CHE242
Presented @ Unilorin
Dr.O.A.Ajayi
1
Terminologies
Isobaric
P
o P = constant
Isochoric
W = P(Vf – Vi)
P
o V = constant
Isothermal
W = 0
o Q=0
V
P
o T = constant
Adiabatic
V
(for an ideal gas)
V
P
(for a monatomic ideal gas)
V
2
Terminologies, contd…
•
•
•
•
•
•
•
Quasi-static means that it occurs slowly enough that a uniform pressure and
temperature exist throughout all regions of the system at all times.
Heat (thermal) reservoir is a sufficiently large system in stable equilibrium to
which and from which finite amounts of heat can be transferred without any
change in its temperature.
A high temperature heat reservoir from which heat is transferred is sometimes
called a heat source. A low temperature heat reservoir to which heat is transferred
is sometimes called a heat sink.
Work reservoir is a sufficiently large system in stable equilibrium to which and
from which finite amounts of work can be transferred adiabatically without any
change in its pressure.
Thermodynamic cycle-A system has completed a thermodynamic cycle when the
system undergoes a series of processes and then returns to its original state, so that
the properties of the system at the end of the cycle are the same as at its beginning.
Heat Engine-A heat engine is a thermodynamic system operating in a
thermodynamic cycle to which net heat is transferred and from which net work is
delivered.
Working substance of the engine, which is the material within the engine that
actually does the work (e.g., the gasoline-air mixture in an automobile engine).
3
•
•
•
•
Preamble
The second law of thermodynamics states that processes occur in a certain
direction, not in just any direction. Physical processes in nature can proceed
toward equilibrium spontaneously:
Water flows down a waterfall.
Gases expand from a high pressure to a low pressure.
Heat flows from a high temperature to a low temperature.
Once it has taken place, a spontaneous process can be reversed, but it will not reverse
itself spontaneously. Some external inputs, energy, must be expended to reverse the
process. As it falls down the waterfall, water can be collected in a water wheel, cause a
shaft to rotate, coil a rope onto the shaft, and lift a weight. So the energy of the falling
water is captured as potential energy increase in the weight, and the first law of
thermodynamics is satisfied. However, there are losses associated with this process
(friction). Allowing the weight to fall, causing the shaft to rotate in the opposite
direction, will not pump all of the water back up the waterfall. Spontaneous processes
can proceed only in a particular direction. The first law of thermodynamics gives no
information about direction; it states only that when one form of energy is converted
into another, identical quantities of energy are involved regardless of the feasibility of
the process. We know by experience that heat flows spontaneously from a high
temperature to a low temperature. But heat flowing from a low temperature to a higher
temperature with no expenditure of energy to cause the process to take place would not
4
violate the first law.
Preamble, contd…
•
The first law is concerned with the conversion of energy from one form to
another. Joule's experiments showed that energy in the form of heat could
not be completely converted into work; however, work energy can be
completely converted into heat energy. Evidently heat and work are not
completely interchangeable forms of energy. Furthermore, when energy is
transferred from one form to another, there is often a degradation of the
supplied energy into a less “useful” form. We shall see that it is the second
law of thermodynamics that controls the direction processes may take and
how much heat is converted into work. A process will not occur unless it
satisfies both the first and the second laws of thermodynamics.
Why does heat only flow
from hot areas to cooler
areas? Can we reverse
this natural process?
5
It will arouse
changes while the
heat transfers from
low temp.substance to
high temp. one.
6
Common character of spontaneous process
Some examples
• Work changes into heat automatically;
• Gas inflates toward vacuum;
• Heat transfers from the high temp. object to the
low temp. one;
• The solution of different concentration can
mixed evenly;
•
The contrary process of these process can not proceed
automatically. When it gets back to the original state by external
force, the effect which can not be wore away will leave.
7
Spontaneous process & its common character
Spontaneous process (change)
Some changes have the trend of
spontaneity, once happens, it can proceed
automatically without the help of outside
force.
Common character
– irreversible, the contrary process of every
spontaneous process is non-spontaneous.
