿ
immediately representing the relationship between the variations of the parameters
intensive.
The number of intensive parameters capable of independent variation is
called the number of thermodynamic degrees of freedom of the system. Thus, components
r have, as we have already pointed out, (+ 2) ÿ 1 r
a simple system with
thermodynamic degrees of freedom.
The whole discussion could have been done in entropic representation; in her the
Gibbs-Duhem relationship stands
Out
ÿ 1T ÿ +
P
Vd
ÿ
T
r
ÿ ÿÿ
.
No
ÿÿµiT
=0
i =1
3.1.3 Summary of the Formal Structure:
We are then left with the following formal structure for our thermodynamics studies:
is was
1. Ttéemmotsoudmaaa Itin
written
action
via in
fundl
taamen
odin
the
energy
â(mnoicrma asolmbreentoesdisetsecmonaheecqiudea)p,oqdueesceorn-
representation as
IN = IN ( S,V,N
)
or, in entropy representation as
S
=
S ( U,V,N
)
.
Machine Translated by Google
40 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
(a) It is important to point out that the fundamental equation needs to be written in
terms of the extensive variables only, without the aid of any intensive variable
(unless, of course, the relationship of the intensive variable with the extensive
parameters is already made explicit). , because in this case it would be a
simple substitution of the occurrences
ratios
of the intensive
variable
its expression
of these
extensive
parameters).
Thisby
becomes
clear if in
weterms
consider,
for example,
in the case of temperature, the equation
=
IN
IN ( T,V,N
)
,
which, given the definition of temperature, can be written as
IN
=
IN ÿ
ÿU
ÿS
ÿ
,V,N
,
IN
which is a partial differential equation in the variable . We must remember
that a partial differential equation generates solutions with
1 Indeterminate functional forms
so that such a solution
may imply a total knowledge of the system, as occurs
with the fundamental equations.
2. Teregmiao)s, likewise, equations of state, given by (representation of en- (((
T
P
=
T U,V,N
P U,V,N
µj U,V,N
=
=
µj
)
,
)
)
also given according to a dependence on the extensive variables.
(a) All equations of state together imply exact knowledge of the thermodynamic
properties of the system (which can be seen from Euler's formula, since
with such equations we can simply rewrite the fundamental equation).
However, we do not always have all these equations and therefore need
methods to
obtain one of these equations in terms of the others. (b)
The equation that gives us the type of dependence between one of the parameters
intensive in terms of the others (and therefore in terms of the extensive
variables) is the Gibbs-Duhem equation.
1For example, the wave equation can be written as the partial equation
2
ÿ f
2
ÿ f
1
=0
ÿ
ÿx
2
in
2
ÿt
2
,
where in is the speed of the wave, you
x the space and t the time. The solutions of this equation, the reader
can check, they are given by any function with the functional dependence
f (x
ÿ
vt
Thus, there is an indeterminacy in the functional form of the solution.
)
.
Machine Translated by Google
41
3.1. FIFTH CLASS: (02/04/2008)
(c) If we are working with a system with a particle type
only, we can also directly integrate the equation
of
T ds
=
VAT,
ÿ
which is the differential form of the fundamental equation in terms of the
intensive parameters.
(d) Integrations of both the Gibbs-Duhem equation and the differential form of
the fundamental equation, however, will only give a result in terms of an
arbitrary constant (coming, of course, from integration).
3.1.4 The Ideal Simple Gas:
Since we have already studied the most basic mathematical elements related to the
fundamental equations, let us now study some specific examples that
characterize thermodynamic systems in an important way.
The ideal simple gas is characterized by the two equations of state
PV
NRT
=
It is
IN = cNRT,
c is a number that depends on the characteristics of the gases: it is 32 for
/
where monatomic gases at relatively low temperatures (stopped cBT little comwith electronic excitation energies), or 52 for some diatomic
gases ; 2 for diatomic
/
cos or even 7 of / gases at higher temperatures (of the order
thousands of Kelvins).
The above equations of state can be used to obtain the fundamental equation in
any of the ways presented above. We can
to write:
1
=
cNR
T
P
cR
=
IN
,
in
=
RN
T
R
=
IN
in
and directly integrate the Gibbs-Duhem equation
d
m
T
S
=
T
+
T
ÿ
ÿ
substituting the result into Euler's
equation
ÿ
1
P
1
out
=
of +
CEO
ÿ
ÿ
P
m
TN.
T
IN
ÿ
,
T
ÿ
We have to
d
ÿ 1T ÿ = ÿ
cRu
ÿ2
,
d
P
ÿ
T ÿ=ÿ
Rv
ÿ2
Machine Translated by Google
42 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
so that
d
m
ÿ Tÿ = ÿ
of
cR
R
ÿ
in
dv
,
in
that can be integrated term-by-term due to its particular characteristic
(very rare, by the way), giving
m
T
ÿ
m0
T
in
cR ln
=ÿ
in
R ln
ÿ
in 0
/0
in0
so that
S
c
IN
Ns 0 + No.
=
ln ÿÿ
IN0 ÿÿ
c
IN
N
IN0 ÿÿ
N 0 ÿÿ
where
s 0 = ( c + 1) R ÿ ÿ ÿ
m
T
(3.3)
,
(+1)ÿ
.
0
Another way to obtain the same result would be to simply integrate the
equation
1
P
ds =
of
ÿ DVD,
T
+ÿ T
giving
s =
s 0 + cR
R
in
in
in
in
0ÿ
(3.4)
,
ln ÿ 0 ÿ +
ln ÿ
which is equivalent to the previous one by a simple definition of shelf. s 0, which is a
A mixture of two or more ideal gases (called a multicomponent simple ideal gas) is
characterized by a fundamental equation that can be written as
S
ÿ
=ÿ
No
j
0+ÿ
ÿÿ
Njjj
j
ÿ
R ln
T
T 0+ÿ
IN
Mr
ln ÿ
j
0ÿ
IN =
,
ÿ
ÿÿ
local unit
ÿ
Njjj
j
RT,
ÿ
where the temperature works as a parameter that, once eliminated (it is a
intensive parameter and cannot appear in the fundamental equation), gives us the fundamental
equation in the form ( Comparison of thisS = S U,V, In }).
equation with equation (3.3) or, more directly, with
theorem of Gibbs
(3.4), indicates the result known as
{
.
Theorem 38
(by Gibbs) The entropy of a mixture of ideal gases from the
O
entropies that each gas would have volumeheValone
occupied T.
at temperature
is the sum
Proof. The proof of Gibbs's theorem can be done through a simple thought experiment,
which the reader should consider from the text of Callen's book.
Machine Translated by Google
43
3.2. SIXTH CLASS: (04/07/2008)
If we write the entropy equation in the form
S
=ÿ
ÿ
ÿ
No
0+
j
Njjj
ÿÿ j
T
IN
R
+ No. ln ÿ
T0
Nv 0 ÿ ÿ
R ln
ÿ
ÿ
j
Nj ln
Nj
N
,
then the last term is called the entropy
entropy mixture
and represents the difference
between that of a mixture of gases and that of a collection of separate gases, each at the same
temperature and density as the original mixture.
Nj/Vj
=
N/V
.
3.2 Sixth Class: (07/04/2008)
3.2.1 O fluido ideal de Van der Waals:
Real gases rarely satisfy the equations of state presented in section
above, except in low density limits, so we have to try to improve the model. One way of doing this
is by appealing to an analysis
statistics (in terms of corpuscles colliding with the walls of the container, etc.) that is from the
outside
field of thermodynamics, but that can,
at this moment, help us to know another characterization
of systems
thermodynamics, for example.
The Van der Waals equation stems from the perception that, in the physical-statistical deduction
of the equation
P
NRT
IN
=
the assumption is made that the particles are all of zero volume (point). If we assume that each of
b
them has a volume then it is reasonable to exchange the total volume
for the corrected volume. In
the same way, when
IN
IN Nb
,
ÿ
we have the particles inside a container, those that are close to the
center of the container suffer forces from all the others that tend, by their
random character, to cancel out; those that are close to the walls, however, are more affected by
the particles that are on the opposite side of the wall and this tends to change the force with which
they collide with the wall (the pressure, therefore). The decrease in pressure must be, in this
perspective, proportional to the 2ÿ , since when
number of
peers of molecules or proportional to ÿ1
/in
the volume goes to infinity, the correction should disappear. So, the equation of
wanted state is
P
=
RT
in
The parameters
a
It is
ÿ
ÿ
b
a2
.
in
b depend, of course, on the specific system considered.
ado. We already know that we must have another equation of state to access the
content of the fundamental equation (unless a constant). In the case of only one mole (molar
equations) it makes sense to look for the equation of state (the one involving temperature).thermal
We cannot use the thermal equation
Machine Translated by Google
44 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
of ideal gases, as it is not compatible with the previous equation. We must, then, look for
the simplest expression that can serve for an equation of
thermal state, compatible with the Van der Waals equation.
The appearance of temperature in the pressure equation implies that we can
solve this problem in the entropy representation, because we have
P
R
=
a2
1
in 2
T
ÿ
T
in
b
ÿ
and, since
ds
P
1
=
T
of +
T
DVD,
we should look for
1
=
fu,(v
T
)
,
between this equation
representing the second equation of state. soughtthe compatibility
ds be
and the equation for pressure comes from the requirement that an exact
differential, so that
ÿ
ÿv ÿ 1T
P
ÿ
=
ÿ
ÿ
ÿu
in
T
ÿ
,
in
since this would imply
ÿ2s
ÿ2s
=
.
ÿvÿu
ÿuÿv
So, we have to
ÿ
ÿv ÿ 1T
=
ÿ
P
ÿ
ÿu
in
ÿ
T
ÿ
=
ÿ
R
ÿ
ÿu
in
a
1
in 2
T
b
ÿ
ÿ
a
ÿ
in
ÿ
=ÿ
ÿu ÿ 1T
in 2
in
ÿ
,
in
a condition that can be written as
in 2
ÿ
ÿ
=
ÿ
ÿv ÿ 1T
a
in
ÿ
ÿu ÿ 1T
ÿ
in
It is like
ÿf
ÿ /(1
v
We have to
=
)
ÿv
ÿf
ÿv ÿ (1 /in )
=
1
ÿf
ÿv ÿ (1 /v/ÿv
)
ÿ
ÿ/in
(1
) ÿ 1T
) ÿ 1T
ÿ u/a
(
in
ÿf
ÿv
1
=ÿ
/in 2
ÿ1
ÿ
by
1
cR
=
in
ÿf
ÿv
,
,
in
so that 1 second /T must be a function that relates to variables 1
u/a the same functional dependency. The simplest way is given
T
in 2
ÿ
=
ÿ
=
+
,
and in
/in
It is
Machine Translated by Google
45
3.2. SIXTH CLASS: (04/07/2008)
in
since this equation tends towards that of the ideal gas if we take ÿ ÿ (which
is the basic assumption of ideal gas). Thus, we have the two equations of state
P
R
=
ÿ
T
in
acR2 +uv
ÿ
of
bcR
1
=
T
+
in
and in
which can be used to obtain the fundamental equation by direct integration of
ds
=
1
T
of +
P
T
dv
cR
=
R
ÿ
ÿ
in
+
and/or widow
in
+ÿ
b
ÿ
acR2 +uv
dv.
of
The integration can be done if we notice that, to have the first term on the right,
we must count on the factor
cR ln ( in +
which, derived with respect to the volume
cR
ÿ
in
+ to/go v2
dv + f
ÿ
()in
and in)
in he is
,
R
dv
ÿ
=ÿ
so that
+ f ()in
in
R ln ( in
f ()in=
b
ÿ
acR2 uv
+
of
ÿ DVD,
b)
ÿ
and so
ds
=
d cR
[
ln ( in +
and in)
+ R ln ( in
ÿ
c
b)] = d R ln[( in +
and in)
{
( in
ÿ
b)]}
,
i.e.
s = R ln[( in + and in)
c
( in
ÿ
b)] + s 0 ,
i.e,
S
=
c
Ns 0 + No. ln[( in +
and in)
( in
ÿ
b)]
.
also
real gases
are not usually represented quantitatively in a
correct form by the Van der Waals equation, since statistical considerations show
that there must be more terms in the expansion, the Van der Waals equation being
der Waals a first approximation. However, the Van der Waals equation
gives good results for many
qualitative
real thermodynamic systems.
3.2.2 Electromagnetic Radiation:
If we consider an empty cavity of material corpuscles, with the walls maintained at
T still be electromagnetic energy in the cavity. This
temperature, then there will
system has empirical equations of state given by the Stefan-Boltzmann law
,
IN
=
bV T
4
Machine Translated by Google
46 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
It is
P
where =b 7
.56 × 10ÿ16
3K
J/m
4
IN
=
,
3 IN
which can be calculated from the principle
most basic outside the scope of thermodynamics.
It should be noted that no mention is made of the number thatN
would be the number of
N
photons. In fact, there is not, in this system, any number that is conserved
2
. All the
and by which we can count the particles inside the cavity
,
fundamental equation must be written only in terms of In the representation IN
It is
IN therefore.
,
of entropy, we must have
1
1//b
4 IN1 4 INÿ1 /
=
T
P
1
=
T
1//b
4 IN34
4
INÿ3
/
4
3
so that
dS
1//b
4 IN1 4 INÿ1
=
/ 4dU +
1
1//b
4 IN3 4 INÿ34 /
dV
3
=
4
1///b
4d
4 IN1
IN3
4
d
1///b
4 IN3 4 IN1
3
3
ÿ
so that the fundamental equation becomes
S
4
=
4ÿ =
1///b
4 IN3 4 IN1
ÿ
4
4ÿ
.
3
3.2.3 The rubber bar:
Thermodynamics can also help to find relatively simple macroscopic equations based on a few
experimental observations to
some systems in which the notion of heat appears. An example of this is the rubber bar; in this
system we have two relevant variables: the length of the bar ( ) and the internal energy ( )
L with it (the extensive parameters)
IN with the 'conjugated' internal variables given by the
associated
t
voltage and the temperature The length is analogous to the volume in usual
systems, while the T
L
voltage ist analogous to pressure ÿ
IN
P
.
Experimentally, it is known that the internal energy
no
depends on the length
length of the rubber bar (other than its initial length) and
is linear with temperature – for temperature orders that do not break the
elastic limit of the bar–, so that
IN = cL 0 T,
2In this system, microscopically, photons are constantly absorbed by the wall and
created by it in such a way that there is only an average number of photons involved, not a
exact (conserved) number of them.
.
Machine Translated by Google
47
3.2. SIXTH CLASS: (04/07/2008)
where c
behave
it is a constant. Tension, also experimentally, is seen if
linearly with the
so that
length
,
L L0
bf ()T L
L0
1ÿ
ÿ
t
=
,
L
whereb is a constant, 1 represents
the elastic limit of the bar, and the f ( T ) is a
function is Tonly restricted by the need for thermodynamic consistency.
So, we have to
dS
=
1
bf (T ) L L 0 dL.
T L1ÿ L0
ÿ
T
of
0
cLdU
IN
=
ÿ
ÿT dL
ÿ
dS is an exact differential, we have
So, for what
ÿ
ÿ 1T
ÿL
ÿ
ÿ
bf (T ) L L 0
T L 1ÿ L0ÿ
ÿ
ÿ
=ÿ
ÿU
IN
L
and since the first member is zero, just put cia (0 f (T ) = T to get consistency
= 0). So we have
L0
L0
ÿ
tT
=
bL
L
1
1
,
T
=
cL 0
IN
ÿ
and we are left with
L0
bL
dL
L 1ÿ L0
b
ÿ
dS
=
0
cLdU
IN
ÿ
=
d cL 0 ln ( U/U (0) ÿ
ÿ
L
ÿ
2 (L 1 ÿ
L 0)2
L
0)ÿ
and so
S
=
S0 + cL 0 ln ( U/U (0) ÿ
b
L
ÿ
2 (L 1 ÿ
L 0)2
L 0)
,
which is the fundamental equation of the system.
3.2.4 Magnetic Systems:
Thermodynamic systems do not always adjust so smoothly to
all
the elements that we have been presenting so far. Some systems may
possess certain idiosyncrasies that prevent one or another immediate application of
certain feature of formalism. Magnetic systems have such a
idiosyncrasy and can be used to exemplify this issue. We are
interested here in para- and diamagnetic systems, which are only magnetized in
the presence of external fields.
In these cases, we must consider our thermodynamic system as consti- and
Coming from a sample placed in an external magnetic field and giving Be respon- .
I Thus, choosing fulfills
magnetic moment
this field an appropriate symmetry,
I the role of an extensive variable and
the variable
Machine Translated by Google
48 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Be fulfills the role of its intensive conjugate variable, so the field in our
fundamental equation must be written as
IN
=
IN ( S,V,I,N
)
(3.5)
,
in terms of the extensive variables only, as we already know, and therefore
ÿU
ÿ
Be
=ÿ
.
ÿI
S,V,N
T magnetic field and the Joule/Tesla
The units are: the tesla () for the
Be ()
for the magnetic moment. Note that our definition of a thermodynamic system
implies that the internal energy refers
INonly to the material system, so
what to energy total
of the system is, in fact, given by
1
OUT =
IN +
2
ÿ1
m0
J/T
2
B eV,
where the second term takes into account the energy stored in the magnetic field in the
volume With a
IN
.
fundamental equation like the one shown in (3.5) we have an equation
by Euler
IN
=
TS
ÿ
PV + At
+
Women
that it is always a relation that involves the product of the extensive variables of the problem
by the conjugate intensive variables; and a Gibbs-Duhem relationship
SdT
ÿ
V dP + IdBe + Ndµ
=0
,
which is an equation that relates the intensive variables of the problem from the way they
are written in terms of the extensive variables.
The specificity of magnetic thermodynamic systems is that there is no restrictive wall
regarding the magnetic moment, so that the magnetic moment is a variable that must always
be considered as non-existent.
restricted. In any case, the ideas we have already presented apply
without major modifications and we can, by way of example, explain a
fundamental equation for paramagnetic systems in the form
IN
=
NRT
S
0 exp ÿ No.
+
I2
N 2 I 02
ÿ
.
3.2.5 Second derivatives and material properties:
The first derivatives of the thermodynamic extensive variables proved to be
important for the constitution of the several equations of state and the consequent definition of
the important intensive variables. We also have a series of second derivatives in these same
parameters that represent experimentally observable properties of the different materials and,
therefore, have great
physical interest. We therefore have:
Machine Translated by Google
49
3.2. SIXTH CLASS: (04/07/2008)
1. The coefficient of thermal expansion, defined as
a
1
=
ÿ
1
=
ÿ
ÿvÿT
in
ÿ
IN
P
ÿV
ÿ
,
ÿT
P
which represents the volume response of the system to an increase or decrease in
temperature, maintained at constant pressure (and the number of moles);
2. Thermal compressibility, defined as
1
kT
ÿV
1
ÿ
=ÿ
ÿ
=ÿ
IN
ÿvÿP T
in
ÿ
ÿ
,
ÿP
T
whichrepresents the volume response of the system to a change in pressure, held at
constant temperature (and the number of moles);
3. The molar heat capacity at constant pressure, defined as
Cp
=
T
ÿ
T
N
=
ÿ
ÿsÿT
P
ÿ
ÿS
Qÿ
dT
1
=
ÿ
ÿT
N
P
ÿ
P
which represents the response of the increase or decrease in the amount of
heat in the system due to changes in its temperature, maintaining
assuming constant pressure (and the number of moles).
There are also many other coefficients that represent important responses of
thermodynamic systems to changes in their variables, however, such as
As will be seen further on, these other coefficients are linked to those presented here by
formal relations. Indeed, by way of example, it is easy to show that it is worth
ÿ
ÿT
ÿP
ÿ
ÿV
S,N
(3.6)
ÿ
= ÿ ÿ ÿS
V.N
since they simply represent the formal identity
ÿ
ÿ 2 IN
ÿ 2 IN
ÿ
ÿSÿV
ÿ
N = ÿ ÿVÿS
N
as one can easily identify from
ÿU
T
ÿS
ÿU
ÿ
V,N
,
P
ÿV
ÿ
S,N
.
=ÿ
=ÿÿ
Note that, in equation (3.6), both terms have physical interpretations
immediate: the right side represents the temperature change associated with an adiabatic
expansion of the volume, while the other represents a
change in pressure when heat is introduced into the system while maintaining the volume
constant.
That these two magnitudes are equal is somethingabsolutely not
trivial e can be considered a first triumph of the theory
.
Machine Translated by Google
50 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
There are, as we have already said, other coefficients that have explicit mathematical
relations with which we have already defined. Thus, to exemplify, since we will deal with this topic
later, it can be shown that, if we define the adiabatic compressibility as being
1
kS
ÿvÿP
in
=ÿ
ÿ
s
ÿ
and the molar heat capacity at constant volume to be
=
Cv
T ÿ
ÿ
,
ÿsÿT
in
then it can be shown that the relations hold
T Va
=
CP
Cv +
T Va
2
NkT
=
kT
,
kS +
2
.
NCP
For now, it is enough to know the three basic coefficients presented in the
beginning of the section and knowing that there are other coefficients, as well as relationships
between them.
3.3 Chapter Exercises:
Exercise 39
a first, as an example. We have
(3.1-1) Let's take
=
S
)1 /
3
,
entropic representation, so that
which suggests the
1
A ( NUV
1
=
T
3
A
( NV )1 /
IN2/ 3
P
3
,
1
=
T
3
A
( NOT )1 /
IN2/ 3
3
m
T
,
1
=
3
A
( UV )1 /
N 2/ 3
3
so that
S
=ÿ1
that once done
mind.
Exercise 40
3
A
3
( NV )1 /
IN2/ 3 ÿ
+ÿ1
3
A
as simplifications, reproduces
=
relationship between
A
S3
NV
,
2
+ÿ1
3
A
3
( UV )1 /
N 2/ 3 ÿ
N
a original expression, evident-
S4
.
NV
T, P
P =2 A
2
µ asobtaining
this, we
have
and extensive variables: doing
T =4 A
IN
a equation of state
(3.2-1) We have
IN
We can find the
eliminating the
( NOT )1 /
IN2/ 3 ÿ
3
IN
S4
NV
,
3
m
=ÿ
state equationsand
A
S4
N 2 IN2
,
Machine Translated by Google
3.3. CHAPTER EXERCISES:
51
or yet
T =4 A
3
s
in
4
in
so that
2
6
sv4 =
A
=ÿ
s
4
in
2 ,
in
1
in 3
16 A 2
=
m
3 ,
s4
2A
P
T
s
P =2 A
2 ,
8A s2
and,
therefore,
T
1
2
A
ÿ
ÿ
64 A
P
4
s
=
in
ÿ2 = ÿ
m.
2
Like this
T
4
64 A P
2
1
m
Exercise 41
=ÿ
.
(3.3-1) We have to
s
T =3 A
2
in
=
P
,
3
As
in
2
,
with To a constant. We can find the
fundamental equation finding, chemical
a
first, sitting is given a function that describes O potential. We have to
a energy as a dependent variable (already
is
with
that the
represent- entropy
a entropy is considered
what
that appears in the equations of state, indicating that
a independent variable), so
=
dµ
T ds
VAT
ÿ
andwe have
dµ
A
=ÿÿ3
s
2
ds
s
A
ÿ
in
3
in
2
dv ÿ = ÿÿ d
As
3
ÿ
in
,
so that
3
m
As in
=ÿ
ÿ1 +
m0 ;
of Euler how
The fundamental equation remains, formula therefore, given by
=
IN
TS
PV + N,
ÿ
or
S
3
S
IN = 3 A NV
3
S
A NV
ÿ
ÿ
3
A NV
+
0
WOMEN
or
IN
=
A
S
3
+
NV
WOMEN0
We could also obtain the molar
fundamental equation
form of the fundamentalequation, given by
of
=
T ds
ÿ
VAT,
.
a from the direct integration
Machine Translated by Google
52 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
giving
of = 3 A
s
2
s
A
ÿ
vds
in
3
s
ÿA
dv = d
2
3
ÿ
in
so that
s
in =
3
A in + in 0
or yet
S3
IN = A
+ Not 0
NV
.
(3.3-2) We have to
Exercise 42
IN = PV,
=
P
BT
2
so that
T
2
IN
=
BV
and,
therefore,
1
T
=ÿ
BV
.
IN
So, we also have to
P
T
=
IN
BV
×ÿ
IN
=ÿ
IN
THIS
IN
,
so that now we have the equations of state written only in
terms of the extensive variables. We can also write its molar form
as
1
T
It is
get the
P
=ÿ
Bvu ,
=ÿ
T
This
in
fundamental equation integrating
ds =
1
T
of +
P
dv
T
for
obterds =
1 2 1 2 ÿ1 2
B / in / in / of
1 2 1 2 ÿ1 2
+ B / in / in / dv
=
1 2 1 2 1
d ÿ 2 B / in / in / 2ÿ
so that
1 2 1 2 1/
s = 2 B / in / v
2
+ s0
yet or,
1 2 1 2 1
S = 2 B /// IN
IN
2
+ Ns 0
.
Machine Translated by Google
3.3. CHAPTER EXERCISES:
(3.3-3) We have to
Exercise 43
P
It is
53
=ÿ
ÿ 2 OF IN
NO NW
we want to find the
1
1
IN // 2 IN
T =2 C
,
2
WATER/N
N ÿ 2 AT
state equation. We write, first
NOT
1
P
=ÿ
IN N ÿ 2 AT
to obtain, by division,
1/ NO
eAU/N
2 ÿÿ1 = ÿ
T = ÿÿ 2
It is
u 1/ 2
2 Cv 3/
CV 3/ 2
P
ÿ
At
It is
2
also
1
1 ÿ 2 At
2 With1/ 2 in 1/
=
T
a
to stay with
At
ÿ
1 2 1 2
ÿin1 ÿ/ 2in and /
2C
At
equation
At
1
ds =
ÿ
It is
2
of
so that
1
in /
2
in 3/
2
u 1/ 2 in
s =
1
At
ÿ
ÿ
dv
It is
C
ÿ=
/
ÿ1
d
ÿ
1 2
in / v
ÿ1
/
2
and
2
At
ÿ
+ s0
It is
C
and,
therefore,
NOT 1/ 2 IN
S =
Exercise 44
/
ÿ1
2
ÿ
It is
C
HAS A + Ns .
0
(3.3-4) Now we have to
3
in =
2
in
Pv,
1/
2
=
1
BT in /
3
so that
1
=
B
T
1
in /
3
in 1/
2
P
,
=
T
2
3
B
1
in /
2
in 2/
3
so
stay =
It is
B
ÿ
1 3
in / u
ÿ1
2
/ of
2
+
3
1 2 ÿ23 /
in / v
dv
1 2 1
This / in /
d
ÿ=ÿ2
like this
1 2 1
s = 2 This / in /
3
+ s 0,
or yet
1 2
S = 2 THIS /// IN
1 3
N
1
6
+ Ns 0
.
3ÿ
ÿ
At
ÿ
,
Machine Translated by Google
54 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Exercise 45
(3.4-2) We have to
NRT
P =
It is
IN = 3 / 2 NRT
,
IN
we also have to
T dS
VAT
VAT,
of an adiabatic compression. Thus, we can see that, in this
=
of
=ÿ
ÿ
case, it is
to write
3
IN =
PV,
2
so that
3
=
of
3
VAT
2
V dP
ÿ
2
VAT
=ÿ
and,
therefore,
5
or,
dv
dP
=3
IN
P
integrating,
5/
PV
3
=
const.
Note further that
s
2 (s
0
3R
in
)
2/
in
3
3
It is
like this
exp ÿ
in
ÿ=
in
Pvu
0ÿ0ÿ
5
Pv /
=
in
3
Pv 5/
3
2
2/
3
3
uv0
0
0ÿ0ÿ
2
3
2/
in
2
=
ÿ
=
3
uv0
/ 3
and
20
2( s
s 0) / 3 R
ÿ
or yet
5
Pv /
3
=
2 3
3
2
Pvv
0 0
2( s
/ 3
and
20
ÿ
s 0) / 3 R
Pv0
=ÿ
/ 3
and
50
ÿ2
s0/ 3R
ÿ
2 s/ 3 R
It is
as presented in the book.
Exercise 46
(3.4-3) We have to (to
T 0 0= 0 C,
The initial pressure
2
moles)
IN0 = 45 × 10ÿ3
3
m
,
Tf
0= ÿ50
C.
It is
P IN = NRT
0
and therefore P =
0
. × 273
2 × 8314
45 × 10ÿ3
The final volume can be given
0
0
.38765 × 10
= 100
a starting from
Pv
5/
3
=
Pv0
/
50
3
.
Well = 010087
MPa.
