Transcript q 2 - q 1

Physical Chemistry
The Second Law Of Thermodynamics
To
Prof. Dr: FAWZY
3.1 INTRODUCTION
 The first law of thermodynamics, when applied
to a change of state of a thermodynamics system,
defines two extreme cases in which either ,w=0 (
isochoric process ) or q = 0 (adiabatic process ) in
which cases , respectively , q = ∆u and w =∆u .
 But if q ≠ 0 and W ≠ 0 , is there a definite
maximum amount of work which the system can do
during its change of state ? the answer to this
question requires an examination of the nature of
process .
 The examination which will be made in this
chapter results in the identifications Of two
classes of processes : reversible and irreversible
processes , and in the introduction of a state
function called the ENTROPY S
 The concept of the entropy will be introduced
from two different stating points : in the first
, the entropy will be introduced and discussed
as the result of a need for quantification of the
degree irreversibility of a processes , and in
the second , the entropy be introduced as a
result of the examination of the properties of
reversibly operated cyclic heat engine .
 This examination leads to a statement known
as the second law of thermodynamics which
together with the first law of thermodynamics
lays the foundation for the development of the
thermodynamic methods of describing the
behavior of matter .
3.2
Spontaneous ,or Natural
processes
A system, left to itself will do one of two things
:
 either it will remain in the state in which it
happen to be (in this case, this state will be
called an equilibrium state ) ;
Or
 it will move , of its own accord, to same other
state ( the initial state in this case will be
named non equilibrium state ) .
 A process which involves the spontaneous
movement of a system from a non equilibrium
state to an equilibrium state is called a
spontaneous or nature process.
 Example of nature process are:
1.the mixing of two gases
2.The equalization of temperature .
3.the spontaneous occurrence of the chemical
reaction : A + B = C + D
In either direction , depending on the initial mixture
of A, B, C, and D until
Reaching the equilibrium state of the system .
 The natural processes can not be reversed
without the application of an external effects,
thus a natural processes would leave a
permanent change in an external environment
.Such a processes is said to be irreversible,
therefore the terms spontaneous nature ,and
irreversible are synonymous in this context
 If a closed system undergoes a spontaneous
process involving the performance of work and
the production of heat, then as the process
continuous, the capacity of the system for further
spontaneous change decreases.
 Finally, once equilibrium is reached, the
capacity of the system for doing further work is
exhausted.
 This means that in the initial non equilibrium
state of an isolated system , some of the
energy of the system is available as useful work
in the system get degraded by convening it to
thermal energy
(Or heat) which a form of energy that is no
longer available for external purpose.
3.3
Quantification of
irreversibility and Entropy
 Since the capacity of the system for
spontaneous change decreases with the
occurrence of the natural process, these should
be a quantitative measure for the degree of
irreversibility ( or the degree of degradation of
the spontaneous process ) of the natural
process.
 The existence of processes which exhibit
differing degrees of irreversibility can be
illustrated by the weight –pulley- paddle wheels
heat reservoir shown in figure(3.1).
Lewis and Randall considered the following three process :
1. the heat reservoir in the weight –heat reservoir system
is at temperature T2 . The weight is then allowed to fall ,
performing work w, and the heat produced ,q , enters the
heat reservoir .
2. the heat reservoir at the temperature T2 is placed in
contact with a heat reservoir at temperature T1 which in
lower than T2 , and the same heat q is allowed to flow from
the reservoir at T2 to the reservoir at T1 .
3. the weight is allowed to fall, performing work w , and
the heat produced ,q , enters the heat reservoir .
 since process (3) is equivalent to processes (1) and (2)
, the degree of irreversibility of process (3) is higher
than the degree of irreversibility of process (1) , and
as T2 is higher , than T1 then (q \ T2 ) will be higher
than (q \ T1 ).
 Thus , when a system undergoes a spontaneous
process which involves absorption of heat q and the
constant temperature T, then the entropy change or the
degree ,ΔS, which can be used as the capacity of
spontaneous change or the degree of irreversibility, can
be expressed as: ΔS = ( q \ T )
 The entropy inverse as a result of the process is
thus a measure of the degree of Irreversibility of the
process
3.4
Reversible Process
 Since the capacity of system for
spontaneous change decreases with the
occurrence of the natural process ( a change
from non equilibrium state to equilibrium state )
until equilibrium is attained at which the
capacity of the system for spontaneous change
will be exhausted ; thus, if the process
proceeds under infinitesimally small driving force
such that , during un process, the system is
never more than infinitesimal distance
From equilibrium the degree of irreversibility
will approach zero value and the process
because a reversible process .