A reversible process is one in which both the system and
its environment can be returned to exactly the states
they were in before the process occurred.
8
Heat Engines
•
•
•
•
•
•
Heat can be converted to work directly
and completely
Converting heat to work requires the
use of a device called a heat engine
Heat engines come in many forms,
pure heat engines (steam power plants)
and semi heat engines (gas turbines)
All have a working fluid
Work can be easily converted
completely to heat and other forms of
energy
Converting other forms of energy to
work is not that easy
9
Heat engine, contd…
•
•
•
•
•
•
Qin=amount of heat supplies to
steam in boiler from high
temperature source (furnace)
Qout=amount of heat rejected from
steam in condenser to a lowtemperature sink
Wout=amount of work delivered by
steam as it expands in turbine
Win = amount of work required to
compress water to boiler pressure
Wnet,out= Wout-Win (kJ)
Wnet,out= Qin-Qout (kJ)
10
The 2nd law of thermodynamics
Clausius saying,
“It will arouse other changes while the heat
is transferred from the low temp. object to the
high temp. one”
Kelvin saying,
“It will arouse other changes while the heat
from the single thermal source is taken out
and is totally changed into work.”
• Ostward expression:
“The 2nd kind of perpetual motion machine can not be
made”。
11
The 2nd law of thermodynamics, ctd…
• Kelvin-Planck Statement:
– It is impossible for any device that operates on
a cycle to receive heat from a single reservoir
and produce a net amount of work.
– No heat engine can have a thermal efficiency
of 100%
– For a power plant to operate, the working fluid
must exchange heat with the environment as
well as the furnace
12
Refrigerators and Heat Pumps
• Heat moves in nature from high
temperatures to lower temperatures, no
devices required
• The reverse process, heat from low temp to
high temp, required special devices called
refrigerators or heat pumps
13
Refrigerators and Heat Pumps, ctd..
14
Heat Engine vs Heat Pump
•
In the refrigeration process, work W is used to remove heat QC from the
cold reservoir and deposit heat QH into the hot reservoir.
15
Working Examples of Heat Pump
•
In a refrigerator, the
interior of the unit is the
cold reservoir, while the
warmer exterior is the hot
reservoir.
•
In a refrigerator, the
interior of the unit is the
cold reservoir, while the
warmer exterior is the hot
reservoir.
16
Performance check
•
•
Coefficient of Performance, COP
The index of performance of a refrigerator or heat pump is expressed
in terms of the coefficient of performance, COP, the ratio of desired
result to input. This measure of performance may be larger than 1,
and we want the COP to be as large as possible.
COPHP = Desired output /Required input = QH/Wnet,in
COPHP = QH/(QH–QL) = 1/(1-(QL/QH))
17
The 2nd law of thermodynamics, ctd…
• Clausius Statement
– It is impossible to construct a device that operates in a
cycle and produces no effect other that the transfer of heat
from a lower-temperature body to a higher-temperature
body
• Perpetual-Motion Machines
– A device that violates the 1st law (creates energy) is a
perpetual-motion machine of the first kind (PMM1)
– A device that violates the 2nd law is a perpetual-motion
machine of the second kind (PMM2)
18
Carnot cycle and Carnot law
• In 1824, French engineer
N.L.S. Carnot designed a
Carnot cycle.
• Ideal gas absorbs heat Qh
from Th thermal source,
exports work W through
ideal thermal machine,
other heat Qc discharged
to Tc thermal sink.
19
Processes in Carnot cycle
A-B=Reversible isothermal expansion,
TH = cont
B-C=Reversible adiabatic expansion,
Q=0
C-D=Reversible isothermal
compression, TL = cont
D-A=Reversible adiabatic compression,
Q=0
20
Processes in Carnot cycle
Process 1
Reversible expansion at Th
from P1,V1 to P2 ,V2 , (A→B)
U1  0
V2
Q2  Qh  W1
W1   pdV  RT2 ln VV12
V1
T2  Th
• The work is showed as the
following area under the
curve AB.
21
Processes in Carnot cycle
Process 2
Adiabatic reversible expansion from P2 ,V2 ,Th
to P3,V3,Tc, (B→C)
Q=0
T1
W2  U 2    CV ,m dT
T2
The work is showed
area under the curve
BC.
22
Processes in Carnot cycle
Process 3
Reversible compress at Tc from P3,V3 to
P4 ,V4 ,(C→D)
U 3  0
V4
Q1  Qc W3   pdV  RT1 ln VV43
V3
T1  Tc
The work is showed as the area under the
curve DC.
23
Processes in Carnot cycle
Process 4
Adiabatic reversible compress from P4 ,V4 ,Tc
to P,V,T, (D→A)
Q0
T2
W4  U 4    CV ,m dT
T1
The work is showed as
the area under the curve
DA.
24
General heat & work
The whole cycle:
U  0
Q  Qh  Qc
Qh >0
Qc <0.
V
V
W= W1  W3  RT2 ln V  RT1 ln V
2
4
1
3
(W2 and W4 can be eliminated)
25
General heat & work, ctd…
According to the formula of the adiabatic
reversible process
Process 2:
Process 4:
ThV2 1  TcV3 1
ThV1 1  TcV4 1
V2 V3

