Chemical Engineering

Thermodynamics
Paper Code: CHC-302
Credits: 3L+1T = 4
Professor (Dr.) Gopinath Halder
Head of the Department
Chemical Engineering Department
National Institute of Technology Durgapur
15-Sep-20 Dr. G. N. Halder- NIT Durgapur 1
 MODULE-I
First Law of Thermodynamics & Its
Application
 MODULE-II

Second Law of Thermodynamics & Its

Application
 MODULE-III

Thermodynamic Property Relations

 MODULE-IV
Solution Thermodynamics
 MODULE-V

Chemical Reaction Equilibrium

 Syllabus in Detail
 Scope of thermodynamics and fundamental concepts
 Microscopic and macroscopic view
 First law of thermodynamics: Applications to batch
and flow systems
 Second and third law of thermodynamics:
Reversibility and irreversibility, Carnot cycle,
entropy, free energies, exergy.
 Real gases: Equations of state, Compressibility
charts, Departure functions
 Thermodynamics of flow processes: Single and
multi-stage compression, expansion through nozzles

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 Refrigeration and Liquefaction of Gases:
 Vapour compression refrigeration cycle

 Absorption and Air/Gas refrigeration cycle

 Choice of refrigerants

 Linde and Claude processes of liquefaction of

gases.
Thermodynamic property relations:
 Maxwell’s relations and thermodynamic
functions of pure substances.
 Residual properties

 Fugacity

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 Syllabus in Detail
 Partial molar properties and its evaluation
 Chemical potential: Effect of Temperature and
Pressure
 Activity coefficient: Effect of Temperature and
Pressure
 Gibbs-Duhem Equation
 Fugacity: Liquid, solid, mixture
 Raoult’s law and ideal solution
 Henry’s law and Dilute solution
 Phase-equilibrium for single and multicomponent
system
 Vapour-liquid equilibrium for binary mixture

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 Syllabus in Detail
 Reaction cordinate
 Equilibrium criterion and constant
 Standard Gibb’s free energy change
 Vant Hoff equation: Effect of Temperature on
equilibrium constant
 Homogeneous gas and liquid phase reaction
 Heterogeneous reaction (Solid-gas/liquid-gas)
equilibria
 Phase rule for reacting system
 Fuel cell

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 Text/Reference Books
1. Chemical Engineering Thermodynamics:
Smith, Van Ness & Abbott, McGraw-Hill
Publications
2. Chemical and Process Thermodynamics: B.
G. Kyle, PHI

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 Continued…
3. Chemical Engineering Thermodynamics:
Y.V. C. Rao, Orient Longman Publication
4. Thermodynamics: An Engineering
Approach: Cengel & Boles, McGraw-Hill
Publications
5. Introduction to Chemical Engineering
Thermodynamics: Halder, Prentice-Hall of
India
 Chemical Engineering Thermodynamics

 Definition:
Thermodynamics is the science which
deals with the energy and its
transformation or quantitative relationship
between heat and work.
 Origin:

The term ‘thermodynamics’ is derived

from the Greek word ‘therme’ means heat
and ‘dynamis’ means power.
 Essence of Thermodynamics
The science of thermodynamics can serve
the following purposes-
 the quantitative estimation of heat and work
requirement for any physical and chemical
process
 determination of equilibrium conditions for
chemical reaction
 the transfer of chemical species between
phases.

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Approaches of Thermodynamics:
 Macroscopic approach
– Classical thermodynamics
 Microscopic approach
– Statistical thermodynamics.

