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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
THERMODYNAMICS
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definitionsdescriptions
no proofs
no violations
Events,Experiments,Observations
Laws of Thermodynamics:0th, 1st, 2nd, 3rd
Applications, Verifications
abstract generalize
dynamics – changesthermo – heat
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Note: does not worry about rate of changes (kinetics) but the states before and after the change not dealing with time
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Classical Thermodynamics
Marcoscopic observables T, P, V, …
Statistical Thermodynamics
Microscopic details dipole moment, molecular size, shape
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Joule’s experiment
T mgh adiabatic wall
(adiabatic process)
U (energy change) = W (work) = mgh
w
h
Thermometer
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w
T time interval of heatingU = q (heat)
Conclusion: work and heat has the same effect to system (internal energy change)
FIRST LAW: U = q + W*
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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
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Positive: heat flows into system work done onto systemNegative: heat flows out of system work done by system
Convention:
Pressure-volume Work
P1 = P2
V2V1
MM
d work F dlext M gdhpiston
weight
AAdh
P Adhext
P dVext work P V Vext 2 1
Mg h
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If weight unknown, but only properties of systemare measured, how can we evaluate work?
PV1V2P
MM
Assume the process is slow and steady,
Pint = Pext
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Free Expansion:
Free expansion occurs when the external pressure is zero, i.e. there is no opposing force
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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
quasi equilibrium process
ò dVPW extò dVPint
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1 2P
V
Example: P1 = 200kPa = P2
V1 = 0.04m3 V2 = 0.1m3
W PdV1 21
2
P V V2 1
200 01 0 04 3kPa m. . 12kJ
what we have consider was isobaric expansion(constant pressure) other types of reversible expansion of a gas: isothermal, adiabatic
*
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Isothermal expansion
remove sand slowlyat the same time maintain temperature by heating slowly
W PdVrev 1
2
nRT
VdV
1
2
nRTdV
V1
2
nRT
V
Vln 2
1
T
VV1 V2
P
area under curve
*
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Ex. V1 = 0.04m3 P1 = 200kPa V2 = 0.1m3
W nRTV
Vrev 1
2
1
ln
PV
V
V1 12
1
ln
200 0 04
01
0 043kPa m. ln
.
. 7 33. kJ
PPV
VkPa2
1 1
2
80
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Adiabatic Reversible Expansion
For this processPV = constant for ideal gas(proved later)
V1
P1 T1
V2
P2 T2
(slightly larger than 1)
C
Cp
v
W PdVPV dV
V 1 1
1
2
1
2
PV V V1 1 21
11
1
P V PV2 2 1 1
1
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Ex. V1 = 0.04m3, P1 = 200kPa, V2 = 0.1m3,
= 1.3
P kPa2
1 3
2000 04
01060 77
.
..
.
W kJ
60 77 01 200 0 04
1 136 41
. . .
..
volume change
|Wa|>|Wb|>|Wc|>|Wd|
W PdV 0
a
db
V
P
200kPaconst
isothermal
adibatic
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State Function Vs Path Function State function: depends only on position in
the x,y plane e.g.: height (elevation)
300200
100mB
AX
Y
1 2
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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
3
2481.13K
192.45K
P
200kPa
V/m30.04 0.1
1
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5 moles of monoatomic gas
C R
U nC T R kJ
W kJ
Q U kJ kJ
W kJ kJ
Q kJ kJ kJ
V
V
3
23
2 289 5 18
12
12 30
7 33 0 7 33
7 33 18 25 33
1 2
1 2
1 3 2
1 3 2
. .
. .
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f x y z w, , , ,
Consider a 2 variable functionor f x y, z f x y , a surface in three
dimensional plot
C
C’
z or
yx
ycxc
Multivariable Calculus
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At point C, there are two slopes orthogonal to each other in constant x or constant y direction
y
zz
x x=xC y=yc
C’C C
C’
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the change in can be calculated as a sum of two parts: change in x direction and change in y direction
z z dx
dxy
dyC Cy xc c
partial derivative partial derivative w.r.t. x w.r.t. y
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Joule’s second experiment
Energy UU(V,T) ???