V1 V4
2 divide 4:
W1  W3  RT2 ln
V2
V1
 RT1 ln
V4
V3
 R(T2  T1 ) ln
V2
V1
26
Efficiency of Carnot Heat Engine
Thermal machine absorbs heat
Qh from Th source, part of heat
is changed into work, other Qc
go back to Tc source.
or
W Qh  Qc


(Qc  0)
Qh
Qh
V2
nR(Th  Tc )ln( )
V1 Th  Tc
Tc


 1
V2
Th
Th
nRTh ln( )
V1
 1
27
Carnot law
Carnot law:
The efficiency of the reversible machine is
the highest.
Deduction of Carnot law: The efficiency of all
reversible machines is the same.
The importance of Carnot law:
(1) In principle inequality (ηI <ηR) has
solved the direction problem of the chemical
reaction;
(2) it solves the limitation problem of the
thermal machine efficiency.
28
Entropy
Conclusion from Carnot cycle
W Qh  Qc Th  Tc



Qh
Qh
Th
Qc
Tc
1
1
Qh
Th
Or:
Qc Qh

0
Tc Th
Qc
Qh

Tc
Th
In qualitative
terms, entropy
can be viewed
as a measure
of randomness
or disorder of
the atoms or
molecules in a
substance.
That is, in the Carnot cycle, the summation
of heat effect and temperature is zero.
29
Thermal temperature quotient of any reversible cycle
The summation of
every reversible cycle
thermal temperature
quotient is zero.
•Adiabatic lines: RS &TU
•ΔPVO= Δ OWQ
•Same work, Δ U, Q
Qi
)R  0
(
i
Ti
or
 (
Q
)R  0
T
30
Thermal temperature quotient
• In many little Carnet cycles,
adiabatic inflation line of the
previous cycle is the adiabatic
reversible compression line of
the next, the work of these
two process counteract.
• The total effect corresponds the
closed curve, so the thermal
temperature quotient addition is
zero. Or the integral of cycle
process is zero.
31
2.4.2 Derivation of entropy
Choose two points A, B from
the reversible cycle curve at
random, the cycle is divided
into two reversible processes
A→B and B → A.
Q
 ( T )R  0
A Q
Q
A ( T )R1  B ( T )R2  0
B
32
Derivation of entropy
Transposition
B Q
Q
A ( T )R1  A ( T )R2
B
The thermal temperature quotient
addition of every reversible process
depends on the initial and final
state, but it has nothing to do with
the reversible approach, this thermal
temperature quotient addition has the
properties of state function.
every reversible
process
33
Definition of entropy
Clausius defined this state function as “entropy”.
Mark: S ; Unit: J.K-1 1
Q
SB  SA  S   ( )R
A
T
B
or
S   (
i
Qi
)R
Ti
For a little change
S   (
i
Qi
)R  0
Ti
Q
dS  (
)R
T
These entropy change formulas are called the
definition of entropy, that is, the entropy change
can be scaled by the thermal temperature quotient
addition of the reversible process.
34
Clausius inequality
Suppose there is a reversible machine
between the high and low thermal source.
Th  Tc
Tc
R 
1
Th
Th
The same, there is an irreversible machine
which have the same temperature with the
previous one.
Qh  Qc
Qc
so： IR 
1
Qh
Qh
35
Thermal temp quotient addition of irreversible process
According to the Carnot law: IR  R
so
Qc Qh

