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THERMODYNAMICS
References





Atkins, P.W., “Physical Chemistry”, Oxford
University Press
Castellan, G.W., “Physical Chemistry”,
Addison Wesley
Levine, I.R., “Physical Chemistry”,
McGraw-Hill
Laidler & Meiser, “Physical Chemistry”,
Houghton Mifflin Co.
Alberty, R.A. and Silbey, R., “Physical
Chemistry”, Wiley
thermo – heat
dynamics – changes
definitions
Events,Experiments,Observations
descriptions
abstract
generalize
no proofs
no violations
Laws of Thermodynamics:
0th, 1st, 2nd, 3rd
Applications, Verifications
Note:


does not worry about rate of changes
(kinetics) but the states before and after
the change
not dealing with time
Classical Thermodynamics
Marcoscopic observables
T, P, V, …
Statistical Thermodynamics
Microscopic details dipole
moment, molecular size, shape
Joule’s experiment
Thermometer
w
adiabatic wall
T  mgh
h
(adiabatic process)
U (energy change) = W (work) = mgh
Page 59
w
T  time interval of
heating
U = q (heat)
Conclusion: work and heat has the same effect
to system (internal energy change)
*
FIRST LAW: U = q + W
U: internal energy is a state function
[ = Kinetic Energy (K.E.)
+ Potential Energy (P.E.) ]
q: energy transfer by temp gradient
W: force  distance
E-potential  charge
surface tension  distance
pressure  volume
First Law:
The internal energy of an isolated system is constant
Convention: Positive: heat flows into system
work done onto system
Negative: heat flows out of system
work done by system
Pressure-volume Work
d  work   Fext dl
  M piston  gdh
weight

 Adh
A
  Pext  Adh
  Pext dV
work   Pext V2  V1 
 Mgh
P1 = P2
M
V1
M
V2
If weight unknown, but only properties of system
are measured, how can we evaluate work?
P
M
V1
M
V2
P
Assume the process is slow and steady,
Pint = Pext
Free Expansion:
Free expansion occurs when the external pressure
is zero, i.e. there is no opposing force
Reversible change: a change that can be
reversed by an infinitesimal modification of a
variable.
Quasi equilibrium process:
Pint = Pext + dP (takes a long time to
complete)
infinitesimal at any time
W   Pext dV
  Pint dV
quasi equilibrium process
Page 64-66
P 1
2
Example:
P1 = 200kPa = P2
V1 = 0.04m3 V2 = 0.1m3
V
*
2

W12    PdV   P V2  V1
1

 200kPa 01
.  0.04m
3
 12kJ
what we have consider was isobaric expansion
(constant pressure) other types of reversible
expansion of a gas: isothermal, adiabatic
Page 65-66
Isothermal expansion
remove sand slowly
at the same time
maintain temperature
by heating slowly
2
Wrev    PdV
21
nRT
 
dV
V
1
2
 nRT 
1
P
*
T
dV
V
 V2 
V1
 nRT ln 
 V1  area under curve
V2
V
Ex. V1 = 0.04m3 P1 = 200kPa V2 = 0.1m3
Wrev
 V2 
 nRT1 ln 
 V1 
 V2 

  PV
1 1 ln
 V1 
 0.1 
 200kPa  0.04m ln

 0.04 
3
 7.33kJ
PV
P2  1 1  80kPa
V2
Adiabatic Reversible Expansion
For this process
PV = constant for ideal gas
(proved later)
 
Cp
(slightly larger
than 1)
Cv

PV
1 1 dV
W    PdV   

V
1
1

1
1
PV
V

V
1 1
2
1

1 
2
2

P2V2  PV
1 1

1 

V2
V1
P1
T1 P2
T2
Ex. V1 = 0.04m3, P1 = 200kPa, V2 =
0.1m3,
 = 1.3
 0.04 
P2  200