V1V2
thermometer
At time zero, open valve
adiabatic wall
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After time zero, V1 V1+V2
T=0 Q=0 W=0 no Pext U=0
thermometer
No temperature change
adiabatic wall
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dTCdVV
UdT
T
UdU
V
U
VTV
T
0
0 (for ideal gas)CV is constant volume heat capacity
U=U(T) Energy is only a function of temperature for ideal gas*
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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
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U
V
T
U
V
ConstantTemperature
VV T
UC
T
U
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Energy is a function of Volume and temperature for real gases
Interaction among molecules
*
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*Enthalpy
Define H = U + PV state function intensive variables locating the state
Enthalpy is also a state function
H = U + PV + VP
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PPTP
CCdPP
HdT
T
HdH
PTHH
0
,
for ideal gases C C R
C
C
P V
P
V
(proved later)
At constant pressureH = U+PV = U - W = QH = QP constant pressure heating
H is expressed as a functional of T and P
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Thermochemistry Heat transferred at constant volume qV = U
Heat transferred at constant pessure qP = H
Exothermic H = -ve
Endothermic H = +ve
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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
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The standard states of a substance at a specified temperature is its pure form at 1 bar
What is the standard state ?*
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25oC, 1 bar: the most stable forms of elements assign “zero enthalpy”
Ho298 = 0 used for chemical reactions
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Standard enthalpy of formation
Standard enthalpy change for the formationof the compound from its elements in their reference states.
Reference state of an element is its most stablestate at the specified temperature & 1 bar
C (s) + 2H2 (g) CH4 (g) Hfo = -75 kJ
289K, 1 atm
From the definition, Hfo for elements 0
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Hess’s Law
The standard enthalpy of an overall reaction is the sum of thestandard enthalpies of the individual reactions into which areaction 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.
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Hess’ Law
R X Y PH H H 2 3 4
H1
H1 = H2 + H3 + H4 state function
Hess’s law is a simple application of the first law of thermodynamics
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e.g. C (s) + 2H2 (g) CH4 (g) H1o = ?
298K, 1 atm
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 H1
o = -393.7 + 2(-285.8) - (-890.4)
= -75 kJ/mole
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Heat of Reaction (Enthalpy of Reaction) Enthalpy change in a reaction, which may be obtained from Hf
o of products and reactants
Reactants Products
products tsreaac
oreactionfR
oprodfp
or HnHnH
tan,,
i fo
iproductsreac ts
Htan
I stoichiometric coefficient, + ve products,
- ve reactantsE.g. CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)
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Hfo/ kJ
CH3Cl -83.7
HCl -92.0
Cl2 0
CH4 -75.3
Hro = (-83.7-92.0) - (-75.3+0) = -100.4 kJ
Reactants Products
elements elements
n HR f Ro , n Hp f p
o ,
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jfo
jo
r HH ,
2A + B 3C + D
0 = 3C + D - 2A - B
Generally,0 = J J J
J denotes substances, J are the stoichiometric numbers
*
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Bond energy (enthalpy)
Assumption – the strength of the bond is independent of the molecular environment in which the atom pair may occur.
C (s,graphite) + 2H2 (g) CH4(g)Ho = -75.4 kJ
H2 (g) 2H (g) Ho = 435.3 kJ
C (s,graphite) C (g) Ho= 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 kJCH Bond enthalpy = 1661.9/4 kJ = 415.5 kJ
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Temperature dependence of Hr
productsreactantsreactants
reaction,products
products, ipiRippipP
T
T
po
rT
r
CnCnCC
dTCHHo
CP,R
CP,P
Hro
P
R
HrT
298 T
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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 H2 = Ho + 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
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For ideal Gases:
H
P
H
TC C R
T PP V
0,
dUU
VdV
U
TdT
dHH
PdP
H
TdT
T V
T P
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dH dU PdV VdP
U
VdV
U
TdT VdP P
V
PdP
V
TdT
U
VdV
U
V
V
PdP
V
TdT
dHU
T
U
V
V
T V T P
T T T P
V T
,
TP
V
TdT V P
V
P
U
V
V
PdP
H
T
U
TP
U
V
V
TC R
P P T T T
P V T PV
0 = R/P for ideal gas
H
PV P
U
V
V
PV P
RT
PV V
T T T
2 0
0 for ideal gas
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Prove PV = constant
for adiabatic reversible expansion of an ideal gas
dVV
UdTCdU
TV
0 for ideal gas
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VCR
V
V
V
V
V
V
T
T
V
V
C
R
T
T
VRdTdCV
dVR
T
dTC
dVV
RTPdVdTC
QWQdU
1
2
1
2
1
2
1
2
0
lnln
lnln
,
Adiabatic expansion
1122
2
1
1
2
1
1
2
1
2
11
22
VPVP
V
V
V
V
P
P
V
V
VP
VP
V
V
C
R
CR
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Reversible vs IrreversibleNon-spontaneous changes vs Spontaneous
changesReversibility vs Spontaneity
First law does not predict direction of changes,cannot tell which process is spontaneous. Only U = Q + W
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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
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Second Law of Thermodynamics
No process is possible in which the sole result is the absorptionof heat from a reservoir and its complete conversion into work
Hot Reservoir
q
w
Engine
* Page 120
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Direction of Spontaneous Change
More Chaotic !!!