 Thus , a reversible process is a process
during which the system is never a way from
equilibrium states ; therefore , the reversible
process is the process which takes a system
from state A to state B along a path of
continuum equilibrium states .
3.5
An Illustration of
irreversible and Reversible
process
 Example: Water evaporation (or
condensation) in a system of water and water
vapor at the uniform temperature T contained in
a cylinder fitted with a frictionless piston
placed in thermal contact with a heat reservoir
at constant temperature T as shown in figure
(3.2) {fig 3.2 ,P .4.2 ) .

Let the system to be initially at the state:
Pevap (saturated water vapor pressure at temperature T = PH2O(T)) = Pext ,V,
Tsystem = Theat reservoir = T; so the system is mechanically and thermally is
in equilibrium state.
If we consider the evaporation process, i.e. if the external
pressure is suddenly decreased by ΔP, a non equilibrium
state is induced, so the piston will move outward causing a
pressure drop inside the cylinder, thus the liquid water
evaporate in a process of establishment of mechanical
equilibrium by equalization of the external and internal
pressure.
The evaporation process leads to a temperature drop
between the temperature of reservoir and the temperature
of water vapor + water system which is lowered because
of the endothermic nature of the evaporation process.
This temperature gradient forces a heat transfer from the
reservoir to the content of the cylinder to reestablish
thermal equilibrium by equalizing the temperature of the
whole system at temperature T.
As the system temperature is reestablished again at
temperature T, the mechanical equilibrium at
Psystem = Pext = P(saturated water vapor at temperature
T – PH2O (T) )
will be reestablished again by increasing the external
pressure back to Pext = PH2O(T) which cause instantaneous
compression of the water vapor raising the system
content temperature to T2 >T which leads to heat flow
from the system to the heat reservoir.
Thus: the work done on the system during this cycle is given by
Pext V – (Pext – Δ P)V = V Δ P
where V is the volume change during the process. Thus, the permanent change
in the external energy of the heat reservoir as a result of the cycle process is
V Δ P.
Now if we consider the condensation process, i.e. if the cyclic process
started by increasing the external pressure, which equals PH2O(T), by
Δ P, it is possible to show that, for a cycle, the permanent change in
the external energy of the heat reservoir is V ΔP.
Thus, as ΔP approaches infinitesimal value, i.e. ΔP→ δP,
reversibility is approached, i.e. when evaporation or condensation
processes are carried out in such as a manner that pressure exerted by
the water vapor is never more than infinitesimally different from its
saturation value at the temperature T, the degree of irreversibility
approached the zero value.
It can also be seen that, as true reversibility is approached
the process became infinitely slow.
3.6
Entropy and Reversible Heat
Consider the evaporation process of water vapor in the
cylinder-piston system of figure (2) which is in contact with
heat reservoir at temperature T: if the process is carried
out in a reversible manner; then, |Wrev| = PH2O (T) V,
and if the process is carried out in an irreversible manner,
then,
|Wirr| = (PH2O (T) – ΔP) V.
Thus, |Wrev| > |Wirr|, i.e. the maximum work alone by the
system will be
|Wmax| = |Wrev|.
Now, applying the first of thermodynamics to the reversible and
irreversible processes we have:
ΔUrev = qrev + Wrev
And;
ΔUirr = qirr + Wirr
Since the initials states and the final states of reversible and
irreversible processes are the same, thus
ΔUrev = ΔUirr
therefore,
|qrev| - |qirr| = |Wrev| - |Wirr| > 0
i.e. |qmax| - |qirr| = |Wmax| - |Wirr| > 0
Thus, if the process is carried out reversibly:
ΔStotal = ΔSsystem + ΔSheat reservoir
= |qrev|/T + (-|qrev|/T) = zero
For irreversible process:
ΔS
heat reservoir
ΔSsystem
= -|qirr| / T
= |qirr | / T + (|qrev| - |qirr|)/T
Where |qirr| is the amount of heat entering the system leaving the heat
reservoir |qrev| - |qirr| represents to amount of work degraded in the system
and enters back to the heat reservoir; thus:
ΔSsystem = |qrev|/T
Therefore:
ΔSsystem + ΔS
heat reservoir
ΔStotal
= |qrev| /T + (-|qirr| /T)
= (|qrev|-|qirr|)/T = ΔSirr

ΔSirr is the entropy created as a result of the irreversible.