 010
. 
W
1.3
 60.77 kPa
 60.77  01
.  200  0.04
1  13
.
d: pressure drop without
volume change
 6.41kJ
200kPa
const
P
isothermal
W    PdV  0
|Wa|>|Wb|>|Wc|>|Wd|
a
b
d
adibatic
V
State Function Vs Path Function
State function: depends only on position in
the x,y plane
e.g.: height (elevation)
300
1
B
2
200
100m
Y
X
A
Path Function: depends on which path is taken
to reach destination
from 1  2, difference of 300m (state function)
but path A will require more effort.
Internal energy is a state function,
heat and work are path functions
P
200kPa 1
2
481.13K
3
0.04
0.1
192.45K
V/m3
5 moles of monoatomic gas
3
CV  R
2
U  nCV T  3 2 R 289  5  18kJ
W1 2  12 kJ
Q1 2  U  12 kJ  30kJ
W1 3 2  7.33kJ  0  7.33kJ
Q1 3 2  7.33kJ  18kJ  25.33kJ
Multivariable Calculus
  f  x, y, z, w,
Consider a 2 variable function
  f  x, y or z  f x, y a surface in three
dimensional plot
z or 
 
C’
C
xc
x
yc
y
At point C, there are two slopes orthogonal
to each other in constant x or constant y
direction
z
C’
C’
z
C
C
x=xC
y=yc
y
x
the change in  can be calculated as a sum
of two parts: change in x direction and
change in y direction
zC  zC 
  
  
 d  
 dx  
 dy
 x  y
 y 
c
partial derivative
w.r.t. x
xc
partial derivative
w.r.t. y
Joule’s second experiment
Energy UU(V,T) ???
thermometer
V1
V2
adiabatic wall
At time zero, open valve
thermometer
No temperature
change
adiabatic wall
After time zero, V1 V1+V2
T=0
 Q=0
W=0 no Pext  U=0
Page 99
U=U(T) Energy is only a function of
temperature for ideal gas *
 U 

 0
 V  T
 U 
 U 
dU  
 dT  
 dV  CV dT
 T V
 V  T
0 (for ideal gas)
CV is constant volume heat
capacity
Kinetic Model for Gases
Qualitatively:
• Gases consists of
spheres of negligible
size, far apart from
one other.
• Particles in ceaseless
random motion; no
interactions except
collisions
U
V
Constant
Temperature
U
V
T
U
 U 
CV  

 T V
T
*
Energy is a function of Volume
and temperature for real gases
Interaction among
molecules
*Enthalpy
Define
Page 101
H = U + PV
state function
intensive variables
locating the state
Enthalpy is also a state function
H = U + PV + VP
At constant pressure
H = U+PV
= U - W
=Q
H = QP constant pressure heating
H  H T , P 
H is expressed as a functional of T and P
 H 
 H 
dH  
 dT  
 dP  C P  0  C P
 T  P
 P  T
C P  CV  R
CP
 
CV
for ideal gases
(proved later)
Thermochemistry
Heat transferred at constant volume qV = U
Heat transferred at constant pessure qP = H
Exothermic
H = -ve
Endothermic H = +ve
Standard states, standard conditions
do not measure energies and enthalpies
absolutely but only the differences, U or H
The choice of standard state is purely a
matter of convenience
Analogy – differences in altitudes between
100 points and their elevation with respect
to sea level
*
What is the standard state ?
The standard states of a substance at a specified
temperature is its pure form at 1 bar
25oC, 1 bar:
the most stable forms of elements assign
“zero enthalpy”
H
o
298
= 0 used for chemical reactions
Standard enthalpy of formation
Standard enthalpy change for the formation
of the compound from its elements in their
reference states.
Reference state of an element is its most stable
state at the specified temperature & 1 bar
C (s) + 2H2 (g)  CH4 (g)
289K, 1 atm
o
Hf
= -75 kJ
From the definition, Hfo for elements  0
Hess’s Law
The standard enthalpy of an overall reaction is the sum of the
standard enthalpies of the individual reactions into which a
reaction may be divided.
Standard reaction enthalpy is the change in enthalpy when the
reactants in their standard states change to products in their
standard states.
Hess’ Law
H2
H3
H4
R 
 X 
Y 
 P
H1
H1 = H2 + H3 + H4
function
state
Hess’s law is a simple application of the first law of thermodynamics
e.g. C (s) + 2H2 (g)  CH4 (g)
298K, 1 atm
H1o = ?
C (s,graphite) + O2 (g)  CO2 (g)
Ho = -393.7 kJ
H2 (g) + ½O2(g)  H2O (l)
Ho = -285.8 kJ
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
Ho = -890.4 kJ
H1o = -393.7 + 2(-285.8) - (-890.4)
= -75 kJ/mole
Heat of Reaction (Enthalpy of Reaction)
Enthalpy change in a reaction, which may be
o
obtained from Hf of products and reactants
Reactants  Products
H r 
o