,
Machine Translated by Google
3.3. CHAPTER EXERCISES:
55
It is
Pv
=
RT
of so that (dividing one equation into another) (note by )
N =2
a divide by
in
in0
5/
2
vf /
3
P 0 in
=
05
/
3
3
100 .8765 × 103 × ÿ45 × 10ÿ32 ÿ/
8 .314 × (273 ÿ 50)
=
RTf
.
= 0097569
and,
therefore,
vf
It is
.
= 0030476
like this
Exercise 47
there
IN
VAT
=ÿÿ
=ÿÿ
IN0
.
a integral of mechanical work:
(3.4-4) Let's do
IN
WM
.0609535 = 61 × 10ÿ3= m 3
0
Vf
IN0
3
Pv500 /
dv
53
ÿ
Pv500
2
=ÿ
in /
/
3
in f
53
in 23/
3
2
Pv00 23 /
3
in /
=ÿ
2
+
3
Pv0
0
in 0
final work of the previous year is
100 .
3
=
WM
/
5/
2
3
3
+
.
1008765
ÿ
so that
O
10ÿ3 /
×
/
/ 2 3ÿÿ
× 45 × 10ÿ3
2ÿ
8765 × 10
2ÿ
.
.
.
(61= ÿ2779620900
= 624960975
+ 3404 10ÿ3)2581875
3
ÿ
10 3ÿÿ 45
×
ÿ
The work per number of moles is given as above. energy
initial can be calculated easily as
IN0
of
3
=
2
NRT
0
= 15. × 2 × 8314. × 273 = 68091660 .
final energy
Uf
=
3
.
= 55620660
2 NRTf
so that
ÿ IN = 55620660
. ÿ 68091660 .= 12471
Total work is NWM to decimal
= 12499. J, being
. J
a difference due to simplifications
places.
Exercise 48
(3.4-5)
There is a relationship
V
T
The work done on
O
IN1
WM
IN0
IN1
=ÿÿ
the
T0
.
IN1 <
gas to compress it to a volume
VAT
=ÿÿ
= ÿ IN0 ÿ
IN0
NRT
IN
dv
=ÿ
No.
T
INh
0
IN1
INthe
0 ÿ IN0
ÿ1
dv
Yes
0 _
NRT h0
=ÿ
the
ÿV 0
2
ÿ
INf
IN
IN0
Machine Translated by Google
56 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
and,
therefore,
WM
the
NRT
=
Vf
0
(3.7)
.
ÿ1ÿÿ
IN0 ÿÿ
the
A de do stays gas energy variation
the
=
IN0
3
2
Uf =
NRT 0 ,
=
3
2 NRTf
3
2
Vf
NRT
IN
0ÿ
0
so that
the
ÿ IN
the heat transferred
to the
3
=
Vf
NRT
2
.
0 ÿÿ 1ÿ
ÿ0 ÿÿ IN
gas is simply
the
ÿ Q =ÿ
=
IN + ÿ IN
so that
ÿ Q =0
integrating directly
with
the
= 23/
-
NRT
0 ÿ 1the
Vf
3 2ÿÿ 1 ÿ ÿ
IN0 ÿÿ
O
We can also calculate ratio
.
,
transferred heat
a
dQ
=
T dS,
T ofin terms
but for that we need to explain
however, that
Tÿ S
=
of
S (or vice versa). We have,
= ÿ IN + ÿ IN
+ VAT
like before.
Exercise 49
of the relationship
(3.4-6) We left
c
c
=
S
Ns 0 + No.
ln ÿÿ
IN
IN
N
IN0 ÿÿ
IN0 ÿÿ
N 0 ÿÿ(+1)ÿ
to get
1
T
P
T
m
ÿS
ÿN
T =
Exercise 50
ÿ
ÿ
=
R
ln ÿ
=
ÿV
c
IN
IN0
ÿ
=
ÿ
ÿ
ÿ
cNR
IN
=
No.
IN
c
N
N
IN
IN0
.
+1)ÿ ÿ
Rc(+ 1)
ÿ
(3.4-7) We have to ÿ
S
= ÿ ÿS
ÿU
= ÿ ÿS
ÿ
ÿ
ÿÿ
0 ÿÿ(
Ns1 1, 0 + Ns2 2 ,0 + ( Nc1 1 + Nc2
2)
R
ln ÿ
T
T0 ÿ +
,
N 1R
ln ÿ
VN
in 0 ÿ +
1
N 2R
ln ÿ
VN
in 0 ÿ
2
with
IN = ( Nc1 1 + Nc2
2)
RT.
Machine Translated by Google
3.3. CHAPTER EXERCISES:
57
A of state, eliminated equation
S
=
O
N 1 s 1, 0 + N 2 s 2, 0 + ( N 1c 1 + N 2 c2) R
N 1R
It is
of
we have the
temperature term,
ln ÿ
VN1 in 0
ÿ+
N 2R
from the parametric
ln ÿ
N
c 1+1+N 2,0 c 2)
IN (1,0(
Nc1
Nc2 2) ÿ
IN0
VN2 in
ln ÿ
+
0ÿ
state equations:
equation of
m1
T
=ÿ
m2
T
=ÿ
1 = N 1 c 1+ N 2 c 2
R
T
IN
P = N 1+ N 2
R
T
IN
T0 ÿ ÿ ln ÿ
s 1, 0 + Rc 1 + R
c 1R
R
ln ÿ T
VN1 in
T0 ÿ ÿ ln ÿ
R
T
VN2 in
ÿ
s 2 , 0 + Rc 2 + R
c2 R
ÿ
ln ÿ
0ÿ
,
0ÿ
Euler
P
IN
T
1
stay =
TU
+
m1 N
m2 N
1ÿ
2
T
T
ÿ
giving
S = ( Nc1 1 + Nc2 2) R + ( N 1 + N 2) R
T0 ÿ ÿ
0 ÿÿÿ
N 1 ÿÿ s 1, 0 + Rc 1 + R c 1 R
R
T
VN1 in
ÿ
ÿ
N2
s 2 , 0 + Rc 2 + R
ln ÿ
c2 R ln TT
ÿ
as simplifications,
stay
ÿÿ
that after
S
=
Ns1
1, 0 + Ns2
RN
is exactly that the
inamic.
Exercise 51
1
ln ÿ
ln ÿ
R ln
ÿ
VN2 in 0
0
ÿ
ÿ
2 , 0 + ( 1 Nc1 +
VN1 in 0
ÿ+
RN 2
ÿÿ
ÿ
2 Nc2) R
VN2 in
ln ÿ
ln ÿ
T
T0 ÿ +
0ÿ
fundamental equation for this type of thermod system
(3.4-8)
V ÿV initial , so that I saw final
There is an expansionof
ÿ
of
ÿV 0 We want to know aaratio between pressure
Vf
ratio between the initial temperature
of
temperature. We have to
=
.
PV
=
pressure final
NRT.
The compression being adiabatic, we already know (from a previous exercise) that
OK
Pv 5/ 3 It is2 s/ 3 R
=
Pv0
/
2s 0 / 3R
3
It is
50
.
The first equation indicates
quePfVf
P0 IN0
Mass
=
ÿPf
P0
T0
s =
expansion being adiabatic, we have
Pf
/
in0
P0 = ÿ vf
Tf
=
ÿ5
.
s 0 , so that
3
=
l
ÿ5
/
3
.
=
of
IN0 ,
Machine Translated by Google
58 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Like this,
Pf
l
=
P0
Exercise 52
ÿ5 /
Tf
3
,
=
T0
l
ÿ2 /
3
.
In each one
tanks.
(3.4-9) We have He in both gas equations
theirs is worth as
P 1 IN1
=
N 1 RT 1
P 2 IN2
=
N 2 RT 2
It is
IN1
3
=
IN2
N 1 RT 1
23
=
2
N 2 RT 2
We can calculate lemma.a initial temperature of each of them with
Like this
T1
=
T2
=
P 1 IN1
ÿ5 × 10 6ÿ × (01). = 60139
.
8314
.
6
(015)
=
R
P 2 IN2
With
R
=ÿ
as temperatures, we calculate
IN1
3
=
you prob data
.524K
. 1427K
.314 = 108251 8
×
10 6ÿ ×
as initial energies:
.
.
58314
× 60139524
= 75 .× 10
J
2
IN2
3
=
.
.
68314
× 1082511427
= 135. × 10
J
2
a valve connecting both
When we open the element of
balance that, in this case, must establish
IN
T1
Remembering that now there is
2
there are tanks, the
=
. × 10
IN1 + IN2 6 = 21
=
T2
. m 3 , we have
IN = 025
moles no interior do volume total
2 . 1 × 106 = 32 2 × 8314
. ×
T
and,
therefore,
T = 84195333
K
.
the final pressure
of the system must be such that
P
=
.314 × 841952.
× 8333
.
025
estab-
. × 10 Well.
6= 56
Machine Translated by Google
3.3. CHAPTER EXERCISES:
Exercise 53
1.
(a)
59
(3.4-10)
O temperatures
case
Now we have initial
of (in the same volumes).
T 1 = 300 K and T 2
350
as initial pressures
We can calculate K
as being
N1=
P 1 IN1
.
.
10 6ÿ
(0 1) = 200 47
RT 1 = ÿ5 8×.314
× 300
N2
so that
=
P 2 IN2
=
=
.
ÿ6 × 10 6ÿ (015)
8 .314 × 350
RT 2
= 309 .29
a final energy stays
= ( N 1 + N 2) RT.
Uf
The are initial energies
IN1
3
=
. × 8314
. × 300 = 75001841 .
20047
2
IN2
3
=
. × 8314
. × 350 = 135000445 .
30929
2
It is
like this
. × 10
IN 6 = 21
It is
like this
T
21. × 10
2
=
3
6
He a
(b) Se o first tank contained
had an idealdiatomic gas with c
already as before,
.
=
.33034314
. × 847 +. 309
(20029)
K him
K
second tank with
300=
/ 25 a 350 K, then we would have
N1
= 200
.47
N2
= 309
.29
the energies
that the result is independent of the constant c.
they would stay
IN1
IN2
so that
=
=
3
5
. × 8314
. × 300 = 750018412 .
20047
. × 8314
. × 350 = 2250734902 .
30929
J
J
a total energy stays
. J
IN = 300075331
It is
like this
3000753 .312 = (
3
. +
20047
3092
5
2
so that
T = 33608
. K.
.29) × 8314
. ×
T
Machine Translated by Google
60 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Exercise 54
P
It is
ideal multi-
ÿS
ÿ
T = ÿ ÿV
It is
one gas simple
a depression
(3.4-11) We have that
given by
component is
ÿ
using
ÿ
S=ÿ
No
ÿ
R ln
Njjj
0+
ÿÿ j
j
ÿ
stay with
P
T
IN
+ No.
T0
ln ÿ Nv
R
0ÿÿ
ÿ
Nj ln
j
No.
=
T
IN
It is like
r
N
Nj,
=ÿ
j =1
stay with
r
P
r
Nj
=ÿ
RT
Pj,
IN = ÿj =1
j =1
are
that the
partial pressures.
Exercise 55
(3.4-12)
as
ÿS
µj
T =ÿÿ
so that
The electrochemical potential is calculated immediately
=
ÿ
ÿNj
local unit0
R
T
=
ÿR
T
ln ÿ T
t is
a explicit function of
f
R (1 + cj ) ÿ cjR
IN ÿ +
ln ÿ
ÿ
(1 + cj ) ÿ cjR
T
ln ÿ T
0 ÿÿ
.
Since
local unit0
Pvv 0
RT
=
IN
,
we can, of course, write
µj
Exercise 56
1.
=
Pvv 0
RT
RT ln
(3.5-1)
ÿ
T
+ RT (1 + cj ) ÿ cjRT ln
ÿ
ÿ
We have:
in
= aPv
,
Pv
2
=
so that, dividing one by the other,
in
bT
=
aPv
Pv
2
=
a
in
bT
.
T0
ÿ
0ÿ
,
Nj
N
,
Machine Translated by Google
3.3. CHAPTER EXERCISES:
61
so that
P
1
=
;
=
T
uncle
T
,
bv2
are our equations of state. We now want to know that they are
thermodynamically compatible. For that, we must remember
what
ds =
of
dv
+
T
ÿsÿu
=ÿ
ÿ
ÿ
ÿv
so that we should
ÿsÿu
=
ÿ
ÿ
wife + ÿ
ÿsÿvudv
ÿ
ÿu
in
ÿ
ÿ
ÿsÿv
ÿ
,
in
P
ÿ
ÿu
ÿ
T
as equations of state alreadyobtained, we have
for compatibility. With
with
ÿ
ÿ
ÿ
ÿv
with verifies,
ÿ
to have
ÿv ÿ 1T ÿ =
uncleÿ
=
ÿ
ÿu
bv2 ÿ
therefore there is no compatibility.
Now we have
in =
2
aPv
Pv
,
so that
1
P
=
T
of
ÿ
ds is exact, we must have e that,
in order for
2.
2
=
=
T
abu ,
bT
bv2
compatibility implies that
ÿ
ÿ
ÿv
ÿ
abuÿ =
ÿ
ÿu
bv2 ÿ
is that satisfied, there is compatibility.
3.
We have
in c + buv
P =
so that
T =
,
in a + buv
in
,
a + buv
as state equationsstay
P
c + buv
=
T
It is
equations
P
1
T
that no
with such
we should have
ÿ
ÿv
ÿ
in
T =
a + buv
ÿ
in
ÿu
ÿ=
doing or, as derivatives
b = b,
implying compatibility.
a + buv
1
,
in
c + buv
ÿ
in
ÿ
,
Machine Translated by Google
62 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Exercise 57
The expansion is
(3.5-2)
isentropic (quasi-static)
It is
adiabatic).
Like this,
P
1
dS = 0 =
of +
T
DVD,
T
or yet
R
1
Tdu
in
b
ÿ
acR2 +uv
ÿ
dv = 0
of
.
ÿ
+ÿ
Now, we know that
a
in
+ cRT
=ÿ
in
so that
a
=
of
dv + cRdT
in 2
stay with
It is
1
ÿ
T
a
in 2
dv + cRdT
with we
that can be simplified
1
R
ÿ+ÿ
ÿ
ÿ
in
acR2 uv
+
b
ÿ
dv = 0
of
,
put
a
T in 2
dv
cR
=
in
+ a/v
.a 2 dv
in
to stay with
cRdT
R
=ÿ
T
in
bdv
ÿ
giving
cR ln T
R ln ( in
=ÿ
ÿ
b) + const.,
or yet
T
c
( in
ÿ
b) = A,
where A is a constant.
Exercise 58
It is
a temperature of
(3.5-3) Two moles of CO are at a volume
of
2
.
adiabatically expanded
It is
statically
m 3 until
its temperature
drops What are the final
45O×gas
10ÿ3
pressures? In this problem, of course, we can immediately usea ÿ50 oC.
It is as
final
a equation we derived in the previous exercise
0
oC
quasivolumes
It is
c
T
deal with
As if
c = 35 .
.
CO
2,
ÿ
b) = A.
.
a = 0401
,
The initial data imply that
(273)7
It is
( in
we have from the table that
/
2
.
/ × 10ÿ3 ÿ 427 × 10ÿ6ÿ
ÿ452
=
like this
A = 7549649773
.
.
A,
b = 42 . 7 × 10ÿ6
It is
Machine Translated by Google
3.3. CHAPTER EXERCISES:
63
No final state, then we will have
/2
(273 ÿ 50)7
vf ÿ 42 . 7 × 10ÿ6ÿ = 7549649
ÿ
. 773
so that
.
vf = 004563167046
and,
therefore,
Vf = 2 ×
.
vf = 009126334092
.
The pressure can be obtained immediately from the equations of state.
Exercise 59
fluent
of
deals with a
and what if
2
T0
IN0 ÿÿ1=
ÿ
/
/2
IN
T
1.
ÿ = ÿ1
(3.5-4) We must assume that
der Waals. We therefore have to
by
(a) The equation of state at pressure becomes
P
so that
a
R
=
ÿ
T
in
2
in T
b
ÿ
3
P =
a2
ÿ
Rv bT
in
ÿ
or yet
R
P =
in
It is
like this,
O
/2
in
b
ÿ
a
T0ÿ
in0
in
ÿÿ1ÿ
2
work is given by
in 1
R
WM
=ÿÿ
in 0
ÿ
in
/2
in
b
ÿ
in0
a
T 0ÿ
dv
in
ÿÿ1ÿ
2ÿ
worth to worth calculating a integral in details; We have
in 1
WM
put
=ÿ
RT
=ÿ
0ÿ
R
in
0 ÿ in 0
ÿ
in ( in
ÿ
b)
dv
a
ÿ
in
v, we are left with
in 1
WM
RT 0ÿ in
=ÿ
0 ÿ in 0
ÿ
2
( in
2
you
ÿ
2 RT
WM
a
in 1
in
in 0
ÿ
b)
ÿ
with to obtain
which can be integrated by parts to
3Note that
;
0ÿ
in0
=ÿ
2b
b ÿ 0 and a ÿ 0 we have the result of ideal gases.
ln
ÿ
in
ÿ
b
a
in 1
ÿ
ÿ
in + b
in
in 0
;
Machine Translated by Google
64 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
or yet
ÿ
WM
RT 0ÿ in
1 ÿ b/ ÿ in
0 ln
b
1 + b/ ÿ in
=ÿ
a
in 1
ÿ
.
in
in 0
Now how
x ÿ 1 x 2 + O (3)
2
x)=ÿ
ln(1 ÿ
x ÿ 1 x 2 + O (3)
2
x)=
ln (1 +
,
we have to second
order in
(up to x)
x ) ÿ ln(1 +
ln(1 ÿ
ÿ0
So for b
,
a ÿ0
,
that theis
ÿÿ2 ÿ
in
2ÿ
in 1
b
=ÿ
b
x.
we would have
RT 0ÿ in0
WM
x ) = ÿ2
= 2 RT 0 ÿ
ÿÿ
in0
in 1
/
ÿ1
,
in
in 0
in 0
= ÿ1
same as we would have if we put
(3.7)
ÿ (b)
/ 2.
The change in energy can be calculated as
IN0 = cRT
a
0ÿ
in0
a
IN = cRT
,
ÿ
=
in
0 ÿÿ1ÿ
in
cR
in
/
2
T0ÿ
a
in
so that
in0
ÿ IN = cRT 0 ÿÿ
ÿ0
with a
I feel that
,
/
2
ÿ1
Exercise 60
ÿ
0ÿ
in
in
.
immediately we will have to
ÿ IN = cRT 0 ÿÿ
that theis
1
ÿ 1ÿ ÿ ÿ 1 a
in
in0
/
2
ÿ1
in
ÿ 1ÿ
,
expected previous result.
(3.5-6)
One mole (N
= 1 ) on one idealmonatomic gas: satisfies
1
as equations
P 1 IN1 = RT 1 ,
One mole of
= 563 .
IN1 =
3
2
RT.
Cÿb2 (N 2 = 1= ) satisfies ) as van equations
10ÿ6 , c 28 .
the Waals (
.
a = 0659
×
P2
We know that
IN2 = 28 . RT
ÿ
INÿ 56
if the
.
0659
R
=
T2
gases are
center of the cylinder volumes
(the
2 ,
T 2 IN2
.3 × 10ÿ6 2
at the same temperature).
We
. 0 have
IN1 = VV
05must
2 =
P1
T1
=
P2
T2
,
.
0659
2ÿ
300 K
.
IN2
O
piston is in
,
Machine Translated by Google
3.3. CHAPTER EXERCISES:
with
T2
=
T 1 = 300 K.
.
8314
65
Like this,
.
8314
=
.
0659
ÿ
05. IN0
05. IN0 ÿ 56 .3 × 10ÿ6
300 × (025.
2
IN0
)
where do we get that
3
.
m
IN0 = 00001430905455715452
like this
. × 10ÿ4 = 0
V 1 = 05 . IN0
and,
therefore,
P1
RT
=
1
.
8314
× 300
=
.
7= 3486
× 10 Well.
. 715452 × 10ÿ4 0
IN1
The universe would have
V 0which is
Exercise 61
(3.6-1)
volume volume
2 0 VAexpansion is knowing
assumed
that the
to universe
be isentropic
(background)
of is
would expand to a
It is
.
a asked at the end of the expansion. expansion is isentropic,Twe
= 27 . K, what temperature is
0 have
energy
With a
to the
=
S0
4
S
1
ÿ=
34
4
IN0 ///bIN10
4
4
=
3
It is,
1
4
3
///b
IN
4
34
IN
3
as
=
IN
bV T
4
,
stay with
3 4
3
b / IN0 T 0
It is
34
b / INT
=
IN = 2 IN0 , we have
so, using that
3
IN0 T 20=
It is
3
IN0 T
3
like this,
/ 3
T = 2ÿ1
Exercise 62
(3.6-2)
T = 27 . K.
perature
T = 21429914200
.
K.
Electromagnetic radiation is in equilibrium at temp. We have to know
a pressure, so we need to state-relate, we
pressure.
a temperature with mind a
Now, from the equations
immediately obtain
that
P
=
IN
bV T
=
3 IN
4
=
3 IN
1
bT
4
,
3
so that
P
=
1
73
Exercise 63
.056 × 10ÿ16 (2134
. 7)4× =10ÿ13
.
(3.7-1) We have to
t
It is,
as ÿ
Well.
=
=
bT
L
L
ÿ
1ÿ
L0
L0
const, we do
ÿ
dt = 0 =
LbÿT
L
1ÿ
L0
L0
+ bT
L
L
1ÿ
L0
a
Machine Translated by Google
66 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
so that
a fractional change as a function of the increase
temperature
in
is
L
ÿT
=ÿ
L
Exercise 64
T
L0
ÿ
.
dS
of
cL 0
IN
=
a out
a heat transfer to
(3.7-2) We want to calculate
dL
of rubber when it is stretcheda shelf length. We have to
ÿ
ÿ
bL
L1ÿ
temperature with-
L0
dL.
L0
a temperature is kept constant from the
It is U just
As
depends on the temperature
length variation), dU
= 0 It is like that,
(no
L0
bL
dL
L 1ÿ L0
ÿ
dS
It is
like that
dS
=
T
=ÿ
What, we have to
L0
b L
dL.
T L1ÿ L0
ÿ
Q
O
This value is evidently dU
=0
=ÿ
additive inverse of the work done, since
.
Exercise 65
(3.7-3) We now have what is worth
IN = cL 0 T 2
of
consistency equation implies that
ÿ
ÿ 1T
ÿL
=
ÿ
ÿ
ÿU
IN
ÿ
ÿÿ
tT
.
L
Now we know that
1
IN
T = ÿ cL 0
which continues a does not dependon the length L. null, so Thus, it
consistency remains
the
that nothing
requirementequationthat
beÿ linear in temperature.
Exercise 66
O
(3.8-1) Knowing that
IN = NRT
I2
S
+
2 2
N
I0
No.
0 expÿ
ÿ
,
we calculate
T
ÿU
=
ÿ
= ÿ ÿS
I,N
T
I2
S
+
2 2
N
I0
No.
0 expÿ
ÿ
left side of
with alters the
Machine Translated by Google
3.3. CHAPTER EXERCISES:
It is
67
also
ÿU
m
=
ÿ
= ÿ ÿN
T
R
I2
S
ÿ2 R
N
N 2 I 02
ÿ
0ÿ
I,S
I2
S
+
N 2 I 02
No.
ÿ exp ÿ
ÿ
It is
IN
=
ÿI
ÿ U,N
2
ÿ
The equation of Euler =
stays
IN
I2
S
No. + N 2 I 2
2RT 0 I
ÿU
Be
0 expÿ
.
0ÿ
I2
S
ÿ×
TS + At + Women
+
2 2
N
I0
No.
= exp ÿ
2
I
2RT I 2
S
ÿ T 0 S + NT
ÿ2 R
R
2 2
N
N I0 ÿ +
IN 020
0ÿ
=
ÿ
ÿ
that gives, after you cancellations
IN = NT 0 R
I2
S
+
N 2 I 02
expÿ No.
ÿ
as expected.
Exercise 67
(3.8-2) Now we have
m0
IN =
2S
2
2NÿI
,
ÿ
expÿ
T
It is
No.
+ Nÿ
so that
ÿU
2
=
ÿ
= ÿ ÿS
ÿR
I,N
S
No.
expÿ 2ÿ
also
ÿU
m
= ÿ ÿN
2
2
2
m0 I R + 2 eNxR
ÿ
=
ÿ
exp ÿ
SNR
ÿ ÿ 4e SN x exp ÿ
2
ÿ
SNR
2
2NÿR
I,S
It is
ÿU
ÿ
Be
It's at equation of
= ÿ ÿI
=
m0 I
No
+
Women
U,N
.
Euler stays
IN = T S +
At
and,
therefore,
IN =
2eS
S
R expÿ 2ÿ+
No.
that after
you
2
m0 I
+
No
2
2
m0 I R + 2 eNxR
cancellations, provides
IN =
as expected.
ÿ
O
result
m0 I 2
+ Nÿ
2No
exp ÿ SNR
2 ÿ ÿ 4e
2NÿR
S
No.
expÿ 2ÿ
,
SN
x exp ÿ 2 ÿ
SNR
Machine Translated by Google
68 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Exercise 68
(3.9-1) For an ideal multicomponent we have gas
Cv
¯
ÿ
ÿS
N ÿ ÿT ÿ
,
.
=
R
N
IN
¯
c R,
coefficient we must calculate ÿ, a
.
P e T
jNjc
=
N 1 c 1 + N 2 c 2 to get the
where c =
betweenV
T
=
relationship
We have
whatPV
=
= ( N 1 + N 2) RT
NRT,
so that
IN
It is
NRT
=
P
like this,
a
1
ÿV
IN ÿ ÿT ÿ
=
No.
1
=
1
=
=
IN P
P
.
T
NR NRT
The isothermal compressibility is given by
1
ÿV
IN ÿ ÿP ÿ
=ÿ
kT
=
T
1
NRT
IN
P
IN
=
1
=
.
P
NRT
2
a constant pressure is
The calculation of molar heat capacity
T
=
CP
made by the formula
ÿS
N
ÿT
ÿ
P
;
ÿ
thus, we must write S of P in T and calculus
to
terms. We have
Ve
(not of
T) to proceed
to the
ÿ
S
No
=ÿ
ÿ
0+
Njjj
ÿÿ
j
j
R ln
ÿ
T
T0 + ÿ
IN
Mr
j
ln ÿ
0ÿ
(3.8)
local unit
and we use that
PV
where N = ÿ
jNj
No
=ÿ
calculate to
ÿ
0+
j
It is
NRT,
to write
ÿ
S
=
ÿÿ
Njjj
j
R ln
ÿ
T
T0 + ÿ
NRT
Mr
j
ln ÿ
local unit0
P ÿ
(3.9)
partial derivative getting
CP = (¯+c 1)
To calculate
garden center P
R
a adiabatic compressibility, we can use
of isentropic volume changes (for
v no case
and multicomponent), once
quekS
=ÿ
1
ÿV
IN ÿ ÿP ÿ
.
S
a relationship that should
O ideal gas
Machine Translated by Google
3.3. CHAPTER EXERCISES:
69
As
NRT
=
IN
,
P
We have to
ÿV
ÿ
ÿ
S
ÿP
=
ÿ
NRT
No.
1
ÿP P
ÿ
S
ÿ
ÿT
P
+
ÿ
S
ÿP
ÿ
;
The first term can be evaluated immediately, giving
NRT
ÿ
2
,
P
O
while
as
second term can be calculated from (3.9) expression
ÿ
dS = 0 =
ÿ
R
Njjj
ÿÿ
ÿ
T
dP
=
j
Mr
dT
+ÿ j
dT
T
ÿÿ
Mr
dP
P
j
so that
No.
P
¯
No. + Nc R
dT,
T
or yet
¯
1
P
It is
(1 + c ) dT
T
=
dP
like this
dT
T
=
¯
(1 + c ) P
dP
,
providing
ÿ
ÿV
It is
NRT
ÿ
+ No.
2
=ÿ
S
ÿP
P
T
2
(1 + ¯)c P
like this
kS
1
=ÿ
IN
is that
O
ÿ
ÿV
ÿP
ÿ
¯
NRT
c
1
2
=
S
¯
P IN ÿ 1 ÿ 1 + c ÿ =
1
¯
1+ cP
,
expected outcome. With these values, we have
2
T Va
CP
=
CV
+
NkT
,
then, replacing
2ÿ
c 1)
(¯+
R
what it is an identity.
=
¯
cR +
T IN
/T
N ÿ1(1
/P
=
)
¯
=
cR +
PV NT
¯
cR + R = (¯+c 1)
R,
Machine Translated by Google
70 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Exercise 69
(3.9-2)
Equation (3.70) tells us that
ÿT
ÿ
It is
ÿ
(3.10)
ÿS
S,N = ÿ ÿ
V,N
we must show that it holds for an ideal multicomponent.
gas first
derivative can be calculated using
a equation by(3.8)
which we have NR
¯
dS
It is
ÿP
ÿ
ÿV
c No.