 Thus:
ΔSsystem = |qirr|/T + ΔSirr

since |qrev| > |qirr|, the value of ΔSirr will be a
positive value.
Consideration of the condensation process shows that
the work done on the system has a minimum value when
the condensation is conducted reversibly , and
correspondingly:
|wrev| < |wirr| ,
and ;

|qrev| < |qirr| ,
|wrev| - |qirr| =
|qrev| - |wirr| < 0
Thus for reversible process ;
ΔSsystem = - |qrev| / T
ΔS heat reservoir = + |qrev| / T
Therefore
ΔSsystem + ΔS heat reservoir = - |qrev| / T + |qrev| / T
=
zero

For irreversible process ;
ΔS heat reservoir = + |qirr| / T
ΔSsystem
= - |qirr| / T a+ ( |qirr| - |qrev| ) / T
= - |qirr| / T + ΔSirr
= - |qrev| / T

Since qirr > qrev , ΔSirr will be appositive value
It represent the entropy in the heat reservoir because
of invisibility.
The last equation indicates that the entropy change
of a system can be measured , or calculated through
a reversible path for which the heat flow in q rev ,
and ∆S in = 0
The important feature to be noted from the previous
equations is that , in going from an initial to a find
state , either by evaporation or condensation , the
entropy difference between the find state and the
initial state ,
( S f – S i ) is independent of whether the process is
conducted reversibly or in reversibly , and thus
independent of the path of the process ; therefore the
entropy of system is a state property .
3.7
The reversible isothermal
compression of a ideal
gas.
 Application of the first law of thermodynamics to the
mentioned isothermal system gives ;
∆u = 0
q + w = 0
q = - w
Thus:
or :
 But for the mentioned reversible system ;
but
Δw = - Pext. dV
= - (P + δp) dV
≈ - P dV
However, for i deal gases;
P = nRT / ν
Thus ;
δw = (- nRT / ν)dν .
By integration, the work done on the system will be
given by;
W = -nRT ln (ν2 / ν1)
Since ν2< ν1 , the work done on the system will be
positive.
 Applying the first law of thermodynamics yields:
q = nRT ln (ν2 / ν1).
Which will be negative value , i .e. the heat will be a
negative value .
 Since the process is isothermal, thus:
Thus ;
∆S sys = │q │/ T = nRT ln (ν2 / ν1).
∆Senv = -│q │/ T.
= -nR ln (ν2 / ν1).
 As ν2< ν1 , the result of the reversible compression
of an ideal gas is decreasing the entropy of the
system and increasing the entropy of the environment
by an equal amount , thus :
∆Sirr = 0
3.8
The Reversible and
Irreversible adiabatic
Expansion of an Ideal Gas
▪ For reversible adiabatic process :
∆S sys = ∆S environment = 0 .
also :
∆Sirr = 0.
▪ For irreversible adiabatic process :
∆S environment = 0.
∆S sys = ∆Sirr > 0.
This is because work degradation occur and
transfer to heat inside the system causing
higher final temperature than the final
temperature obtained in the case of reversible
process.
For irreversible adiabatic processes , the greater the
heat produced in the gas (system) due to degradation ,
i.e. the greater the degree of irreversibility , the
higher the final temperature and the internal energy ,
and the greater the entropy increase.
3.9
summary statements
▪ From the discussion as far, three points have emerged :
When a system under gases a spontaneous ( irreversible )
process , the entropy of the system increase.
When a system under gases a reversible process ,
no entropy is created , i.e. . ∆Sirr = 0 , entropy is
simply transferred from one part of the system to
another part .
Entropy is a state function.
3.10
The Properties Of Heat
Engines.
Thus , during an irreversible expansion ,
the work done by the gas still equals the
decrease in the internal energy of the
system (gas) ; however , the decrease in
U is less than ∆u of the reversible
expansion due to the heat appearing in
the gas as the result of degradation.
▪ A heat engines is a device which converts heat into work.
