Important Thermodynamic
equations
dU  TdS  pdV
dH  TdS  Vdp
dS sys  dS surr  0
dA   SdT  pdV
d (G ) P ,T  0
dG  Vdp  SdT
Conditions for equilibrium between Phases
(1)
dS U ,V
 0

For isolated system,
dS  dS   dS   0
 dq
dq

 0
T
T
T  T 
(Thermal Equil.)

If the temperatu re is not same, heat will
flow from one phase to another.
(2)
dAV ,T
 0
At constant v olume and temperatu re,
p  dV  p  dV  0
p  p 
(Mechanica l Equil.)
At constant V and T, the pressure of the two
phases should be the same. If not, one phase
would increase in volume
other.
at the expense of
(3)
dG T , P
 0
For an open system,
G  f
T , P, n 
dG   SdT  Vdp 

i
dni
At constant t emperature and Pressure,
dG T , P


i
dni  0
 i dni   i dni  0
 i   i
(Chemical Equil.)
At constant Temp. and Pressure, the chemical
potential of a component should be the same
in two phases. If not, the transfer of that
component will take place from one phase to
other.
Various Equilibrium conditions
Capacity
Factor
S
Intensity
Factor
T
Equilibrium
Condition
T  T 
At const. U,V
V
P
p  p 
At const. V,T
ni
i
i  i
At const. P,T
Chemical Potential
dG   SdT  Vdp    i dni
 G 

 n  T , P ,n j
i  
dU  TdS  pdV    i dni
 U 
 H 
 A 
i  

 


 n  S ,V ,n j  n  S , p ,n j  n  T ,V ,n j
Intergration of function
dU  TdS  PdV    i dni
Here, U, S, V and n all are extensive function,
U  TS  PV    i ni
G  H  TS  U  PV  TS
G    i ni
Phase changes of pure component
solid
G 
 liquid
 vapour
N

i 1
i
ni
For one component system,
G  n
 
G
 molar Gibb' s Function
n
For a spontaneou s reaction,
dG T , P  0
i dni   i dni  0


i



  i  0
i  i
A species diffuses spontaneou sly from the
phase where its chemical potential is higher to
a phase where its chemical potential is lower.
Physical meaning of Chemical
potential
• Tendency of a system to give particles.
Temperature dependence of phase stability
dG   SdT  Vdp
 G 
 S


 T  p
 Gm 
 Sm


 T  p
  
 Sm



T

p
(negative
S m (g)  S m (l)  S m (s)
quantity)
Ice-skating
• As we glide across
the ice, we exert
pressure on the thin
blade, and are
therefore creating a
small stream of water
in our path by melting
that ice. The water
between the blade
and the ice is what we
really glide across.
Response of melting temperature to the applied pressure
  
   Vm
 p T
Relationship between equilibrium
temperature and pressure
e.g. Solid-liquid equilibria

Solid
Liquid
Suppose at Pressure P , the melting temperatu re is T
1
1
So at T and P , the chemical potential of component is same in both phases
1
1
s  
(i)
l
Now on changing the Pressure to P , the melting temperatu re changes to T
2
2
s'   '
(ii)
l
(ii) -(i)
s'  s   '  
l
l
ds  d
l
 S dT V dP  S dT V dP
L
L
S
S
(S  S )dT  (V V )dP
L S
L S
dP  S
dT V
Relation between dp and dT
a ( p , T )   a ( p , T )
At b, the chemical potentials
changes but
are still equal.
b ( p , T )   b ( p , T )
 d  d 
 S , m dT  V , m dp   S  , m dT  V , m dp
V
,m
 V , m dT 
 trs S
dp

dT
 trsV
S 
,m
 S , m dp
(Clapeyron equation)
The solid –liquid boundary
 ve
 dp   fus H

 
 dT  T fusV  ve but small
  ve, large
Steep curve
Liquid –Vapour boundary
dp  vap H
 ve


dT T vapV  ve, large
  ve, small
 vapV  Vm (g)
dp

dT
 vap H
 RT 
T
 p 



 vap H
d ln p

dT
RT 2
(Clausius Clapeyron Equation)
Solid-Vapour Boundary
dp  sub H  vap H
 ve


dT T subV T vapV  ve, large
  ve
 Steeper th an liquid - vap
1 bar
Elevation of boiling point
• 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
Elevation of boiling point
• An application of this effect is the
pressure cooker. A pressure
cooker traps steam inside it,
raising the pressure to about
twice the atmospheric pressure.
As a result, the water boils at
120 °C. Food cooks quicker at
the higher temperature. There is
a safety valve on the cover which
opens if the pressure is too high.
d ln p  vap H