=
dv
IN
S e N are kept constant, we have
as, in this derivative,
ÿT
ÿ
T
ÿ
¯
=ÿ
ÿV
To calculate to
dT +
T
.
c IN
S,N
a expression (3.8)
another partial derivative, we must use
a temperature as a function of pressure
a condition to eliminate
It is
with
do volume.
Thus, we are left with
ÿ
ÿ
S
No
=ÿ
0 + ÿÿ
Njjj
j
j
so that
+ÿ
PV NRT
0
Mr
ln ÿ
j
0ÿ
local unit
¯
c No.
=
dS
It is
IN
R ln
ÿ
dP
P
like this
ÿ
ÿP
there
P
=
ÿ
ÿS
NRT
=
¯
V,N
T
=
¯
c No.
¯
c NRV
c IN
result (3.10) is established.
Exercise 70
you coefficientsÿ, kT in terms of P and v for Waals.
(3.9-3) We want a
a van equation
of
the
system that obeys
The coefficient a It is defined
as
a
1
=
ÿ
ÿ
O
To calculate it, we must express
It is of
the pressure. We have to
P
P
volume molar v
ÿ
b
in
aTv
P is constant
so that, differentiating (with
2 ,
ÿ
R
dP = 0 =
It is
RT
ÿ
bdT
( in
ÿ
after multiplying by
dv 2+
ÿ
in
in terms of temperature
R
=
T
.
ÿvÿT
in
b)2
a
in
or it is
R
2a
RT
ÿ
in
ÿ
bdT
=ÿ
( in
ÿ
b)2
in
3ÿ
dv
3
DVD,
T),
Machine Translated by Google
3.3. CHAPTER EXERCISES:
71
and,
therefore,
a
1
=
ÿ
It is
replacing
R
=
ÿ
ÿvÿT
in
garden
ÿ
P
( in
b)
ÿ
RT
in
ÿ
in
b)
ÿ
2
a
P
b =
ÿ
2 of (
ÿ
+ in 2
,
stay with
a
worth
O
you
even for
Exercise 71
Rv 2
=
3
[in P
+
of
ÿ
.
b
]
other coefficients.
(3.9-5) Show that
=
CPCv kTkS .
We have to
=
CP
T Va
Cv +
2
NkT
so that
TV a
=1+
2
.
NkT Cv
CPCv
In the same way
TV a
=1+
kTkS
2
NkSCP
so that, subtracting, we are left with
=
ÿ
CPCv
kTkS
TV a
TV a
2
NkTCv
TV a
N
2
=
ÿ
NkSCP
2
1
ÿ
ÿ
ÿ kTCv
1
kSCP
or yet
=
ÿ
CPCv kTkS
TV a
2
ÿ
ÿ
ÿ
,
CPCv kTkS
NCPkT
or yet
TV a
k
ÿ
TkSÿÿ 1 ÿ
of so, how can we not
It is
2
ÿ
CPCv
to have
N)
ÿ=0
NCPkT
always (for any temperature, volume
TV a
2
,
ÿ1ÿ
NCPkT
ÿ=0
we should have
=
CPCv
kTkS .
Machine Translated by Google
72 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
Exercise 72
IN
(3.9-10) We will be keeping derivations,
It is
N always constant in
so that
of
=
T dS + BedI
since we must have dU as an exact differential, we must have e,
ÿT
ÿBe
ÿ
ÿS
ÿ
I
V
where we are always assuming that
ÿ
= ÿ ÿI
It is
,
S
are not kept constant. Like this,
more fully,
ÿBe
ÿS
ÿ
Doing the
ÿT
ÿ
I,V,N
.
S,V,N
even in the representation of entropy, we get
ÿS
T
Exercise 73
ÿ
ÿBe
=
ÿ
T
ÿI
ÿ
I,V,N
ÿT
ÿ
.
S,V,N
(3.9-11) We know that
ÿU
ÿI
Be
=ÿ
It is
ÿ
= ÿ ÿI
ÿ
S,V,N
also that
P
ÿU
ÿ
= ÿ ÿ ÿV
.
S,I,N
Now how
of
It is
=
T dS
ÿ
VAT + BedI + µdN
a entropy
as we are always considering
of
there
fact of
=ÿ
N constants, we have
VAT + BedI
dU being an exact differential implies that
ÿP
ÿI
With
there
definitions of
Be
It is
is true, indicating that
ÿBe
ÿV
.
P, we are left with
ÿ
ÿI
=ÿ
ÿU
ÿÿ ÿ
ÿV ÿÿ = ÿ
ÿ
ÿV
ÿU
ÿ
ÿI
ÿ
a expression contained in the exercise as well
O
It is.
Machine Translated by Google
3.3. CHAPTER EXERCISES:
Exercise 74
73
(3.9-13) We have to
m0
N
=
h
so we must express
ÿI
ÿ
ÿ
ÿBe
,
T
in terms
be
And from
It is
T . For the model, we have
2
I
S
0ÿ
+
2 2
N
I
No.
0 exp ÿ
IN = NRT
and,
therefore,
Be
I2
S
+
N 2 I 02
expÿ No.
2RT I
=
02
IN
0
ÿ
It is
T
T
=
I2
S
+
N 2 I 02
0 expÿ No.
ÿ
,
or it is
=
Be
t constant),
so that, differentiating (with
2RT
=
dBe
2
IN
or it is
h
=
m0
N
ÿI
ÿ
T
2
m0 I 0
2RT
=
,
a T, as desired.
(3.9-14) We have to
ÿS
It is
Of,
0
2
m0 IN 0
N 2 RT
=
ÿ
ÿBe
is inversely proportional that
Exercise 75
2 RI
T
IN 02
=
m0
N
ÿ
ÿI
ÿ
ÿBe
S
we already calculated that
2RI
T,
IN 02
=
Be
so that
=
dBe
2RT
2
IN
0
Of +
2 RI
2
IN
T depending on
We calculate dT to from the equation to
=
2IT
Of
N 2 I 02
2RT
2I
2
ÿ 1 + IN
dT
dT.
0
so that
dBe
=
IN
0
2
2
0
ÿ
Of
Se I
,
giving
Machine Translated by Google
74 CHAPTER 3. FORMAL RELATIONS AND EXEMPLARY SYSTEMS
It is
then
ÿI
ÿ
IN
=
ÿ
ÿBe
2I
2
0
ÿ1+
2RT
S
IN
2
2
ÿÿ1
0
and,
therefore,
2I
m0 I 0
2RT
2
=
ÿS
Exercise 76
ÿ1
2
IN
2
ÿ1+
.
0ÿ
(3.9-16) We must show that
ÿT
ÿ
ÿT
ÿ
ÿBe
=ÿ
ÿI
S
knowing that
IN =
ÿS
ÿ
=ÿ
ÿS
ÿ
ÿI
S
=ÿ
=0
ÿ
ÿBe
T
,
T
S
m0 I 2
+ Nÿ
2No
expÿ 2ÿNo.
so that
T
so that is
ÿU
=ÿ
ÿS
2
=
ÿ
ÿR
ÿ
S
ÿU
Be = ÿ
,
expÿ 2ÿNo.
ÿI
easy to see that
ÿ
ÿT
ÿI
ÿT
ÿ
=ÿ
S
ÿBe
=0
ÿ
S
as other equations being obtained in an analogous way.
,
=
ÿ
ÿ
m0 I
No
Machine Translated by Google
Chapter 4
reversible processes and the
Content. of Work Max.
4.1 Seventh Class (04/16/2008):
4.1.1 Possible processes and impossible processes: Physics
has several situations in which certain phenomena become blocked as
to their occurrence in nature because, once they occur, they would violate a
or more boundaries and/or physical laws. Thus, we have not encountered phenomena (so
far, at least) involving speeds greater than that of light, nor
nor phenomena that are associated with negative temperatures (except
in specific situations where the very concept of temperature is changed.)
These are limits that physics imposes on the occurrence of any natural phenomenon.
In addition to the imposed limits, there are also restrictions given by the characteristics
of the physics.
with himThere is, therefore, no natural phenomenon that violates the principle of
conservation of energy.
Thermodynamics also has a principle that selects, among all those phenomena,
those . This principle says
that any thermodynamic
phenomenon that implies an increase
possible
imaginable
or maintenance of the
,
value
of entropyphysical
(assuming
thatissuch
a phenomenon does not violate other nonthermodynamic
laws)
permissible.
Thus, when machines of such machines were built (the Middle Ages are prodigal of
perpetual mutual this), they did not work, as they would violate, precisely, the principle
of conservation of energy. Likewise, an engineer can
build a machine involving thermal parts and it just doesn't work simply because it fails to
enforce the principle of entropy maximization.
,
75
Machine Translated by Google
76CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
4.1.2 Quasi-static and reversible processes:
The principle of entropy maximization is particularized in several other more specific theorems as
we consider different classes of
therefore
It is interesting to analyze how this happens. Before, however, it is im- processes.
introduce appropriate tools to facilitate our analyzes of
thermodynamic phenomena.
of terspace
One of these tools is the definition of a
dynamic
. This definition closely follows that given in mechanics for the
mechanical phase space . As, in mechanics, the relevant variables are positions and moments, a
phase space is defined in the mechanical field in
that the axes are precisely the positions of each of the particles involved in the problem, in
addition to defining other axes that contain the momentum of these particles. Once this is done,
the mechanical state of each of the particles is known when we know the point associated with the
particle in the space of
phase (to know all the points, therefore, is to know the system completely).
This happens, in mechanics, because knowing how the moments vary with the positions, we have
a complete description of the physical system.
Example 77
No case of a particle equation
AND =
p
O
moving on a background plane
2
1
+
2m
It is
with
a of energy given by
a harmonic force, we have
2
mÿ
2
xx
2
1
+
2
2
mÿ
2
yy
as surfaces like of constant energy in this space are
this,
total momentum p) as shownin the figure
a follow,
where we consider only energy,
for each value of
O
first quadrant (obviously,
(considering
the curves
ellipsoids in three dimensions). they are,
Thus, as we said, analogously in thermodynamics, we define a
thermodynamic configuration space (better not to mention thermodynamic phase space , since the
name "phase" has a particular definition in this field),
Machine Translated by Google
77
4.1. SEVENTH CLASS (04/16/2008):
in which the extensive variables of the problem in question appear, considering all the
subsystems that are present. In this space, therefore, (a fundamental equation can be
surface
as a surface, in the case of many dimensions) that connects the
represented
various values of the variables
thermodynamics.
Look the case of ideal simple monatomic gases. we have a
fundamental equation given by
Example 78
in
0 ÿ + ln ÿR
s = s 0 + cR
ln ÿ
in
in
0ÿ
in
,
so that a surface of constant entropy can be represented as in figure
a follow,
it is very similar to the one shown earlier for that mechanic.
phase to space
from
thermodynamics
As we are within the
in which we consider the equations
fundamental representing equilibrium states, each point of the space of
configuration represents a state of equilibrium. One in this space defines acurve
quasi-static
process and can be seen as a dense succession of equilibrium states of the system. This
is a mathematical abstraction, since in usual situations, a thermodynamic system will pass
through several non-equilibrium states in the passage between two equilibrium states.
Moreover, as we do not consider these non-equilibrium states, the analysis made by this
,
,
thermodynamic configuration space leaves out important concepts, such as rates of change
of thermodynamic variables, relaxation times, etc.
In fact, a quasi-static process is nothing more than an orderly succession of equilibrium
states, while a real thermodynamic process is a
successiontemporal
of equilibrium and non-equilibrium states.
In any case, it is possible to make a system, in its process of
go from a state of equilibrium A to another state of equilibrium B, pass
Machine Translated by Google
78CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
through as many points as we want from a certain quasi-static path (on a surface that
defines a fundamental equation). To do this, we just need to control the various links
(walls) of the system, controlling the way in which
are permissible or not to the various thermodynamic variables. With this, we make the
system follow any curve we want on the surface almoststatic.
The importance of this space appears, among other reasons, because the
such
mechanical
work
andfor quasi-static
T dS processes.
identification of ÿVAT
as the
heatastransfer
are only
valid
The only restriction so that we can make the quasi-static approximation of the real
process is that we keep always obeying the entropy maximization principle. In cases
where the final equilibrium state has an entropy greater than the initial one, we will have
irreversible
a transformation if the entropy does not change in the process, so we
reversible
have an almost static process (curve on a surface) that can be approx.
;
.
locus
So, a
imagined by a real process in a closed system only if the entropy is monotonically
non-decreasing over the
locus
.
4.1.3 Relaxation and irreversibility times:
Normally, the constraints of the thermodynamic system of
paused way, but in a . What do we meanlevel
by 'slow', however? The denomination 'slowly'
is actually correlated with the
specific system we are considering: it is a velocity with
that we change the constraints that allow the system to find its new equilibrium situation
before the thermodynamic variables (the point in the thermodynamic configuration
space has moved too far from the one where the first observation was made).
In the case of an adiabatic expansion of a gas, if we pull the piston from
very quickly, so we generate a series of inhomogeneities in the system and
turbulences that prevent the system from passing through several states of equilibrium
when changing. Such a change is neither quasi-static nor reversible.
However, if we pull the piston very slowly, we give the gas that is there
time inside the cavity to adjust its homogeneity to the new volume, so that we never
move away from the situation of homogeneity
(balance, considering the volume variable). The situation of homogeneity
ity occurs
due
to the
the to
gaseous
particles
carry
out amongwith
themselves
and
with the
walls,
inshocks
order tothat
return
a situation
of total
homogeneity
some
speed–something that is related to the average linear size of the cavity and the speed
of sound, that is, with a relaxation time of the order of
t
where c is the speed of sound.
=
IN1//3c,
Machine Translated by Google
79
4.1. SEVENTH CLASS (04/16/2008):
4.1.4 Heat flow: coupled systems and heat reversal
Law Suit:
As an example, let's consider the quasi-static transfer situation of
heat between two systems.
1. In the first case we have a heat transfer from a system 1
for
a system 2 that are at the same temperature. Thus, the process is evidently
reversible, since the amount of heat flowing in the 1 ÿ 2 direction is the same as
that flowing in the 2 ÿ 1 direction, with an increase in the entropy of a system,
but a decrease in entropy on the other, so that = 0;
a total entropy
S
remains the same, that is, ÿ
2. In the second case, each subsystem initially has a temperature of 1 ( so that T 10
T 20,
T 10 T< 20. If we know the heat capacities ) ), then we
C2know
(T
C T that the entropy change in system 1 is given
It is
It is
by
dT
Q =
T
C 1 (1)
,
T
T
the same goes for system 2. The total change in energy and entropy is
dS
1
=
1
1
Tf
T 10
ÿ IN =
ÿ S
ÿ
Tf
T 20
C 1 (T )dT +
Tf
C 1( T )
T 10
T
=
ÿ
1
dT +
C 2 (TdT
)
Tf
Tf
=
dT
T 20
T
ÿ
ÿ much C
C 1 how
2
Considering the case where both
temperature, we have, doing the first integration
=0
.
C 2( T )
are independent of
C 1T 10 + C 2 T20
C 1 + C2
and so, for the change in entropy,
ÿ S
=
Tf
C
1
ln ÿ T 10 ÿ +
To show that this variation on this
expression to get
ÿ S = ln
C2
Tf
.
ln ÿ T 20 ÿ
it's always positive just replace
Tf
,
T C 1+
C2
f
C
T 1 10 T 2C20
,
so that
ÿ S = ( C 1 + C 2)ln
ÿ
ÿ
ÿ
C 1T 10 + C 2 T 20
C 1 + C2
1
C1
ÿ
T 10 T
C2
20 ÿ1
ÿ
ÿ
.
/ ( C 1+ C 2)
ÿ
Since the arithmetic mean (weighted in the case) is always greater than the
geometric mean (equal only when the temperatures are equal), we have that the
argument of the logarithm is always greater than one, and therefore ÿ
S > 0.
Machine Translated by Google
80CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
These analyzes show us that:
• Even when the process is irreversible, it can be represented by a
quasi-static process;
The process can be associated with a flow of heatspontaneous
to a
•
hotter system to a colder one, provided that (a) this flow is sufficiently slow
(conditions that can be adjusted by choosing a wall that has sufficiently high thermal
resistivity) and (b) that the wall separating the two systems do not change the
thermodynamic properties of the two systems.
• Finally, we can see that it is always possible to decrease the entropy of a particular
system, provided we use another system, coupled to it, whose entropy is increased
in such a way as to make the total entropy stay the same or be increased.
4.2 Eighth Class: (04/23/2008):
4.2.1 The maximum work theorem:
The principle by which the entropy of a system must always increase can be
be used to build mechanisms that do work. Refrigerators, air conditioners, steam engines
are examples of such use.
In order to efficiently build these mechanisms, we must
have three different interconnected subsystems:
1. a main system, which must have its thermodynamic state changed from
A for B
,
2. a reversible repository of work, defined as a boundary system
formed by impermeable (does not change the number of moles) and adiabatic
(does not change the temperature) walls that has sufficiently large relaxation times
for us to consider the thermodynamic process
as quasi-static;
3. a reversible heat repository, consisting of rigid walls (does not change the volume)
and impermeable (does not change the number of moles) characterized by
relaxation times large enough for the processes to
thermodynamics occurring there can be considered quasi-static.
By the definition in question, we have that it must hold, by the principle of
conservation of energy
of + QRHS + WRWS = 0 ,
(4.1)
reversible
QRHS is the change in heat in the reversible heat store ( ) and
heat source
WRWS is the variation of work in the reversible repository of
reversible work source ). This expression must hold because the variations
work (
where
Machine Translated by Google
81
4.2. EIGHTH CLASS: (04/23/2008):
RHS
of energy inside the RRC ( ) occur in the
sense of changing only the
heat (by definition) and energy changes inside the RRT() occur
in the sense of changing only the job (by definition).
From the point of view of entropy variation, we have that
dStot
=
dS +
QRHS TRHS
ÿ0
R HS
(4.2)
,
since the entropy variation in the RRT must be zero (isentropic variations). But then, joining
(4.1) and (4.2), we must have
WRWS
ÿ
TRHSdS
ÿ
of
(4.3)
so that the greatest possible work corresponds to the equal sign in
previous equation. But in that case, we also have an equality in (4.2), so that
dStot = 0
,
and the process will be reversible. This proves the following theorem:
Theorem 79
For all
you processes
taking a system from an initial state releasing
a heat release is
a
work
reversible process. Furthermore, heat release) identical for all reversible
a an end state,
a maximum (and is minimal) work to a
(It is
you
It is
processes.
We can, in fact, calculate the maximum work released. Once we have
of
=
Q +
IN,
we get (using the maximum value presented in expression (4.3),
TRHS
dWRWS
=ÿ
T
ÿ
Q
ÿ
TRHS
of
=ÿ1ÿ
T
ÿ (ÿ
Q ) + (ÿ
IN ) ,
indicating that in an infinitesimal process, the maximum work that can be
released to the RRT is a sum of
• the work (ÿ
IN ) directly extracted from the subsystem;
• a fraction (1 ÿ ) of the TRHS/T
heat (ÿ system (this
Q ) directly extracted from the subfraction of heat that can be converted into work in
an infinitesimal process is called ).
thermod machine efficiency
inactive
4.2.2 Machine, cooler and heat pump coefficients:
We have already seen that for a system where there is a hot (primary) subsystem ( ), a coldh
c (
RWS ), we must have, infinitesimally,
RRC () and an RRT
( Qh +
Wh ) +
Qc +
WRWS = 0
Machine Translated by Google
82CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
It is
Sh +
Qc
=0
Tc
so the work released is algebraically maximum. With this we can classify several types of
useful thermodynamic machines:
1. A is thermodynamic
one where a hotmachine
primary system
has part of the generated heat used to produce some kind of work
(performed by mechanical machinery) and is linked to a repository of
heat responsible for cooling it. Examples of this type of application are the
furnaces, heaters (thermal type), etc. The measure of performance
Qh
is made by the fraction of heat (ÿ ) removed
from the hot (primary) system
Wh
which is converted into heat. Putting
= 0 in the first
WRWS
previous equation it is easy to see that we must have
machine efficiency
thermodynamics ÿe
one given by
eh
=
WRWS
(ÿ
Qh
=1ÿ
TcTh ,
)
showing that the efficiency of a thermodynamic engine increases as the RRC
temperature decreases. This is predictable, since a low temperature of the RRC,
compared to the primary system,
Theat
transfer fromeh
second
to = 1 and all
iompprliimcaeinruom. aEavltidaecnatpemaceindtaed,esdee
= 0 then
c
the heat produced by the primary system can be converted into heat (since ) = 0).
= T
Qc
T (ÿ Qh
c
h
2. One isfridge
a thermodynamic machine operated in reverse mode
to the previous item. The idea is to remove heat from the cold system and, with
entry of a minimum amount of work, inject this heat into the
hot system. Thus, the previous equations remain valid, but the
should be defined as the reason
refrigerator performance coefficient
between the heat removed from the refrigerator (the cold system) and the work
that must be done by the machinery, i.e.
ÿr
(ÿ
=
Qc )
Tc
=
;
(ÿ
WRWS
)
Th
ÿ
Tc
= Tc
evidently if the temperatures then the efficiency
is infinite, since no work is required
Th
to transfer heat from one system to another. It is intuitive that efficiency decreases
as we make the RRC colder relative to the hot system.
,
(primary), since heat will have to be transferred using increasing mechanical
force to counteract the direction of spontaneous heat flow;
pump that is used to heat a system already
3. A is heat
a system
hot, extracting heat from a cold system and also extracting some work from an RRT
(removing the door of a refrigerator and placing it in front of
Machine Translated by Google
83
4.2. EIGHTH CLASS: (04/23/2008):
to a window produces such an effect, as the refrigerator will try to "warm" the
atmosphere (a thermal bath, in this case) causing the heat removed from the atmosphere,
together with the energy obtained from the electric company, to be
thrown directly into the room by the cooling coils at the back of the
heat
pump performance coefficientÿp
refrigerator. the is the
reason
ednatnrdeoo the heat released to the hot system and the work extracted from the RRT,
Q
Th
=
=
ÿp
.
(ÿ
WRWS
)
Th
ÿ
Tc
4.2.3 The Carnot cycle:
In all the previous sections we have not specified the processes by which heat and
work are effectively used to modify the states of the primary system.
However, we know that we must use auxiliary systems for this transfer to take place (devices,
machines, etc.). in such a way that such values do not enter into the general sum of changes in
energy and/or entropy. Thus, it is necessary to carry out one (with respect to thermodynamic
variables) in these auxiliary systems: this cycle we call the
cycle
cycle
Carnot
.
Let us consider a particular case where the hot (primary) system and the
cold system (RRT) are thermal baths (cold and hot, respectively). With
this we will be able to do an analysis that will not go into details of integration, etc and that will
1
.O
take place according to finite transfers, instead of infinitesimal ones
Carnot cycle is in four steps:
1. The auxiliary system that is initially at the same temperature as the
primary system (the hot reservoir) is placed in contact with this
reservoir and the RRT. This auxiliary system is then passed through a
isothermal process (an isothermal expansion of a piston, for example) in order to vary its
entropy. In this process there is a flow of heat between the hot reservoir to the auxiliary
system and a transfer of
heat occurs from the auxiliary system to the RRT;
2. Remove contact between the auxiliary system and maintain contact with the
RRT by making this auxiliary system expand adiabatically until
its temperature drops to that of the cold reservoir. In this case, more work is transferred
from the auxiliary system to the RRT. The entropy of the system
auxiliary does not change, the process being isentropic (occurring over a
adiabatic curve);
3. The auxiliary system is brought into contact with the cold system and the RRT and
is compressed isothermally so that the entropy of the
1If the hot and cold systems are RHS, rather than reservoirs (baths), the Carnot cycle
must be considered from infinitesimal steps and the above considerations must be
be made from integrations over the thermal capacities.
Machine Translated by Google
84CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
auxiliary system return to original value. In the process there is new transfer of work,
but now it is a transfer of work from RRT to
the auxiliary system, with a heat transfer from the system
auxiliary for the cold reservoir;
4. Finally, the auxiliary system is adiabatically compressed and receives work from the
RRT, causing the auxiliary system to return to its normal state.
original in the various thermodynamic variables and completing the cycle. Such
process is done isentropically.
In this whole process we have that: The
heat removed from the primary (hot) system in process (1) is the area under
S work transferred to the system
the graph, given opr ÿ while
Th the
,
cold during the process (3) is given by the area under the graph, given by The
Tc ÿ S
difference () ÿ isTh
the net
TcworkStransferred to the RRT in the cycle
complete. The coefficient of performance of this type of process is given by the ratio
totalsystem,
between the net work transfer from the RRT and the heat transfer from the primary
giving
.
ÿ
Th
eh =
Tc
ÿ
=1ÿ
Th
TcTh ,
as we have already seen.
Real machines never reach an ideal of thermodynamic efficiency, so the coefficients
already described are only maximum values. In general, 30 or 40% of the thermodynamic
efficiency is reached due to mechanical friction.
internal, etc.
4.3 Chapter exercises:
Exercise 80
It is
volume
temperatureTb
A) at temperature ls
(4.1-1) We have an ideal mole (gas
of system
fluid
we have a mole
E of der Ta Va.
By
It is (system
B) has
It is
volume Vb.
in case I
question, we know that
ÿ IN = ÿ IN1 + ÿ IN2 = 0
so that, as
ÿ IN1
=
cR ( Tb
ÿ Ub
),
Facing
=
cR (Tf
ÿ
O
Tb)
ÿ
which implies that
Tf
=
Facing,
or that is, there is a temperature inversion. For that to be
possible, we have to have S ÿsuchÿthat
0,
ÿ S
=
cRÿ
ÿ + ln
cR
ln ÿ
TbTa
the process
RTa
RTb
ÿ
ÿ
a/vb
a/v
ÿÿ0
whether physically
Machine Translated by Google
4.3. CHAPTER EXERCISES:
O
85
what gives, finally
RTa
Tb
ln ÿ
Facing
a/vb
ÿ
((RTb
)
a/va )ÿ ÿ 0
ÿ
and,
therefore,
Tb/vb
Exercise 81
Ta/wa.
ÿ
(4.1-3) We have
that ()T=
DT
+ const.,
S
n
so we calculate
IN
=ÿ
n +1
DT
=
C ( TdT
)
n +1
C ( )
TdT
T
=ÿ
=
DT
+ const.
nn
showing that we have
IN =
n +1) /n
D
nS
ÿ(+ 1 ÿ
D
n
n ( n +1)
f ( IN) =
/n
D
n +1
ÿ1
n
S ( +1) /n f ( IN)
/n
a
since we are always considering Well, f is constant. a heat capacity volume
homogeneous Vwe
, that
have
is, to
SVwrite linear
( ) so that U is ÿV must
ÿS e
imply U ÿU.
ÿ
then
a function
But
ÿ
n 1+1 /n
IN =
D
n +1
ÿ1
/n
ÿ1
S 1+1 /n IN
/n
,
ÿ
.
To calculate the
maximum work that can be released, leaving subjects at the you two sis of
to have ÿ S = 0 ,
same temperature, just remember that we must so that
ÿ S
of
D ÿ2
=
n
ÿ
n
T 10
ÿ
n
T 10
n
T 20ÿ
=0ÿ
Tf
n
T 20
+
/n
ÿ1
2
=ÿ
energy stays
ÿ IN
there
T fn
n
T 10
D
=
n
+
+1ÿ2ÿ
n
ÿ(
ÿ
T 10+1
n
=
wrrt
than for
n =2
n
D
Tn10+1
+
Tn20 +1
( n +1) /n
n
T 10 + T 20
2
ÿ
ÿ
ÿ2ÿ
+ 1 ÿ3
,
D
3
3
=
T 10
+ T 20
implies WRRT
ÿ
as wished.
n
T 20+1ÿ
ÿ
u stay
= ÿÿ
work, given by WRRT
n +1) /n
n
T 20
2
1
ÿ
ÿÿ 2
2
/
T 10 + T 2 20ÿ3
2ÿ
,
Machine Translated by Google
86CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
Exercise 82
ÿ in = 0
and,
so,
(4.2-3) We have a free expansion
.
Of this
way
is and like a
P
=
dS
dv
T
gas
monoatômico ideal PV
=
NRT
so that
P
No.
=
T
It is
IN
like this
No.
=
dS
dv
IN
and
by integration,
INf
ÿ S
= ÿ INi
Exercise 83
No.
IN
=
dv
Vf
No.
ln ÿ We
.
ÿ
(4.2-4) We now have
T
Of 2 /s,
=
=
It is
We want to find
easily that
P = ÿ2
Of ln ( s/s
Tf ( To , vo , vf ) .