0
Tc Th
If an irreversible process contacts with
many thermal source:
Qi
(
)IR  0
i Ti
36
Thermal temp quotient addition of irreversible process
An irreversible cycle has two parts.
A Q
Q
( )IR,AB   ( )R  0
B
T
T
i

A
B
Q
Q
( )R  SA  SB SB  SA  ( )IR,A B
T
T
i
Q
SAB  ( )IR,AB  0
T
i
If A→B is a reversible process:
Q
SAB  ( )R,AB  0
T
i
37
Clausius inequality
or
Q
SAB  ( )IR,AB  0
T
i
SAB  (
i
Q
)R,AB  0
T
Combine the two formulas
together:
SAB
Q
 ( ) A  B  0
T
i
38
Clausius inequality
For a little change:
or
Q
dS 
0
T
Q
dS 
T
All these are called Clausius inequality,
the math expression of the second law of
thermodynamics.
39
Importance of Clausius inequality
The sign of inequality can be used as
the criterion of changing direction and
limitation in thermodynamics.
Q
dS 
T
“>” means irreversible process
“=” means reversible process
dS iso  0
“>” means spontaneous process
“=” means equilibrium state
40
Importance of Clausius inequality
Sometimes we conclude the surrounding
which has the close relationship of the system
to judge the spontaneity of the process, that is:
ΔSiso= ΔS(system)+ΔS(surrounding)≥0
“>” means spontaneous process
“=” means reversible process
41
Principle of entropy increasing
For adiabatic system,
dS  0
• Under the adiabatic condition, the process
which approaches equilibrium makes the
entropy increased.
• Under the adiabatic condition, the process
in which its entropy decreases can not happen.
• As for an isolated system the principle can
be stated as: the entropy of the isolated
system never decrease.
42
Isothermal process
(1) Ideal gas
QR
S 
T
V2
p
S  nR ln( )  nR ln( 1 )
V1
p2
(2) the isothermal isobaric reversible change
S ( phase change) 
H ( phase change)
T ( phase change)
(3) the mixed process of the ideal gas.
mix S   R nB ln xB
B
Example A
43
Non-isothermal process
The isometric non-isothermal process
of the fixed substantial quantity
T nCV , m dT
S  T
T
The isobaric non-isothermal process
of the fixed substantial quantity
T nC p , m dT
S  T
T
2
1
2
1
44
Non-isothermal process
(3) the fixed physical quantity change from
p1,V1,T1 to p2,V2,T2.
Isothermal first and then isometric
T2 nCV ,m dT
V2
S  nR ln( )  
T1
V1
T
Isothermal first and then isobaric
T2 nC p ,m dT
p1
S  nR ln(
)
T1
p2
T
Isobaric first and then isometric
V2
p2
S  nC p ,m ln( )  nCV ,m ln( )
V1
p1
45
Chemical reaction
(1) under the standard pressure and
298.15K, every substance standard molar
entropy can be found out in the table,
we can work out the entropy of the
reaction.
 r S   B S (B)
$
m
$
m
B
46
Chemical reaction
(2) Under the standard pressure, work out
the entropy of the reaction temperature
T.
The entropy’s change at 298.15K can
be found out from the table:
 r S (T )   r S (298.15K)  
$
m
$
m
T
298.15K