dT
RT 2
p 2  vap H
ln

p1
R
1 1
  
 T1 T2 
(Clausius Clapeyron Equation)
Autoclave
• Sterilization of both materials
and contaminated media is an
essential process in a
laboratory for in vitro cultures.
• This sterilization is usually
done in a device called an
autoclave.
• Essentially, an autoclave is a
container that exposes the
material to be sterilized to
temperatures above the boiling
point of water, which is
achieved by increasing
pressure.
H fus  6010
Q. It has been suggested that the surface melting of ice plays a
role in enabling speed skaters to achieve peak performance.
Carry out the following calculation to test this hypothesis. At 1
atm pressure, ice melts at 273.15 K, ΔHf =6010 J mol-1, the
density of ice is 920 Kg m-3, and the density of liquid water is
997 kg m-3.
(a) What pressure is required to lower the melting temperature by
5.0 ºC ?
(b) Assume the width of the skate in contact with ice has been
reduced by sharpening to 25*10-3 cm, and that the length of
the contact area is 15 cm. If the skater of mass 85 kg is
balanced on one skate, what pressure is exerted at the
interface of the skate and the ice ?
©What is the melting point of ice under this pressure?
(d) If the temperature of ice is -5 ºC, do you expect melting of ice
at ice-skate interface to occur ?
H of vaporization of water is 539.4 cal/g at the
normal boiling point. Many bacteria can survive
at 100 C by forming spores. Hence autoclaves
used to sterlize
medical and laboratory
instruments are pressurized to raise the boiling
point of water to 120 C. What will be the
pressure required to increase the boiling point
of water to 120 ºC ?
Q. Ar has normal melting and boiling points of 83.8
and 87.3 K; its triple point is at 83.8 K and 0.7 atm,
and its critical temperature and pressure are 151 K
and 48 atm. State whether Ar is a solid, liquid, or gas
under each of the following conditions (a) 0.9 atm and
90 K (b) 0.7 atm and 80 K (c) 0.8 atm and 88 K (d) 0.8
atm and 84 K (e) 1.2 atm and 83.5 K.
Q. Dry ice is frozen carbondioxide. A block of dry ice has a surface
temperature of -78.5 C. If you want to send something frozen across
the country, you can pack it in dry ice. It will be frozen when it
reaches its destination, and there will be no messy liquid left over
like you would have with normal ice. Explain this phenomenon with
the help of a phase diagram?
The vapor pressure of zinc varies with temperature as
log p (mmHg)= -6850/T – 0.755log T + 11.24
and that of liquid Zn as
log p (mmHg) = -6620/T -1.255 log T + 12.34
calculate
(a) boiling pt of Zn.
(b) the triple point.
© the heat of evaporation at the boiling point.
(d) heat of fusion.
(e) the difference in Cps of solid and liquid Zn.
The effect of applied pressure on vapor
pressure
At equilibriu m,
 (l )   ( g )
For any change that preserves equili.
d (l)
 d
Vm (l) dP 
(v)
RT
dp
p
Vm (l) P  RT ln
p2
p1
When pressure is applied to a condensed
phase, its vapour pressure rises.
Phase Rule
F=C-P+2
C = number of component
Degree of Freedom
• Number of independent intensive variables
needed to specify its intensive state.
T, P and mole fraction in each phase.
Total number of intensive variable = pc +2
Mole fraction of components in each phase must be
one
Number of relation =number of phases= p
For chemical equilibrium,
Chemical potential of individual component will be
same in each of the phases.
1  1  1  1  .......
In each of the component
(p-1) relation for each of the component
Total number of relation for chemical equilibrium =
c(p-1)
Degree of Freedom = pc + 2 – p - c(p-1)
=c–p+2
What happens to boiling point or freezing
point of a liquid in the presence of a solute?
What happens to chemical potential of a liquid in the
presence of solute?
For pure solvent
 A (l)   A (g)


 RT ln p 
 A
A
A
In presence of a solute
 A (l)   A (g)

 RT ln p A
A  A


 RT ln
A  A

p
p
A

 RT ln x A
A
(Raoult' s law
 Chemical potential of a liquid
on addition of non - volatile
p  p*
A xA )
decreases
solute.
There is elevation of boiling point upon addition of
a solute.
There is depression of freezing point upon addition of
a solute.