From the first equation we find
2
T0
=
0)
.
sfs
0
Tfv
fv02
To find
a expression of sf/s 0 in terms of variables
the fact that we are considering a free expansion, for the
ds
in 0 , vf,
we use the
which one should we have
P
=
T
DVD,
so that
ds
2 Of ln ( s/s
Of 2 /s
=ÿ
0)
dv
or yet
ds
2
=ÿ
vdv.
s ln ( s/s 0)
Integrating, we have
ÿ
d s/ss/
(0)
.
( s/s 0)ln( s
0)
= ln ÿ
vfv
0 ÿÿ2
The first integral can be solved easily by ÿ substitution: dÿ / s/s so that
=
d s/s
(
0)
(
0) ,
ÿ
dÿ
X
= ln X = ln(ln (
in0
sf/s
0)) = ln ÿ
vf ÿ2
= ln ( s/s 0) ,
Machine Translated by Google
4.3. CHAPTER EXERCISES:
87
giving
in
sfs
0
= exp ÿ
in
2
02 ÿ
f
so that we are finally left with
in 2
T0
in02
Tfvfv 0
2
f ÿ=
expÿ
2
and,
therefore,
Exercise 84
T 0 exp x
=
Tf
T
ÿ IN = ÿ T f C 1 (TdT
)
10
A ( Tf
ÿ
)
+ ÿ T 20 C 2 ( TdT
T 20) + 1 // 2 B
ÿ
1 2B
T
=
T
2
ÿ
f
T2
ÿ2 f
in0/vf
A + BT
=
(4.4-1) We have to C
T
x =(
,
x
ÿ
T 10) + 1 / 2B
ÿ
A (2 Tf
2 20ÿ =
.
We immediately calculate
.
A ( Tf
2
T 10
)2
T 10 ÿ
ÿ
ÿ
T2
f
T 2 10ÿ+
ÿ
T 20) +
.
T
ÿ
2 20ÿ
We also calculate
Tf
ÿ S
=
T
A
ln ÿ
Tf
B Tf
10 ÿ +
A ln ÿ
T
T 10) + A
B Tf
B (2 Tf ln ÿT 10 20
ÿ ÿT+(20)
ÿ
T( 2f
T 10 T 20 ÿ +
ÿ
T 20) =
.
ÿ
we have to demand ÿ IN = 0 by conservation of energy. Like this,
A (2 Tf
ÿ IN = 0 =
ÿ
T 20) + 1 / 2B ÿ2T 2
f
T 10 ÿ
ÿ
2
T 10
ÿ
2
T 20ÿ
.
Substituting values, we have
2 × 10ÿ2 T f2 + 16 Tf ÿ ÿ8 (400 + 200) + 10ÿ2 (16000 + 4000)ÿ = 0
whose solutionis
Tf = 3071 . K.
Like this,
.
2 3071
ÿ S = 8 ln
Exercise 85
given by
+ 2 × 10ÿ2 (6142 ÿ. 600) = 1
400
200
.6
×
(4.4-2)
ÿ
The first two bodies have heat capacity
ÿ
C
=
A + BT,
so that, placed in contact, they imply
Tf
Tf
CdT
ÿ IN
= ÿ T 10
BT
CdT
+ ÿ T 20
=ÿ
1
2
f + 2 ATf
ÿ
2
B
ÿ
2
2
T 10
+ T 20ÿ ÿ A ( T 10 + T
.
20)ÿ
Machine Translated by Google
88CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
A =8
We have to
It is
B = 002
.
2
ÿ IN = 002
. T f + 16 Tf
solutionwhose
the body
T
T
2
It is
.
Like this,
2
+ 200 2ÿ ÿ 8 (600) = 0
,
.
K.
= 3071067812
3,
putting in contact with the body
T
20
( A + BT dT
)
ÿ IN
=ÿ
ÿ 0 .01 ÿ400
T 20020
=
It is
It is
Tf
Taking we
have
T 10 = 400
Furthermore,
.
20
BTdT
BT
=
+ ÿ T 30
f
1
2
20ÿ
BT f
2
1
2
C3
that have
ÿ
2
BT
2
=
A ( T 20 ÿ
30+
BT
,
Tf ) = 0
so that
2
.01T 30 + 8(200 ÿ 30711). = 0
. ÿ 001(307
.
0 .02 (200)2 ÿ 011)2
It is
like this
.
ÿ1000045521
ÿ 001
2
T 30 = 0
.
30 able to
which has no real solution, so there is no temperature T to return to the initial temperature.
be
2
of
O body
Exercise 86
(4.4-3)
Already done in the body of the text.
Exercise 87
(4.4-4)
The two bodies have equal thermal capacities that are independent of
they are
a temperature made
temperature. Thus, from equilibrium, by calculations
in the text, is
Tf
=
C 1T 10 + C 2 T 20
C (T 10 + T 20)
2C
=
C 1 + C2
T 10 + T 20
=
,
2
as wished.
Exercise 88
(4.4-5) We have to
C
=
A/T.
Like this,
ÿ IN
=ÿ
C ()TdT
=
A
Tf
ln ÿ T 0 ÿ
T 10
If we have two such systems, with temperatures
ÿ IN = ÿ IN1 + ÿ IN2
so energy)
=
A
Tf
ln ÿ T
Tf
= ÿ T 10 T 20
T 20 ,
Tf
Aÿ
10 ÿ + ln
a stands, equilibriumtemperature
It is
.
T 20 ÿ =
by
.
so we are left with
T f2
A
ln ÿ
ÿ IN = 0
T 10 T
20 ÿ
(conservation
and
Machine Translated by Google
4.3. CHAPTER EXERCISES:
Exercise 89
89
3
ÿ IN =
like the
IN = (32)/
(4.5-1) We have to
2
R ÿ T,
ÿ S
/
2
ÿ
IN0 ÿ3
IN
.
IN
0 ÿÿ
T = 400 K, we're afraid
they are the same
final
Like this,
and,
ÿ S
V0
if they appear
It is
It is
= 300 K.
Three
It is
IN
R
=
ln ÿÿ
initial temperatures therefore,
ÿ IN = 0
RT
= 10ÿ3 m 3
It is
Vf
R
=
R ln 2 ,
ln ÿ IN0 ÿ =
= 2 × 10ÿ3 m 3
Vf
.
exactly this quantity multiplied by the
= Three ÿ
subsystem
wrrt
Exercise 90
The work that can be released
temperature because RRC: Tres,
=
ÿ ÿ Usubsystem
Three ÿ subsystem
= 300 R ln 2
.
= It's RT IN = cRT
, for a mole. We have
(4.5-2) We have that PV
= 10ÿ3 m 3 It is Vf = 2 × 10ÿ3 m 3 It is T 0 = 400 K. We know that
also that V
0
S
s0 + R
=
c
IN
IN0 ÿÿ
T 0 = 400 K
a change
Thus, in the isentropic expansion, in which it occurs, we have the
300 K,
IN
.
ln ÿÿ IN0 ÿÿ
ÿ
Tf
=
change in energy given by
(1)
ÿ IN
to the
cR (300 ÿ 400) = ÿ100
=
cR
same time we have
c
INf (1)
ÿ S =0ÿ
like this ÿ
IN(2) = 0
,
It is
what
IN(1)
=
the f
IN0(2)
.
then temperature O gas is expanded
a
is maintained
a same. We have
but
ÿ S
already
U cR.
= 100
W (1) = ÿÿ
The work of this step is
in contact with reservoir O
IN0(1)
= ÿ 400 300ÿ
,
as
=
R
INf (2)
0 ÿ = ln ÿ
(2)
ln ÿ IN
It isobvious.
INf (1)
ÿ
,
The work done is
INf (2)
(2)
IN
INf (2)
R
= TresR
Vf (1)
In the adiabatic phase (until reaching the third stage we have
ln aÿ gas compression
ÿ;
2 × 10ÿ2ÿ
Vf at =
temperature
m3
S =0
400
K), have, again, implying that
Tf so =that
we must
It is
c
R
ln ÿÿ 400 300ÿ
2 × 10ÿ3
INf (2)
ÿ=0
to the
volume
Machine Translated by Google
90CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
giving
c
INf (2)
c
2 × 10ÿ3
Vf
= ÿ 400 300ÿ
.
= ÿ 400 300ÿ
given by U is ÿcR. = cR cR,
so ÿthat
three
processes, we have
(400
300)
= 100
joining the
The change in work
O energy is IN (3) =
100
ÿ
WTotal
=
IN (1) +
IN (3) = 100 cR + TresR ln(2) ÿ 100
IN (2) +
cR = 300 R ln 2 ,
as expected.
Exercise 91
ÿ IN = 0
(4.5-3) As it is a free expansion, we have
ÿ S
=
Vf
0ÿÿ
R
= Threeÿ
IN
ln ÿ IN
S = 300 R ln 2
. K. m 3
Exercise 92
in0 = 0001
(4.5-5) Initial volume initial temperatureT
3
It is
.
m
400
=
0002
Final state final temperature
Tf K.
vf
It is
like this
.
It is
0
= 400
=
the variation
of energy is given by
ÿ usys
=
cR ( Tf
T 0) = 0
ÿ
.
we still have
Tf
ÿ sRRC +ÿ ssys
=ÿ
300
T
R
Tf
T
= ÿ 300 ÿ 2 + 150ÿ
equation for
a
ln ÿ 300ÿ+
RdT = 2 ( Tf ÿ 300) +
ÿ
300
ÿ 300)+ R ln2 = 0
Tf
(150
1
Finally, QRRC
using to
R
Tf
dT = 2 R
T ÿ 2 + 150ÿ
2
T f ÿ 300 2ÿ
entropy, we have to
= 25033.
Tf
,
.
which, substituted into the following equation,=gives1904585549
QRRC
ÿ
Of
= ÿÿ
so that wrrt
Exercise 93
usys
(4.5-9) We have to C
ÿ
QRRC
1 ()T =
.
= 190458555
C 2 ()T=
.
C. Like this,
T2
ÿ IN = C (2 Tf
ÿ
T 10 ÿ
T 20)
,
ÿ S
=
f
C
ln ÿ
T 10 T 20 ÿ
.
.
Machine Translated by Google
4.3. CHAPTER EXERCISES:
91
Does there is an RRC present,
eradonot
se relate to
so what conditionO maximum work possible
ÿ S =0ÿ
to be lib-
= ÿ T 10 T 20
Tf
and,
therefore,
ÿ IN = C ÿ 2ÿ T 10 T
T
20 ÿ
T
10 ÿ
20ÿ
there
work stays
Se have
we =
8 J/K,
IN = C
= ÿÿ
wrrt
OC
T 10 = 100
and
ÿÿ
T
20 ÿ ÿ T 10ÿ2
.
T 20 = 0 oC, so
Tfmin = ÿ 373 × 273 = 461 . o C
we take out Tfmax on the condition that in
which we know that WRRT
It is
ÿ IN = 0 (withouttaking into account
ÿ S =0
,
case
the maximum). Like this,
Tfmax
100 + 0
=
= 50
o C.
2
Finally, we have
max
IN RRT
Exercise 94
C 1 (TdT
)
=ÿ
It is
( ) = a/T. Like this,
(4.5-10) We have to CT
ÿ IN
.3 J.
= 8 ÿÿ 373 ÿ ÿ 273 ÿ = 62
C 2 ( TdT
)
+ÿ
=
T f2
a
ln ÿ
T 10 T 20 ÿ
also
C 1 ()
TdT
T
ÿ S
=ÿ
We want work (not O any RRC
energy from the
present,
a system itself)
system 2– which has to
ÿ S
C 2 ()
T
+ÿ
has
It is and the
ÿ
ÿ Tf
2
ÿ
1
ÿ
T 10
ÿ Tf
2
T 20 ÿ
total entropy – of system 1 added to that of
Like this,
1
a
1
a
=ÿ
.
reversible to, precisely, generate either process O maximum
so we have to take all
with cancel).
=ÿ
TdT
1
ÿ
T 10
T 20 ÿ = 0
,
Tf
2T 10 T 20
=
T 10 + T 20
But then we have to
ÿ IN = a ln
2
2
4T 10
T 20
.
(T 10 + T 20)2 (T 10 T 20)
of
Machine Translated by Google
92CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
with we
Now,
have T 10 = T 20 , so is
T 20 = 2 T 10, then
obvious that ÿ IN = 0
as expected.
,
with by
topics
other side,
ÿ IN
=
4
16 T 10
a ln
2
×
9 T 210
=
T
8
a ln
2
9
10
there
job can be written as WRWS = ÿÿ
Exercise 95
=
a ln
9
.
9
8
aT e We want
1 (TCT
) =an RRT.
2( )=2
(4.5-11) We have C two
bodies involved possible. Like
8
a ln
=ÿ
Usubsystem
bT e
there are only
the greatest work
to release
It is
this,
C1 (
ÿ S
T
=ÿ
C2 (
)
TdT
T
+ÿ
)
=
TdT
so that
aT 10 +
=
Tf
there
a (
Tf
T 10) + 2( b
ÿ
bT 20
a 2+ 2b
maximum work released comes from the calculation of
ÿ IN
=
C 1 (TdT
)
1
=
C 2 ( TdT
)
+
2
a T
T
2
ÿ
f
+ b T
2
ÿ
so that
ÿ
ÿ IN
one
2
10
=ÿ
ÿ
ÿ
one
Exercise 96
aT 20 ÿ 2 aT 10 T 20 ÿ
a 2+ 2b
=1
(4.5-12) We fear that N
.
c
ÿ s
=
Tf
R
whose solution is
vf
ÿ
Of
ln ÿÿ
we
cR
ÿ
Tf
vf
Of
we
b
R TC
b
TC
f
ÿ
ÿ
ÿ
f
ÿ
we
T C + cR
=ÿ
we
b
ÿ
vf
T cR T C
i
0
ÿ
= ÿÿ
vf
b
/cR
(
R
ÿ
=0
R
b
ÿ
b
ÿ
T0
=1
like this,
Tf
Tf
0
ÿ
and,
therefore,
C ln
bÿ +
ÿ
bT 2 + 2 bT
20
We know that
b
ÿ
T cR T C
i
ÿ0 ÿ1
2
20ÿ
2
+ aT 20 ÿ 2 aT 10 T 20 ÿ 2 bT 20
+ 2 bT
20ÿ
a +2 b
2
10 +
T
ÿ
work
ficaWRRT =
It is
2
f
10ÿ
ÿ
ÿ
there
T 20) = 0
ÿ
Tf
+ C)
.
20ÿ
.
Machine Translated by Google
4.3. CHAPTER EXERCISES:
Exercise 97
(4.5-13)
93
The heat capacity is constantC feels. Like this
ÿ IN = C ( Tf
T 0)
ÿ
ÿ S
,
C
=
there is a pre-RRC
Tf
ln ÿ T 0 ÿ
of heat to transfer work to the reservoir
One of them O
It is
.
reservoir
O
of work.
In this case we have
thatWRRT = Three ÿ subsystem
ÿ ÿ Usubsystem.
so we stay
comWRRT = Three C
Exercise 98
that is
90
O
Three
C ( Three
ln ÿ T 0 ÿ ÿ
T 0)
ÿ
is (4.5-14) anItapplication of the previous exercise.
O
K. system, water, is from K.
a 5 C The
reservoir
= 278
C = 363
.
The environmentis
a a temperature
= 75
CJ/moleK.
The molar heat capacity given by is
.6 J
So for a mole, WRRT
= 278 x 75ln ÿ 278 363ÿ ÿ 75 (278 ÿ 363) = 812
1kg of water
We have, however,
pasta
of water: we have
It is
we need to know how many moles are in this
1mole molecules
ÿ 18
n
.
n = 5556
so that we have
(4.5-16)
ÿ IN
=
g
Like this, O total work stays
.
= 8126. × 5556.= 45148 12 .
wrrt
Exercise 99
g/mol
ÿ 1000
If we put the
C ( Tf
three bodies together, we have to
T 3) + C ( Tf
ÿ
J
ÿ
T 2) + C ( Tf
ÿ
T 1)
It is
ÿ S
=
T f3
C
ln ÿ
T 3T 2 T
,
1ÿ=0
so that
T 1T 2 T 3
Tf
It is
like this,
wrrt
Exercise 100
=
= 3ÿ
C ÿ3ÿ3 T 1T 2 T 3 ÿ ( T 1 + T 2 + T
.
3)ÿ
(4.5-17) We have two bodies with given heat capacities
by
C
these bodies are
a temperatures
=
A + 2 BT.
T 10 = 200 K e
T 20 = 400 K.
Machine Translated by Google
94CHAPTER 4. REVERSIBLE PROCESSES AND GRADE. OF WORK MAX.
1.
a equilibrium
Ask yourself
taken. To do this, just calculate
minimum temperature
has
which they can be
T2
ÿ S
f
A
=
T 10 T
ln ÿ
A = 8 J/K
Knowing that
It is
B (2 Tf
B = 2 × 10ÿ2
T
ÿ S
ÿÿ
IN = A (2 Tf
Exercise 101
.
2
f
ÿ 24 = 0
25 Tf
2
T 20) + B ÿ2T f
T 10 ÿ
stay with
,
2
RRT is given by
to the
ÿ
T 20) = 0
K.
= 29297.
The biggest job provided 2.
2
J/K
= 0 ÿ 8 ln ÿ 80000ÿ +
whose solution provides Tf
T 10 ÿ
ÿ
20 ÿ + 2
2
T 10
ÿ
2
T 20ÿ = 679 .36 J.
ÿ
The state equations are
(4.5-18)
T
=
1/
As/v
2
P
,
where A is a constant. initial state T
2
T
=
T2
IN1 Final state
It is
1
1 2
4 Of // ,
.
It is
IN2 DRR to
.
2
(Tc < T ) with heat capacity
temperaturaTc
1/
BT
=
CV
2
.
The equations of state allow us to calculate
of
=
T ds
ÿ
VAT
=
A
ÿ
ÿ1
sv
1
2
/ds
2 ÿ3
s in
ÿ
/ 2dv
d
ÿ=ÿ1
4
2
As
so that
in =
With this, we can calculate
in 0
2 ÿ1 /
2) s in
+ ( A/
2
.
a change in energy of the system as being
ÿ in
A
=
ÿ
2
ÿ1
2
s 2 in2
/
2
2
ÿ1
s 1 in 1
ÿ
/
2
ÿ
,
where
s1 = A
ÿ1
Tv1
1/ 2
,
s2 = A
ÿ1
Tv2
12/
2
,
so that
1
ÿ usys
With
2A
ÿ
2
T2
2
T1
in2 ÿ
in
1ÿ
a change in entropy in the subsystem
data, we can calculate
ÿ ssys
=
=
A
ÿ1 ÿ
Tv2
/
12
2
ÿ
Tv1
1 2
1/ÿ
2 ÿ1 /
in
2ÿ
,
Machine Translated by Google
4.3. CHAPTER EXERCISES:
It is
with
95
a heat capacity, as
ÿ sRRC
possible to evaluate a variation
gives entropy not RRC
It is
=
B ÿ
T f T 1/
2
Tc
B
dT = ÿ2
T
ÿ
T 1/
2
T 1/
2
T c1/
ÿ
f
2
ÿ
so that
=
ÿ stotal
A ÿ1 ÿ T 2 IN 1/
2
T 1 IN11/ÿ2 ÿ 2 ÿ
2
ÿ
T 1/ 2 ÿ = 0
c
ÿ
f
,
so that
1
= ÿ Tc + 2 AB
Tf
ÿ
T 2 IN /
T 1 IN 1/ 2
1 ÿÿ2
2
ÿ
12
.
For the
Tf
T 1/ 2 dT
=
B ÿ
RRC, we still have QRRC
2B
=
ÿ
T 3/
2
ÿ
T 3/
2
c
f
3
Tc
ÿ
.
The least amount of
= ÿÿ
work will be WRRT
Exercise 102
QRRC.
ÿ
usys
(4.7-1) We have to
s = R ln[( in
consider the
ÿ
b)( in +
and in)
c
in =
] + s0 ,
O
process indicated in figure 4.1: For
cRT
ÿ
process
and in
AB, we have to
Figure 4.1:
ÿ T = 0 and,
therefore,
ÿ sab
=
ÿ
in =
R ln[( vb
ÿ
a/vb
to/go
ÿ
=
b) T ch ] ÿ R ln[(
a ( vb
and
ÿ
b)
ÿ
and)
/vbva. We also have
T ch ]
vb
= R ln ÿ
and
The work released in the process
WAB
=
Th ÿ s ÿ ÿ in
It is
=
b
ÿ
ÿ
given, therefore, as
vb
ThR
ln ÿ
and
b
ÿ
ÿ
bÿ ÿ
to/go + a/vb.
bÿ
.
Machine Translated by Google
96CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
No pathBC we are on an isentropic, so that
ÿ s =0
implying
,
what
c
ÿ
vBC
f
ÿ
c
bÿ T c = ( vb
=
ÿ uBC
while
b)
ÿ
cR ( Th
T ch
Tc )
= ÿÿ
uBC
No trajetoCD we fear thatT
a/vBC
f
=
cR ( Th
Tc
It is
b)ÿ
ÿ
ThTc
+ b
work stays
+ a/vb.
Tc ) + a/vBC
f
ÿ
a/vb.
ÿ
ÿ uCD
we quickly obtain that
+
a/vCD
f
=ÿ
a line DA is a
The volume can
vCD
be calculated remembering that it is isentropic,
.
ÿ
f
ÿ
=ÿ
= ( vb
a/vBC
ÿ
WBC
vBC
f
ÿ
so that in it, we must have
c
vCD
f
It is
=(
and
b)ÿ
ÿ
ThTc
+ b
ÿ
like this,
=
ÿ sCD
and
R
ln ÿ (( vb
there
work
c
b)
ÿ
ÿ
Th/Tc
)
b() ( Th/Tc
)
and
ÿ = lnRÿ
c
vb
b
ÿ
bÿ
ÿ
he is
= TcR
given by WCD
goes
b
b
ÿ
+ a/vfCD
a/vfBC
.
ln ÿ
Finally, on the way
DA, we have adiabatic compression with a/vCD ÿ s = 0
O work
+ f
Th de
,
he is
cR ( Tc
a/vaque
) ÿ modo
ÿÿÿ
ÿ thighs
,
=
ÿ
WDA
cR ( Tc
=ÿ
Th ) + to/go
ÿ
a/vCD
f
ÿ
.
The total work is
IN
=
WAB
ÿÿ cR ( Th
ÿ
+ WBC + WCD + WDA
Tc ) + a/vBC
f
ÿ
ÿÿ cR ( Tc
Th ) + to/go
ÿ
b
ÿ
b
in bÿ
ThR
in aÿ
= ÿln ÿ
ÿ
a/vb ÿ + ÿlnTcR
in aÿ
in bÿ
b
bÿ
ÿ to/go + a/vba/
ÿ+
ÿ to/vBC
+ vCDf
f
ÿ+
ÿ
a/vCD
f
ÿ
giving
b
vb
IN = ( Th
Note now that
ÿ sab
=
R
in bÿ
ln ÿ
in aÿ
ÿ
Tc
)
R
and
b
therefore, QAB
b ÿ and,
IN
Q
Carnot efficiency.
ÿ
=
Th
ÿ
Tc
,
Th
b
.
ÿ
ln ÿ
So, we have to
is
that the
ÿ
= ThR
ln[( vb
ÿ
b) / ( and
ÿ
b)]
.
Machine Translated by Google
4.3. CHAPTER EXERCISES:
Exercise 103
97
(4.7-2) We now have a cylinder with electromagnetic radiation.
The equations are
4
IN = bV T
4
S =
,
3
bV T
,
3
we put
it in terms to
of make
the that
already
temperature
AND
In the first part of the cycle
ÿ IN = bT h4 ( VB
a simpler analysis.
VV
ÿ
Th did
,
AND)
ÿ
We have
4
ÿ S =
,
bT h3 ( VB
3
AND)
ÿ
of
=
so wab
Th ÿ S ÿ ÿ
IN
b
=ÿ
3
second part of the cycle we have tropic VB VC .
=0
So, S implying that ÿ
Already
3
there
It is
ÿ
4
3
bVCT
Th
AND)
ÿ
Tc,
ÿ
.
but
It is
free change
4
=
c
T h4 ( VB
h3ÿ
bVBT
3
VB
VC
=ÿ
ThTc ÿ3
work
bVCT
bVBT
ÿ
c
VC
In the third part of the cycle, we have
CEO
ÿ
ÿ U b=
4
VC ) T c
ÿ
3
bVDT
c
3
=
3
h ( Tc
Th )
ÿ
.
fixed tc , giving
,
( CEO
Nonetheless, VD is in the same adiabat as
4
4 =
bVBT
h
4
IN =
ficaWBC = ÿÿ
.
VA , so that
4
bVAT
3
h3ÿ
CEO
=
AND ÿ
ThTc ÿ3
so, and,
ÿ IN =
3
h ( AND
bTcT
ÿ
VB )
.
The entropy variation is in this part
4
ÿ S =
3
so that
O
bT c3 ( CEO
4
ÿ
VC ) =
3
bT h3 ( AND
VB
ÿ
)
work in the third part of the cycle becomes
b
=
WCD
Tc ÿ s ÿ ÿ
IN
3 Th3 Tc ( AND
=ÿ
VD
Finally, in the last part of the cycle, we have
ÿ IN =
It is
ÿ S =0
.
The work
bVAT
4
h
4
bVDT
ÿ
c
=
ÿ
AND
bVA ( Th
ÿ
he is
WDA
=
bVA ( Th
ÿ
3
Tc ) T h
.
VB
ÿ
It is
Tc
)
ÿ
3
Tc ) T h
.
Th, so that
Machine Translated by Google
98CHAPTER 4. REVERSIBLE PROCESSES AND THE CONTENT. OF WORK MAX.
the total work
bT 4
h VB
b T 33
+ ÿÿ 3 hTc ((AND
ficaW =
ÿÿ
3
h ( Tc
AND)ÿ + ÿ)ÿ + bVBT
ÿ
VB ÿ
ÿ
bVA ( Th
ÿ
Th )ÿ
Tc ) T h3
ÿ
ÿ
giving
IN = 4 bT h3 ( Tc
3
ÿ
Th )(
VB
ÿ
AND)
so that
IN
ÿ
as wished.
Q
=
43
/bT
3
h Tc
(ÿ43
/bT
ÿ
Th )(
4
h ( VB
VB
ÿ
ÿ
AND)
AND)
=
Th
ÿ
Tc
,
Th
Machine Translated by Google
Chapter 5
Alternative formulations
5.1 Ninth Chamber (30/04/2008):
5.1.1 The Principle of Minimum Energy:
The maximum entropy principle can be used to obtain a number of important
results in thermodynamic theory. However, thermodynamics
characterizes, more than other areas of physics, by having its problems greatly
simplified due to the particular choice of important variables.
Thus, a description in terms of the principle of maximum entropy, while adequate
in principle, may be practically useless, as the problem may become so complex
that its solution is unavailable.
We have already mentioned that the principle of maximum entropy is equivalent
to the principle of minimum energy. So eventually a problem
formulated from the principle of minimum energy can become much more
simpler than if formulated from the principle of maximum entropy, even if these
theoretically equivalent
principles are
We have already studied the thermodynamic configuration space and we know
that we must write it from the extensive thermodynamic variables. So, for example, if
we have a problem in which the fundamental equation holds
1
.
in =
Bs
5/ 2 ÿ1
in
/2
,
then we can express this equation as the surface shown in figure a
B = 1), where we also present the plane = 20 of in
follow (where we
made constant energy. The curve that is the result of the intersection of the plane
= 20
in
1Something similar happens when we try to solve certain problems in partial differential
equations; if the problem naturally has spherical symmetry and we try to solve it in a
Cartesian coordinate system, the solution can become so complex that we cannot solve it. In
classical mechanics we also have a similar situation: we can make use of the Lagrange, Hamilton or
Newtonian formulations, depending on which variables
(generalized coordinates, generalized coordinates and moment or force diagram) are
the most simplistic. This phenomenon, in thermodynamics, reaches much greater degrees.
99
Machine Translated by Google
100
CHAPTER 5. ALTERNATIVE FORMULATIONS
Figure 5.1: Plot of the energy surface for the fundamental equation presented in
in
the text. Also shown is the plane that defines free energy = 20
constant.
with the surface shown in the graph is given by (
s = 20
2/ 5 1/
v
B
= 1)
5
and can be represented as in the following figure. This last graph tells us that,
Figure 5.2: The curve representing the intersection between the con- = 20 energy
standing in plane and the energy surface for the fundamental equation presented in
the text.
in
for an energy = 20 (constant),
and volume varying between 0 and 10, the highest
entropy value is given for = 10.in
This same problem could have been solved by looking at the equation
Machine Translated by Google
101
5.1. NINTH ROOM (30/04/2008):
fundamental taking entropy as the dependent variable, so that
s =
B 2/ 5 in 25/ v
1/
5
which gives a surface like the one shown in the following figure, in which also
we present the plan
. this graph, for a constant and equal entropy
s = 2 5.In
Figure 5.3: Entropy surface graph for the fundamental equation presented in the text. Also shown
is the plane that defines entropy = 25
.
s
constant.
a 2 .5, we have the curve ( B = 1)
in 5= 25.