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Entropy (S) is a measurement of the randomness of the system, and is a state function!
S Q S 1 / T
Spontaneous change is usually accompanied by a dispersal of energy into a disorder form, and its consequence is equivalent to heating
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Expansion into a vacuum
irreversible reversible
Vac
V1
V2
V1
V2
W P dV
nRTV
V
q W
rev ext
rev rev
ln 2
1
W P dV
U Q
V V W nRTV
V
irr ext
irr
0
0
2 12
1
, lnmin
qrev > qirr
-Wrev > -Wirr Wirr = 0 = qirr = U
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T
dqdS rev
Entropy S
For a reversible process, the change of entropy is defined as
Another expression of the Second Law:
The entropy of an isolated systems increases in the course of aspontaneous change:
Stot > 0
where Stot is the total entropy of the isolated system
(thermodynamic definition of the entropy)
*
Page 122
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Entropy S
The entropy of an isolated systems increases in the course of areversible change:
Stot = 0
where Stot is the total entropy of the isolated system
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V1
V2
revrev
extrev
revrev
Wq
V
VnRT
dVPW
V
VnR
T
qdqS T
1
2
1
21
ln
ln
Entropy Change for an isothermal expansion of a perfect gas
Entropy is a state function
Depend only on V
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Carnot’s Theoretical Heat Engine
Heat flows from a high temperature reservoirto a low temperature body. The heat can be utilized to generate work.
e.g. steam engine.
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For the cycle,-Wnet = -(W1+W2+W3+W4)
= qnet = q1+q3
Ucycle=0
q1 positive
q3 negative
qnet positive
|q3| < |q1|
Consider the sequence of reversible processes
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Net effect of the gas going through a cycle
|q1| qH
|q3| qL
|Wnet| W
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Efficiency of theoretical heat engine
1
3
1
3
1
31
1
11q
q
q
q
q
q
W
q
W
heatavailable
W net
H
netnet
Carnot theorem: Engines operating between two temperature TH, TL have
the same efficiency
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H
Lth
A
B
C
D
A
B
D
C
A
B
A
D
B
C
C
D
A
B
T
T
T
T
VV
nRT
VV
nRTVV
nRT
q
V
V
V
V
V
V
T
T
V
V
T
T
UqV
VnRTW
UqV
VnRTW
1
1
0
0
1
3
1
31
1
31
1
1
3
1
1
3
333
111
ln
lnln
,
,ln
,ln
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limiting thermal efficiency of a heat engineIn actual cases, heat engine have much lowerefficient. 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)
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Show that 2nd law and Kelvin’s principle areequivalent
Examine an adiabatic irreversible process. We want to evaluate the entropy change for the process by an reversible path BCDA
D C B C adiabatic
compression C D isothermal A compression B D A adiabatic Q=0 expansion
Wnet
irreversible
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1st law for the cycle, U = 0 0-(WAB + WBA) = QBA + QAB
-Wnet = QBA
By Kelvin’s principle, -Wnet must be negative,
otherwise a 100% efficient heat engine QBA = -Wnet must be negative
QBA 0
SQ
Trev
Q
T
A BA B
B A
H
0
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Evaluation of Entropy Changesisothermal expansion:TdS dq dU PdV
Sdq
T
PdV
T
nRdV
VnR
V
VnR
P
P
rev
rev
ln ln2
1
1
2