B
C p ,m (B)dT
B
T
47
Electrochemical reaction
Work out the entropy from the
thermal effect of the reversible battery
or the temperature change rate of the
electromotive force.
QR
 r Sm 
T
E
 r Sm  zF ( ) p
T
48
Surrounding
(1) the changing surrounding entropy of
every reversible changes
dS ( surrounding )  QR ( system) / T ( surrounding )
If the system heat effect is irreversible,
however, because the surrounding is large
enough, as to the surrounding, it can
be considered as reversible heat effect
dS (surrounding )  Q(system) / T (surrounding )
49
The other calculation
According to the definition of the Gibbs
free energy
G  H  TS
For every isothermal changing process
G  H  TS
S  (H  G ) / T
Apply this method to every thermodynamics equilibrium system.
50
The nature of the 2nd thermodynamics
• Every spontaneous process is irreversible.
• Every irreversible process proceeds
toward the confusion increasing, while
entropy function can be considered as
a measurement of the system confusion,
this is the essence of the irreversible
process.
51
Helmholz free energy and
Gibbs free energy
Why new function is necessary?
The state function of thermodynamics
energy is deduced by the 1st thermodynamics
law. In dealing with the problem of the
thermochemistry, the enthalpy is also
defined. The state function of entropy is
deduced by the 2nd thermodynamics law.
However, when the entropy is used as
criterion, the system must be isolated , that
is, the entropy changes of both system and
surrounding must be considered at the same
52
time, but it is not so convenient.
Helmholz free energy
So it is necessary to bring in the new
thermodynamics function of System
itself to judge the direction and
limitation of the spontaneous change.
Helmholz define a state function:
F  U  TS
def
F is called Helmholz free energy .
53
Helmholz free energy
dF  dU  TdS  SdT
 δQ  δW  TdS  SdT (dU  δQ  δW )
 δWmax
(Isothermal, reversible
or
δQ  TdS)
(  dF )T,R  δW
max
54
Helmholz free energy
That is: during the isothermal, reversible
process, the most work which export to
the surrounding by the system is equal
to the decreasing value of the system
Helmholz free energy, so F is called work
function. If it is an irreversible process,
the work which is done by the system
is less than the decreasing value of F.
55
Helmholz free energy
If the system is under the isothermal,
isobaric condition and do not export
other works
or
(  dF )T,V,W f
 0
0
( dF )T,V,W f
 0
0
“=” : reversible process
“>” : irreversible spontaneous process
The spontaneous change always
proceeds toward the direction that
Helmholz free energy decreases.
56
Gibbs free energy
Gibbs J.W., 1839-1903 define a state
function
G
def
H  TS
G is called Gibbs free energy, it is a
state function, it has capacity properties.
57
Gibbs free energy
dG  dH  TdS  SdT
because dH  dU  d ( pV )
 δQ  δWe  δWf  pdV  Vdp (We  pdV )
 δQ  δWf  Vdp
dG  δQ  δWf  Vdp  TdS  SdT
 δW f, max (dT  0, dp  0, reversible
or
(dG)T,p,R  δWf, max)
58
Gibbs free energy
That is: in the isothermal, isobaric
and reversible process, the most noninflation work which exports to the
surrounding by the system equals to
the decrease value of system Gibbs
free energy. If it is the irreversible
process, the work which is done
by
the system less than the decrease
value of system Gibbs free energy .
59
Gibbs free energy
If the system is in the condition of isothermal,
isobaric and exports non-inflation work,
(dG)T , p ,W 0  0
f
or
(dG)T , p ,W 0  0
f
“=”: reversible process
“<”: irreversible spontaneous process.
The spontaneous change always happens toward
decreasing of Gibbs free energy.
dG is also called isothermal, isobaric potential.
60
Gibbs free energy
In the isothermal, isobaric and reversible
battery reaction
rG   W f, max   nEF
n is the substantial quantity of electron
in the battery reaction,E is the electromotive