/ v2
ÿ1
/
2
which is represented graphically as in the figure below. Thus, for a constant entropy equal to 2 5,
the minimum energy that our system
. can have, for the molar volume in the interval (010] is
in
obtained at = 10.
,
In the first case we will know that the equilibrium will be given by the maximum point of the
curve that gives the entropy in terms of volume (constant), while in theinsecond case we will know
that the thermodynamic equilibrium will be given by the curve that gives the energy in terms of
Minimum volume. molar volume (constant point). Thus, the following two principles s
are fully
equivalent:
Theorem 104
(Principle of
Maximum Entropy)
any unrestricted internal parameter given value
of total internal energy.
The balance value of
such is that it maximizes a entropy for a
The balance value of
(Principle of Least Energy) any unrestricted
internal parameter is such that it minimizes a given value of total entropy. a energy for a
Theorem 105
Machine Translated by Google
102
CHAPTER 5. ALTERNATIVE FORMULATIONS
Figure 5.4: The curve representing the intersection between the constant entropy
s = 2 .plane and the entropy surface for the presented fundamental equation
constant
not text.
Proof. (informal) So far, however, we have not demonstrated the equivalence of
these two fundamental principles. To do so, consider a situation where
the principle of maximum entropy holds no
and that the system is in equilibrium, but
(by contradiction) the internal energy is minimal. We could then remove energy from
the system in the form of work keeping the entropy constant and then return this
energy to the system in the form of heat. The entropy of the system would
increase, as the system would return to
Q = T dS
its original energy situation. However, the system in the final position would have a
since we bigger than that of the initial position; well, this is a contradiction, entropy
assumed that the system was in equilibrium and that the principle of maximum
entropy was valid. Thus, the principle of maximum entropy implies that of minimum
energy. We still need to show that the principle of minimum energy
implies maximum entropy, which we will do in the exercises.
Proof. (formal) From a formal point of view, consider that the principle of maximum
entropy is valid, that is, that it is valid for the equilibrium of thermodynamic systems
ÿ 2S
ÿS
ÿV
ÿ
IN = 0
ÿV
IN < 0 .
ÿ
and ÿ
2ÿ
It is worth remembering that the meaning of the derivative
()
ÿS/ÿV
IN represents various
entropy action with respect to the extensive variable that is free to vary, keeping
the energy fixed, which is exactly the content of the principle of maximum entropy.
Now we have to compute the derivative of energy as a function of (the constant
IN
entropy,
as stated in the principle of least energy) for
Machine Translated by Google
103
5.1. NINTH ROOM (30/04/2008):
2
show that it must represent a minimum point
ÿ
ÿU
ÿ
ÿV
ÿS
ÿX
IN
ÿS ÿÿ
ÿU
IN
=ÿ
ÿ
ÿ
S
ÿS
T ÿ
=ÿ
=0
ÿ
ÿV
. Like this,
,
IN
showing you have IN in fact, it has an extremum exactly at the point where
it.
To show that this point is a minimum, we calculate the second derivative )]
P = ( ÿU/ÿV
of IN to get [we did
S
,
ÿ 2 IN
ÿ
ÿV
ÿP
2ÿ
=ÿ
ÿP
ÿ
ÿV
S
=ÿ
ÿU
ÿ
ÿU
S
ÿ
ÿP
ÿ
ÿV
IN
O
+ÿ
S
ÿ
ÿV
IN
or yet
ÿ 2 IN
ÿ
ÿV
=
2ÿ
P ÿ
S
ÿ 2 IN
ÿV
ÿ
2ÿ
=ÿ
S
ÿV
ÿP
ÿ
+ÿ
ÿP
ÿ
ÿV
IN
=ÿ
ÿV
IN
ÿ
,
IN
P = 0. But then
since we have, at the point,
ÿ
ÿP
ÿU
ÿÿ ÿ
ÿ
2
ÿS
ÿX
IN
ÿS ÿÿ ÿÿ
ÿU
IN
ÿ S
ÿV 2
ÿS
ÿU
=ÿ
S
2
ÿ S
ÿVÿU
ÿS
+
ÿV
ÿ
T
=ÿ
ÿS
ÿU ÿ2
ÿ 2S
ÿV
2
> 0,
IN it is minimal.
P at which It is important to
so that = 0 is a point
note that these two principles provide two different ways of reaching equilibrium. We must
always remember, however, that however equilibrium is obtained, the final equilibrium state of
the system satisfies both principles.
In a previous chapter, we showed that thermal equilibrium between two systems must occur
when both are at the same temperature. We did this, however, using the entropy representation.
We can do the same using
2 The relationship
ÿU
ÿ
ÿX
ÿ
ÿ S
ÿX
ÿ
ÿS
ÿU
IN
=ÿ
ÿ
S
ÿÿ
X
can be demonstrated in the following way: we have, by the assumption of the minimum theorem
energy, that the function S = S ( U,V,N ) is constant. So, we have to
dS = ÿ
and putting
dN
ÿS
ÿU
ÿS
of
ÿ
+ÿ
ÿV
V,N
dv
ÿ
ÿ S
+ÿ
=0
ÿ
ÿN
U,N
U,V dN
= 0, we arrived
ÿU
ÿS
aÿ
ÿU
ÿ
ÿ
IN
ÿV
ÿS
ÿ
=ÿÿ
ÿ
ÿV
S
,
IN
where the term ( isÿU/ÿV
entropy constant
because this has been our assumption since the
)S
start. We are then left with the final expression
ÿU
ÿ
ÿV
ÿ
ÿS
ÿV
ÿ
ÿS
ÿU
IN
=ÿ
ÿ
S
ÿÿ
IN
.
Machine Translated by Google
104
CHAPTER 5. ALTERNATIVE FORMULATIONS
the representation of energy. So, we have to
IN = IN
(1) ÿ
S (1) , IN(1) , ÿ N j
(1)
IN
ÿÿ +
(2) ÿ
S (2) , IN(2) , ÿ N j
(2)
ÿÿ
.
We know the values of all volumes and moles and they do not change since the
also the
systems are rigid and impermeable; the total energy does not change, since
system is closed. Thus, the variable to be computed is the
entropy. We can do this by assuming that the equilibrium state minimizes the
energy, so that any infinitesimal variation of it will give zero, that is, (1) +
of
=
T
(1)
(2)
T
dS
dS (2) = 0
,
as long as we have a fixed total entropy (this is the energy minimization principle).
But then, we have (1) +
S
=
S (2)
S
It is
dS = 0
,
so that
dS (1)
dS (2)
=ÿ
and therefore,
=ÿ T
of
(1)
T
ÿ
dS (1) = 0
(2)ÿ
giving
(1)
T
(2)
T
=
,
just like when we used the entropy representation.
5.1.2 Legendre Transformations:
Both in the representation of energy and in the representation of entropy, the
independent variables are always taken as being the extensive variables of the
problem. Intensive parameters are always conceived as derived concepts. This
situation is in conflict with laboratory practices in
thermodynamics, in which it is generally easier to control the variables
intensive than extensive ones (in fact, instruments for measuring entropy do not
exist, while instruments for controlling temperature
are usual).
Thus, we are interested in being able to rewrite the entire thermodynamic
formalism using one or more intensive parameters as independent variables. Let's
see how this can be done.
Example 106
consider the fundamental s/r equation
in
( s, v ) = Of
ÿ2
exp(
)
all of it from
written
Let'sinassume
terms ofthat
extensive variables we
,
in
It is
s.
want to write this equation in terms of the intensive variable P (pressure).
if we write
ÿ s,P
(
)=
in
ÿ (ÿ Pv ) = Of
ÿ2
exp( s/R ) + Pv,
(5.1)
Machine Translated by Google
105
5.1. NINTH ROOM (30/04/2008):
in
with we eliminate
we immediately observe that, arriving
to the our result.
Oh, we know that
of this expression, we can
ÿu
P,
=ÿ
ÿv
so we are left with
P = 2 Of
ÿ3
exp( s/R ) = 2 u/v
We then have two equations
ÿ s,P
(
=
in
Of
ÿ2
3
=
+ Pv
in
)=
2
Pv
exp( s/R
)
But then
in
A exp( s/R
P
=ÿ2
)
/
3
ÿ1
,
a
we arrived
3
ÿ s,P
(
)=
2
P 2/
3
A
)1(2
ÿÿ
ÿP
waiting
as a way to eliminate what we
3
exp( s/ 3 R )
,
s e P. I feel that
is that just a function of
how it would beofif
/
of
v?
=
in,
(ÿ pls
expression (5.1). Why do we use variable O term
A is simple answer
with remember
)
have
of
Like this,
with
=
ÿ
VAT.
we put
=
p
It is
T ds
in
+ Pv
differentiate, we get
dÿ
=
indicating that
of + VAT + vdP = ( Tds
VAT ) + VAT + vdP
ÿ
=
T ds + vdP,
a new independent variable is P, no longer v.
From the previous
example,
therefore, we conclude that if we have a function that can
be expressed
in differential
form as
dY
FjdXj,
=ÿ
j
where
ÿY
Fj
ÿ
= ÿ ÿXj
,
X
k
AND
the (
X 1 , X 2 ,...
)
Machine Translated by Google
106
CHAPTER 5. ALTERNATIVE FORMULATIONS
so if we want to write our formalism in terms of the example variable, we just
write
2
K,...
p (X 1 , X,...,F
)=
AND
FK
,
by
(X 1, X 2 ,... ) ÿ FKXK,
since, in this way, we obtain, by differentiation
dÿ
=
dY
FKdXK
ÿ
XKdFK jFjdXj
=ÿ
= ÿ j ÿ= K FjdXj
ÿ
ÿ
ÿ
FKdXK
XKdFK
ÿ
XKdFK
so the independent
p is now written in terms of the parameter
variable function. Note further that
ÿ
ÿÿ
ÿFk ÿ = ÿ
FK
as
XK.
These transformations allow us to define a series of important functions
in thermodynamics, calls that we will study
thermodynamic
next. One ispotentials
left, however, with the impression
that there is an arbitrary character in the definition of the function obtained by adding one
or more terms to the
p
AND so that we can ask whether this function
equation relative to the function
AND
is representative of thermodynamic systems in the same way as the function
,
,
,
supposedly was. In the same way that we can pass from the representation in the
energy for the representation of entropy, with a rewriting of the principle for the new variable
(which passes from “minimum energy” to “maximum entropy”), also in the general case of
a Legendre transformation, to the same extent that there is a change in the thermodynamic
step by
step
function, there is a change in the principle that corresponds to it (minimum,
maximum,
etc.)
so that the function and the principle associated with it maintains the validity of the
more
description
thermodynamics that the new function entails–this topic will be discussed in
throughout Chapter 6. Just as an illustration, we can define the potential function
of Helmholtz from the replacement of entropy by temperature in the equation
fundamental to free energy, that is, from the equation
F
IN
=
TS,
ÿ
so that
dF
=
of
TdS
ÿ
ÿ
SdT = ( TdS
VAT +
SdT
VAT + µdNµdN
=ÿ
ÿ
ÿ
TdS
)
ÿ
SdT
ÿ
O balance value
and, for this potential, we have the minimization principle:
of any unconstrained internal parameter in a system in di-potential contact of
athermal with a thermal reservoir minimizes state set
O
Helmholtz about
O for vacuum
you which T = T r , being
do
T r a temperature travel
. See, therefore, that the content of the principle has changed together
with changing the definition of the function.
Machine Translated by Google
107
5.2. TENTH CLASS (05/05/2008):
5.2 Tenth Class (05/05/2008):
5.2.1 The Thermodynamic Potentials:
The thermodynamic potentials can then be defined in terms of which choice we make for the
exchange of the extensive variable(s) for the corresponding intensive(s). Note that we must always
replace the extensive variables with the
T the
P Nj
S from
INprevious
corresponding intensives, that is, . This can be seen
developments,
since we are
µj
ÿ
ÿ
,
ÿ
,
always looking to change the form of the differential term (it is variable intensive, it is variable
extensive), so we must always include the term ÿ ) and thus move
from
My hands
Ij
ijej
not
in order to obtain the differential ÿ (
ÿ
ejd
+
ejd
My hands
My hands
for
.
We therefore have:
1. Energy Representation:
T giving
H [T,V,N
,
IN [ S,V,N ] of
: replacement in
S
by
];
IN [S,V,N ]
enthalpy : replacement in
(b) A
(c) A
Helmholtz
free energy of
(a) A
IN by
free energy of
by P giving G [T,P,N
,
P giving
,
IN [ S,V,N ] of S
Gibbs from : replacement
F [ S,P,N ];
IN
by T
It is
into ];
2. Entropy Representation:
Massieu
function of
(a) A
IN
for 1
(b) A
by
/T
,
giving
S [U,V,N ] of
in temperature: substitution in
IN
[1/T,V,N
];
in pressure: substitution in [ ];
function of Massieu
P/T giving MV U,P/T,N
S [U,V,N ] of
,
(c) A [ function of
Massieu
in pressure and temperature: replacement in
S U,V,N ] de IN for 1 giving
/T it's from IN by
P/T
MOV [1];/T,P/T,N
,
It is easy to show that there is dependence between some Massieu functions and the potentials
defined in the energy representation. So, for the Massieu function on temperature, we have
]= S
IN [1/T,V,N
=
U/T
ÿ
ST
IN
ÿ
=ÿ
T
F
T
while for the Massieu function on pressure and temperature, we have
S
MOV T ,PT , N
ÿ1
ÿ=
1
ÿ
ÿ
TU
P
IN
T
IN
=ÿ
ÿ
ST + PV
T
=ÿ
G
T
.
all
Also note that if you try to replace extensive variables
with variables
intensive, the solution is identically zero, due to Euler's equation. In fact, when exchanging all the
extensive variables, we would have the Legendre transformation:
f
=
IN
ÿ
TS + PV
ÿ
Women ÿ 0
.
IN
Machine Translated by Google
108
CHAPTER 5. ALTERNATIVE FORMULATIONS
5.3 Chapter Exercises:
monatomic gas
(5.3-1) To find in a fundamental equation of a Gibbs
It is
In the representation of
the Hemlholtz representation, entropy enthalpy.
we have
Exercise 107
queS
=
N
IN
N
IN0 ÿÿ 0 ÿÿ(+1)ÿ
IN c
0 ÿÿ
IN
Ns 0 + No.
ln ÿÿ
Note, however, that we have to have the
So, we invert this equation to obtain
1.
.
equation in representation
c +1) /c
N
IN = INÿ(0 ÿ 0
N
c
ÿ
IN0 /c
IN ÿ1expÿ
S S0
cNR
power.
ÿ
ÿ
.
From this energy representation, we have that the Helmholtz a repre(Hemlholtz)
sentencing is given by
F
=
IN [T ] = IN
TS
ÿ
It is like
T
ÿU
=
=
ÿS
IN
cNR
of so we are left with:
c
F
=
cNRT
TNs
ÿ
NRT
0ÿ
ln ÿÿ
cNRT
0 ÿÿ
cN 0 RT
N
N 0 ÿÿ
IN
0 ÿÿ
IN
c
(+1)ÿ
giving
c
F
2.
=
Nf 0 + NRT c(
T
0 ÿÿ
T
ÿ 1)ln ÿÿ
IN
0 ÿÿ
IN
N0
ÿÿ
N
.
(Enthalpy) Now we have
H
=
IN [ P ] = IN + PV
with
P
ÿU
=ÿ
ÿV
=
IN
cV
so that
H [S,P,N ] = cPV + PV
we have to eliminate
we have
It is
a variable
c
ÿ
N0
N
IN
0 ÿÿ
IN
Using
V. _
c +1)
a expression for
S S0
expÿ cNR ÿ =
ÿ
IN0
IN
or yet
c
ÿ
PcV
0 ÿÿ
PcV
0
N0
ÿ(ÿ(
N
c +1)
S S0
expÿ cNR ÿ =
ÿ
IN0
IN
a energy,
Machine Translated by Google
5.3. CHAPTER EXERCISES:
109
so that
c
P
ÿ
It is
eliminating
c +1
N0
P 0 ÿÿ
S
ÿ
N
H [S,P,N
]
IN0
ÿ=ÿ
cNR
expÿ
we found
S0
ÿ
IN
c +1
ÿ
.
(5.3-2) Van der Waals e given by
Exercise 108
s = R ln[( in
c
b)( in +
ÿ
and in)
] + s0
so that
( in
we have to f ( T, in ) =
in
ÿ
T =ÿÿ
Now,
ÿu
=ÿ
in
in
=
ÿ
ÿs
+
s 0) /cR ] ÿ
ÿ
a
and in
in
ÿ
=ÿ
cR
in
in
+ ( c ÿ 1) RT ln[( in
(5.3-3) For
Exercise 109
exp[( s
b)1 /c
ÿ
in
;
Ts, with
a
f
a
1
in =
c
b)( cRT
ÿ
a
radiation we have
S
=
4
+ cRT.
1///b
1
4 34
IN
IN
)
.
]
4
3
so that
/
IN = ÿ 3 4ÿ4
1.
3
b
ÿ1
// 3 S 43 IN
ÿ1
/
3
=
ÿS
43
/
IN
ÿ1
/
3
.
To the representation in enthalpy, we have (no dependence on
=
H
N):
H [S,P ] = IN + PV
with
ÿU
P
=ÿ
.
ÿV
Like this,
a
P
=
4/
S
IN
3ÿ
so that
IN
3
= ÿ a ÿ3/
,
ÿ
4
P
ÿ34 /
S.
3
With this result, we can write
IN
=
ÿS
4/ 3 ÿ1 /
IN
3
=
ÿS
a
4/
3ÿ
3
/
ÿÿ1
4
ÿ1
P 1/ 4 S /
3
4
1= 3 / a
4
3//P
1 4
S.
Machine Translated by Google
110
CHAPTER 5. ALTERNATIVE FORMULATIONS
Thus, we are finally left with
4
H [S,P ] =
3 // 4 P 1 4
(5.3.5) We have
3
= a NV IN /
queS
(
)1
It is
S,
as other representations.
similar calculations apply to
Exercise 110
/ 4a
13
3
we want to write the
result in the representation of Gibbs. So, we have to
first write
IN
It is
S3
=
b
NV
remember that
G
=
G [T, P ] = IN
ST + PV,
ÿ
with
ÿU
=
T
ÿS
P
,
ÿU
=ÿ
.
ÿV
Thus, we are left with
S2
S3
T = 3 b NV NVV
P =these
b
,
with
It is
U,S
we have to eliminate
P
T
=
It is
equations. note that
N
1
9b S
2
2
ÿ
S
NT
=
2
IN
,
9ÿP
=
NT
3
81 ÿP
2
so that
IN
It is
NT
=
3
9ÿP
like this,
G
We still need to calculate the
a ( T, P
)
=
NT
N T
3
NT
3
+
ÿ
9ÿP
9ÿP
NT
3
=
81ÿP
thermodynamic coefficients. To calculate VT (and P) to do a
we have to find a relationship between
It is
derivation
a ( T, P ) = 1
.
IN ÿ ÿVT ÿ
P
Now, such a relation, so that we have already found
1
a ( T, P ) =
O
even for other coefficients.
3
.
81 ÿP
NT
IN 27ÿP
2
2
,
=
a
Machine Translated by Google
5.3. CHAPTER EXERCISES:
Exercise 111
111
H [S,P,N
(5.3-6) We have to
S
ÿ
A
=
] = IN + PV
2ÿ1/
NOT + BV
with
,
2
It is
2
ÿS
2
bV
N
IN =
ÿ
,
with
ÿU
P
=ÿÿ
=
ÿ
ÿV
2bV
N
S,N
PN
IN =
ÿ
Like this,
H
is that
=
2
ÿS
bV
ÿ
2
+ PV
N
P 2N
2
ÿS
=
+
N
;
2b
4b
desired result. We compute V by simply re-
knowing that
IN
as
ÿH
PN
=
ÿ
= ÿ ÿP
,
2b
S,N
had already been obtained.
Exercise 112
(5.3-7) We have to
2
H
1
AS N
=
ÿ
ln ( P/P
0)
we want to calculate Cv T, P (A good) representation,
therefore, would be that
Pe
Gibbs, which is alreadygiven in terms of these T variables.
Thus, we must go to
It is
.
this representation by doing
G
H
ST,
ÿ
S e h using
eliminating
T
=
ÿH
= ÿ ÿS
ÿ1
= 2 ASN
ÿ
ln ( P/P 0) = 2 H/S
S
ÿ
NT
=
P,N
2A ln ( P/P
Thus, we are left with
G [T,P,N
]=ÿ H
AS 2 N
=ÿ
NT
ÿ1
ln ( P/P 0) = ÿ
4 A ln ( P/P
now we know
ÿ 2G
whatCp
= ÿ ÿT
We can also calculate ÿT
= ÿ ÿT
and a.
=ÿ
IN
ÿ
ÿ
=ÿ
.
2A ln ( P/P
P
In fact,
ÿV
1
Mr. T
N
ÿS
2ÿ P
ÿP
ÿ 2G
1
ÿ
=ÿ
T
IN
ÿ
ÿP
2
2ÿ T
0)
;
0)
0)
Machine Translated by Google
112
CHAPTER 5. ALTERNATIVE FORMULATIONS
It is
ÿV
1
=
a
ÿ
IN
ÿ
1
=
ÿ
ÿT
ÿ
IN
P
ÿG
ÿ
ÿT
ÿ
ÿP
.
ÿ
T
P
With these results, we can use
2
Cv =
already obtained in chapter 3.
Cp
T INa
NÿT
ÿ
So we have
N
=ÿ
Cp
,
2
NT
=
Mr. T
,
2 A ln ( P/P
4P
0)
[2 + ln (( P/P 0)]
2 OF
ln
3
P/P
0)
It is
a
NT
=
2 FIG
2
ln
( P/P
0)
so that
N +2 T
Cv
Exercise 113
=ÿ
.
2 A ln ( P/P
0)
(5.3-8) We have to
F
˜
A kBT ln
=
ÿ
It is
and we want the
calculationof
b
ÿ
in +
It is
b
d
ÿ
ÿ
ÿ
O
representation in terms of entropy. For that, consider
ÿ ÿF
( ÿÿ)
=
ÿF
F + b
;
ÿÿ
where b = 1 /kBT. we have to
ÿF
b
=
ÿÿ
ÿF dT
b
ÿF
b
ÿF
1
=ÿ
=ÿ
ÿT dÿ
kBÿ
2
ÿT
=ÿ
ÿT
ÿkB
T
ÿF
=
T S.
ÿT
Like this
F + b
ÿF
=
=
F + TS
IN,
ÿÿ
by definition. Like this,
ÿ
=
ÿ
in + hey hey
wow _
˜
IN
N
d
.
ÿ
ÿ
It is
b in +
ÿ
It is
b
d
ÿ
a fundamental equation of one of ideal
(5.3-10) We must find
gases in the Helmholtz representation. Thus, for a mixture, we have
Exercise 114
ÿ
S
ÿ
0+
=ÿ
So-called
j
ÿÿ
Njjj
j
R
T
ln ÿ T
0ÿ+ÿ
No.
IN
ln ÿ Nv
0ÿÿ
R
ÿ
j
Nj
ln Nj
N
Machine Translated by Google
5.3. CHAPTER EXERCISES:
113
It is
ÿ
ÿ
=
IN
ÿÿ
RT.
Njjj
ÿ
j
S por T as an independent variable:
we must replace
F [ T, V,
UT, N,eliminated
Note that we already have
TS. as a functionof so that is already
a variable U of the new variables. On the
other hand, S also already in
given in terms of
jNjs
0 + ÿÿ
ÿ
{
=}
jNj
ln Nj
so we are simply left with
T,
F [ T,V,
T ÿÿ
terms is
IN
]
Nj
Nj }] = ÿÿ
{
T
0
T
ÿ ln
Rÿ
jNjc
jNjcj jNjR
ÿ+ÿ
RT
ÿ
0ÿÿ R
ÿ ln ÿ
ÿ
VNv
N
which can be written as
F [ T,V,
{
Nj
}] = ÿ
NJRT
T
cj ÿ 1 ÿ ln ÿ T 0 ÿÿ ÿ
ÿ
j
and,
therefore,
F [ T,V,
{
}] = ÿ
Nj
IN
s 0/R + ln ÿ
0 ÿÿ
local unit
F [ T,V,N j
.
]
j
Exercise 115
A system follows the
(5.3-12)
(s
We need to calculate the
s 0)4 =
ÿ
Gibbs
potential of
in =
ÿ1
Bv
/
fundamental relationship
2
He had
.
G [T,P,N
2
(s
ÿ
]
We have to
.
s 0)2
and,
therefore,
ÿ1
T = 2 Bv
The potential of
2
=
s 0)
ÿ
P = 1 / 2 Bv
,
in
ÿ( s
ÿ3
/
2
(s
ÿ
s 0)2
s 0) T + Pv + s 0 T
ÿ
T
and u and in terms s, v
we need to eliminate from
P
T2
It is
(s
Gibbs fica
g
It is
/
=
1
in
8B
ÿ1
/ 2ÿ
P.
I feel that
T4
64 B 2 P
in =
2
like this
(s
ÿ
T
s 0) =
ÿ1
2 Bv
T
=
/
T
T3
2
=
2 B 8 BP
2
16 B 2 P
Finally
in =
B
8 BP
T
2
T
6
16 2 B 4 P
=
2
T4
32 B 2 P
.
ÿ
Machine Translated by Google
114
CHAPTER 5. ALTERNATIVE FORMULATIONS
so we are left with
g
T4
=
T4
32 B
2P
16 B
T4
+ P
ÿ
2P
16 B
2P
T4
=
32 B 2 P
2
and,
therefore,
Exercise 116
T 4N
32 B 2 P
=
G
(5.3-13) We have to
.
in
= (32)/
=
AvT
Pv
P
It is
=
4
AvT
Thus, con-
.
we concluded that
Of
1
T = ÿ 32
in
/
2
ÿ1
4
P
,
T
3
=
Of
Of
in
2
ÿ3
/
2
4
ÿÿ3
so that
/
ds
4
1 2
in / u
A
ÿ1= ÿ23
where a
It is
ÿ1
// 4
4
/ of
A
1 4
ÿ1
in
/ u2
3/ 4
dv
=
ÿd
ÿ
1 2
in / u
3/ 4ÿ
+ ÿ 3 2ÿÿ3
some constant irrelevant to our interests. Like this
s =
1/ 2
u
ÿv
34/
and,
therefore,
ÿ2
43
/
in = ÿs / in
3
,
this being the fundamental equation. To obtain the Gibbs
a representation in terms of
potential, we remember
= in
queg
Ts + Pv
ÿ
with
P
ÿu
=ÿ
we have to eliminate
=
ÿv
2
3
s, in
P
T
ÿs
by u.
=
4 / 3 ÿ53 /
in
T
,
ÿu
=
4
=
ÿs
3
ÿs
1/ 3 ÿ2 /
in
3
.
Taking
4// 3
ÿs
(23)
in
4 4 3
s /v
4
(4 /b3)4
ÿ5
/
3
3
=
ÿ8
/
3
3
2 7b
3
in
or it is,
in
b
= 2 ÿ 4 3ÿ3
3
P
.
T4
Then
s
= ÿ 3 4b ÿ3
T 3 in
2
= ÿ 3 4ÿ3 1b
3
T
b
34 ÿ 4 3ÿ6
6
P2
T
8
b
= 4 ÿ 4 3ÿ3
3
P2
.
T
5
Machine Translated by Google
5.3. CHAPTER EXERCISES:
115
Like this,
in =
b ÿ 4 ÿ 4 3ÿ3 b
so that
g
Exercise 117
64
=
3
b
9
P
2
T
4
P
3
T
/
2
P
3
b
ÿ 2 ÿ 4 3ÿ3
T
5 ÿ4
256
3
b
ÿ
27
/
3
P
2
T
4
128
+
3
b
27
3
9
4 ÿÿ2
P
2
T
4
64
=
27
=
+ ahave
(5.3.14) We have to f YT / We
( in still
)
to
Cv T()
=
1
b ()in
2
64
=
3
3
b
P
2
T
4
b
P
2
T
4
.
where f
,
=
f [T,V ]
Oh, we know that
.