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Continued
 Macroscopic approach
In the macroscopic approach to the study
of thermodynamics, the behaviour of
individual molecules is not required to be
taken into consideration. This approach is
adopted in classical thermodynamics.
For example, state of the certain amount of
gas contained in a vessel.
 Microscopic approach:
In the microscopic approach, it is
necessary to discuss the average behaviour
of large group of individual molecules
with significance. Because, the molecular
kinetic theory has been incorporated as a
supplement to thermodynamics for the
modern development of this subject. This
approach is adopted in statistical
thermodynamics.
Example: Statistical explanation of entropy
 Macroscopic & Microscopic Approaches-
At a Glance
Macroscopic Approach
 a few number of variables to explain the
thermodynamic state of the system
 measurable variables
 no consideration of behaviour or structure of the
individual molecules.
Microscopic Approach
 a large number of variables to explain the
thermodynamic state of the system
 variables can not be measured
 it is necessary to consider the behaviour or structure
of the individual molecules.
IMPORTANT QUANTITIES:
 Force:
Force is a dimension. Its unit is derived from
Newton’s second law of motion which can be
mathematically expressed as
Force ∞ mass x acceleration
or, F ∞ ma
 Unit of Force: SI (International System of Unit)
unit of force is ‘Newton’. One Newton is defined
as the force required to accelerate a mass of 1 kg
at a rate of 1meter/second2
i.e., 1N = 1 kg-m/sec2
The English unit of force is pound-force (lbf).

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 Problem:
Assuming that the radius of the Moon and
the Earth are 0.3 and 1 respectively
Calculate the weight of 1 kg mass on
the Moon when the mass of the Moon
is 0.013 related to the Earth’s mass.

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Solution:
According to Newton’s law of gravitation,
Gm1 m 2
F 
r2
For Earth, Gm e m 2 For Moon, Gmm m2
Fe  Fm 
re 2 rm2
On the moon, the weight of the mass is equal to
the force acting on the mass on the moon and
comparing with the earth, we get
Fe me  rm2  m e  rm 
2
1
 0.3
2
   
Fm mm  re2 
= =
m m  re  = 6.92 ≈ 7.0
0.013
Hence, the mass of 1 kg will weigh
approximately 1/7 kg on the moon.
15-Sep-20 Dr. G. N. Halder- NIT Durgapur 18
Pressure:
Pressure is defined as normal force exerted
by the fluid per unit area. It is a scalar
quantity.
Mathematically, p  F
A
Where, F = Force normal to the area
A = Area of the surface
P = Pressure

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Pressure:
Unit of Pressure:
In SI system: Newton per square meter
(N/m2) i.e., one Newton force acting on one
square meter area. This is also called pascal
(Pa).
1 Pa = 1 N/m2
1 bar = 105 Pa = 105 N/m2
Again, 1 bar = 105 Pa = 0.1 MPa = 100 kPa
Pressure:
In F.P.S system: Pound force per square inch
(lbf/inch2).
1 lbf/inch2 = 144 lbf/ft2
In M.K.S system: kgf/cm2
1 kgf/cm2 = 104 kgf/m2
 1 standard atmosphere (atm) = 1.01325 bar =
1.01325 x105 Pa = 760 mm Hg = 29.92 in Hg
 Absolute Pressure = Gauge Pressure +

Atmospheric Pressure
Pabsolute = Pgauge + Patmospheric
Work
 Work is a form of energy. It is defined as
the product of the force and the
displacement in the direction of the applied
force. In other way, work performed
whenever a force acts through a distance.
The work-done can be mathematically
expressed as
dW  Fdl
where, F = applied force and
dl = displacement

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UNITS OF WORK:
 SI unit: Newton-meter.

1 N-m is equivalent to 1 Joule,

i.e., 1 N-m = 1 J
1 kJ = 103 J
 English unit : ft-lbf
SIGN CONVENTION OF WORK:
 Work-done by the system is treated as
positive quantity because of the
displacement in the same direction
Work done = +Ve quantity
 While the work done on the system by
external agent is regarded as negative
quantity.
Work done = -Ve quantity
Consider a common example on compression
or expansion of a gas in a cylinder-piston
assembly. The gas is enclosed by a
frictionless piston in a cylinder. The piston
moves along the length of the cylinder.

The total force acted upon the piston is

F = PA
Where, P = Pressure of the gas and
A = Area of the piston
 The displacement of the piston in the
direction of force is dV
dl 
A

where, dV = Change in volume of the gas

Putting the value of ‘dl’ and ‘F’, we get
dV
dW  Fdl = PA.  PdV
A
Integrating the equation with integral limits
V = V1 and V = V2 as the volume of the gas
V
changes and we get 2

W   PdV
V1
15-Sep-20 Dr. G. N. Halder- NIT Durgapur 27
 A gas is enclosed by a movable piston in a
cylinder. The gas expands from initial volume
of 3 m3 to a final volume of 5 m3 as a
result of 210 kJ of work-done on the system
by an external source. The pressure of the
system remains constant at 560 kPa. Estimate
the net work-done by the system.