force of the reversible battery, F is the
Faraday constant.
61
Equilibrium condition
Qi
S   (
)R  0
Ti
i
ΔSiso= ΔS(system)+ΔS(surrounding)≥0
( dF )T,V,W f
( dG )T,p,W f
 0
 0
0
0
“=” means reversible,
equilibrium
“<” means irreversible, spontaneity
“=” means reversible,
equilibrium
“<” means irreversible, spontaneity
62
∆G calculation
G of isothermal physical change
(1) the ∆G of the change phase of the
isothermal, isobaric and reversible reaction.
G  F  pV
Because it do not export non-inflation work
dF  δWe
dG  dF  pdV  Vdp
 We  pdV  Vdp (We  pdV , dp  0)
0
63
∆G calculation
∆G of isothermal physical change
(2) Under the same temperature, the system
changes from P1,V1 to P2,V2,
suppose Wf=0
dG  We  pdV  Vdp (We  pdV )
p2
G   Vdp
 Vdp
p
For the ideal gas:
1
p2
V1
G  nRT ln  nRT ln
p1
V2
64
∆G calculation
∆G of isothermal reaction change
dD  eE  fF  gG
ΔG1
ΔrG
ΔG3
dD  eE  fF  gG
ΔrG2=0
pD'
pE'
pG
pF
G1  dRT ln
 eRT ln
G 3  fRT ln '  gRT ln '
pD
pE
pF
pG
Δ rG =ΔG1+ ΔG3
pFf pGg
rGm   RT ln K p  RT ln d e  RT ln K p  RT ln Qp
pD pE
65
van’t Hoff isotherm
r Gm   RT ln K p  RT ln Qp
This is called as van’t Hoff isotherm,
or chemical reaction isotherm. Kp is the
equilibrium constant Qp is the pressure
ratio.
66
van’t Hoff isotherm, ctd…
r Gm   RT ln K p  RT ln Qp
• When QP<KP, ∆rGm<0, Reaction proceed
towards the positive way.
• When QP=KP, ∆rGm=0, Reaction is in the
equilibrium state.
• When QP > KP, ∆rGm>0, Reaction proceed
towards the opposite way.
67
The relation among thermodynamics
functions
Definition of thermodynamics function
H  U  pV
F  U  TS
G  H  TS
So G  F  pV
68
Figure relationship among
thermodynamics function
H
H  U  pV
U
TS
TS
pV
G  H  TS
 F  pV
pV
F  U  TS
G
F
函数间关系的图示式
69
Four basic formulas
1
because
dU  TdS  pdV
dU  Q  pdV
• This is the combine formula of the first and
second law of thermodynamics, it can be used
in the closed system which has invariable
component and do not expert non-inflation
work.
• Formula (1) is the most basic formulas among
four.
70
Four basic formulas
(2)
dH  TdS  Vdp
because H  U  pV
dH  dU  pdV  Vdp
dU  TdS  pdV
therefore dH  TdS  Vdp
71
Four basic formulas
(3)
dF   SdT  pdV
because F  U  TS
dF  dU  TdS  SdT
dU  TdS  pdV
therefore dF  SdT  pdV
72
Four basic formulas
(4)
dG  SdT  Vdp
because G  H  TS
dG  dH  TdS  SdT
dH  TdS  Vdp
therefore dG  SdT  Vdp
73
Derivative formulas
(1)
(2)
dU  TdS  pdV
dH  TdS  Vdp
(3)
(4)
dF  SdT  pdV
dG  SdT  Vdp
Deduce from formulas of (1), (2)
U
H
T (
)V  (
)p
S
S
Deduce from formulas of (1), (3)
U
F
p  ( )S  ( )T
V
V
74
Derivative formulas
(1)
(2)
dU  TdS  pdV
(3)
dF   SdT  pdV
dH  TdS  Vdp
(4)
dG   SdT  Vdp
• Deduce from formulas of (2), (4)
H
G
V  ( ) S  ( )T
p
p
•Deduce from formulas of (3), (4)
F
G
S  ( )V  ( ) p
T
T
75
Character functions
Character function and their character
variables.
G(T , p)
H (S , p)
(1)
(2)
F (T ,V )
S (H , p)
U (S ,V )
dU  TdS  pdV
dH  TdS  Vdp
(3)
(4)
dF   SdT  pdV
dG   SdT  Vdp
76
Maxwell formulas
Review of full differential coefficient
z  z ( x, y)
dy
dx
dZ  ( x  dx)( y  dy)  xy
 xdy  ydx  dxdy  xdy  ydx
(Z ) x  xdy
(Z ) y  ydx
z
z
dz  ( ) y dx  ( ) x dy  Mdx  Ndy
x
y
77
Maxwell formulas
M and N are the x, y functions
M
2 z
(
)x 
,
y
xy
M
so
(
)x
y
(1) dU  TdS  pdV
(2) dH  TdS  Vdp
N
2 z
( )y 
x
xy
N
 ( )y
x
p
T
(
)S  ( )V
V
S
T
V
( )S  ( ) p
p
S
78
Maxwell formulas
S
p
)T ( )V
V
T
(3) dF  SdT  pdV
(
(4) dG  SdT  Vdp
S
V
( )T  ( ) p
p
T
So, we can use these partial differential
coefficient which can be measured in the
experiments instead of those which can
not be measured directly.
79
so
Application of Maxwell
formulas
Relationship between U and V
According to 1st basic formula,
dU  TdS  pdV
Work out the partial differential coefficient
of V at the same temperature
U
S
(
)T  T ( )T  p
V
V
(
S
p
( )T  ( ) V
V
T
U
p
)T  T (
)V  p
V
T
80
Why U=f (T) for ideal gas?
As we know,
pV  nRT
p  nRT /V
p
nR
( )V 
T
V