▪ It is interesting to note that the first steam engine ( which is a
devices that transfer heat into work using steam as medium ) was
built in 1769 and was operational for considerable number of years
before introducing the principles of reversible and irreversible
processes.
Heat
reservoir
at high
q2
Heat
q1
reservoir
at low
▪ Based
on trials
, the principle
of impossibility
ofHeat
creating
virtual
Temperature
t2 of the second type
engine
temperature
t2
motion
machine
(i.e. the heat engine
which is
capable of absorbing an amount of heat from a heat reservoir and
transferring it fully into useful work have been emerged.
Work w
▪ Thus , a heat engine can only be created by working a medium (
say ideal gas ) through a cyclic process ; the simplest heat engine
as shown in figure 3 p:50. In this device operation the heat engine
withdraw an amount of heat , q2 from a heat reservoir at high
temperature t2 , and same of this heat is converted to work , w2
and the remainder of heat , q1 is transferred to a lowtemperature heat reservoir at t1 , thus :
│q2│ > │q1│
and
t2 > t 1
▪ A typical example of heat engine is the familiar simple
steam engine where superheated steam is passed from a
boiler ( the high-temperature heat reservoir ) into a
cylinder where it performs work by expanding against an
piston (the engine ) ; as a result , the steam
temperature decreases and at the completion of the
piston strake , the spent steam is exhausted to the
atmosphere ( the low- temperature heat reservoir ). A
flywheel then returns the piston to it's original position
; thus completing the cycle and preparing for the next
working stroke.
The efficiency of engine is defined as :
Efficiency = work obtained = w
Heat input
q2
Car not in 1824 explained the factor
governing the efficiency of an engine by considering a
cycle that consist of 4- reversible steps shown in figure
(4) . This cycle consist of two expansion steps : an
isothermal one (A → B) and an adiabatic step (B → C)
and two compression steps: an isothermal one (C → D)
and an a diabetic step (D → A) .
▪ For this cycle , which is known as Car not cycle , w is
.given by :
w = q2 - q1
Where q2 is the heat absorber by the engine fro the
heat- temperature heat reservoir at t2 during the
reversible expansion step (A → B) and q1 is the heat
transferred fromThusthe
thermodynamics
substance , heat
, for Can
not cycle :
engine , to the low- temperature heat reservoir at
t1during the reversible compression step (C → D)
Thus , for Car not cycle :
Efficiency = ( q2 - q1 ) / q2
= 1 - ( q2 - q 1 )
Let now question the following :
Is it possible to have another
more efficiency heat engine
working between t2 and t1
following Car not cycle ????????
▪ Let us propose that this is possible , thus assume that
we have two devices the first has the parameters q1 ,
q2 ,and w , and the second has the parameters q′1 , q′2
, and w′ such that
q2 = q′2 and w > w′ ,so q1 <
q′1 ; then the first device is more efficient than the
second .
Let the first engine to sun in the forward direction of
Car not cycle
thus it performs as heat engine , then :
w = q 2 - q1 .
Let the second engine to sun in the reverse direction of
Car not cycle , thus it function as heat pump , then ;
thus ;
-w′ = - q′2
+ q′1 = - q2 + q′1
w - w′ =
q′1 - q1
this mean that an amount of work (w - w′ ) has been
obtained from quantity of that ( q′1- q1 ) by absorbing
it from the low- temperature heat reservoir without
leaving aching in any other thermodynamic system .
Though this is in agreement with the first law of
thermodynamics , it conflict with human experience fact
of impossibility of certain of a perpetual motion machine
of the second type . Thus , the two machine must be of
the same efficiency .
▪The above discussion gives rise to a preliminary
formulation of the second low of Thermodynamics which
is known as the principle of Thomson : the principle of
Thomson states that :
It is impossible , by means of cyclic
process , to take heat from a
reservoir and convert it into work
without , in the same operation ,
transferring heat to a cold reservoir.”
“
▪Consider that we have two other engines ; the
operating parameters of the first one q2 , q1 ,w and
the second operating of the second one q′2 , q′1 , w′
such that :
w = w′
thus ;
q2 > q′2
q1 > q′1
This mean that the second engine is more efficiency than
the first . Let the second engine to sun in the forward
direction of Car not cycle , thus it performs as heat
engine , then ;
w = q′2 - q′1
Let the second device to sun in the reverse direction of Car not cycle
, thus it performs as heat pump , thus ;
therefore ;
-W = - q2 + q1
q2 - q′2 = q1 - q′1 = q.
which means that an amount of heat q is pumped from the lowtemperature heat reservoir to the high- temperature heat reservoir .