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Tm: melting pt., Tb: boiling pt.,
If CP = constant,
S CT
TPln 2
1
H/S VL
S
Tm Tb T2
isobaric heating:
2
1
T
T
gasp
b
v
T
T
liqP
m
m
T
T
solidP
b
b
m
m
dTT
C
T
HdT
T
C
T
HdT
T
CS ,,,
T1
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change of temperature and volume (pressure)
TdS dU PdVU
TdT PdV nC dT PdV
S nC
TdT
P
TdV
nCT
TnR
V
V
nCT
TnR
P
P
VV
V
V
P
1
2
1
2
2
1
2
1
2
1
1
2
ln ln
ln ln 1
2P
V
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The efficiencies of heat engines
Hot Reservoir
q
w
Engine
S = - |q|/Th < 0 not possible! contrary to the second law*
Page 142
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The efficiencies of heat engines
Hot Reservoir
qh
w
Cold Reservoir
qc
Sh = - |qh|/Th
Sc = + |qc|/Tc
S = - |qh|/Th + |qc|/Tc 0
|wmax| = |qh|- |qc,min| = (1- Tc /Th) |qh|
Maximum efficiency:rev= |wmax|/|qh|= 1- Tc /Th
rev 1 as Tc 0 or Th
Page 130
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Energetics of Refrigeration
Hot Reservoir
Cold Reservoir
Sh = + |qc|/Th
Sc = - |qc|/Tc
S = - |qc|/Tc + |qc|/Th < 0
not possible!|qc|
Page 144
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The energetics of refrigeration
Hot Reservoir
qh
w
Cold Reservoir
qc
Sh = + |qh|/Th
Sc = - |qc|/Tc
S = |qh|/Th - |qc|/Tc 0
|wmin| = |qh,min|- |qc,| = (Th /Tc-1) |qc|
Maximum efficiency of performance:crev= |qc|/ |wmin| = Tc /(Th- Tc)
qh
Page 144
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The energetics of refrigeration
Hot Reservoir
qh
w
Cold Reservoir
qc
Sh = + |qh|/Th
Sc = - |qc|/Tc
How to keep it cool?dqc/dt = A(Th -Tc)
d|w|/dt = (1/ crev) dqc/dt = A (Th -Tc)2 / Tc
qh
Page 145
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The Nernst Theorem
The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero.
S 0 as T0
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Third Law of Thermodynamics
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.
Crystalline form: complete ordered, minimum entropy
Implication: all perfect materials have the same entropy (S=0) at absolute zero temperature
* Page 147
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Statistical Interpretation of S
S = 0 at 0K for perfect crystals S = k ln
Boltzmann number of arrangements postulate of entropy Boltzmann constant
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Entropy Change of Mixing
one distinguish arrangement
S k S kA B ln!
!, ln
!
!
4
40
4
40
A B
A B
A B
A B
N N
N N
S k
!
! !
ln
8 7 6 5
4 3 2 170
70
number of arrangement increased
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In general mixing NA, NB
!!
!ln
BA
BAmix NN
NNkS
Entropy change of mixingStirling’s approximation: ln N! N ln N + 0(N)
for large N
0
BBAABA
BBAABABA
BBAABABAmix
XXXXNNk
NXNXNNNNk
NNNNNNNNkS
lnln
lnlnln
lnlnln
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From classical thermodynamics, isothermal reversible expansion of gases A & B
0
BBAAT
A
BAB
A
BAAmix
B
BAB
B
rev
A
BAA
A
BAA
A
rev
A
rev
XXXXRn
V
VVRn
V
VVRnS
V
VVRn
T
q
V
VVRn
V
VV
T
RTn
T
W
T
q
lnln
lnln
ln
ln
ln
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Assignment (due on 06/12/1999)
2.4, 2.5, 2.37, 3.5, 3.23, 4.10, 4.29
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Extensions of 2nd Law
TdS dq Clausius Inequality
For adiabatic process,
TdS or dS 0 0
Entropy will always attain maximum in adiabatic processes.