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 Heat is defined as the form of energy. It can
be transferred from higher temperature region to
lower temperature region whenever the
temperature difference exists between them. The
flow of heat continues until the difference in
temperature is equalized. Heat is the energy in
transit. Like other forms of energy, heat can not
be completely converted into work. There are
three modes of heat transmission.
These are-
 Conduction
 Convection
 Radiation
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 Units of Heat
 In SI system, the unit of heat is Joule
1 Joule = 1 N-m
1 calorie = 4.2 Joule
 In English engineering system of unit,
the unit of heat is ft-lb
1 ft-lb = 1.3558 J
1 BTU = 1055.04 Joule

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 Energy can be defined as one property which
can be transformed into or produced from work.
The total energy ‘E’ of a system comprises of
the following two forms. These are-
 Macroscopic form and
 Microscopic form
Macroscopic Form of Energy: It includes the
energies those a system possesses with respect
to outside reference plane, such as, Kinetic
energy and potential energy. This mode of
energy is concerned with motion and influence
of some external effects viz. electricity,
magnetism, surface tension, gravity etc.

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Macroscopic Forms of Energy
Kinetic Energy: The energy possessed
by a system by virtue of its motion
is called kinetic energy.
1
EK  mV 2

2
Potential Energy: The energy possessed by
a system by virtue of its elevation with
reference to an arbitrary reference plane is
called potential energy.
E P  mgZ
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Microscopic Forms of Energy
Internal Energy: Every system has a certain
amount of energy within itself. This is known
as internal energy. The internal energy is
nothing but the sum of all the microscopic
forms of energy.
Other Forms of Energies: These are electrical,
magnetic, surface energy etc. These energies
can be neglected apart from some special
cases.
Hence, the total energy ‘E’ of a system can
be represented as
E = E K + EP + U
Macro Micro
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 Power is defined as the rate at which work is
done or ratio of work done to unit time.
It can be expressed as,
Work-done Joule
Power = ---------- = ---------- = Watt
Time Second
SI unit of power is kJ/sec or kW.
The most familiar unit of power is horse
power (hp)
1 hp = 745.7 W

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 PROBLEM
Determine the power required to
accelerate a car weighing 1200 kg from a
velocity 10 km/hour to 100 km/hour in
just 1 minute on a level road.

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The extent of work needed to change in
kinetic energy of the body to accelerate is
W  m V22  V12 
1
2
V1 = 10,000 m/hour = 10, 000m
 2.777 m / sec
3600sec
V2 = 100,000 m/hour = 100, 000m  27.777m / sec
3600sec

Substitutingthe values of m, V1 and V2 in the above

equation, we get 1  2
m  m 
2

W x1200kg  27.777    2.777  

2  sec   sec  

15-Sep-20 Dr. G. N. Halder- NIT Durgapur 36

 Continued……..
1  
2
m   m  
2

W  x1200 kg  27.777    2.777  

2  sec   sec  
= 458310 N-m
= 458310 J
= 458.31 kJ
Power required = Work-done/Time
= 458.31 kJ/60 sec
= 7.63 kW