This corresponds to the spontaneous heat flow from cold bodies to hot
bodies which opposes the human experience. Thus the two devices
must have the same efficiency. The above discuss have
laid to the formulation of the principle of clauses which
is one of statements of the second law of
thermodynamics ; it states that:
“ It is impossible to
transfer heat from a cold
reservoir to hot reservoir
without , in the same
process, converting a
certain amount of work into
heat .”
▪
3.11
The Thermodynamic
temperature scale
▪The above results suggest that all reversible Car not
cycle operating between the same upper and lower
temperature must have the same efficiency , namely the
maximum possible .
▪This maximum efficiency is independent of the working
substance and is a function only of the working
temperature t1 and t2 ; thus :
Efficiency
= (
q2 - q1 ) / q2
= 1 - (
q2 - q1
Thus :
q1 / q2 = ƒ( t1 , t2 )
▪ Consider the Car not cycle shown in figure 5. The figurer
identify 3cycle , the first operating between t1 , t2 , the
second is operating between t2 , t3 , and the third is
operating between t1 , t3 ; thus :
q1 / q2 = ƒ( t1 , t2 )
q2 / q3 = ƒ( t2 , t3 )
so ;
q1 / q3 = ƒ(t1 , t3 )
q1 / q3 * q3 / q2 = ƒ(t1 , t3 ) = q1 / q2 = ƒ( t1 , t2 )
ƒ( t2 , t3 )
▪As ƒ( t1 , t2 ) is independent of t3 , then ƒ(t1 , t3 ) ,
and ƒ( t2 , t3 ) must be of the form :
And ;
Then :
ƒ ( t2 , t3 ) = ƒ (t2) / ƒ (t3 )
ƒ ( t1 , t2 ) = ƒ (t1) / ƒ (t2)
q1 / q2 = ƒ ( t1 ) / ƒ ( t2 )
▪ Kelvin took these function to here the simplest possible form
, namely T1 , T2 : thus :
q1 / q2 = T1 / T2
then the
Efficiency is given by ;
( q2 - q1 ) /q2 = T2 – T1 / T2
This defines the absolute thermodynamics scale of
temperature which is independent of the working
substance .
▪The zero temperature of this scale is that 
temperature of the cold reservoir that makes the
Efficiency of the Car not cycle equal to unity
The absolute thermodynamics scale is identical to the
ideal gas temperature scale .
This can be demonstrated by considering the working substance in
Car not cycle to be 1 mole of an ideal gas ; by reefer to the
figure 5 , and applying the first law of thermodynamics and the ideal
gas laws for reversible and adiabatic processes and on the whole cycle
, it can be shown that :
VB / VA = VC / VD ,
q2 = R T2 ln (VB / VA)
and
:
w = R (T2 – T1 ) ln (VB / VA)
thus ,
the Efficiency
= w / q2 = T2 – T1 / T2 .
which is identical to Kelvin's equation derived previously :
( q2 - q1 ) / q2 = T2 – T1 / T2
3.12
The Second Law Of
Thermodynamics
The pervious equation can be written as :
q2 / T2 - q1 / T1 = 0
i.e.
∑q/T=0
taking into consideration that the heat entering to the
system is positive and the heat leaving the system is
negative .
By taking Avery general cycle represent by the loop
ABA such that shown in figure 6 p: 54 into
consideration ; this cycle can be broken down into a
number of Car not cycle as shown ; for this cycle :
∑q/T=0
By making Car not cycle smaller and smaller , the zigzag
path of the Car not cycle will approach the general loop
ABA ; and in the limit of coincidence , the summation can be
replaced by acyclic
Integral ; i.e.
Ф δq / T = 0
Thus :
B
∫A ﴾δq / T ﴿
+
A
∫B ﴾δq / T﴿ = 0
This mean that these integrals are perfect
differential of semi function of state of the system .
Let this function to be S and is called the entropy ;
thus :
dS = δq / T
since the derivation of this equation is based on Car
not cycle which is based on reversible processes ;
this equation must be written as :
dS = δq rev / T
•The Second Law Of Thermodynamics can thus be stated as :
I.