2
ÿf
ÿT
s ÿÿ
=ÿ
Like this,
ÿ f
ÿT 2 ÿ
in
b ()v
T
=ÿÿ
ÿ
Cv/T.
=ÿ
ÿsÿT
in
2
ÿ f
ÿT 2 ÿ
ÿ
=ÿ
=
once we know that f
d
2
AND
dT
ÿ1
1
/ 2ÿ
b ()
v =
in
in
=
T
2
ÿ1
+
,a
( T ) / ( in + a ) But then we're left with
AND
.
/ 2ÿ
AND
4
()T=
3
3
T /
2
+ dT +
It is,
is
and where
d It are
arbitrary constants. To calculate remember O work released, just
that
1
ÿf
ÿ 2 T 1/ 2
s
s =
+ d ÿ + const
=ÿ
ÿ
ÿT
1
2T /
2
2 T 10/
+ d
f
=
ÿ s
2
+ d
.
ÿ
+ a
vf
Exercise 118
+ a
in
so that
a
in0 +
(5.4-2) We have to
1
S
T
We can calculate remembering that S
1
=
T
3
4
ÿU
ÿ1
S
ÿ1 ÿ=
ÿ
TU.
1
34
// ÿU
IN
=
// 4 IN 1
4
ÿ
1
IN /
4
4
=
so that
3
4
ÿT
1
IN /
4
.
Like this,
S
=
3 3// 4 1
IN
b ÿ 3 4ÿ3 T IN
3
4
ÿ
4
ÿU
ÿ1
// 4 UV 1
4
=
1
4 ÿ 3 4ÿ3 4
or it is,
1
S
ÿ1
T
,
INÿ =
ÿ 3 4ÿ3 ÿÿ3 4bÿ 1
T
IN.
4
3
b T IN
.
Machine Translated by Google
116
CHAPTER 5. ALTERNATIVE FORMULATIONS
Machine Translated by Google
Chapter 6
Extreme Principles
Transformed:
6.1 Eleventh Class (05/07/2008):
6.1.1 The minimization principles for potentials:
We haveonly
already
seen that
thechange
Legendre
are with
thermodynamically
relevant
because,
as we
the transformations
function associated
the equation
fundamental introducing one or more new terms, we must also have the alteration of
the principle of maximi-minimization. Thus, for each of the Legendre transformations,
that is, for each of the thermodynamic potentials,
we're going to have a particular minimization or maximization principle.
Helmholtz Free Energy:
In the Helmholtz free energy we consider
F [ T,V,N
] = IN [ S,V,N
] ÿ TS,
S , V,N
to move from a representation in terms of to one in terms
of . Now, the only gain that
we can have in this type of mathematical artifice
happens if we are in a situation where the variable to which we changed
T ) is
(in this case the temperature
. Such
constant
a situation is one in which
T,V,N
1
we have our composite system (of two subsystems, for example) connected to a
thermal reservoir. In fact, in this case, the principle of conservation of
energy gives us
IN + INr
=
const.
1
This is fully equivalent to the case of Legendre transformations in classical mechanics.
There we try to find a Legendre transformation that takes us to the situation where we have a
maximum number that
motion constants
thus reducing the number of integrations
we must do to obtain the equations of motion.
,
117
Machine Translated by Google
CHAPTER 6. TRANSFORMED EXTREME PRINCIPLES:
118
so that
d ( IN + INr ) = 0
(6.1)
and the principle of least energy implies that
d 2 ( IN + INr ) = d 2 IN = 0
,
where we are assuming that the reservoir is large enough not to undergo second-order
changes in the process. Remembering that the principle of least energy is valid only for
situations in which entropy is being maintained
constant, we also have:
d (S + S r ) = 0
.
The expression (6.1) implies the terms (in addition to those referring to the other
extensive thermodynamic variables)
T
(1)
dS (1) + T
(2)
dS (2) + T r dS r
T
=
(1)
(2)
dS (1) + T
dS (2)ÿ
Tr
ÿ
dS (1) + dS
(2)ÿ = 0
so that we get
(1) =
T
(2) =
T
T
r
representing the thermal equilibrium of the composite system with the reservoir.
Thus, we can write expression (6.1) as
d ( IN + INr ) = of
r
r
+ T dS
=
of
T r dS = 0
ÿ
which almost the expression of the Helmholtz free energy in differential format
T r and
Tin general).
it is (being that it differs from this one by the appearance of the temperature
no However, as we are in a situation where this temperature is kept fixed, exactly by using
a thermal reservoir, we can write
d ( IN + T r S ) = 0
,
since we are only
which can be considered the Helmholtz free energy,
considering the states associated with temperature
T r . So, we arrived
the expression
dF
=
d ( IN
TS ) = 0
ÿ
,
which is the expression of a critical point. However, we know that
d 2 IN = d 2 ( IN
ÿ
T r S>)
0,
since the terms related to the reservoir do not influence this last result. So we also have
d2 F >
0, implying that we must
tpeorteunmciamlídneimHoe.lmOhroelstuz.lt of this analysis is the minimization principle for the
The balancevalue of any unconstrained internal parameter
in a system in diathermal contact with a thermal reservoir min set of states for
the streets O potential of
Helmholtz of this system on osquais O
T = T r .2
Theorem 119
2We use the complement “of this system” to explain the fact that the potential of
Helmholtz
thermal reservoir to
does not represent a relative part
.
Machine Translated by Google
119
6.1. ELEVENTH CLASS (05/07/2008):
We can easily understand this result. From equation (6.1) we have that the change in
energy in the composite system must be compensated by the change () and the energy term
dF = d inINthe reservoir.
TS
However, we have that (ÿ ) represents the change in
T r energy
Tr = T
dS of the reservoir (since ). Thus, with the term (assuming the
dS r =
dS existence of a reservoir),
TS
ÿ
It is
ÿ
eoss teaumcoasr“arteetr irmanndiom”oa. part related to the energy reservoir, still maintaining
Minimum Enthalpy:
All the above considerations can be equally extended to the other thermodynamic
potentials under identical conditions. In the case of enthalpy, we consider one (which will
keep the pressure constant).
pressure reservoir
We have, therefore,
d ( IN + INr ) = of
ÿ
P r dv
r =
of
r
+ P dv
=
d ( IN + P r V ) = 0
so that
dH = 0
.
r
Yet, since we must have P is a constant and
d2H
=
d 2 ( IN +
IN is an independent variable,
IN) = d 2 In > 0 ,
Pr
so the principle holds:
Theorem 120
The balance value of any unconstrained internal parameter
in a system in contact with pressure vessel minimizes which
P = Pr
of this system on the
you
set of states for
a enthalpy
.
Minimum Enthalpy:
Now we have a reservoir that keeps the temperature and pressure fixed (a
thermodynamic system in contact with the atmosphere, for example)
d ( IN + INr ) = d ( IN + T r dS r
P r dv
ÿ
r
) = d ( IN
ÿ
TrS
P r V ) = dG = 0
ÿ
,
so that
d2 G
=
d 2 ( IN
ÿ
TrS
ÿ
PrV
2
) = d In >
0
and we are left with the result:
Theorem 121
The equilibrium value of any unrestricted internal parameter in a system
in contact with a temperature reservoir
It is
a free energy of
Gibbs of this system about which set
pressure minimizes
T = T r It is P = P r
of states for
you
.
Machine Translated by Google
120
CHAPTER 6. TRANSFORMED EXTREME PRINCIPLES:
6.1.2 The Helmholtz Potential:
For a composite system in contact with a thermal reservoir, the equilibrium state minimizes the Helmholtz
potential for states that have
constant temperature equal to that of the reservoir. There are many practical situations in which
thermodynamic processes occur under situations related to thermal reservoirs. Thus, processes that
take place in rigid and diathermic containers at room temperature are particularly suited to the use of
Helmholtz free energy. As this potential is a natural function of and { }, the condition that variables is a
T problem
T variables { }. In a problem where there must
IN constant
variables in the
and becomes
reduces the
a function
numberofofjust the
Nj
,
be equalization of pressures between F
two
IN
Nj
It is
r
T sought
subsystems (both connected to a thermal reservoir at temperature ) we have that the equations
are
P
(1) ÿ
T r , IN(1) , ÿ N (1) ÿÿ = (2)Pÿ
j
T r , IN(2) , ÿ N (2) ÿÿ
j
r
which is an equation with one less variable (namely, which is constant)– T
,
,
in fact, only one variable, since the number of moles is assumed to be constant
also. The other equation represents the restriction on the total volume, and is given by
(1)
IN
(2)
+ IN
IN = const.
=
IN(1)
so that we can find the complete solution in terms of
It is
IN(2)
.
Note that pressure can only be obtained already in terms of temperature because
we are using the Helmholtz potential representation, for which
ÿF
P ()j
ÿVj
=ÿÿ
with
ÿ
,
{ N k}
F [ T r , IN,{ Nj }].
Another important feature is that the Helmholtz potential allows us to appreciate only the
composite system, without taking into account details.
relating to the thermal reservoir. Furthermore, this potential, as we shall see later, impressively simplifies
statistical mechanics calculations.
For a system in contact with a thermal reservoir, the Helmholtz potential can be interpreted as
the work available at constant temperature. Indeed, for a system capable of doing work that is in
contact with a thermal reservoir, we have
WRWS
=ÿ
of
ÿ
of
r
=ÿ
of
ÿ
T r dS r
=ÿ
d ( IN
ÿ
T r S) = ÿ
dF,
so that the work released in a reversible process by a system in
contact with a thermal reservoir is equal to the decrease in the Helmholtz free energy of the system
(hence the name “free energy”).
Machine Translated by Google
121
6.1. ELEVENTH CLASS (05/07/2008):
6.1.3 At Enthalpy:
For a composite system in interaction with a pressure vessel, the
equilibrium state minimizes the enthalpy over the set of states that have
the same constant pressure as the reservoir. It is rarer to find, in practical situations,
thermodynamic conditions involving constant pressure (something that would
presuppose, in most cases, a system in contact with the environment) without
there is heat exchange.
Enthalpy can be interpreted as a “potential for heat”. From the
differential form
dH = T dS + INdP
µjdNj,
+ÿ
j
it becomes evident that a system in contact with a pressure vessel and that
be waterproof will imply
dH
T dS
=
=
Q
so that: heat added to the system at constant pressure and with the other extensive
S IN
system parameters (not included) appears as a
increase in enthalpy.
It is
6.1.4 The Gibbs Potential:
For a composite system in interaction with a reservoir of both pressure and
as temperature, the equilibrium state minimizes the Gibbs potential over the set of states
at constant temperature and pressure equal to those of the
reservoirs. }
T,P, { Nj
The Gibbs
potential
is a natural function of variables that are
and
in
T P constant, reducing the problems of obtaining { }. cases in which
there are
Nj
innumerable processes that occur both at pressure and at temperature
constant.
It is
The Gibbs potential of a many component system is related
to the chemical potentials of the individual components. In fact,
G
=
IN
ÿ
TS
PV
ÿ
µjNj
=ÿ
j
so that, for a single-component system, we have
=
GN
m.
As we saw in Chapter II, chemical reactions are a natural application of the Gibbs
potential, precisely because they generally take place in contact (pressure
and temperature) with the environment. In these reactions, we have
ÿjAj,
0ÿÿ
j
Machine Translated by Google
122
CHAPTER 6. TRANSFORMED EXTREME PRINCIPLES:
n j are the stoichiometric coefficients of the reaction. How the changes in
where numbers of moles must be proportional to the stoichiometric coefficients, we
have that
dN
1
dN
=
n1
2
dN
=
n2
so that
3
=···=
n3
=
dNj
a
n yes
In a chemical reaction conducted at constant pressure and temperature, we are
with
dG
=
a
ÿ
=0
njµj
,
j
that is, we immediately have an equation
ÿ
=0
njµj
(6.2)
.
j
0
Nj ÿ as the initial values of the numbers of moles, so in
If we write ÿ
new values will be
N f = N
0
The chemical potentials
µj
+ yeah
j
j
are functions of T, P and the number of moles, that is,
a The solution
a system; the only
the singleforparameter
determines
the equilibrium
of equation
composition
(6.2) for of the
criterion
applying the
procedure
N f becomes negative (since it is not possible to
have none of the quantities ÿ3
j
moles of 2, for Hexample). To represent this condition or criterion for which
application, the notion of
degreeof reaction . The maximum value of a
.
all ÿ
N f
j
ÿ remain positive defines the maximum extent of the reaction. da ÿ
N f
a
likewise, the minimum value of
j remain
for which all ÿ
positives represents the maximum extent of the reverse reaction. The real value a
of an equilibrium situation must lie somewhere between these two extremes.
of reaction
The
is defined
as represented by
e
degree
in
,
,
e
a
a min
ÿ
.
ÿ
a max
ÿ
a min
Thus, it is possible that an algebraic solution of (??) gives or less a which is greater
a max than
a than min; in these cases, the process is terminated by the value
associated
with the
end of
one will
of the
components
ends
first).
The
of
a physically
a min. In this
relevant
value
then
be a max or(whichever
case,
even
though the
equation (??) is not satisfied (since the left term never reaches zero), it certainly
reaches the smallest absolute value accessible to the system.
Machine Translated by Google
Chapter 7
Maxwell's Relations:
7.1 Thirteenth Class (06/09/2008):
7.1.1 Maxwell's Relations:
We have already seen that there are several physical parameters of interest represented by first derivatives
between extensive variables and intensive parameters, of intensive parameters among themselves, etc. Such derivatives
always have the form
ÿX
ÿ
ÿ
ÿY
Z, W
IN S
and, given the large number of thermodynamic potentials (extensive variables () and
,
,
,
T
S variables
IN N is enormous.
intensive variables ( numberIN
of such
,
,
pÿ
,
,
F
H
,
,
G ), var ),
the respective
Fortunately, the number of such derivatives that can be considered independent is quite restricted, being, in
fact, only three in number. Chosen a set of three independents, all others can be obtained from these.
The way to obtain all the others as a function of these three chosen independents is precisely what we will now
investigate.
Many of these relationships are due to the imposition of continuity of the derivatives
mondays. For example: we know that the continuity and good behavior of
free energy implies IN
that
ÿ 2 IN
ÿSÿV
ÿ 2 IN
=
ÿVÿS
,
But such a relationship also implies that
ÿ
ÿV
ÿU
ÿ
ÿS
ÿT
ÿ=ÿ
ÿV
ÿP
ÿ
=ÿÿ
S,N
ÿS
=
ÿ
V,N
where the indices representing the variables that remain fixed result from the
fact that, in the free energy representation, the independent variables are,
123
ÿ
ÿS
ÿU
ÿ
ÿV
ÿ
,
Machine Translated by Google
124
CHAPTER 7. MAXWELL'S RELATIONS:
N
S IN
precisely
and the outermost derivative (being partial) determines the
independent variables being held fixed. This last relation is, in fact, a prototype of sorelations of Maxwell
called equalities. Such relations can be obtained
directly from the equations for the potentials
thermodynamics. In fact, consider a thermodynamic potential ÿ any
It is
,
.
written in abstract form as ÿ =
d
ÿdA + ÿdB
+ c dC,
we immediately have to
ÿÿ
ÿÿ
ÿA
ÿ
ÿB
ÿ=ÿ
ÿÿ
ÿ
,
ÿC
ÿ
ÿÿ
ÿA
ÿ=ÿ
,
ÿ
ÿ
ÿÿ
ÿC
ÿÿ
ÿB
ÿ=ÿ
ÿ
obtained by the second derivative in exactly the same way as we did before. We can see
this, for example, for equality of the middle, by doing
ÿ 2S
ÿÿ
= ÿ ÿA
ÿAÿC
ÿ 2S
ÿÿ
ÿCÿA
= ÿ ÿC
=
ÿ
B,C
.
ÿ
A,B
Thus, knowing the differential form of the thermodynamic potential, we can
he However, these relationships
all
immediately get the relationships it generates.
are equally valid regardless of the representation (of which potential
thermodynamic) we are considering, since, in the calculation process, the potential
itself is eliminated from the final result.
mentally
The reader is therefore invited to
the necessary operations
to obtain all the Maxwell relations that we present below for each of the thermodynamic
potentials.
IN [ S,
T dS
=
of
IN,N
ÿ
S, IN
]
VAT + µdN
ÿ
S,N
IN,N
F [ÿ T,
dF
SdT
=ÿ
ÿ
ÿ
IN,N
VAT + µdN
H [S,P,N
dH
=
=ÿ
]
T dS + VAT + µdN
G [ÿ T,P,N
of
]
]
SdT + V dP + µdN
ÿ
ÿT
ÿV
S,N
ÿT ÿÿÿ
ÿN
S.V
ÿ
ÿÿ
T,V
ÿP
ÿN
T,N
ÿÿ
IN,N
ÿÿ
S,P
S,N
P, N
ÿÿ
ÿ
S.V
S,N
ÿT
ÿN
S,P
ÿV
ÿN
ÿÿÿ
S,P
ÿÿ
ÿS
ÿP
T,N
ÿÿ
ÿS
ÿN
ÿ
=ÿ
ÿP ÿ
ÿT
V,N
ÿV
ÿN
S,N
=ÿ
ÿµ
ÿ
ÿT
=ÿ
ÿµ
ÿ
ÿV
ÿV
ÿS
=
T, P
P, N
ÿµ
ÿ
ÿV
T,V
ÿT
ÿP
V,N
=ÿ
T,N
ÿS ÿÿÿ
ÿN
T,V
ÿP
ÿN
ÿµ
ÿ
ÿS
=ÿ
ÿS
ÿV
ÿ
ÿP ÿ
ÿS
V,N
=ÿÿ
ÿ
=ÿ
ÿ= ÿ
ÿ
P,N
ÿµ
ÿ
ÿP
S,N
T,N
ÿV
ÿT
ÿÿ
T,P
=ÿ
ÿµ
ÿ
ÿT
T,P
=ÿ
T,N
P,N
ÿµ
ÿS
=ÿ
ÿ
V,N
ÿµ
ÿ
ÿP
ÿ
P,N
P,N
T,N
Machine Translated by Google
125
7.1. THIRTEENTH CLASS (06/09/2008):
Exercise 122
you same calculations for
Do it now
cos (there is no specific name for them):
IN [ S,
IN, m
ÿ
IN [ÿ T,
ÿ
]
,
you thermodynamic
IN, m
ÿ
IN [ S,P,
ÿ
]
,
ÿ
potentials
m
]
queHseájatondeoceusmsármioecaanre isamlizoaçmãoedmeôcnái lccoulpoas r.aEes st teaébeoletce emr aesdtaasppraósxsiamgaens seç sãeom.
7.1.2 Thermodynamic Diagram:
In the textbook (Callen) a mnemonic method for obtaining Maxwell's relations is presented. This
diagram is based on a square that takes into
it counts the variables S
but
the variables (by dimensionality:
N them into account, you
,T,notV,P
mto take
It is
,
would need something like a cube, geometrically speaking). The method is interesting and should
be studied carefully by the student. Here we present an auxiliary method, which we consider
simpler for thermodynamic variables (extensive and intensive). and which involves The method
all
consists of writing the thermodynamic potential in the form
a
A
ÿ=ÿ
b
B
c
C
ÿ
where, in the upper line, we present the dependent variables and in the lower line the
independent ones (the conjugated
independent ones must be presented with
the respective signs, exactly as in the expressions already calculated). Thus, we would have, for
example,
F [ÿ T,
ÿ
IN,N
]=ÿ
ÿ
S
T
ÿ
P
IN
m
N
ÿ
so that we immediately get
ÿ
ÿP
ÿS
)ÿ = ÿ
ÿ (ÿ IN
ÿ (ÿ T
ÿS
ÿP
)ÿ
i.e,
=
.
ÿV
ÿT
T,N
V,N
ÿ areÿ determined
ÿ by the
ÿ independent variables (bottom row):
Note that the variables held constant
as the left derivative is in terms of the constant independent variable, as the right derivative
IN we keep
,
T
It is
N
T know that all IN
is in terms of the independent variable we keep we
,
It is
N constants. As
the representations to be used have as reference to
representation of free energy, it is always very easy to know what are the variables
independent, dependent and signs.
Machine Translated by Google
126
CHAPTER 7. MAXWELL'S RELATIONS:
Example 123
a derivative
Imagine that we want to know how to relate
ÿT
ÿ
ÿV
ÿ
S,N
one
croelmaçãoolguma another derivative. So, such a derivative must have come from PV
T
S
ÿ
m
N
ÿ
,
T with respect like a
so that we can initially do (S, V and pending). Note, a derived from
keep the
variables
as inde- own
S e N constants
however, are not , this, variables
It is a representation of energy
that it is free, so that we must, in fact, write
ÿ
as
T
S
ÿ
P
m
IN N
ÿ
,
a derivative we are looking for is
of V it is clear that
exchanged. So, sign
ÿT
ÿÿ
Ve
ÿP
ÿ
ÿV
S,N
= ÿ ÿS
ÿ
,
V,N
of so we are left with
ÿT
ÿ
ÿP
ÿ
ÿV
Note, however, that this is not
S,N
and the
= ÿ ÿ ÿS
ÿ
.
V,N
only possibility, we could also
then
to have
ÿ
T
S
ÿ
P
IN
N
ÿ
m
ÿ
(and no other), where we notice the fact that we are O negative sign of µ which of free
a representation
always considering what determines all
energy as
as others. For this configuration, we are left with
ÿT
ÿÿ
ÿV
ÿS
S, m = ÿ
ties of
Consider the
ÿ
V, m
previous problem, but now we look for
Maxwell associated with
ÿV
ÿ
ÿT
ÿ
.
P,N
to the
a
ÿP
ÿ
obtained
from
the previous
that could µ, since none
of these
variables
were ter
being
onlyvaried by the exchange
derivatives.
Example 124
with he must
N
as re
Machine Translated by Google
127
7.1. THIRTEENTH CLASS (06/09/2008):
In this case, we would have
ÿ
ÿ
S
T
IN
P
m
ÿ
N
P with O positive for
It is
minus sign
(in which we know that T must comeas
a paradU expression: whenever you switch
the variables
immediate comparison with
O
conjugates
sign must change with respect to the expression of sodU)
that
ÿV
ÿÿ
ÿT
ÿS
=ÿ
ÿ
ÿ
ÿP
P,N
T,N
immediately, and
ÿV
ÿÿ
ÿT
ÿS
=ÿ
ÿ
ÿ
ÿP
P, m
.
T, m
The reader, of course, must choose, after careful study, which of these methods is easier
for him to use: the one presented here or the one that can be obtained from the textbook.
A procedure for reducing derivatives:
Since the derivatives we are interested in are mutually dependent, it is
necessary to use a method which enables us to obtain the one in terms of the other. One
possibility is as follows:
Criterion 125
of all
independent
first-order derivatives, only three can be
any derivative can be expressed in terms of a set
three chosen as basic. Conventionally, we choose arbitrary from
It is
O
set as being
ÿS
=ÿ
ÿ
CPT
ÿT
ÿ
Va =ÿ
,
P
ÿV
ÿT
ÿ
,
ÿ
V ÿT = ÿ
P
ÿV
ÿP
ÿ
.
T
All
as first derivatives (involving both intensive and extensive
parameters) can be written in terms of the second derivatives of the Gibbs potential, they
constitute a complete set
CP , a e kT
you numbers of
independent (maintained
moles constant).
a representation of Gibbs, for
a which
Proof. The proof can be given using
Proposition 126
we fear
[ÿ T,P,N
]
so that ÿ
2
=
ÿ
CPT
ÿT
G
2
,
INa =
ÿ 2G
ÿTÿP
,
ÿ
V ÿT =
ÿ 2G
ÿP
2
are as only possibilities (kept
you numbers of constant moles) of
what second derivatives in this representation. Not, like the representations are equivalent
with functions are independent in the Gibbs representation, also
alents, these
O are in all
other representations.
Machine Translated by Google
128
CHAPTER 7. MAXWELL'S RELATIONS:
The strategy for reducing derivatives is based on the following
ÿX
ÿ
=ÿ
ÿY
identities:ÿ
ÿY
ÿÿ1
ÿX
WITH
It is
ÿ
ÿY
ÿ
ÿW
WITH
WITH
ÿZ
÷ÿ
ÿ
ÿY
WITH
÷ÿ
ÿ
ÿW
WITH
ÿZ
=ÿÿ
ÿ
ÿX
=ÿ
ÿ
ÿY
WITH
ÿX
ÿY
ÿX
ÿ
,
ÿ
.
ÿX
X
AND
in order
The method then consists of the following three steps to be performed
in which they appear
:
1. If the derivative contains some of the potentials, bring them one by one to the numerator and
eliminate using the techniques in the preceding section;
2. If the derivative contains the chemical potential, take it to the numerator and
eliminate it using the Gibbs-Duhem relationship
dµ
=ÿ
sdT + vdP
;
3. If the derivative contains entropy, bring it into the numerator. If any of Maxwell's relations eliminates
it, use it. If the Maxwell relations (use equation 2 above
cannot delete it, put one over the ). The numerator
ÿT will be
with
ÿS
IN = T expressible as one of the specific heats.
fics (or
CP
or Cv );
4. Bring the volume to the numerator. The remaining derivatives will be expressible in terms of
a
It is
.
Mr. T
5. The original derivative has now been expressed in terms of the four quantities . Eliminate using the
equation
Cv
CP Cv a
Mr. T
,
,
It is
Cv
=
CP
Tvÿ
ÿ
2
/kT.
7.2 Chapter exercises:
Exercise 127
(7.2-1) we know that
in =
It is
we want to calculate is
in 0 +
a (T
ÿ
T 0) + b ( P
a heat transfer
ÿ
P 0)
Q system
constant
fortemperatureT
s and at
changed by a small increment dv Now, Q dS to calculate= in
=
T we must
temperature.
calculate That
as dsisconstant
0
in
ÿ
0
volume molar
.
,
with relates
to dv à
done simply like
dS = ÿ
ÿS
ÿ
ÿV
TdV.
However, we do not know how S varies with V equation of S.
, simply because we don't have
a
a relationship between v, T e Q, so that
We have, however,
0
Machine Translated by Google
7.2. CHAPTER EXERCISES:
as relations of Maxwell to modify
a relation of S with
It is P .
We have like this, using aspreviously studied techniques, which
we can try to use
V for some betweenV
ÿ
ÿ
129
P
S
T
ÿS
m
N
IN
ÿ
ÿÿÿÿ
ÿP
ÿ
ÿV
so that
ÿ
ÿT
=ÿÿ
T
IN
ÿP
dS
ÿ
ÿT
=ÿ
VdV.
To calculate the partial derivative above, just implicitly differentiate
original, remembering that V is being held constant, that is,
=
ÿv
a
equation
ÿv 0 + aÿT + bÿP,
giving
ÿP
ÿ
a
ÿ
ÿT
=ÿ
.
b
IN
Thus, we are left with
=
Q
Exercise 128
0
aTdV. b
=ÿ
(7.2-3) We know that
a =1
to the
T dS
IN
ÿ
=
ÿVT ÿ
ÿ1
T
P
same time as
=
CP
T
ÿS
ÿ
N
ÿ
ÿT
.
P
We want to show that
ÿCP
ÿ
ÿ
1
=
ÿ
ÿP
ÿÿ
N
T
ÿP
ÿ
T
ÿS
=0
ÿ
ÿT
P
.
T
we use
S
T
ÿ
ÿ
IN
P
m
N
ÿS
ÿÿÿ
ÿV
ÿ
ÿP
=ÿÿ
T
ÿT
ÿ
.
P
So that
=
T
ÿCPÿP
or yet,
ÿ
ÿ
1
N
ÿP
ÿ
ÿ
P ÿ
T
1
N
1
N ÿÿ
ÿÿ ÿÿÿ
1
N
as if
1
N
=
ÿCÿP
=
P
ÿT
ÿ
ÿ
ÿS
T
ÿT
ÿ
ÿT
ÿÿ
wanted to demonstrate.
ÿ
ÿ ÿÿ ÿ
ÿT
T
ÿÿ
T
ÿ
()ÿ
ÿT
T
1
ÿ
ÿT S
N
ÿP
ÿT
ÿ
ÿ
ÿÿ
ÿT S
ÿP
T
ÿS ÿÿÿ
ÿP
T
P
ÿV ÿÿÿ
+
ÿT
P
P
1 ÿ
IN P + N
ÿ
ÿÿ
1 ÿ ÿS ÿ
N
ÿP
T
ÿS
=
ÿP
T
1 ÿ ÿV ÿÿ
=
N
ÿT
P
ÿV ÿ
=
0
ÿT
P
ÿ
P
1
N
ÿ
S
P
ÿ
=
T
ÿ
ÿÿ
Machine Translated by Google
130
CHAPTER 7. MAXWELL'S RELATIONS:
as
(7.3-1) Derive
Exercise 129
equations:
TdS
=
NCvdT
TdS
=
NCP dT
+ ( T/a K T ) dv
a
TVdP
ÿ
to derive the
ÿS
first, we write
TdS
= T
ÿ
+ T
ÿ
ÿT
ÿS
ÿ
ÿ
ÿV
VdT
TdV
where
ÿ
ÿ
S
T
ÿ
P
IN
m
N
ÿS
ÿÿÿÿ
ÿP
ÿ
ÿV
ÿV/ÿT
=
ÿ
ÿT
T =ÿÿ
IN
INa
P =
((ÿV/ÿP
))T
of
so that TOS
=
+ TTdS
NCvdT
T
=
ÿS
ÿ
ÿ
ÿT
+
VdT
Ta
ÿTdV.