  H
n p H f , prod 
o
products

i
 nR H f ,reaction
o
reaactan ts
o
f
i
products
 reac tan ts
I stoichiometric coefficient, + ve products,
- ve reactants
E.g. CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)
Hfo/ kJ
CH3Cl
-83.7
HCl
-92.0
Cl2
0
CH4
-75.3
Hro = (-83.7-92.0) - (-75.3+0) = -100.4 kJ
 n
Reactants
R
H f , R o
elements
Products
 n H
elements
p
o
f ,p
Page 83
2A + B  3C + D
0 = 3C + D - 2A - B
Generally,
0 =  J J J
J denotes substances,
J are the stoichiometric numbers
*
H r   j H
o
o
f ,j
Bond energy (enthalpy)
Assumption – the strength of the bond is
independent of the molecular environment
in which the atom pair may occur.
o
C (s,graphite) + 2H2 (g)  CH4(g)H = -75.4 kJ
H2 (g)  2H (g)
Ho = 435.3 kJ
o
C (s,graphite)  C (g)
H = 715.8 kJ
 C (s,graphite) + 2H2 (g)  C (g) + 4H (g)
Ho = 2(435.3)+715.8 = 1586.5 kJ
C (g) + 4H (g)  CH4 (g) H = -75.4-1586.5
= -1661.9 kJ
CH Bond enthalpy = 1661.9/4 kJ = 415.5 kJ
Temperature dependence of Hr
T
H r  H r   C p dT
T
o
To
C P 


C p i ,productsn p 
products
reactants
P
HrT
CP,P
Hro
R
CP,R
298
C p i ,reaction nR 
T
 C
i
reactants
 products
pi
What is the enthalpy change for vaporization
(enthalpy of vaporization) of water at 0oC?
H2O (l)  H2O (g)
Ho = -241.93 - (-286.1) = 44.01 kJmol-1
o
H2 = H + CP(T2-T1) assume CP,i
constant wrt T
H (273) = Ho(298) + CP(H2O,g) - CP (H2O,l)(273-298)
= 44.10 - (33.59-75.33)(-25)
= 43.0 kJ/mole
For ideal Gases:
 H 
 H 
   0,    CP  CV  R
 P  T
 T  P
 U 
 U 
dU    dV    dT
 V  T
 T  V
 H 
 H 
dH    dP    dT
 P  T
 T  P
dH  dU  PdV  VdP
 U 
   dV
 V  T
 V 

 U 
 V 
   dT  VdP  P   dP    dT ,
 T  V
 T  P 
 P  T

 U 
 U   V 
 V 
  dV      dP    dT 
 V  T
 V  T  P  T
 T  P 
 U   U   V 

 V  
 V   U   V  
 dH          P  dT  V  P       dP
 T  P 
 P  T  V  T  P  T 
 T  V  V  T  T  P

  U   V 
 H 
 U 
        P       CV  R
 T  P  T  V   V  T  T  P
0
= R/P for ideal gas

