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Entropy changes in irreversible Processes
To obtain the change in entropy in an irreversible process we have to calculate S along a reversible path between the initial state and the final state.
Freezing of water below its freezing point
H2O( l , -10 °C) H2O( s , -10 °C)
H2O( l , 0°C) H2O( s , 0 °C)
Irrev
273
263
263
273ln ice
crysliq C
T
HCS
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A sample of 1.00 mole of monoatomic perfect gas with Cv,m = 1.5 R, initially at 298 K and 10 L, is expanded, with the surroundings maintained at 298 K, to a final volume of 20 L, in three ways (a) isothermally and reversibly (b) isothermally against a constant external pressure of 0.5 atm (c) adiabatically against a constant external pressure of 0.5 atm . Calculate ΔS and ΔSsurr, ΔH and ΔT for every path.
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Question?
• A 50 gm mass of Cu at a temperature of 393 K is placed in contact with a 100 g mass of Cu at a temperature of 303 K in a thermally insulated container. Calculate q and ΔStotal for the reversible process. Use a value of 0.4184 Jg-1K-1 for specific heat capacity of Cu.
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Absolute entropy of a substance
)(
)(
)()0()(
0
T
T
p
b
vT
T
p
f
fT p
b
b
f
f
T
dTgC
T
H
T
dTlC
T
H
T
dTsCSTS
Third law of thermodynamics:
The entropy of each pure element or substance in a perfectly crystalline form is zero at absolute zero.
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energy. free helmholtz called isA
0)(
0)(
)(
T and Vconstant At
0
or 0
0
0
workadditional no olume,constant vAt
0
0
,
,
,
,
TV
TV
VS
VU
v
sys
surrsys
surrsys
Ad
TSUd
TSddUTdSdU
dU
dS
TdSdU
TdSdq
TdSdqT
dqdS
dSdS
dSdS
Spontaneous process
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energy. free sGibb' called isG
0)(
0)(
)(
T and Pconstant At
0
or 0
0
0
workadditional no pressure,constant At
0
0
,
,
,
,
TP
TP
pS
pH
p
sys
surrsys
surrsys
Gd
TSHd
TSddHTdSdH
dH
dS
TdSdH
TdSdq
TdSdqT
dqdS
dSdS
dSdS
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Enthalpy/ Entropy driven process
A-Enthalpy drivenB-Entropy driven
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Hydrophobic Effect in Protein Hydrophobic Effect in Protein FoldingFolding
More Hydrocarbon-Water Interfacial Area
Less Hydrocarbon-WaterInterfacial Area
HOH
HOH+
Folded Unfolded
S=+
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Hydrophobic forcesHydrophobic forces
• Hydrophobic interactions are considered to make a major contribution to stabilizing the native structures of proteins in aqueous environment.
• The hydrophobic effect is a unique organizing force based on repulsion by the solvent instead of attractive forces at the site of organization.
• Thermodynamics of protein unfolding can be explained on the basis of transfer of non-polar groups from organic solvent to water.
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Enthalpy and EntropyEnthalpy and Entropy
Go = Ho - TSo
•Typically, G and H are measurable and S calculated
•H and S Provide Mechanistic Insight
•dH = CpdT Gives Thermal Dependence of K
•In very rough generalities:H related to bond formation/breakingS related to configurational freedom and water ordering
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O
H
H
O H
HWater molecules seek
favorable positions and partially freeze their
thermal motion.
Extreme ordering of waters caused by hydrophobic molecule
Hydrophobic bond
• Net effect is entropic rather than energetic in nature just because the energy of H-bonds is extremely high and waters would prefer to become frozen and sacrifice a part of their freedom than to lose the large energy of a hydrogen bond.
Reason for negative Reason for negative SS
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A simple binding process
++
Binding can be enthalpy or entropy driven
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Enthalpic contribution to Free energy of binding
• Ionic and hydrogen bonds.
• Electrostatic interaction.
• Van der Waal interaction
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entropy of bindingentropy of binding
• Binding is favoured if it leads to a net increase in disorder or entropy.
