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Transcript
Entropy changes in irreversible Processes
To obtain the change in entropy in an irreversible
process we have to calculate DS along a reversible
path between the initial state and the final state.
Freezing of water below its freezing point
H2O( l , -10 °C)
Irrev
H2O( s , -10 °C)
H2O( s , 0 °C)
H2O( l , 0°C)
DS  Cliq ln
273 DH crys
263

 Cice
263
T
273
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.
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.
Absolute entropy of a substance

S (T )  S (0) 

Tf
0


Tb
Tf
C p ( s ) dT
T
C p (l ) dT
T

T
C p ( g ) dT
Tb
T


D
f
H
Tf
Dv H
Tb
Third law of thermodynamics:
The entropy of each pure element or substance in a
perfectly crystalline form is zero at absolute zero.
Spontaneous process
dS sys  dS surr  0
dS sys   dS surr
dq
T
dq  TdS  0
At constant v olume, no additional
dqv  TdS  0
dS sys 
dU  TdS  0
dSU ,V  0 or
dU S ,V  0
At constant V and T
dU  TdS  dU  d (TS )
 d (U  TS )V ,T  0
 d ( A)V ,T  0
A is called helmholtz
free energy.
work
dS sys  dS surr  0
dS sys   dS surr
dq
T
dq  TdS  0
At constant pressure, no additional work
dS sys 
dq p  TdS  0
dH  TdS  0
dS H , p  0 or
dH S , p  0
At constant P and T
dH  TdS  dH  d (TS )
 d ( H  TS ) P ,T  0
 d (G ) P ,T  0
G is called Gibb' s free energy.
Enthalpy/ Entropy driven process
A-Enthalpy driven
B-Entropy driven
Hydrophobic Effect in Protein
Folding
DS=+
+
HOH
HOH
Folded
More Hydrocarbon-Water
Interfacial Area
Unfolded
Less Hydrocarbon-Water
Interfacial Area
Hydrophobic 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.
Enthalpy and Entropy
DGo = DHo - TDSo
•Typically, DG and DH are measurable and DS calculated
•DH and DS Provide Mechanistic Insight
•In very rough generalities:
DH related to bond formation/breaking
DS related to configurational freedom and water ordering
•dDH = DCpdT Gives Thermal Dependence of K
Reason for negative DS
H
O
H
O
H
H
Water molecules seek
favorable positions and
partially freeze their
thermal motion.
Hydrophobic bond
Extreme ordering of waters caused by hydrophobic molecule
• 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.
A simple binding process
Binding can be enthalpy or entropy driven
+
+
Enthalpic contribution to Free
energy of binding
• Ionic and hydrogen bonds.
• Electrostatic interaction.
• Van der Waal interaction
entropy 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 DS
– 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 DH
Combining the First and Second law
dU  TdS  pdV
U  f (S ,V )
 U 
 U 

 T

  p

S

V

V

s
 T 
 p 
    
 V  s
 S V
dH  dU  pdV  Vdp
dH  TdS  pdV  pdV  Vdp
dH  TdS  Vdp
 H 

 T

S

p
 H 

  V

p

S
 T 
 V 
    
 S  p
 p  s
H
S
U
P
G
V
A
 S 
 p 

 
 

V
  T  T V
 S 
 V 
    
 T  p
 p  T
 T   V 
    
 p  s  S  p
 T 
 p 


 
 

V
 s
 S V
T
Good people have studied under very
able teachers.
dA  d U  TS 
dA  dU  TdS  SdT
dA  TdS  pdV  TdS  SdT
dA   SdT  pdV
 S   p 
   
 V T  T V
dG  dH  TdS  SdT
dG  TdS  Vdp  TdS  SdT
dG  Vdp  SdT
 S 
 V 
    
 T  p
 p T
The Thermodynamic Square
S
H
U
V
P
G
A
dU  TdS  pdV
dH  TdS  Vdp
dA   SdT  pdV
dG  Vdp  SdT
T
Thermodynamic equation of state
 U 
 p 

  T
 P
 V  T
 T V
 H 
 V 

  T 
 V
 P  T
 T  P
 U 


 V  T
Prove it
 V 
 CP 

  T  2 
 P T
 T  P
2
Derive the expression for for a gas following the virial
equation with Z=1+B/V
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
When extra (useful) work is
involved.
Important Thermodynamic
equations
dU  TdS  pdV  dwe
dH  TdS  Vdp  dwe
dS sys  dS surr  0
dA   SdT  pdV  dwe
d (G ) P ,T  0
dG  Vdp  SdT  dwe
Helmholtz free energy
dS sys  dS surr  0
dS sys  dS surr
dq
T
dq  TdS  0
For Reversible process
dU  dw  TdS  0
dw  dU  TdS
dwmax  dU  TdS  dA (At constant Temperatur e)
dS sys 
Maximum work done (including expansion work) can be calculated by
measuring the change in A.
Gibb’s Free Energy
In the presence of extra work,
dG  -SdT  VdP  dw add, rev
At constant t emperature and pressure,
E.M.F. work
dG   SdT  pdV  dwe
DG P ,T  0
dGP ,T  dwe  nFdE
 nFDE P ,T  0
DGP ,T  nFDE
DE P ,T  0
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) .
DE P ,T  ve  1.09V
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.
DG   SdT  VdP  mgdh
DGT  VdP  mgdh
At equilibriu m
VdP  mgdh
dP   gdh
 PM 
dP  
 gdh
 RT 
dP
 Mg 
 
dh
P
 RT 
P
ln  2
 P1
  Mg 
  
(h1  h2 )
  RT 
The Thermodynamics of a Rubber
band
dU  TdS  pdV  fdL
 f 
 H 

  f T

 T  p , L
 L  T , p
• 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).
The variation of Gibb’s energy with temperature.
H
  G / T  

 2
T
 T 
Variation of the Gibb’s energy with pressure
dG  Vdp  SdT
At constant t emperature ,
dG  Vdp
(a) Liquids and solids
Vm  constant
DGm  V
 p2
 p1 
(b)Gases
Ideal Gas
DGm 

pi  p 
pf
RT
dp  RT ln
p
pi
(standard pressure at 1 bar)
DGm  RT ln
pf
p
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 DVm = 2.75
cm3. At what pressure would aragonite become
the stable form at 298 K?.
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