To the second equation, we have
TdS
=
T
ÿS
ÿ
+ T
ÿ
ÿT
PdT
S
T
IN
P
ÿ
ÿS
=
ÿ
ÿP
+ T
NCPdT
ÿ
ÿS
ÿ
ÿP
TdP
TdP.
Now
ÿ
ÿ
m
N
ÿS
ÿÿÿ
ÿV
ÿ
ÿP
T
=ÿÿ
ÿT
Va
=
ÿ
P
so that
TdS
=
NCP dT
TV a dP,
ÿ
as wished.
(7.3-2) We know from the previous problem that
Exercise 130
TdS
=
NCP dT
ÿ
a
TVdP
so is
T
ÿ
ÿS
ÿT
=
ÿ
NCP
ÿ
TV a
ÿ
in
or yet
NCv
=
ÿP
IN
Va
NCP + TV a
ÿ
=
NCP
V ÿT
Tvÿ
2
ÿ
,
Mr. T
as wished.
P
))T
T Va2
ÿ
Mr. T
which, dividing by No, provides
Cv = CP
ÿV/ÿT
((ÿV/ÿP
NCP + TV a
=
ÿ
ÿT
a
=ÿ
V ÿT
ÿ
Mr. T
Machine Translated by Google
7.2. CHAPTER EXERCISES:
131
(7.3-3) Standard
calculations C P, T ÿT, e a,
ÿ
ÿH
=
ÿ
ÿV
derivative ( ÿH/ÿV of) T,N
P. in terms of quan
We have to
a
Exercise 131
T
ÿS
ÿ
T,N
+ IN
ÿ
ÿV
ÿP
ÿ
ÿ
.
ÿV
T,N
T,N
The first derivative can be reduced using
S
T
ÿ
ÿ
It is
ÿ
P
IN
m
N
ÿS
ÿÿÿÿ
ÿP
ÿ
ÿV
T,N
ÿ
ÿT
=ÿÿ
V,N
this last one can be reduced using
ÿ
ÿP
ÿÿ
=
ÿ
ÿT
ÿ
V,N
ÿV
ÿT
P,N
ÿV ÿÿ
ÿP
T,N
a
Va
=
=ÿ
.
V ÿT
ÿ
Mr. T
We are like this with
ÿH
ÿ
It is
a
we can reduce
=
ÿ
ÿV
a
T
+ IN
ÿP
ÿ
T,N
ÿ
ÿV
ÿT
T,N
last derivative easily, using simply that
ÿ
ÿP
ÿV
)ÿ1
ÿV
= (ÿ V ÿT
ÿP ÿÿ1
T,N
ÿ
T,N
=ÿ
.
Finally, we have
ÿH
ÿ
= ( Ta ÿ 1) /kT.
ÿ
ÿV
Exercise 132
T,N
a
(7.3-4) We want to reduce
ÿ
derivative
ÿ
ÿvÿs
P
which can be reduced by doing
ÿ
T
in
s
P
ÿT
ÿ
ÿÿÿÿ
ÿ
ÿvÿs
ÿ
ÿs
P =ÿ
P
=1 /
ÿ
=
ÿ
ÿsÿT
so that
ÿ
Exercise 133
=
ÿ
ÿvÿs
P
T
.
CP
(7.3-6) We want to reduce
ÿ
=
ÿ
ÿsÿf
P
1
( ÿf/ÿs
1
=
)P
[ÿ S ( ÿT/ÿs
)P
ÿ
P ( ÿV/ÿs
T
CP
P
)] P
Machine Translated by Google
132
CHAPTER 7. MAXWELL'S RELATIONS:
and,
therefore,
1
ÿ
=ÿ
ÿ
ÿsÿf
ST/C
P
.
+ P ÿV/ÿs
(
P
)P
to reduce the
as relations of
last derivative in the denominator, we could use
a derivative. so we do
Maxwell, but they don't reduce
( ÿV/ÿs
) P = ( ÿV/ÿT
) P ( ÿT/ÿs
= IN T a
/CP
)P
so we are left with
1
ÿ
ÿsÿf
1
=ÿ
ÿ
=ÿ
ST/C
P
+ PVT a /C
P
a derivative
(7.3-7) We want to reduce
Exercise 134
ÿ
ÿ
ÿh
ÿ
ÿvÿhÿ
ÿs
ÿ
s
=ÿ
ÿ
ÿsÿv
.
CPT S + PV a
P
h
.
in
The derivativeof the numerator can be written
ÿh
ÿ
ÿv
derivatives of the numerator
AceMaxwell
of
(try it!), so we use
P
= in
ÿ
ÿv
s
ÿ
ÿ
ÿP
= in ÿ
ÿ
.
s
ÿsÿvÿsÿP
in
ÿ
ÿT
ÿ
ÿ
ÿ
ÿsÿv
ÿ
denominator cannot be reduced by relations
It is
ÿsÿT
=ÿ
P
ÿ
CP
=
ÿ
ÿv
P
vÿT
P
It is
ÿ
ÿ
ÿsÿP
=ÿ
ÿsÿT
ÿT
ÿP
ÿ
ÿ
in
in
C in
T
ÿ
ÿ
C in
T
=
in
T
vÿ
=ÿ
ÿT
ÿP
ÿ
ÿ
( Ta
CvÿT/
va
C in ( ÿv
ÿP )T
T
=
in
=
( ÿv
ÿT )P
)
so that
P
ÿh
ÿ
ÿv
CvÿT
= in
ÿ
ÿ
s
ÿh
( Tds
=
ÿs
+ vdP
)
ÿh
ÿs
in
ÿs
in
ÿ
ÿ
in
ÿ
ÿ
in
= T +
ÿ
ÿP
= T + in
ÿs
in
ÿ
ÿ
.
CvÿT
)
a derivative
We still have to solve
and,
therefore,
CP
=
( Ta
CvÿT/
=
( ÿs/ÿP
) in
vT ÿ
T +
ÿ
= T
CvÿT
We are like this with
ÿ
CP
=
ÿ
ÿsÿv
h
T ÿTCv
(
ÿ
ÿ
ÿÿ 1 ÿ vá ÿTCv
va
)
.
Machine Translated by Google
7.2. CHAPTER EXERCISES:
133
from where can we use
=
Cv
CP
Tvÿ 2 /kT
ÿ
to finally get
CP
=
h
ÿsÿv
ÿ
ÿ
T ÿT
[
( CP
CP
=
Tvÿ 2 /ÿT ) ÿ va
ÿ
.
T ÿT
(
ÿ
]
CP
Tvÿ
ÿ
ÿ
Exercise 135
(7.4-2) We have to
ÿP
in ÿ
=ÿ
ÿS
ÿP
IN
=ÿ
ÿ
ÿv
ÿ
ÿ
ÿV
s
.
S,N
Like this,
IN
=ÿ
ÿS
ÿ
ÿP
ÿ
=
ÿ
ÿV
ÿ
S,N
ÿS
ÿV
P
;
ÿS ÿÿ
ÿP
IN
according to some results of problem (7.3-7), we have
ÿS
ÿ
ÿV
CP
vÿT
=
ÿ
P
ÿS
ÿ
,
=ÿ
ÿ
ÿP
( Ta
CvÿT/
)
IN
so we are left with
=
ÿS
CP
vCvÿT
ÿ
.
So, we have to
ÿ
=
vÿ SÿT
=
kTkS
CPCv
giving
1
ÿS
Exercise 136
.
vÿS
(7.4-3) We must, in order to
dT
For a gas
=ÿ
ÿT
= ÿ ÿV
free expansion,
P
dv
ÿ
a
Ta
ÿ
ÿ
= ÿ NCv
U,N
simple ideal, we have (same number
=
Ns 0 + No.
moles)
c
T
S
IN
T
IN
ÿ
ln ÿÿ
This implies that
ÿS
ÿ
ÿT
=
ÿ
0 ÿÿ
0ÿ
cNR
T
in
so that
=
CvT
cNR
T
ÿ
Cv
dv.
NCvÿT
=
cNR.
.
2ÿ
va
)
Machine Translated by Google
134
CHAPTER 7. MAXWELL'S RELATIONS:
For another side, we have
P
S
ÿ
ÿ
T
ÿ
ÿS
ÿ
ÿÿÿÿÿ
IN
ÿ
ÿP
ÿ
ÿV
=ÿÿ
T
=
ÿ
ÿT
ÿ
IN
ÿV
ÿT
P
ÿV ÿÿ
ÿP
T
INa
=
a
=ÿ
,
V ÿT
ÿ
Mr. T
so we are left with
ÿS
ÿ
No.
=
ÿ
ÿV
a
=
IN
T
or it is,
ÿV
=
Mr. T
Mr. T
.
No.
Substituting these results into the initial expression, we get
P
cN 2 R
dT = ÿ
Ta
cN 2 R (a V/NR
ÿ
T
dv
)ÿ
=ÿ
T
ÿ
cNV
cNV
ÿ
dv
=0
.
Exercise 137 1%
in
.
O volume of a system adiabatically
(7.4-5) The variation of
we must find the
the chemical potential was reduced. We have, therefore,
what
ÿµ
ÿ
ÿV
dµ = ÿ
a starting from
The derivative can be reduced
ÿµ
ÿ
ÿ
ÿV
= ÿÿ
S
+ V/NdP
S/NÿT
ÿV
=ÿ
ÿ
ÿV
S,N
dv.
S,N
ÿ
N
S,N
ÿT
+
ÿ
S,N
Such derivatives need to be further reduced:
ÿ
ÿV
ÿ
=ÿ
ÿ
ÿT
ÿ
S,N
ÿS
ÿT
V,N
ÿS ÿÿ
ÿV
T,N
Cv/T
=ÿ
ÿ
ÿS ÿ
ÿV
T,N
where, using
P
S
ÿ
ÿ
It is
T
ÿ
ÿS
ÿ
ÿÿÿÿ
IN
yet
ÿP
=
ÿT
so that
ÿ ÿÿ
V,N
ÿ
ÿV
ÿV
ÿP
T,N
ÿV
ÿT
ÿÿP,N
ÿP
ÿ
=ÿ
T,N
ÿ
=
V ÿT
ÿ
ÿT
V,N
=
;
ÿV
ÿ
Mrs
ÿ
ÿV
ÿ
ÿT
=ÿ
ÿ
S,N
Now
ÿ
ÿP
ÿ
ÿV
=ÿ
ÿ
ÿ
S,N
ÿCv
.
ÿTT
ÿS
ÿV
P.N
ÿS ÿÿ
ÿP
V,N
=
CP
.
INCvÿT
IN
N
ÿP
ÿ
ÿV
.
ÿ
S,N
Machine Translated by Google
7.2. CHAPTER EXERCISES:
135
Stay with
ÿµ
ÿV
ÿ
S ÿCv
=
ÿ
S,N
IN
+
N ÿTT
CP
N INCvÿT
so that
ÿSCv
1
=
dµ
C
+
T
NÿT
dv.
PCv
ÿ
Exercise 138
ÿ
(7.4-8) We have to
ÿ
ÿCP
ÿ
ÿ
ÿ
N
T
ÿ
.
ÿT
P
P constant, we are left with
T
=
ÿ
ÿP
ÿS
ÿ
N
to how
T
Differentiating with respect
T
=
CP
ÿP
ÿS
ÿ
P
T
=
ÿ
ÿT
ÿ
ÿS
ÿT
ÿP
ÿÿ
N
T
ÿ
ÿ
T
P
which can be written as
ÿCP
ÿ
=
ÿ
ÿP
The derivative ( ÿS/ÿP
ÿ
ÿ
T
T
ÿ
ÿS
ÿT
ÿP
ÿÿ
N
IN
P
ÿ
ÿ
T
.
P
Maxwell
a relation of
) T can be reduced with
S
T
ÿ
V a.
=ÿ
ÿSPÿ T
ÿÿÿÿ
ÿVT ÿ P
=ÿÿ
So we have
ÿ
ÿCP
ÿP
=
ÿ
T
ÿ
N
T
ÿ (ÿ V a
ÿT
)
T
ÿ
ÿV
ÿÿa
=ÿ
N
P
+ IN
ÿ
ÿT
ÿ
what from the
ÿCP
ÿ
ÿ
Tv
=ÿ
ÿP
a
ÿÿ
2
ÿ
+ ÿ ÿT
T
ÿ
.
P
So, having to
P
ÿ in
A
+
P constant, we arrive
we have, first, that, keeping
a
=
2A
1
P
ÿvÿT
in
ÿ
so that
RT
ÿ2 ÿ =
T
P
ÿvÿT
ÿ
ÿ
P
ÿÿ
ÿ
va ÿ 2 A/T 3ÿ = R
or
from where with can
calculate
ÿ
ÿ
a
=
R
2A
+
PvvT
a desired ratio.
=
3
P
3
T
ÿ
a
R
ÿÿ
ÿT
P
ÿ
ÿ
P
Machine Translated by Google
136
CHAPTER 7. MAXWELL'S RELATIONS:
Machine Translated by Google
Chapter 8
Systems Stability
Thermodynamics:
8.1 Fourteenth Class (06/18/2008):
8.1.1 Intrinsic stability of thermodynamic systems:
The basic principle of thermodynamics says that
2
d S<
dS = 0 ,
(8.1)
0
so that the equilibrium states are those represented by a maximum entropy. So far we have
only considered the condition = 0. However, the second dS
condition, which says that it is, in fact, a maximum, is associated with
to the notion of stability
equilibrium systems. We intend, therefore, to consider the second inequality
above to investigate the conditions under which a thermodynamic system is stable.
Consider two identical systems that have the same fundamental equation given by
2
S = ( NOT)1/exp
ÿÿ
bUV + 05. IN2
N2
(8.2)
ÿ
which is not a valid fundamental thermodynamic equation. This function has a graph like the
one shown in figure 8.1. Note that if we remove a
amount of energy ÿ from one IN
of the subsystems and take it to the other, then
S ()U,V,N
which was worth 2ÿ
would now be worth the entropy,
S ( IN + ÿ U,V,N ) + S ( IN ÿ ÿ U,V,N
)
which, by the graph, is greater than zero! Thus, in the passage of energy from a
subsystem to identical
the other for the latter there would be an entropy increase,
implying that the whole system could not be in equilibrium. But the subsystems, being
identical, can be considered as a single system and not
137
Machine Translated by Google
138 CHAPTER 8. STABILITY OF THERMODYNAMIC SYSTEMS:
Figure 8.1: Graph of entropy in terms of energy (para = 5) of the
N = 1, IN equation presented in the text.
N
IN constant
It is
it makes sense to say, at the same time, that the system was in thermodynamic equilibrium
and that entropy increased in a process that should maintain the state of equilibrium. In fact,
these internal inhomogeneities are a basic characteristic of phase transitions and precisely
identify non-equilibrium states. Equation (8.2) is a fundamental thermodynamic equation
it is notfor which
precisely because it presents this region
S ( IN + ÿ U,V,N
) + S ( IN ÿ ÿ U,V,N
)
> 2 S ( U,V,N
)
,
precisely because such a relationship would imply
[S ( IN + ÿ
U,V,N
or (dividing by ÿ
) ÿ S ( U,V,N
)] ÿ [ S ( U,V,N
IN2 and taking ÿ
) ÿ S ( IN ÿ ÿ U,V,N
)]
>
0
IN ÿ 0)
d2 S > 0,
and not the correct expression for the thermodynamic equilibrium state presented
in (8.1).
Now, the condition that must be satisfied, therefore, is
S ( IN + ÿ U,V,N
) + S ( IN ÿ ÿ U,V,N
which, in differential terms, implies
ÿ 2S
ÿ
ÿU 2 ÿ
V,N
ÿ0
.
) ÿ 2 S ( U,V,N
)
(8.3)
Machine Translated by Google
139
8.1. FOURTEENTH CLASS (06/18/2008):
Figure 8.2: Construction of a thermodynamically valid fundamental equation from an equation that violates
the thermodynamic equilibrium condition.
Note, however, that this condition is less restrictive than condition (8.3), which must hold for any ÿ ÿ 0.
IN instead of only valid for ÿ
IN
extensive
variables
The same condition must hold for all other thermodynamics (we will talk about
the intensive
ones next, but it
is clear that there must be similar conditions, although not identical, since we know that equilibrium
conditions exist for the other thermodynamic potentials that can, we know, be expressed in terms of one or
more intensive variables.)
From this fundamental equation (8.2), which is not stable at all points, a stable thermodynamic
fundamental equation can be obtained by a construction like the one shown in figure 8.2, where the tangents
to the curve are taken to construct a new curve ( envelope of the first) for which the condition (8.3) always
applies. It is interesting to note that, in Figure 8.2, only the hatched region implies
non-observance of the principle of maximum entropy in the differential form, while the entire region that is below
the straight line constructed from the
tangent to the curve violates the non-differential (or global) condition. The region between the figure is
It is
B
,
said to be locally stable but globally unstable. of the fundamental relation (of the straight
A
A point in the portion between
It is
C
line) corresponds to a separation of phases in which
part of the system is in the state as will be seen in detail in the next chapter.
A
and part of it is in the state
C
,
of
Example 139 To the van's fundamental fluid equation fundamental
a thermodynamic equation
gas of hydrogen molecules, constructively acceptable
astext. In this case we have
according to specifications seen in the
waals for one
A
Machine Translated by Google
140 CHAPTER 8. STABILITY OF THERMODYNAMIC SYSTEMS:
Figure 8.3: Envelope construction for a van der Waals fluid representing a gas of
H 2 with in = 130.
a equation
s = R ln[( in + v/a
where, for molecules of
thus, fixing
c = 25.
.
s ()in= 8
in
c
( in
)
H 2 worth the values
= 130 , we find
.
314ln ÿ (130 +
ÿ
b)]
.
a = 00248
,
b = 26 .6 × 10ÿ6
,
a function
in/ 0 .0248)2
.5 ÿ
in ÿ 26
.
6 × 10ÿ6ÿÿ
In the figure we also show a grafted (no
whose graph is shown in figure 8.3.
O from
behavior
frame) second derivative
of this
function for
s
indicate on which interval
its second derivative so that we
to have
must
with
()in
makes it greater than zero. No
2
ÿ s
ÿ
ÿv
ÿ0
2ÿ
,
in
we come to the conclusion that it Ois interval
a thermodynamically
, 5 × 10ÿ4ÿ interval for The
ÿ8 × 10ÿ5unacceptable
a function.
construction by envelopes generates a curve
represented by a darker (black) line.
a
Machine Translated by Google
141
8.1. FOURTEENTH CLASS (06/18/2008):
Evidently, for the general case in 3 dimensions (in which we let
IN as
IN ) the requirement that must be met is given by
so much
S ( IN + ÿ U,V + ÿ IN, N ) + S ( IN ÿ ÿ U,V ÿ ÿ
IN, N ) ÿ 2 S ( U,V,N
)
which, in differential form, is now given by
2
2
2
ÿ S ÿ S
ÿU
2
ÿÿ
2
ÿV
ÿ S
.
ÿUÿV
ÿÿ0
This last condition is more general than simply requiring that
2
2
ÿ S
ÿ
ÿ0
ÿU 2 ÿ V,N
,
ÿ S
ÿ
ÿV 2 ÿ U,N
ÿ0
,
as it also prevents the
IN
IN there is violation of
out of coordinate axes
principle of maximization (known as ”fluting”).
In physical terms, local stability conditions ensure that inhomogeneities in extensive
variables do not increase entropy. In the section that
Next, we will deal with these same conditions for other thermodynamic potentials.
It is
Before continuing our exposition, it is interesting to note that these conditions that we
have been studying imply restrictions on the signs of the functions that characterize
the thermodynamic system (thermal capacity, compressibility, etc.). Indeed, we can easily
show that
2
ÿ
ÿ S
ÿT
1
=ÿ
T 2 ÿ ÿU
ÿU 2 ÿ V,N
1
ÿ
=ÿ
NT
V,N
2
Cv
ÿ0
,
indicating that the molar capacity at constant volume must be positive in a
stable system (analogous conditions valid for other physical observables).
8.1.2 Stability conditions for thermodynamic potentials:
The passage to the valid conditions for the other thermodynamic potentials is straightforward.
In the case of energy, for example, we know that it must be a minimum, so it must be
IN ( S + ÿ S, IN + ÿ IN, N ) + IN ( S ÿ ÿ S, IN ÿ ÿ
IN, N ) ÿ 2 IN ( S,V,N
)
,
where, of course, we must respect the variables in terms of which said potential is written.
The local convexity conditions become
2
ÿ IN
ÿS
2
ÿT
=
2
ÿ IN
ÿ0
ÿS
,
2
or, in the general case, for joint variations of
2
ÿS
2
ÿV
2
ÿ0
ÿV
S
It is
IN
2
2
ÿ IN ÿ IN
ÿP
=ÿ
ÿV
ÿÿ
ÿ IN
.
ÿSÿV
ÿ2 ÿ 0
Machine Translated by Google
142 CHAPTER 8. STABILITY OF THERMODYNAMIC SYSTEMS:
This passage to the representation of energy is interesting because, in
our developments, we define all thermodynamic potentials in terms- ], we have
F [ T,V,N energy
terms. So, for the case of
ÿF
S
so that
2
ÿ IN
ÿS
1
=
( ÿS/ÿT
2
ÿ(
)
,
ÿ F/ÿT
2)
2
2
ÿ F
ÿ IN
ÿ
ÿS
1
=
ÿS
indicating that if
ÿU
=
T
,
ÿT
ÿT
=
2
=ÿ
ÿS 2 ÿ V,N
ÿ0ÿÿ
ÿ0
ÿT 2 ÿ V,N
while we remain with
2
ÿ
ÿ F
ÿ0
ÿV 2 ÿ T,N
,
F does not change the relationship with IN
once the function
The same method applied to the other thermodynamic potentials implies that
.
2
ÿ IN
ÿS
2
ÿT
=
1
=
ÿS
1
=
( ÿS/ÿT
ÿ(
)
2
ÿ G/ÿT
2)
It is
2
ÿ IN
2
ÿV
ÿP
ÿV
=ÿ
so that
1
=ÿ
=
( ÿV/ÿP
1
2
ÿ(
ÿ G/ÿP
2)
2
2
ÿ
)
ÿ G
ÿT 2 ÿ P,N
ÿ0
,
ÿ
ÿ G
ÿP 2 ÿ T,N
ÿ0
.
If we remember the convention by which we represent the various potentials in the
form (for e.g.) G
,
G
ÿ
=ÿ
S
T
IN
P
m ÿ
N
,
so we get the rules for the second derivative right away, just by looking
the sign of the second line. In fact, where the sign is different from that found in the
energy representation, the relation in the second derivative changes the sign, where
is equal, it keeps. So, immediately,
H
=
ÿ
immediately implies that
T
IN
m
S
P
N
ÿ
ÿ H
ÿS 2 ÿ P,N
ÿ
2
2
ÿ0
,
ÿ
ÿ H
ÿP 2 ÿ S,N
ÿ0
.
Another way to remember these results is simply to note that:
thermodynamic potentials are convex functions of their extensive variables and
concave functions of their intensive variables
.
you
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143
8.1. FOURTEENTH CLASS (06/18/2008):
8.1.3 Physical consequences of stability:
We can now return to the issue involving the constraints that the conditions of
a
Mr.
T appeal to the different
Cv CP obtained
stability impose on quantities. These conditions can be immediately
if we
,
It is
,
representations via thermodynamic potentials. In fact, remembering that
ÿV
1
=ÿ
Mr. T
IN
ÿ
ÿ
ÿP
T
and knowing that the representation in which the partial derivative above can appear
T
is easily the one where we have a function of ), we write
ÿ
ÿ 2F
ÿP
ÿV 2 ÿ T = ÿ ÿ
ÿV
IN
(It is
N
1
=
ÿ
,
V ÿT
T
ÿ0
,
so that we get
kT > 0
.
Other similar identities can be obtained for any coefficient
CPforathis to
Mr.use
T
physical, not only being Cv
enough
the appropriate thermodynamic potential, with, however,
,
,
It is
,
overlapping results. To theyes, for example
ÿ 2G
ÿ
ÿV
ÿ
ÿP 2 ÿ T = ÿ ÿP
V ÿT ÿ 0
=ÿ
T
so that again we get
Mr. T
ÿ0
.
ÿ
Cv
ÿ0
It is possible to show, in fact, that
CP
is that
ÿS
Mr. T ÿ
ÿ0
,
just use the relations
CP
Cv
ÿ
=
Tvÿ
,
Mr. T
because we know that
Mr. T
is positive, and yet
=
Mr. T
2
CvCP .
,
Machine Translated by Google
144 CHAPTER 8. STABILITY OF THERMODYNAMIC SYSTEMS:
8.1.4 Le Chatelier's principle:
The subject of second derivatives allows us to access a principle, called Le Chatelier's principle,
which provides the physical content of the concept of stability of thermodynamic systems. Le
Chatelier's principle says the following:
Theorem 140
Any inhomogeneity that somehow develops in
a
a system must induce a process that tends to eradicate this inhomogeneity.
with
In a container with fluid in equilibrium, if a certain region of the
system is heated in some way, there is local heating that works
as an inhomogeneity. Le Chatelier's principle then guarantees that there will be a process of heat
flow between this warmer region and the colder regions in order to homogenize the temperature of
the system as a whole.
The variable that guarantees this process is the specific heat, which is positive.
In the same way, the appearance of a longitudinal wave in a fluid
induces regions with higher and lower densities. Regions with higher density, and therefore greater
pressure, tend to expand, while regions of lower density, with less pressure, tend to contract; the
positivity of the compressibility guarantees that the system responds to the appearance of the wave
in order to make the pressure of the system to homogenize again.
In fact, apart from external causes, any physical system will always
has local inhomogeneities. In fact, from the point of view of statistical mechanics, all systems
continuously develop local fluctuations.
The state of equilibrium, static from the point of view of classical thermodynamics, is incessantly
dynamic. In this sense, thermodynamics always deals with
with variables that represent average values (which is the concept correlated to
"macroscopic").
8.1.5 The Le Chatelier-Braun principle:
In addition to changes in the system that are immediately caused by the appearance of
inhomogeneity, there are also other changes, only indirectly.
caused, which also contribute to the resumption of balance. This is, from
Indeed, the content of Le Chatelier-Braun's principle:
Theorem 141
with
Any inhomogeneity that somehow
develop into
a eradicate this
A system must directly induce a process that tends to
ianeormraodgiecnaerideastdaeienotmamogbeénme, iddeir.etly, other processes that also tend to
Proof. To demonstrate this principle, consider a spontaneous fluctuation occurring in a
fdX 1 composite system. Directly, there will be a change in the intensive parameter so that
P 1, linked to X 1
dP 1f
=
1
ÿPfdX
ÿX 1
1;
(8.4)
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145
8.1. FOURTEENTH CLASS (06/18/2008):
P
the variation since such fdX 1 will also change the intensive parameter 2 (one however,
parameters also depend, in general, on the variable 1.) Thus, we will also haveX
2
ÿPfdX
ÿX 1
f =
dP
2
(8.5)
.
1
r
Then, the system
appear,
which
we constant
will call j signals
here are
determined
by theresponses
minimization
of energy
(with
total entropy).
Like this,
d ( IN + INres ) = ( P 1 ÿ
P res
dX
)
r
P 2res)
+ ( P2 ÿ
1
dX
r
2
dX i
with-
ÿ0
so that
dP
As
Dr
1 and
Dr
2
f
1
dX
r
1
r
+ dP 2f dX
ÿ0
2
.
are independent, we have
dP
f
1
Dr
1
f
Dr
ÿ0
,
dP
ÿ0
,
2
ÿPfdX
ÿX 1
2
ÿ0
2
.
Using (8.4) and (8.5) we get
1
ÿPfdX
ÿX 1
r
dX
1
1
1
dX
r
2
ÿ0
.