15-Sep-20 Dr. G. N. Halder- NIT Durgapur 37

 A gas is enclosed by a frictionless piston in a 0.3 m
diameter cylinder and a metal block is placed on the
piston. The weight of the piston and the block rested
on it equal to 100 kg. The atmospheric pressure and
the acceleration due to the gravity ‘g’ are 1.013 bar
and 9.792 m/s2 respectively. Calculate:
a) the force exerted by the atmosphere, the piston and
the weight on the gas.
b) the pressure of the gas in kPa
c) if the gas expands on application of heat, the piston
moves upward along with the weight by 0.5 m, then
what is the work-done by the gas in KJ?
d) What is the change in potential energy of the piston
and the weight after the expansion of gas
15-Sep-20 Dr. G. N. Halder- NIT Durgapur 38
Solution: a) From the definition of force, we have
F = PA = Pressure x Area
Here, the force exerted by the atmosphere can be
calculated as
F = 1.013 x 105 x п/4 x (0.3)2 m2
= 7160.47 N
Force exerted by the piston and the metal block
= mg = 100 kg x 9.792 m/s2
kg  m
= 979.2 2
= 979.2 N
s
Hence, the total force acting upon the gas
= 7160.47 N + 979.2 N = 8139.67 N
15-Sep-20 Dr. G. N. Halder- NIT Durgapur 39
b) The pressure of the gas can be estimated,
as we know Force 8139.67
Pressure = Force =  2 = 
 0.3m 
2
Area d
 
= 1.152 x 105 N/m2
c) The gas expands on application of heat and
the volume of the gas goes on increasing
and the piston moves upward.
The Work done due to expansion of gas is
given by
W = Force x Displacement
= F dl = 8139.67 N x 0.5 m
 = 4069.835 N-m = 4069.835 J
15-Sep-20 Dr. G. N. Halder- NIT Durgapur 40
 d) Change in potential energy of the piston
and the weight after expansion process,
EP = mgZ
= 100 kg x 9.792 m/s2 x 0.5 m
= 489.6 kg-m2/s2
= 489.6 N-m = 489.6 J.

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 Terminology and Fundamental
Concepts
 System: It is defined as the quantity of matter
or a region in a space upon which the
attention is concentrated in the analysis of a
problem.
 In other words, a system is defined as any portion
of the universe separated from the surroundings
by real or imaginary boundary.
 For example, a piston, a solution in a test tube,
a living organism, a planet etc.

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 A typical thermodynamic system is illustrated
in the figure

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 Surroundings:
The area or region external to the system
is known as the surroundings.
 Universe: The system and surroundings
constitute an universe.
 Boundary:
The system is separated from the
surrounding by a real or imaginary
surface, which is known as boundary. A
boundary is basically of no thickness and
it can neither contain any mass nor occupy
any volume in space. A boundary is not
necessarily rigid.
15-Sep-20 Dr. G. N. Halder- NIT Durgapur 44
 The system is separated from the
surrounding by a real or imaginary surface,
which is known as boundary. A boundary is
basically of no thickness and it can neither
contain any mass nor occupy any volume in
space. A boundary is not necessarily rigid.
The boundary may be broadly categorized
into two ways-
1. Fixed boundary
2. Moving boundary

15-Sep-20 Dr. G. N. Halder- NIT Durgapur 45

 Consider a piston-cylinder assembly. The gas is
confined in the cylinder having a closely fitted
piston. Now, if the gas is heated, gas will start
expanding and the piston will move upward.
Here, cylinder is the fixed boundary and piston is
the moving boundary.

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 A system is said to be open if it can exchange
both mass and energy with the surroundings.

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EXAMPLE OF OPEN SYSTEM:
Most of the engineering devices such as air
compressor, turbine, nozzle etc belong to open
system. In case of air compressor, air enters at
low pressure and leaves at high pressure,
both mass and energy are crossing the system
boundary.

CONTROL VOLUME
Open system is often called control volume.
Actually the region in the space selected for
the analysis of an open system is known as
control volume.
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Consider a compressor in which mass flows
into and out of the device, can be analyzed as
control volume. In this device, an arbitrary
region in space can be identified as control
volume.

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 A system is said to be closed if it can
exchange energy only with its surroundings, not
the mass.
A closed system is basically of fixed mass, in
which there is no mass transfer across the
boundary. Only the energy can be transferred into
or out of the system
Example of Closed System:
i) a certain quantity of gas confined in a cylinder
bounded by a frictionless piston.
ii) hot liquid taken in a closed metallic flask
iii) reaction in a batch reactor

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A closed system can be depicted as

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 A system is said to be isolated one, if it can
neither exchange energy nor mass with its
surroundings. In any isolated system, there
is no interaction between the system and
the surroundings. The system is of constant
mass and energy.