 H 
 U   V 
 RT 
   V  P  
    V  P   2   V  V  0
 P  T
 V  T  P  T
 P 

0 for ideal gas
Prove PV = constant
for adiabatic reversible expansion of an ideal
gas
 U 
dU  CV dT  
 dV
 V T
0 for ideal gas
Adiabatic expansion
dU  Q  W , Q  0
RT
CV dT   PdV  
dV
V
dT
dV
 CV
 R
T
V
CV d ln T    Rd ln V 
T2
R V2
ln  
ln
T1
CV V1
 T2   V2 
    
 T1   V1 
R
CV
P2V2  V2 
  
P1V1  V1 
1
P2  V1 
  
P1  V2 

R
R
CV
CV
 V1 
  
 V2 

 P2V2  P1V1

Reversible vs Irreversible
Non-spontaneous changes vs Spontaneous
changes
Reversibility vs Spontaneity
First law does not predict direction of changes,
cannot tell which process is spontaneous.
Only U = Q + W
Second Law of Thermodynamics
•Origin of the driving force of physical and chemical change
•The driving force: Entropy
•Application of Entropy:
•
Heat Engines & Refrigerators
•
Spontaneous Chemical Reactions
•Free Energy
*
Page 120
Second Law of Thermodynamics
No process is possible in which the sole result is the absorption
of heat from a reservoir and its complete conversion into work
Hot Reservoir
q
w
Engine
Direction of Spontaneous Change
More Chaotic !!!
Spontaneous change is usually accompanied by a
dispersal of energy into a disorder form, and its
consequence is equivalent to heating
Entropy (S) is a measurement of the randomness
of the system, and is a state function!
SQ
S 1/T
Expansion into a vacuum
irreversible
reversible
Vac
V2
V2
V1
V1
Wirr    Pext dV  0
U  Qirr  0
V2  V1 ,Wmin
V2
 nRT ln
V1
Wirr = 0 = qirr = U
Wrev    Pext dV
V2
  nRT ln
V1
q rev  Wrev
qrev > qirr
-Wrev > -Wirr
Page 122
Entropy S
For a reversible process, the change of entropy is defined as
dqrev
dS 
T
(thermodynamic definition of the entropy)
*
Another expression of the Second Law:
The entropy of an isolated systems increases in the course of a
spontaneous change:
Stot > 0
where Stot is the total entropy of the isolated system
Entropy S
The entropy of an isolated systems increases in the course of a
reversible change:
Stot = 0
where Stot is the total entropy of the isolated system
Entropy Change for an isothermal expansion of a perfect gas
V2
V1
S  1
T
 dq
rev
q rev
V2

 nR ln
T
V1
W rev    Pext dV
Depend only on V
V2
  nRT ln
V1
q rev  W rev
Entropy is a state function
Carnot’s Theoretical Heat Engine
Heat flows from a high temperature reservoir
to a low temperature body. The heat can be
utilized to generate work.
e.g. steam engine.
Consider the sequence of reversible processes
For the cycle,
-Wnet = (W1+W2+W3+W4)
= qnet = q1+q3
Ucycle=0
q1 positive
q3 negative
qnet positive
 |q3| < |q1|
Net effect of the gas going through a cycle
|q1|
qH
|Wnet|
W
|q3|
qL
Efficiency of theoretical heat engine
 Wnet
 Wnet  Wnet



available heat
qH
q1
q3
q1  q3
q3

 1  1
q1
q1
q1
Carnot theorem: Engines operating between
two temperature TH, TL have
the same efficiency
VB
 W1  nRT1 ln
 q1 , U  0
VA
 W3  nRT3 ln
 1
VD
 q3 , U  0
VC
T3  VC 
T3  VD 
 ,

 
 