A similar function for other processes?
*Page 133
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Define Helmholtz free energyA = U - TS Thermodynamic
State Function dA = dU - TdS - SdTSubstitute into Clausius Inequality
0
0
0
dq TdS
dq dA dU SdT
PdV dA SdTfor isothermal, isochoric (constant volume)process,
0dA
*
*
Page 149
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A will tend to a minimum value
isothermal, isochoric process,dA 0dA 0
equilibriumspontaneous
If only isothermal, 0
dA PdV
PdV dA
W dAisothermal reversible expansion
W RTV
VA ln 2
1
*
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change in Helmholtz free energy = maximum isothermal work
Example of isothermal, isochoric process: combustion in a bomb calorimeter
O2 +fuel
Temp. bath
O2, CO2,H2O
Higher P heat givenout
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Gibbs Free Energy
Define Gibbs free energy G = H - TS Thermodynamic
state function = U + PV -TS dG = dU +PdV + VdP -TdS -SdT
substitute into Clausius inequality
SdTVdPdG
SdTVdPdG
TdSPdVdU
TdSdq
0
0
0
* Page 149-155
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constant pressure, constant temperature
dG0
G will tend to a minimum value equilibrium spontaneous change
dG 0dG 0
More applications since most processes areisothermal, isobaric
chemical reactions at constant T, PReactants Productsendothermic H is positiveexothermic H is negative
time
G
*
*Page 154
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H is not a criteria for spontaneityS only isolated system G = H - TS
R
P
G
rxn. Co-ord.
S H G
+ ve - ve - ve spontaneous dissociation of unstable compounds
- ve + ve + ve non-spontaneous
forming unstable compounds
+ ve + ve ? ? dissociation of a strong compound
- ve - ve ? ? recombination reactione.g. H + H H2
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Change of Gibbs free energy with temperature(constant pressure)
ST
G
SdTGG
VdPSdTdG
P
12
0
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oo
oo
P
fRTGG
P
PRTGG
f
fRT
P
PRT
VdPGG
SdTVdPdG
ln
ln
ln
ln
1
2
1
2
12
0
Ideal gas
Real gas fugacity
Change of Gibbs free energy with pressure(constant temperature)
Ideal gas
Real gas
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Gibbs free energy
oio
oio
i
P
fRTG
P
pRTGG
ln
ln Ideal Gases
Real Gases
Total Gibbs free energy of a mixture of gases
BBAAiBBiAA
oB
BoBBo
AA
oAA
BBAA
XRTXXRTXGXGXn
P
pRTnGn
P
pRTnGn
GnGnG
lnln
lnln
,,
÷øöçèæ
÷øöçèæ
*
Page 168-174
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0
mix
iiimixmix
iii
oii
H
XRTXSTG
XRTXGXn
ln
ln
for ideal gases
Chemical Equilibria
aA + bB cC + dDconsider G for a pass of the reaction at constant T & P
G = cGC + dGD - aGA - bGB
molar Gibbs free energy
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ioii fRTGG ln
pure state fugacity
i
ioii
bB
aA
dD
cCo
BADCoB
oA
oD
oC
fRTG
ff
ffRTG
fbRTfaRTfdRTfcRTbGaGdGcGG
ln
ln
lnlnlnln
At equilibrium, G = 0
Kff
ff
RT
Gb
Ba
A
dD
cC
o
lnln
equilibrium constant
fi in the unit of bar
0 = JJ J
*
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for ideal gaseous mixture fi = pi
nxb
Ta
T
cT
dT
bB
aA
cC
dD
p
T
ii
nc
n
aA
bB
cC
dD
p
ii
i
p
p
o
bB
aA
dD
cC
PKPP
PP
XX
XXK
p
pX
RTKRTcc
ccK
RTcV
RTnp
constK
constK
RT
G
pp
pp
÷÷øöççèæ.