The entropy S , defined by the equation
is the function of state ,and
ds = δq rev / T ,
II. The entropy of system in adiabatic enclosure can never
decreases ; it remains constant in a reversible process,
and increases in an irreversible processes .
Thus :
dS system
+ dS environment ≥ 0
There for , ∑ dSi = 0 , for reversible processes ,
where i refer to the parts of the system .
If any of these i parts behaves irreversibly ; thus :
∑ dSi = dS irr
•where the value of , entropy created on the degree of
irreversibility and is a measure of that degree of
irreversibility .
Thus the value of dS irr can vary between zero for
reversible processes to increasingly positive values for
increasingly irreversible processes
3.13
Maximum Work Obtained
From a Thermodynamics
System.
For a system change from state A to state B , the
application of the first law of thermodynamics gives :
usys = uB – uA = q + w .
Thus , for infinitesimal change :
( - δw ) = δq – du
system
where ( - δw ) is the work done by the system .
The Second Law Of Thermodynamics gives ;
dS system = δq / T + dS irr
thus ;
δq = T * dS system - T * dS irr
Combining the mathematical statements of the First
and Second Laws Of Thermodynamics yields
( - δw ) = T * dS system - du
since
dS irr
> 0
system
- T * dS irr
thus ;
( - δw ) ≤ T * dS system - du
system
,
Integration from state A to state B gives :
(-w) ≤ T ( SB – SA ) – (UB – UA )
and as S and
U
are state function , then
(-w)max = T ( SB – SA ) – (UB – UA )
thus , for reversible processes , the work done by the
system is given by
(-w)max
; for irreversible processes
, the degraded work is given by :
(-w)max - (-w) .
Application of these principle on a system that
expands isothermally in a reversible manner from state
A to state B yields :
(-w)max = qrev = R*T ln (VB / VA)
If the process is irreversible and the expansion occurs
freely as in joule's experiment , the work done by the
system equals zero
the degraded work = wmax - w = wmax – 0 = wmax = qrev
i.e. the degraded work = R*T ln (VB / VA) ,
and
∆Sirr = R ln (VB / VA)
3.14
Entropy and The Criterion
of Equilibrium
in an Isolated System
 Consider the chemical isolated system that contain the
chemical constituents A, B, C, and D and let this system
to be of constant internal energy U of constant volume V .
Equilibrium
 By considering that this system attain , equilibrium at
entropy
specific composition of A, B, C, and D, thus, starting with
A and B .The reaction A + B = C + D will spontaneously
occur from this non equilibrium state until the attainment
of the equilibrium composition .
 Since
process , entropy of the system
( A this
+ B ) is a spontaneous
composition
(C+D)
increases until the equilibrium point is reached as shown in
figure ( 7 )
Also, if we started by C and D , the reaction will
produced from left to right , and again the entropy of the
system will be increasing as shown in the figure .
 Thus, the system entropy will be maximized at the
point of equilibrium , at which the system composition
can not more either to the right or to the left
spontaneously because the process will be accompanied
by decrease in entropy which does not agree with the
second law of thermodynamics .
3.15
the combined statement of
the first and
the second laws of
thermodynamics
 For a reversible process , we can write :
but
and
thus
or
dUsystem = δqrev + δwrev
δqrev = TdS ,
δwrev = - PdV
dUsystem = TdSsystem – Pdvsystem
dU = TdS – PdV
This equation in valid for closed systems and for the cases
where the only form of work performed in the work due to
volume change , i.e. the P-V type of work .
 Based on the pervious equation we can write the
following a function:
U = U( S, V )
The total differential of which is :
dU = (∂U / ∂S)v dS + (∂U / ∂V)s dV .
thus
T = (∂U / ∂S)v dS
,
P = (∂U / ∂V)s dV
Also we can write :
S = S ( U , V )
thus ;
dS = (∂S / ∂U)v dU + (∂S / ∂V)u dV .
but
dS = dU / T + ( P / T ) dV
therefore ;
(∂S / ∂V)u = P / T .
Equilibrium in a system of constant U and V occurs
when the entropy of the system is maximized and in
system of constant S and V occurs when U is
minimized.
The further development of classical thermodynamics
results from the fact that S and V are inconvenient
pair of independent variables; i.e., in considering a real
system, considerable difficulty would be experience in
arranging the state of the system such that
simultaneously it would have the required entropy and
would occupy the required volume.