T1  VB 
T1  V A 
VB VC


V A VD
 1
VB
VD
nRT1 ln
 nRT3 ln
q1  q3
VA
VC
T3


 1
VB
q1
T1
nRT1 ln
VA
th
TL
 1
TH
limiting thermal efficiency of a heat engine
In actual cases, heat engine have much lower
efficient.
 irreversible processes, friction losses, etc..
Kelvin: It is impossible, by means of a cycle to
take heat from q reservoir and convent
it to work without at the same time
transferring heat from a hot to cold
reservoir. (We cannot have a 100%
efficient heat engine)
Show that 2nd law and Kelvin’s principle are
equivalent
Examine an adiabatic irreversible process.
We want to evaluate the entropy change for
the process by an reversible path BCDA
D
C
Wnet
A
B
Q=0
irreversible
B  C adiabatic
compression
C  D isothermal
compression
D  A adiabatic
expansion
1st law for the cycle, U = 0
-(WAB + WBA) = QBA + QAB
-Wnet = QBA
0
By Kelvin’s principle, -Wnet must be negative,
otherwise a 100% efficient heat engine
 QBA = -Wnet must be negative
QBA  0
S A B
Q A B

rev
T
QB  A

TH
0
Evaluation of Entropy Changes
isothermal expansion:
TdS  dq rev  dU  PdV
dq rev
PdV
S  

T
T
 V2 
 P1 
nRdV

 nR ln   nR ln 
V
 V1 
 P2 
isobaric heating:
S 
Tm

T1
C P , solid
T
H m b C P ,liq
H v 2 C p , gas
dT 

dT 

dT
Tm Tm T
Tb Tb T
T
T
Tm: melting pt., Tb: boiling pt.,
If CP = constant,
 T2 
S  C P ln 
 T1 
H/S
V
L
S
T1 Tm
Tb
T2
change of temperature and volume (pressure)
 U 
TdS  dU  PdV  
 dT  PdV  nCV dT  PdV
 T  V
2
2
CV
P
S  n 
dT   dV
T
T
1
1
2
P
T2
V2
 nCV ln  nR ln
T1
V1
T2
P1
 nC P ln  nR ln
T1
P2
1
V
Page 142
The efficiencies of heat engines
Hot Reservoir
q
w
Engine
*
S = - |q|/Th < 0
not possible! contrary to the second law
Page 130
The efficiencies of heat engines
Sh = - |qh|/Th
Hot Reservoir
S = - |qh|/Th + |qc|/Tc  0
qh
w
qc
Cold Reservoir
|wmax| = |qh|- |qc,min|
= (1- Tc /Th) |qh|
Maximum efficiency:
erev = |wmax|/|qh|= 1- Tc /Th
Sc = + |qc|/Tc
erev 1 as Tc  0 or Th 
Page 144
Energetics of Refrigeration
Hot Reservoir
Sh = + |qc|/Th
S = - |qc|/Tc + |qc|/Th < 0
|qc|
Cold Reservoir
not possible!
Sc = - |qc|/Tc
Page 144
The energetics of refrigeration
Sh = + |qh|/Th
S = |qh|/Th - |qc|/Tc  0
Hot Reservoir
qh
qh
w
qc
Cold Reservoir
|wmin| = |qh,min|- |qc,|
= (Th /Tc-1) |qc|
Maximum efficiency of
performance:
crev = |qc|/ |wmin|
= Tc /(Th- Tc)
Sc = - |qc|/Tc
Page 145
The energetics of refrigeration
Sh = + |qh|/Th
Hot Reservoir
How to keep it cool?
dqc/dt = A(Th -Tc)
qh
qh
w
d|w|/dt = (1/ crev)  dqc/dt
= A (Th -Tc)2 / Tc
qc
Cold Reservoir
Sc = - |qc|/Tc
The Nernst Theorem
The entropy change accompanying any physical or
chemical transformation approaches zero as the
temperature approaches zero.
S  0 as T0
* Third Law of Thermodynamics
Page 147
If the entropy of each element in its most
state is taken as zero at the absolute zero
of temperature, every substance has a
positive entropy. But at 0K, the entropy of
substance may equals to 0, and does
become zero in perfect crystalline solids.
Implication: all perfect materials have the same
entropy (S=0) at absolute zero temperature
Crystalline form: complete ordered,
minimum entropy
Statistical Interpretation of S
S=0
at 0K for perfect crystals
S = k ln 
Boltzmann
postulate of
entropy
number of arrangements
Boltzmann constant
Entropy Change of Mixing
one distinguish
arrangement
4!
4!
S A  k ln
 0, S B  k ln
0
4!
4!
 A B
N
 N B !
number of
arrangement
increased
87  65