Example of Isolated System: thermoflask

15-Sep-20 Dr. G. N. Halder- NIT Durgapur 52

Illustration:

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Homogeneous System:
A system is considered to be
homogeneous when the properties are
uniform throughout.
A homogeneous system contains only one
phase necessarily.

For instance, liquid water in a container.

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Heterogeneous System:
A system is referred to as heterogeneous when
the properties are not uniform throughout the
system. It consists of more than one phase. It
requires the presence of at least two distinct
homogeneous phases to proceed.

Example: presence of water in liquid and

vapour phase in a container, a mixture of
water and toluene is an interesting example of
heterogeneous system. Here, toluene and water
are two immiscible liquids.
The term ‘ state’ refers to the condition in which
the system exists. The state of a system at any
given condition can be better characterized by
certain observable properties of a system. A
system in order to be in a state must have
definite values assigned to its properties such as
temperature, pressure, volume, composition etc.
Consider a horizontal piston-cylinder arrangement contains
certain amount of gas. The piston is held in place by latches.
This stage is designated to be state-I. Now if the restraining
latches are removed, the gas inside the cylinder will start
expanding to an appreciable extent. This is state-II. It is
clearly understood from the figure that the system has
different set of properties in two different states. Thus the
same gas can exist in a number of different states.
 The state of a system interacting with the
surroundings is said to be steady if it does
not vary with time. Simply, the change of
state is zero, but the interactions are non-
zero.
In this context, one thing is to be kept in
our mind, steady and uniform conditions are
not the same matter. Uniform means no
change with location over a specified
region.

15-Sep-20 Dr. G. N. Halder, NIT Durgapur 58

 The term ‘equilibrium’ is used to imply that a
state of balance of a system.
If there is no change in the state of the
system, then the system is considered to be in a
state of thermodynamic equilibrium.
Hence, a system is said to be in equilibrium
state, if the system has no tendency to
undergo any further change. i.e., the properties
like pressure, temperature, composition are
uniform in magnitude throughout the system.

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 The equilibrium may be of many kinds. A
system in order to be in complete
thermodynamic equilibrium, it must satisfy
the conditions of following three relevant
types of equilibrium.

These are, i) Thermal equilibrium

ii) Mechanical equilibrium
iii) Chemical equilibrium

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 Thermal Equilibrium
A system is said to be in thermal equilibrium if
the temperature is uniform throughout the entire
system.
Consider two different systems A and B. If the
thermal equilibrium is attained then there is no
difference in temperature between the interacting
systems A and B. In other words, no temperature
gradient exists between two systems. Here, driving
force is temperature.
At this condition, TA = TB , where, TA and TB are
the temperatures of system A and B respectively.

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 A system is said to be in mechanical
equilibrium if the pressure is uniform
throughout the entire system.
Consider two different systems A and B. If there
is no imbalance of forces between the different
parts of the interacting systems A and B, then
the system is in mechanical equilibrium. Here,
driving force is pressure. Thus mechanical
equilibrium implies the uniformity of pressure.
At this condition, PA = PB , where, PA and PB are
the temperatures of system A and B respectively.

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A system is said to be in chemical equilibrium
if the composition of the system does not
change with time.
For two different systems A and B in chemical
equilibrium, there is no tendency for the
chemical reaction to occur between the
interacting systems A and B. Here, driving
force is chemical potential. At this condition,
i i , for i = 1,2,3,4, ……n.
A B