The first inequality implies Le Chatelier's principle, since it multiplies
1
ÿP/ÿX
Typing
the equation
convexivity,
we get by
1
,
dP
f
1
which is positive because of the criterion of
r
dP (1) ÿ 01
,
indicating that the system response tends to make a change in the parameter
intensive that is opposite to the one that caused the inhomogeneity. The second inequality
can be rewritten using Maxwell's relation
2
1
=
ÿPÿX
1
ÿPÿX
2
in the form
ÿP
fdX
1
ÿX 2 ÿ
1ÿ
dX
r
2
ÿ0
,
ÿP/ÿX
1
so that, multiplying by the positive quantity
1
ÿPfdX
1
ÿPdXr
1
or
ÿX
1
ÿÿ
ÿ
dP
f
1
ÿX
1, we arrive at
0
2
2
r
dP 1 (2) ÿ 0
ÿÿ
,
which shows that the change in the parameter P 1 due to system response with
with respect to other variables also occurs in the sense of eliminating inhomogeneities. It
can be easily shown that we must also have
dP 2f
r
dP (2) ÿ 02
,
Machine Translated by Google
146 CHAPTER 8. STABILITY OF THERMODYNAMIC SYSTEMS:
P2, referring to the other variables,
so that the parameters will also be
changes that will tend to eliminate the inhomogeneity that was indirectly
caused in them by the variation fdX1 .
Machine Translated by Google
Chapter 9
Phase Transitions from
First Order:
9.1 Fifteenth Class (06/23/2008):
9.1.1 One-component systems:
A phase transition takes place in a thermodynamic system when, depending on the variation of one
of its pertinent variables, its properties are abruptly modified. Thus, the water solidifies below the
its structural properties
.
K
and it vaporizes above temperature 373 changing.15
temperature
27315 so
K
,
are evident.
Each phase transition can be seen as represented by the linear region shown in figure ??
BHF
of the previous chapter. There, we saw that the fundamental equation was not thermodynamically
acceptable precisely because it had a
region in which the principle of maximum entropy was not valid. The solution
was to build a thermodynamically acceptable fundamental equation by the tangent envelopes.
thermomechanical in which there may be a phase
As an example of system
IN of different materials
transition, we can consider an inverted-shaped pipe in which two metallic spheres
and, therefore, of different coefficients of thermal expansion are placed at its ends. The barrel also
has a piston capable of sliding due to the thermodynamic properties of the system. At temperatures
the sphere on the right is larger, pushing the piston to the left; this piston descends a height and is
T gas
> Tc
balanced by the pressure of the
contained in the pipe below it in the left part. At temperatures
however, the sphere on the left is larger, and pushes the piston
Heto the right; it descends to a height
and is balanced by the pressure of the gas contained in the pipe
Hd
below the piston. We therefore have two equilibrium situations that depend on
system temperature (a thermodynamic parameter).
If the two spheres are different in size, then the two minimum points
147
T Tc<
,
Machine Translated by Google
148
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
are located at different heights. In fact, the smallest minimum point will be on the side of the smallest
sphere. These two minimum points can be represented in the Helmholtz potential, for example,
remembering that the potential energy
gravitational must be represented in thermodynamics.IN as well as the terms themselves
,
or if you are a small person, you will be loaded with turmocaatdraenusmiçãloaddoe pfaas readopur timroediroa coir lidnedmro,
temperature-induced (but could be any change in some other thermodynamic parameter). The two
thermodynamic states between which there is a first-order phase shift occur in separate regions of the
thermodynamic configuration space.
On the other hand, if we consider the same problem, but now with spheres
identical, then we would still have two stable equilibrium points, each one of a
side of the cylinder (taking the highest point as the center). the heights would be
identical. As we increase the temperature of the system, its pressure
also increases and the two equilibrium heights approach the highest (central) point until, above a certain
Tcr
temperature, the piston is at
in equilibrium only at the center point. If we now reverse the process, we see clearly that by decreasing
or increasing the temperature from a single equilibrium state into two symmetrical equilibrium
states. This one
Tcr
,
O
bifurcate
is an example of a second order phase transition and the temperature is called the critical
temperature. The states between which a second-second transition
order occurs are contiguous states in thermodynamic configuration space.
Figure 9.1: Thermodynamic potential with multiple minima.
Returning to the case of dissimilar spheres, it may happen that the piston is
at the left local minimum, which is a local-only minimum point, not
global (figure 9.1. After a while that the system remains in this state,
Tcr
Machine Translated by Google
9.1. FIFTEENTH CLASS (06/23/2008):
149
the occurrence of fluctuations can generate a fluctuation with sufficient amplitude to
cause the piston to overcome the gravitational potential barrier of the barrel
in
INand move to the other end of the pipe, assuming a state of global equilibrium.
At that point, the system will stay until a new fluctuation sends it back to the local
equilibrium point; however, the fluctuation in this case has to be very
mficaairar memuitaímssipmliotumdeais ,teamsspimo ,nampuoitsoiçmãoendoesmpírnoivmáovegll .obDalesdsoe
qmuoednoa, poos si içstãeomdae local minimum. Thermodynamics, as we have been studying
up to this point, is not concerned with the states referring to local minima, being interested only
in the system occupying its global minimum, since that is the state that it will assume in the
absolute majority of the time.
However, it is worth considering that in some systems there may be a barsuch a high potential line between the minimum points that fluctuations with
magnitude sufficient to break such a barrier will be too infrequent to statistically
expect the system to change state effectively. In this case, the system, even if it is
a local minimum, should
effectively be considered to be in a stable equilibrium (although, theoretically, it is a
metastable equilibrium).
A thermodynamic example could be the case of a vessel with steam
of water at a pressure of 1 atm and a temperature slightly above 37315 Taking.a K
small subsystem as a sphere of variable volume that, at any instant, contains one
gram of water, we have two subsystems (the
container and the 'bubble') in which the first serves as a thermal reservoir for the
second. The equilibrium condition is that the Gibbs potential, ) is a minimum.
G (T,P,N The
two variables that are fixed by the equilibrium conditions are
the free energy and volume IN
of the subsystem. In this subsystem there can be
fluctuations in volume (and they do occur continuously and spontaneously).
The slope of the curve that represents the thermodynamic potential in terms of the
extensive variable represents an intensive parameter (in this case the pressure,
since the extensive parameter is the volume), which acts as a restoring force that
takes the system back to its original homogeneity. density according to
Le Chatelier's law. Even so, there may be larger fluctuations that cause the
system to contract considerably in order to greatly increase its density; in these
cases, we have the appearance of small drops of liquid that live for a few moments
to soon disappear again.
If, in this system, we gradually decrease the temperature, the diagram of minima
of the Gibbs potential varies as shown in figure 9.2, in such a way 4 the two minima
T are equal. at this point we have a lock, at temperature
.
phase division, since both phases, represented by the different minimum points,
are equally possible and will coexist in the system. If the steam were
cooled very gradually, then the system could have remained at the minimum that
was once global, but which became local due to the change in temperature. The
system would then still be in the previous thermodynamic state. Being a point (now)
of metastable equilibrium, any spontaneous fluctuation (internally) or caused by
external disturbances will cause the system to overcome the energy barrier and
go to the new state of stable equilibrium. AND
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150
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
Figure 9.2: Schematic variation of the Gibbs potential with volume (or inverse density) for
T 1 T<T<2temperature
T(the
< <T3
4
various temperatures ( 5). Temperature 4 is the transition
minima coincide).
stop
TThe high density phase (left minimum) is stable below the transition temperature.
This is precisely what happens with liquids that we cool in the freezer of our refrigerators
and that, when we take them in our hands, still show up in black and white.
liquid; we clearly noticed that when touching the casing, the liquid starts
immediately to freeze, changing its state to the new equilibrium state
stable (solid). The same occurs in the liquid-vapour transition, but the velocity
with which the phenomenon occurs does not allow us to visualize it.
Figure 9.3: Gibbs potential minima as a function of temperature
T
.
This whole analysis in terms of the Gibbs potential is interesting because it allows us to
easily understand the issue of phase coexistence in the first-order phase transition process. In
such a process, the Gibbs potential has the form shown in figure 9.3, where its values are
shown.
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151
9.1. FIFTEENTH CLASS (06/23/2008):
ores for the most diverse phases. Note, however, that the "physical" Gibbs potential is the
one that is built with the minimum points presented there, so that its curve presents points in
which the phase is solid, passing later to liquid and then to gas, with successive points of
discontinuity in the derivative. At each of these points there is a phase transition, of course.
In addition,
also
be ait discontinuity
in the thermodynamic
parameters
represent
the slopethere
of themust
curve
(since
changes discontinuously).
Since the
graph is a that
graph,
versus
T first-order phase transition points. In
there must be a discontinuity in entropy at G
each
of these
,
fact, in general, there is always a discontinuity in all other thermodynamic potentials,
unless by coincidence, this very property being the one that defines a first-order transition.
Considering the graphs ??, we see that as we increase successively, at what
D
the temperature, we pass through the points
A,B,C
where there is coexistence of the phases that are separated by the curve in question.
However, on the way, looking at the figure on the right, we see that the minimum points
D
are collapsing one on top of the other, so that at the point there is only one minimum.
This
behavior is totally similar to the one already discussed for
the case of the inverted pipe, representing
IN
a second-order phase transition. The point
D is so called
critical point
It is
.
9.1.2 The entropy discontinuity: We saw in the previous
section that, on the phase change curves, there is always a discontinuity in the entropy
value, which is the slope of the curve that
G by temperature. About any point of a
represents the variation of
phase shift curve, both phases coexist and the two phases have
exactly the same value for the Gibbs potential.
In our example in question, water, if we have it in the region that represents
”ice” (solid), so we can gradually increase the temperature by introducing heat into the
system. When the temperature reaches that of melting ice, we are precisely on the line of
coexistence of the two phases. We can continue
putting heat into the system, but that heat should now melt ice and also
heat liquid water that already exists in the system. Inserted heat does not increase
the temperature and what we see is the appearance of more and more water in a state
liquid, maintained at the same temperature, until all the ice has melted
(and the system traversed its path along a line of coexistence, or a
solid-liquid
transitionthe
line).
Once all the ice has melted, more heat introduced into the
system
will phase
again increase
temperature.
temperature, until a new phase transition point is reached.
The amount of heat required to melt one mole of solid is called and is related to the
of latent heat of fusion
change in entropy between
the liquid and solid phases in the form
ÿLS =
T
( ) ÿ Ss(
ÿ Ls
)ÿ
,
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152
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
where T is the (constant) melting temperature (at a given pressure). In
in general, the latent heat in any first-order transition is given by
ÿ
=
Tÿ
s.
heat
If the phase transition is between the liquid and gas phases, then we have the
latent sublimation vaporization
,
if it is between the solid and gaseous phases, we have the
latent heat
.
It is important to note that the method by which the transition is induced is irrelevant–
latent heat is independent of this. We could heat ice under pressure
constant or increase pressure at constant temperature; anyway
we would have the same latent heat involved, that is, the same amount of heat would be withdrawn from the
thermal reservoir.
9.1.3 The slope of the coexistence curves:
The coexistence curves shown in figure ??, for example, are not as
arbitrary as it may seem; its slope, given by is totally determined by the two coexisting
phases and such slope
dP/dT
has great physical interest. Consider the four states shown in figure 9.4, so that the states are on the coexistence
curve but correspond to different phases at a different temperature.
A
It is
A
ÿ
. In the same way we have the states
FACING
B
It is
B
TB . There is a difference on
the curve
coexistence
but
at (such
a temperature
assumeeven
to beholding
infinitesimal
Pcurve,
ÿ
B
Well
the
is
as ÿ ) thewe
pressure
for the temperature
difference
slope of
PB ÿ The
Well
ÿ
TB
dP/dT
.
Figure 9.4: Four coexisting states.
Phase balancing now requires that
gA = gA
ÿ
,
gB = gB
ÿ
ÿ
FACING
=
dT
.
ÿ
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153
9.2. SIXTEENTH CLASS (06/25/2008):
so that
gB
=
gA
ÿ
gB
ÿÿ
gA
ÿ
.
But
gB
ÿ
sdT + vdP
=ÿ
gA
ÿ
at the same time as
gB
ÿÿ
gA
ÿ
s dT + in dP
ÿ=ÿ
so we arrive at the result
dP
ÿ
s
=
dT
s
ÿ
in
ÿ
ÿ s
=
ÿ in
in
ÿ
,
where ÿ and
s ÿ are
in discontinuities in molar entropy and molar volume associated with the phase transition.
As we have
ÿ
=
T ÿ s,
we arrived at the result
dP
1
=
ÿ
T ÿ s
dT
,
which is the Clapeyron equation. The Clapeyron equation includes Le Chatelier's principle.
9.2 Sixteenth Class (06/25/2008):
9.2.1 Unstable isotherms and first-order transitions:
So far we have focused our discussion on the Gibbs potential which, in fact, is a natural potential to
deal with the problem, given its characteristic of continuity in the passage between phases. However, a
description in terms of isotherms is much more common in thermodynamics. Thus, in this section, we
will deal with first-order phase transitions using the formalism of
unstable isotherms. Our main tool will be the van der Waals equation which, at least qualitatively,
represents very well the main characteristics of these transition processes – all semi-empirical equations
of state have similar behavior.
For a van der Waals fluid, we can write
RT
P
=
( in
a
b) ÿ
ÿ
in 2
(9.1)
,
H 2), we have
.
a = 00248,
=
T 28 5, .
so that, for the realistic case of the hydrogen molecule ( the curves shown in figure
9.5. The parameters used were: 6×10ÿ6 and the temperatures used in the various
.graphs were: = 26
b
T 2 = 295 . K T = 3053
. K
,
,
T
= 3154. K
,
T
= 3255. K
,
T
. K
= 3356
It is
T
. K
1 = 3457
1For simplicity of notation, we will not place the units for the variables, assuming them
known in each pass.
1
.
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154
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
Figure 9.5: van der Waals isotherms for the data given in the text.
These isotherms have a region of instability, where the second derivative conditions are not
valid, here represented by the equivalent condition
kT > 0, that is,
ÿP
ÿ
ÿV
0
ÿ
.
T<
If we consider just one of these isotherms in isolation, as in Fig.
9.6, we can see in more detail that the violation region is given by This violation of the stability F KM
FKM
principle implies that this portion of the
isotherm must be considered non-physical so there must be a phase transition in it.
The Gibbs potential is determined by the shape of the isotherm (less than
an arbitrary function on
T ), then
dg
=
dµ
=ÿ
sdT + vdP,
by the Gibbs-Duhem condition. Thus, integrating at constant temperature, we are left with
g =ÿ
in ()PdP
+
Phi()T
,
P it makes more
sense Phi()Tis the arbitrary function of temperature. Since () is the integrand,inwhere
to consider the curve shown in 9.6(b) than the curve
in × P
shown in 9.6(a). If we put an arbitrary value to
,
P ×
in the potential of we can
,
Gibbs (equal to the chemical potential) at some point, such as the point
A
,
.
Machine Translated by Google
155
9.2. SIXTEENTH CLASS (06/25/2008):
Figure 9.6: A particular isotherm of the van der Waals equation.
calculate the chemical potential
as in
B using
m at any other point of the same isotherm,
,
B
in ()PdP.
ÿ
B
µA
=ÿ A
Another possibility is to make a change of variables (or integration by parts) in order to
obtain
ÿ
B
µA
B
B
vP | A ÿ ÿ
=
P v()
DVD,
A
which can be more interesting depending on the form of the function
Consider the equation (9.1)
Example 142
spondent. We have to
RT
P =
b)
( in
against ao
equation
a
ÿ
,
in 2
ÿ
in order to write
in
3
ÿÿ
b+
RT
P
ÿ in
2
ab
+
ÿ
aP v
whose solutions are, putting
a
=
b+
b
=
c
=
RTP
bP
abP
P
=0
,
P ()inor
in
(P
in ().P
)
corre-
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156
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
It is
3
ab + 108 c + 8 a + 12ÿ b
X = ÿ36
ÿ 3 a 2b 2 ÿ 54 a B C + 81 c
3
2
+ 12 a 3c,
as following:
1/3
in2 =
in3 =
a
29
a
in 1 =
3+ X
3 ÿ 6 ÿ
(1+ i ÿ 3) 1/ 6
ÿ
x
3
12
a
(1ÿ i ÿ 3)
ÿ
x
12
3
ÿ
i
+ 3 ÿ1 ÿ
1/
3
a
b3 ÿ
X
ÿ1
b
ÿ 3 ÿÿ
/
3
a
ÿ
b
+ 3 ÿ1 ÿ i ÿ 3 ÿÿ 3
a
ÿ
ÿ1
/
3
ÿ1
/
3
X
29 ÿ
3
,
X
29 ÿ
we look at
that are, obviously three different equations, as is obvious from figure 9.6(b).
consider the
Example 143
of
equation (9.1)
isotherm shown in the figure
T K.= 305 .
Tick the
µ value
= 1 for the
a temperature
9.6(b) for
a
=
b+
.
=
= 00248
ÿ aP abP
P
=
= 0 .65968×10ÿ6
c
and the
results should be given by
ÿ
B
µA
=
P v()
ÿÿ
dv
A
Now we have to know what points are
× 10 Pa, which has associated
.
Well 7= 011
ÿ
and A
=
so, of course, just ).
Now to
O point
=
B
vP | A
To e B.
a she
ÿ
vB
RT
ln ÿ
ÿ
which also implies only one acceptable result. But then, for this
pressure, we will have
.
(012)
For a pressure given by
ÿ
µA
PB
=
.
= ÿ8600978443
. × 10 Pa, we are with
7= 009
ÿ
.
(009)
inB
and A
a B
b
b
ÿ
ÿÿ
.
in A
The point A is given by the volume
pressure
.
000004702626122
.
.
i
0000095
ÿ
.
.+ 0000050
i
00000500000095
ÿ
mB
ÿ
ÿ
O third result can be used pressure
(The =
we use a
P 7= 012
. × 10 ) we have
B (if
ÿ
.
(012)
inB
=
.
.
i
ÿ00042
ÿ
.
.+
i ,
00075ÿ00042
.
000750008790866116
ÿ
ÿ
.
00088
,
P
B
B
vP | A
.
2535770
P
= 0 .266 × 10ÿ4 +
RTP
point
P 10 7=
. 009 × Well It is
P 7= 011
. × 10 Calculations Pa.O chemical potential at points
P 7= 012
. × 10 Well. We have that our parameters are:
ÿ
ÿ
.
000005025853956
.
000008356120333
.
00001745324793
,
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157
9.2. SIXTEENTH CLASS (06/25/2008):
so we are left with three possible outcomes
.
mB(009)
.
mB(009)
.
(009)
mB
ÿ
µA
ÿ
µA
ÿ
µA
.
= ÿ8615511998
.
= ÿ8611684053
.
.
= ÿ8617125577
Thus, according to the previous example, we can consider the system in a
A contact with a pressure reservoir and a heat reservoir. If we increase the pressure of the
state and in
pressure vessel quasi-statically, so
T
to keep the temperature constant, so our system traverses the 305 isotherm in our example. For
. K less than
pressures
PB
,
=
we see that the volume of
the system (in this isotherm) is univalued. As the pressure increases
P toT the
above , however,
PB three states of the same but different make it accessible
,.
system. Assume
It is
in
,
with
N
C Lrepresent
the triplet of values shown in the graph; Of these three values, it is unstable, but both
It is
,
L
C
It is
,
N
minima of the Gibbs potential. Which of these two minimum points does the system
choose will depend on which one represents a global minimum of the Gibbs potential (or the chemical
potential) (for this temperature).
Exercise 144
Construct a Pµ curve for varyingfrom
P () P Pa Pa. O the previousofexample system table,
the steps
We fear
built
from the previous example:
P (10
there
P 015×10
. 7=
.
7= 007×10
and second
m
7
)
.
0070
.
ÿ8661232235
.
0075
.
.
ÿ8623285354
ÿ8622885724 ÿ8648373102
.
0080
.
.
ÿ8620631355
ÿ8619418980 ÿ8636798359
.
0085
.
.
ÿ8618046424
ÿ8615695918 ÿ8626403344
.
0090
.
.
ÿ8615511998
ÿ8611684053 ÿ8617125577
.
0095
.
.
ÿ8613017956
ÿ8607297182 ÿ8608966285
.
0100
.
ÿ8610557769
.
0105
.
ÿ8608126823
.
0110
.
ÿ8605721664
.
0115
.
ÿ8603339593
.
0120
.
ÿ8600978443
.
.
.
.
.
graph is given by figure9.7
As the pressure is increased further, the system finds the point at which the curve intersects itself.
pressure the
D
,
m point comes from the other branch of the curve. So at
At that point we have single
=
up of the points
minimum ofmwhich is greater than the physical state is . Below system is made
PQ
ON
PD
PD the physical state of
Q
between . But then,
,
B
significant
Exercise 145
system
It is
is the one that connects all these dots
build the
examples.
A
a physically isotherm
.
significant isotherm for two previous O graphic
P
× in
using the
,
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158
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
Figure 9.7: The path taken by the system from the potential analysis
chemical (Gibbs potential) and the pressure.
The isotherm obtained in the previous example is a fundamental thermodynamic
namically acceptable. To know how to determine the branch point, just note D relationship of change
mD
r that (cf. figure 9.8) so that
O
D
ÿ
=ÿ
=
O
(PdP
)
= 0,
FROM _
remembering that the integral is taken along the hypothetical isotherm. but this
integral can be rearranged in the form
K
ÿFD vdP
+ÿF
M
vdP
vdP
+ÿK
+ÿ
vdP = 0
ABOUT
or yet
vdP
FK
FD
ÿ
vdP
KM
ÿÿ
ÿ
and therefore the condition tells us that
´
area
vdP
=
I
=
´
area
ÿÿ
ABOUT CEO,
II.
It is only after we have truncated the non-monotonic isotherm with this equal-area relation
that the isotherm thus constructed represents a physically acceptable isotherm.
Machine Translated by Google
159
9.2. SIXTEENTH CLASS (06/25/2008):
Figure 9.8: The thermodynamically acceptable isotherm.
Note that there is a non-zero change in the molar volume at the point of the phase
transition, that is, there is a discontinuity in the volume. In the previous examples, 7
D is given approximately by pressure = 00916PD
Well. The
point
× 10 .
volumes for this pressure are:
0 .500 × 10ÿ4
,
00.859 × 10ÿ4
,
. 168 × 10ÿ3
,
where the middle volume represents the unstable value. Thus, we have a change in
D
volume at the point given by
ÿ in = 0 . 118 × 10ÿ3
.
Find the
point PD a from the chart by mere inspection
is too rude. So, using the
make a program in computer algebra, value of
bipartition method, which finds automatically with a preset O
PD
you vD values
by vO tolerance.
Example 146
The program can be written as:
> a:=0.0248; b:=26.6*10^(-6);R:=8.314;T:=30.5;
> eq:=v^3-(b+R*T/P)*v^2+a/P*va*b/P=0;
> P0:=0.11*10^7;
> Tol:=0.1*10^(-7);stepsize:=-0.005*10^7;flag:=false;
> for i from 1 to 100 by 1 while flag<>true do
>
vB1:=solve(subs(P=P0,eq));
>
if Im(vB1[1])<>0 or Im( vB1[2])<>0 or Im (vB1[3])<>0 then
>
P0:=P0+stepsize:
> else
>
>
Pf:=P0+stepsize:
vB2:=solve(subs(P=Pf,eq)):
It is
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160
>
>
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
comp1:=(vB1[3]-vB1[1])*P0R*T*ln( (vB1[3]-b)/(vB1[1]-b) )-
>
a/vB1[3]+a/vB1[1]:
>
comp2:=(vB2[3]-vB2[1])*PfR*T*ln( (vB2[3]-b)/(vB2[1]-b) )-
>
>
>
>
a/vB2[3]+a/vB2[1]:
if comp1*comp2<0 then
Pmid:=(P0+Pf)/2:
>
vB3:=solve(subs(P=Pmid,eq)):
>
comp3:=(vB3[3]-vB3[1])*PfR*T*ln( (vB3[3]-b)/(vB3[1]-b) )-
>
a/vB3[3]+a/vB3[1]:
>
>
if comp1*comp3<0 then
>
stepsize:=stepsize/2:
else
>
P0:=Pmid:
>
>
>
>
stepsize:=stepsize/2:
be;
else
>
>
P0:=P0+stepsize:
be;
> be;
>
if -stepsize/10^7<2*Tol then flag:=true: fi:
> from;
> Pprocurado:=P0;vB1:=solve(subs(P=P0,eq));
> test:=(vB1[3]-vB1[1])*Pprocurado>
R*T*ln( (vB1[3]-b)/(vB1[1]-b) )>
a/vB1[3]+a/vB1[1];
The result, for the values we've been using so far, is:
Pwanted=913288.8793 (compare with 916000.0 000 per inspection)
vD=0.5005060328e-4, vK=0.8559177063e-4, vO=0.1686101911e-3
test = 0.167e-4
(the closer to zero, the better)
Variations in other variables can also be obtained. The change in
entropy can be calculated from the integration of
ds
ÿvÿs
ÿ T
dv
omkfd = ÿ
on a hypothetical isotherm
or, using Maxwell's relations,
,
ÿ s
ÿP
=
sD
ÿ
sO
= ÿ omkfd
ÿ
ÿT
ÿ
vdv,
which can be interpreted graphically as the area between two neighboring
isotherms, as shown in figure 9.9. In this case, as the system
Machine Translated by Google
161
9.2. SIXTEENTH CLASS (06/25/2008):
is transformed at fixed temperature and pressure going from phase
pure, for the pure
D phase, it absorbs an amount of heat per mole equal to ÿ
s
The ÿDO = T
.
Figure 9.9: The discontinuity in molar entropy. The area between adjacent isotherms is
related to entropy discontinuity and therefore heat
latent.
vary
Calculate, for O same system as the previous examples, D.
To calculate
actionon entropy there latent heat involved in the transition
O
this quantity, we must note that
Example 147
ÿ
ÿ
ÿP
R
=
ÿ
ÿT
in
in
b
ÿ
,
so that
R
ÿ s
in
= ÿ omkfd
ÿ
=
bdv
so that
R ln
ÿ
CEO
vO
ÿ
b
=
R
b
.
168 × 10ÿ3 ÿ
ln ÿ 0 .
b
ÿ
b
0500 × 10ÿ4 ÿ
ÿ s = 1493760176
.
and,
therefore,
ÿ
=
T ÿ s = 305 .× 1493760176
.
= 455597
.
K.
Having calculated this difference and also the difference in volume, it is easy to
calculate the difference in energy to be
ÿ in
=
out
ÿ
uO = T ÿ s
ÿ
P ÿ in.
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162
CHAPTER 9. FIRST ORDER PHASE TRANSITIONS:
With this method, we reach the end of our calculations with the classification of
SO liquid phase, the region is in the pure gas
the whole thermodynamic process: the region is in the pure
AND phase. The flattened region corresponds to a
OKD
mixture of the two phases. The region of mixing of liquid and gas phases is limited by a curve that
resembles a parabola joining the ends of the
flattened regions of each isotherm, as shown in figure 9.10.
P
Figure 9.10: Classification of phases in the plan
in
ÿ
.
All that remains is to know how to calculate the amount of each phase in the region
where they coexist, that is, in the flattened regions of the graph. Knowing that we must always have
IN
=
=
Nv
+ Nxgvg,
Nxÿvÿ
D fraction
where isvgthe volume of the point is the mole
vÿ isofthe
xg from which
volume
of mole
the point
gas
and the
fraction O
of liquid,
It is
,
It is
xÿ
we derive the identity
xÿ =
vg
vg
ÿ
ÿ
in
vÿ
,
xg =
in
vÿ
ÿ
.
vg
ÿ
vÿ
This is called the level rule.
Exercise 148
liquid
It is
when
as molar percentages of the phases
In our examples, calculate gaseous
.
in = 00001
.
O volume is
Thus, at the end of these calculations, we can say that: a van der fluid
Waals, representing a gas of hydrogen molecules, being held for
. K state
the whole process at temperature = 305 isTin pure liquid
,
Machine Translated by Google
9.2. SIXTEENTH CLASS (06/25/2008):
163
7
to the pure
at
P = is
.091328×10
0 found
CEO = 0 . 5005×10ÿ4 at pressure , and
gaseous state volume from the phase
. 1686 × 10ÿ3 at the same pressure. A
vO =
transition volume implies a change in volume of ÿ Between
011855 these
× 10ÿ3 . in= 0 .
two volumes, there is a phase transition, at pressure and temperature
constants in which the liquid and gas phases coexist. When, during the
3
= 58%
of
in = and
0 . 1×10ÿ
tion,system
the volume
the of
, so
wesystem
have ainpercentage
of state.
=
x g 42%
the
in theisliquid
the
the gaseous
Inxÿthe
process
phase transition state, the latent heat of vaporization is (for the temperature used) ÿ 456.
Note the ÿvapor
very low volumes and the low temperature at which we can find liquefied hydrogen
H
gas ( 2 ) .