 70
NA!NB !
4  3 2 1
A
S A B  k ln 70
In general mixing NA, NB
S mix

N A  N B !
 k ln
N A! N B !
Entropy change of mixing
Stirling’s approximation: ln N!  N ln N + 0(N)
for large N
S mix  k  N A  N B  ln  N A  N B   N A ln N A  N B ln N B 
 k  N A  N B ln  N A  N B   X A ln N A  X B ln N B 
 k  N A  N B  X A ln X A  X B ln X B   0
From classical thermodynamics, isothermal
reversible expansion of gases A & B
n A RT  V A  VB
 qrev 
  W rev 
ln 

 
 
T
 T A  T A
 VA
 V A  VB 

 n A R ln 
 VA 
 V A  VB
 qrev 

  nB R ln 
 T B
 VB






 V A  VB 
 V A  VB
  nB R ln 
S mix  n A R ln 
 VA 
 VA
 nT R X A ln X A  X B ln X B 
0



Assignment (due on 06/12/1999)
2.4, 2.5, 2.37, 3.5, 3.23, 4.10, 4.29
Page 133
* Extensions of 2nd Law
TdS dq
Clausius Inequality
For adiabatic process,
TdS  0 or dS  0
Entropy will always attain maximum in
adiabatic processes.
A similar function for other processes?
*
*
Page 149
Define Helmholtz free energy
A = U - TS
Thermodynamic
State Function
dA = dU - TdS - SdT
Substitute into Clausius Inequality
0  dq  TdS
0  dq  dA  dU  SdT
0  PdV  dA  SdT
for isothermal, isochoric (constant volume)
process,
0  dA
* A will tend to a minimum value
isothermal, isochoric process,
dA  0 equilibrium
dA  0 spontaneous
If only isothermal,
0  dA  PdV
 PdV  dA
W  dA
isothermal reversible expansion
V2
W   RT ln
 A
V1
change in Helmholtz free energy = maximum
isothermal
work
Example of isothermal, isochoric process:
combustion in a bomb calorimeter
O2 +
fuel
Temp. bath
heat given
out
O2, CO2,
H2O
Higher P
*
Gibbs Free Energy
Define Gibbs free energy
G = H - TS
Page 149-155
Thermodynamic
state function
= U + PV -TS
dG = dU +PdV + VdP -TdS -SdT
substitute into Clausius inequality
0  dq  TdS
0  dU  PdV  TdS
0  dG  VdP  SdT
 dG  VdP  SdT
*
constant pressure, constant temperature
0  dG
*
G will tend to a minimum value
dG  0 equilibrium
dG  0 spontaneous change
G
Page 154
time
More applications since most processes are
isothermal, isobaric
chemical reactions at constant T, P
Reactants  Products
endothermic H is positive
exothermic
H is negative
H is not a criteria for spontaneity
S only isolated system
G
G = H - TS
S
H
G
+ ve
- ve
- ve
spontaneous dissociation of unstable
compounds
- ve
+ ve
+ ve
nonspontaneous
+ ve
+ ve
?
?
dissociation of a strong
compound
- ve
- ve
?
?
recombination reaction
e.g. H + H  H2
R
forming unstable
compounds
P
rxn. Co-ord.
Change of Gibbs free energy with temperature
(constant pressure)
dG   SdT VdP  0
G2  G1    SdT
 G 

  S
 T  P
Change of Gibbs free energy with pressure
(constant temperature)
dG  VdP SdT  0 
G2  G1   VdP