 A
where, i and ,  i B are the chemical potentials of
component ‘i’ in system A and B respectively.
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Boundary Wall of a System:
 Permeable Wall: It is defined as the wall that
allows the passage of both matter and energy.
 Rigid Wall: The wall whose shape and
position are fixed.
 Adiabatic Wall: The wall which prevents the
passage of matter or energy. It does not permit
the flow of heat into and out of the system.
 Diathermal Wall: The wall prevents the passage
of matter but allows the transmission of energy
 Isothermal Process: The process in which the
temperature remains constant throughout the
system. Here, T = constant, or, dT = 0
 Isobaric Process: The process during which the
pressure remains constant throughout the system.
Here, P = constant, or, dP = 0
 Isochoric Process: The process in which the
volume remains constant throughout the system.
Here, V = constant, or, dV = 0
 Adiabatic Process: The process in which there is
no exchange of heat between the system and the
surroundings. Here, Q = constant, or, dQ = 0
 Cyclic Process: It is defined as the
process in which a system having
undergone a change returns to its initial
state at the end of the process. Here, the
path of the process is called a cycle and
for a cycle, initial and final states are
identical.
 Quasi-static Process: It is defined as the
process which takes place very slowly and
with infinitesimal driving forces.
The characteristics which are experimentally
measurable and which enable us to define a
system are called property. The properties can
describe properly the state of a system.
The properties may broadly be categorized
into two ways-
a) Extensive property and
b) Intensive property
 Extensive Property: The properties
dependent upon the extent or quantity of
the mass of a system are known as
Extensive properties. Mass, volume,
length, surface area, internal energy,
enthalpy etc are extensive properties.
 Intensive Property: The properties
independent of the extent or quantity of
the mass of a system are known as
Intensive properties. Temperature, pressure,
density, specific volume, specific heat
etc are intensive properties.
 This law was coined and formulated by R.
H. Fowler in 1931. The law states that ‘If the
two bodies are in thermal equilibrium with
a third body, they are also in thermal
equilibrium with each other’.

 Suppose two bodies A and B are brought into thermal
contact with each other. Then heat will flow from the
body at higher temperature to the same at lower
temperature. The process continues until the thermal
equilibrium is established between A and B. Now if
the another body C is in thermal equilibrium with A,
then B and C should also be in thermal equilibrium.
Hence, it can be concluded that A, B and C should
also be in thermal equilibrium.
 Mathematically, if TA = TB and TC = TA
then TA = TB = TC
Phase Rule:
 A phase is defined as a physically distinct but
homogeneous part of a system separated from other
parts by boundary surface . The phase rule is
qualitative treatment of systems in equilibrium. It
was formulated by J. W. Gibbs in 1876. The
number of independent variables in a multi
component and multiphase system is clearly
indicated by the phase rule and given by
 F = C – P +2
 Where, F = degrees of freedom
 C = number of components
 P = number of phases
Continued….
 In the above formulation, 2 is taken into
consideration for the fact that two variables T and
P must be specified to describe the state of
equilibrium. This rule is widely used for the study
of different types of systems in equilibrium.
 Example: water system
Ice Liquid water Water vapour
Here, P = number of phases = 3,
C = number of component = 1
Then, F = 1-3+2 = 0 i.e., ice, liquid water and water
vapour are in equilibrium.
Reversible Process:
A process is said to be reversible if both the
system and the surroundings can be restored to
their respective original states by reversing the
direction of the process.
A reversible process is one in which the
properties of the system at every instant remain
uniform when the process takes place.
Example: i) Vapourization of a liquid in a
closed vessel at a constant temperature.
ii) Chemical reaction in a galvanic
cell
WCompression = WExpansion
Characteristics of a Reversible Process:
 A reversible process proceeds with no driving
force through a succession of equilibrium steps
 A reversible process can be restored to its
original state at the end of the reverse process
 The process undergoes an infinitesimal change
towards completion.
 During the reversible process properties of the
system remain uniform at every instant.
 In reversible process, the magnitude of the
thermodynamic quantities at different stage will be
same as in the forward direction but opposite in
direction.
Irreversible Process
A process which does not satisfy the criterion
of reversible process is known as irreversible
process. In this process, the change occurs very
rapidly and there is no chance of attainment
of equilibrium of the system. All spontaneous
processes occurring in nature are irreversible.
They can not be reversed by the help of any
external agency.
For example, flow of heat from one body to
another, free expansion of a gas, rusting of iron
in presence of atmospheric oxygen
Factors Responsible for Irreversible Process
The following factors are responsible for the
irreversibility of the process-
 Friction

 Mixing of two fluids

 Unrestrained expansion

 Heat transfer across a finite temperature

difference
 Electric resistance

 Inelastic deformation of solids

 Chemical reaction