.ln
ln
ci = concentration
varies withtotal pressure
n = c+d-a-b
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For liquids, use activity
G
RT
a a
a aK
oC
cD
d
Aa
Bb eqln ln
Gibbs - Helmholtz Equation
G VdP SdT
G
TS
P
Also, G H TS H TG
T P
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G
T
H
T
G
T
GTT
G
T T
G
T
P
P
P
2
1
GTT
H
T
orG
T
TH
P
P
2
1
Gibbs - Helmholtz Equation
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For a reaction
GT
T
H
T
GT
TH
P
P
2
1
Similarly,
AT
TU
P
1
1/T
G/TSlope=H
G2
H
G1
1/T
G/T
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Change of equilibrium constant with temperature
G RT K
G
TR K
To
To
ln
ln
Gibbs Helmholtz equation
GT
T
H
T
RK
T
H
T
O
To
To
2
2
ln
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ln K
T
H
RTTo
2Van’t Hoff Equation
If constant (i.e. HP -HR is constant)Hrxno
ln tanKH
RTcons t
o
endothermic, H + ve, K with Texothermic, H - ve, K with T
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Relation between Thermodynamic functions
dU = TdS-PdV 1st lawdH = TdS+VdP, H = U+PVdA = -SdT-PdV, A = U-TSdG = -SdT+VdP, G = H-TS
From multivariable differential calculus dz = M dx + N dy total differential, i.e.
z depends on x & y
M
y
N
xx y
*
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Maxwell Relations
T
V
P
S
T
P
V
S
S
V
P
T
S
P
V
T
S V S P
T V T P
,
,
Phase equilibrium
Clapeyron equationmolar Gibbs free energy
G=0G=G-G
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G G
dG dG
V dP S dT V dP S dT
dP
dT
S S
V V
S
V
H
T V
P
T
liq
vaps
Clausius- Clapeyron equationFor vaporization and sublimation
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dP
dT
H
T V
H
TV
P H
RT
dP
Pd P
H
RTdT
PH
RTconst
P
P
H T T
RT T
vap
vap
vap
vap
vap
vap
sat vap
sat
vap
2
2
2
1
2 1
1 2
ln
ln .
ln1/T
In P
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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
dP
dT
H
T V V
cal mol l atmcal
K l mol
atm K
torr K
dT
dPK torr
sat
sat
vap
v l
9725 0 04129
37315 30180
0 03566
2710
0 0369
1 1
1
1
1
1
.
. .
.
.
.
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Example: Calculate the change in pressurerequired 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
V V l g
P
T
H
T V V
cal g l atmcal
K l gatm K
l s
sat
sat
fus
l s
1
0 9998
1
0 91689 06 10
79 7 0 04129
2731 9 06 10133
5 1
1 1
5 1
1
. ..
. .
. .
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liq
vapsolid
P
T
50kg
skate blade
iceliq. H2O
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Example 2.1A 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 2.2Calculate the work done when 50 g of iron reacts with hydrochloricacid in: (a) a closed vessel of fixed volume; (b) an open beaker at25oC.
Example 2.3The internal energy change when 1.0 mole CaCO3 in the form ofcalcite converts to aragonite is 0.21 kJ. Calculate the difference between the enthalpy change and the change in internal energy.
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Example 2.4The 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 2.5Water 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 sthrough a resistance in thermal contact with it, it is found that 0.798 g of water is vaporized. Calculate the molar internal energyand enthalpy changes at the boiling point (373.15 K).
Exercise 2.6At very low temperatures the heat capacity of a solid is proportionalto T3, and we can write Cv=aT3. What is the change in enthalpy ofsuch a substance when it is heated from 0 to T?
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Example 3.6A sample of argon at 1.0 atm pressure and 25oC expands reversiblyand adiabatically from 0.50 L to 1.00 L. Calculate its final tempera-ture, the work done during the expansion, and the change in internalenergy. The molar heat capacity of argon at constant volume is 12.48 JK-1 mol-1.
Example 4.1Calculate the entropy change in the surroundings when 1.00 molH2O(l) is formed from its elements under standard conditions at298.15 K.
Example 4.4Calculate 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.
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Example 5.4The pressure deep inside the Earth is probably greater than 3×103 kbar,and the temperature is around 4×103 oC. Estimate the change in Gon 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 5.5Calculate the change in molar Gibbs energy when water vaporizes at1 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 5.6Suppose 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.