 RT ln 


 RT ln 




f2 

f1 
P2
P1
 P 
G  G o  RT ln  o 
P 
 f 
G  G o  RT ln  o 
P 
Ideal gas
Real gas fugacity
Ideal gas
Real gas
Page 168-174
Gibbs free energy
*G  G o  RT ln  pi 
i
o
P 
 fi 
o
 G  RT ln  o 
P 
Ideal Gases
Real Gases
Total Gibbs free energy of a mixture of gases
G  nAG A  nBGB
pA
pB
o
 n AG  n A RT ln o  nBGB  nB RT ln o
P
P
o
A

 n X AG A,i  X BGB ,i  X A RT ln X A  X B RT ln X B


 n X i G  X i RT ln X i
o
i

i
Gmix  TS mix   X i RT ln X i 
i
H mix  0
for ideal gases
Chemical Equilibria
aA + bB  cC + dD
consider G for a pass of the reaction at
constant T & P
G = cGC + dGD - aGA - bGB
molar Gibbs free energy
Gi  G  RT ln f i
o
i
pure state
fi in the unit of bar
fugacity
G  cGCo  dG Do  aG Ao  bGBo  cRT ln f C  dRT ln f D  aRT ln f A  bRT ln f B
c
d
fC f D
 G  RT ln a b
f A fB
o

  i G  RT ln  f i
o
i
i

At equilibrium, G = 0
* G o
 f cf d

 ln  C a D b   ln K
RT
 f A fB 
0 = SJ J J
equilibrium constant
for ideal gaseous mixture
ln
pC p D
c
d
a
b
p A pB
G o

RT
fi = pi
ln K p  const .
K p  const .
ni RT
pi 
 c i RT
V
Kp 
c D cC
d
c
b
a
cB c A
RT 
n
ci = concentration
 K c  RT 
n
pi
Xi 
pT
n = c+d-a-b
varies with
total pressure
d
c
d
c
X D X C PT PT
 n
Kp 

K
P
x
a
b
a
b
X A X B PT PT
For liquids, use activity
aC a D
G

 ln a b  ln Keq
RT
a A aB
o
c
d
Gibbs - Helmholtz Equation
G  VdP  SdT
 G 

  S
 T  P
 G 
Also, G  H  TS  H  T  
 T  P
G
H  G 



T
T  T  P
 
 GT
T
P
 
 GT
T
P
G
1  G 
 2  

T  T  P
T
H
 2
T
 G 
T

 H
or
1 

TP
Gibbs - Helmholtz Equation
For a reaction
  G 
T    H

2
 T 
T

P
G/T
  G 
T

 H
 1 

T P
Similarly,
Slope
=H
1/T
G2
G/T
  A 
T   U

 1 

T P
H
G1
1/T
Change of equilibrium constant with temperature
GT   RT ln K
o
GT o
  R ln K
T
Gibbs Helmholtz equation
o
 G T
H T
 2
T
T
o
H T
 ln K
 R
 2
T
T
O
 ln K H To

T
RT 2
Van’t Hoff Equation
o
If H rxn
constant (i.e. HP -HR is constant)
H o
ln K  
 cons tan t
RT
endothermic, H + ve, K
exothermic, H - ve, K
with T
with T
*
Relation between Thermodynamic functions
dU = TdS-PdV
dH = TdS+VdP,
dA = -SdT-PdV,
dG = -SdT+VdP,
1st law
H = U+PV
A = U-TS
G = H-TS
Page 149
From multivariable differential calculus
dz = M dx + N dy total differential, i.e.
z depends on x & y
 M 
 N 


 
 y  x  x  y
Maxwell Relations
 T 
 P 
 T 
 V 
     ,
   
 V  S
 S  V
 P  S  S  P
 S 
 P 
    ,
 V  T  T  V
Page 163
 S 
 V 
    
 P  T
 T  P
Phase equilibrium
Clapeyron equation
G=0
molar Gibbs free energy G=G-G