• This includes entropy of– the system (interacting molecules and solvent)
• represented as change in entropy or S – the environment (everything else)
• as the system releases or absorbs heat it changes the entropy of the surroundings
• heat release is measured as change in enthalpy or H
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Combining the First and Second law
ps
Sp
Vs
sV
S
V
p
T
Vp
HT
S
H
VdpTdSdH
VdppdVpdVTdSdH
VdppdVdUdH
S
p
V
T
pV
UT
S
U
VSfU
pdVTdSdU
),(
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P
V
S
T
H
U G
A
Vs
ps
Tp
VT
S
p
V
T
S
V
p
T
p
S
T
V
T
p
V
S
Good people have studied under very able teachers.
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Tp
VT
p
S
T
V
SdTVdpdG
SdTTdSVdpTdSdG
SdTTdSdHdG
T
p
V
S
pdVSdTdA
SdTTdSpdVTdSdA
SdTTdSdUdA
TSUddA
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The Thermodynamic Square
P
V
S
T
H
U G
A
SdTVdpdG
pdVSdTdA
VdpTdSdH
pdVTdSdU
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Thermodynamic equation of state
VT
VT
P
H
PT
pT
V
U
PT
VT
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Prove it
PT
P
T
VT
P
C
2
2
Derive the expression for for a gas following the virial equation with Z=1+B/V
TV
U
TV
U
TV
U
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True or False ??
No heat transfer occurs when liquid water is reversibly and isothermally compressed.
During an adiabatic, reversible process at constant volume the internal energy can never increase.
The internal energy of a system and its surroundings is never conserved during an irreversible process, but is conserved for reversible processes.
The work done by a closed system can exceed the decrease in the system’s internal energy
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When extra (useful) work is involved.
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Important Thermodynamic equations
e
e
e
e
dwSdTVdpdG
dwpdVSdTdA
dwVdpTdSdH
dwpdVTdSdU
0)(
0
,
TP
surrsys
Gd
dSdS
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Helmholtz free energy
e)Temperaturconstant (At
0
process ReversibleFor
0
0
max dATdSdUdw
TdSdUdw
TdSdwdU
TdSdqT
dqdS
dSdS
dSdS
sys
surrsys
surrsys
Maximum work done (including expansion work) can be calculated by measuring the change in A.
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Gibb’s Free Energy
revadddwdG ,
rev add,
pressure, and emperatureconstant tAt
dwVdP-SdTdG
work,extra of presence In the
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E.M.F. work
EnFG
nFdEdwdG
dwpdVSdTdG
TP
eTP
e
,
,
0
0
0
,
,
,
TP
TP
TP
E
EnF
G
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Displacement of metal from metal salts
• A more active metal can "displace" a less active one from a solution of its salt.
• “Active" metals are all "attacked by acids“.
• Zn(s) + Cu2+ → Zn2+ + Cu(s) .
VveE TP 09.1,
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Pressure as a function of height
• An increase in height will correspond to a decrease in pressure.
• Boiling point of water is raised if the pressure above water is increased; it is lowered if the pressure is reduced. This explains why water boils at 70 °C up in the Himalaya.
)(ln
mequilibriuAt
211
2 hhRT
Mg
P
P
dhRT
Mg
P
dP
gdhRT
PMdP
gdhdP
mgdhVdP
mgdhVdPG
mgdhVdPSdTG
T
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The Thermodynamics of a Rubber band
LppT T
fTf
L
H
fdLpdVTdSdU
,,
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• Q.Experiments show that the retractive force f of polymeric elastomers as a function of temperature and expansion L is given by f(T,L) = aT(L-L0) where a and L0 are constants.
• (a)Use Maxwell relations to determine the entropy and enthalpy at constant T and p.
• (b) If you adiabatically stretch a rubber band by small amount, its temperature increases but volume does not change. Derive an expression for its temperature as a function of L, L0, a and C (heat capacity).
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The variation of Gibb’s energy with temperature.
2
/
T
H
T
TG
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Variation of the Gibb’s energy with pressure
p
pRTG
pp
p
pRTdp
p
RTG
ppVG
VdpdG
SdTVdpdG
fm
i
i
fm
m
ln
bar) 1at pressure (standard
ln
Gas Ideal
(b)Gases
constantV
solids and Liquids (a)
,emperatureconstant tAt
12
m
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Question ?
• Calculate ΔG for the conversion of 3 mol of liquid benzene at 80 C (normal boiling point) to vapour at same temperature and a pressure of 0.66 bar? Consider the vapour as an ideal gas.
• In the transition CaCO3 (aragonite) to CaCO3 (calcite), ΔGm (298) = -800 J and Vm = 2.75 cm3. At what pressure would aragonite become the stable form at 298 K?.