p

U
( )T  T ( )V  p
V
T
nR
T 
 p 0
V
So, Ideal gas’s U is only the function of temperature.
81
Relationship between H and p
According to 2nd basic formula,
dH  TdS  Vdp
Work out the partial differential coefficient of
p at the same temperature
S
V
H
S
( )T  ( ) p
(
)T  T ( )T  V
p
T
p
p
H
V
(
)T  V  T (
)p
p
T
(for ideal gas)=V-T(nR/p)=V-V=0
82
Gibbs-Helmholtz equation
Relationship of temperature and ∆rG, ∆rF.
(1)
(3)
(G)
G  H
[
]p 
T
T
 (F ) F  U
[
]
T V
T
G
)
H
T
[
]p   2
T
T
(
(2)
(4)
F
(
)
U
[ T ]V 
T
T2
83
Gibbs-Helmholtz equation
The deducing of formula (1)
(G)
G  H
(1) [
]p 
T
T
According to the basis formula:
dG  SdT  Vdp
G
( ) p  S
T
(G )
[
] p  S
T
84
Gibbs-Helmholtz equation
According to the definition:
G  H  TS
Under certain T:
G  H  T S
So
therefore
S 
G  H
T
(G)
G  H
[
]p 
T
T
85
Gibbs-Helmholtz equation
G
( )
H
T
(2) [
]p   2
T
T
1
(1)two sides T
1 (G)
G  H
[
]p 
T T
T2
1 (G)
G
H
[
]p  2   2
T T
T
T
(
[
G
)
T ]   H
p
T
T2
 d(
G
H
) p    2 dT
T
T
86
Clausius-Clapeyron equation
Clapeyron equation
Under certain T and p, when two phases of
pure substances are in equilibrium, the change
rate of vapor pressure which changes with
temp. can be figure out by the following
equation:
dp / dT= ∆H / T∆V
87
Deducing of Clapeyron equation
Phase A and Phase B are balanceable in T and
P. Ga=Gb. Changing dT and dP, and the system
is balanceable again. △G=0.
Phase A
Phase B
T
P
Ga
Gb
T+dT
P+dP
Ga +dGa
Gb +dGb
Ga +dGa =Gb +dGb;
dGa =dGb
 S1dT  V1dP   S 2 dT  V2 dP
dP S 2  S1

H


dT V2  V1 TV
88
Clausius-Clapeyron equation
As to the gas-liquid equilibrium
of two phases, 1 mol ideal gas,
neglecting the liquid volume,
 vap H m
dp  vap H m


dT TVm (g) T ( RT / p)
d ln p  vap H m

dT
RT 2
If ∆vapHm has nothing to do with temp,
p2  vap H m 1 1
ln

(  )
p1
R
T1 T2
89
The 3rd thermodynamics law
Thermometric scale of thermodynamics
In the thermodynamics temp scale, temperature
of three-phase-point of water is thought as
273.16 K. Choose 1/273.16 as the unit of thermodynamics temperature, which is called one
Kelvin degree. Every system thermodynamics
temperature is the value compared to it.
Qc
T  273.16K 
Qh
90
The 3rd thermodynamics law
• The temperature of the substance can
not drop to 0 K by finite ways
• When the temp is approaching 0 K, the
entropy of system is unchangeable.
• At 0 K, entropy of every integrity crystals
is zero.
91
The relationship among ∆H, ∆G and
T in gathering system
In 1902, T.W. Richard researched the
relationship of T with ∆G or ∆H, which
were measured by some battery reaction.
∆G and H are almost the same size at
low temperature.
lim(G  H )  0
T 0
T  0K ,
H  G
92
E-mail:
Tel.
segeaj@gmail.com
08033175209;08092979494
93