G  G
dG  dG
P
liq
V dP  S dT  V dP  S  dT
dP S   S
S
H



dT V  V
V TV
s
Clausius- Clapeyron equation
For vaporization and sublimation
vap
T
H vap PH vap
dP H vap



dT TVvap
TVvap
RT 2
H vap
dP

 d  ln P  
2 dT
P
RT
H vap
sat
 ln P  
sat  const .
RT
P2 H vap T2  T1 
ln

P1
RT1T2
In P
1/T
Example: What is the change in the boiling
point of water at 100oC per torr
change in atmospheric pressure?
Hvap = 9725 cal mol-1
Vliq = 0.019 l mol-1
Vvap = 30.199 l mol-1
sat
dP
dT sat
9725 cal mol 0.04129 l atmcal 


T V  V 
37315
. K 30180
.
l mol 
H vap
1
1
1
v
l
 0.03566 atm K 1
 27.10 torr K 1
dT
 0.0369 K torr 1
dP
Example: Calculate the change in pressure
required to change the freezing
point of water 1oC.
At 0oC, the heat of the fusion of ice is 79.7cal g-1,
the density of water is 0.9998 g cm-3 and that of
ice is 0.9168 g
1 
 1
Vl  Vs  

  9.06  10 5 l g 1
 0.9998 0.9168 
H fus
P

sat 
T
T Vl  Vs 
sat
79.7 cal g 1 0.04129 l atmcal 1 
2731
. K   9.06  10 5 l g 1 
 133 atm K 1
50kg
P
liq
solid
skate blade
vap
T
ice
liq. H2O
Example 2.1
A certain electric motor produced 15 kJ of energy each second as
mechanical work and lost 2 kJ as heat to the surroundings. What
was the change in the internal energy of the motor and its power
supply each second?
Example 2.2
Calculate the work done when 50 g of iron reacts with hydrochloric
acid in: (a) a closed vessel of fixed volume; (b) an open beaker at
25oC.
Example 2.3
The internal energy change when 1.0 mole CaCO3 in the form of
calcite converts to aragonite is 0.21 kJ. Calculate the difference
between the enthalpy change and the change in internal energy.
Example 2.4
The enthalpy change accompanying the formation of 1.00 mole
NH3(g) from its elements at 298 K is -46.1 kJ. Estimate the
change in internal energy.
Example 2.5
Water is heated to boiling under pressure of 1.0 atm. When an
electric current of 0.50 A from 12-V supply is passed for 300 s
through a resistance in thermal contact with it, it is found that
0.798 g of water is vaporized. Calculate the molar internal energy
and enthalpy changes at the boiling point (373.15 K).
Exercise 2.6
At very low temperatures the heat capacity of a solid is proportional
to T3, and we can write Cv=aT3. What is the change in enthalpy of
such a substance when it is heated from 0 to T?
Example 3.6
A sample of argon at 1.0 atm pressure and 25oC expands reversibly
and adiabatically from 0.50 L to 1.00 L. Calculate its final temperature, the work done during the expansion, and the change in internal
energy. The molar heat capacity of argon at constant volume is
12.48 JK-1 mol-1.
Example 4.1
Calculate the entropy change in the surroundings when 1.00 mol
H2O(l) is formed from its elements under standard conditions at
298.15 K.
Example 4.4
Calculate the entropy change when argon at 25oC and 1.00 atm in
a container of volume 500 cm3 is allowed to expand to 1000 cm3
and is simultaneously heated to 100oC.
Example 5.4
The pressure deep inside the Earth is probably greater than 3×103 kbar,
and the temperature is around 4×103 oC. Estimate the change in G
on going from crust to core for a process in which V=1.0 cm3 mol-1
and S = 2.1 JK-1 mol-1.
Example 5.5
Calculate the change in molar Gibbs energy when water vaporizes at
1 bar and 25 oC. Note that the molar Gibbs energies of formation are
-237.13 and -228.57 kJ mol-1 for water and its vapor, respectively.
Example 5.6
Suppose that the attractive interactions between gas particles can be
neglected and find an expression for the fugacity of a van der Waals
gas in terms of the pressure.