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UNIVERSITY
OF NAIROBI
DISTANCE LEARNNG STUDY
MODULE
Nyongesa, F. W., Ph.D
University of Nairobi
Reviewer
Awour, J. B., Ph.D
University of Nairobi
 CODL, University of Nairobi
SPH 302: Thermodynamics
DISTANCE LEARNNG STUDY MODULE
Dr. Nyongesa F. Wanjala, Ph.D., MSc., UoN.
Department of Physics
University of Nairobi
Reviewer
Dr. Awuor John Buers, Ph.D., MSc., UoN.
Department of Physics
University of Nairobi
 CODL, University of Nairobi, 2015
1
SPH 302: Thermodynamics
Forward
The study of thermodynamics was inaugurated by 19 -century engineers, who
th
wanted to know the ultimate limitations the laws of physics impose on the operation of
steam engines and other devices that convert heat to mechanical energy and vice
versa. Today the scope of thermodynamics has immensely increased with
innumerable applications in chemistry and engineering.
Thermodynamics is the study of relationships involving heat, mechanical work and
other aspects of energy transfer that takes place in devices such as refrigerators,
heat pumps, internal combustion engines etc. These relationships are governed by
the four laws of thermodynamics which are now some of the most important
fundamental laws in nature.
This module attempts to provide a clear and modern view of the essential Principles
of Thermodynamics and its applications, relevant to Science and Engineering. The
module is organized in 6 lectures. Lecture 1 begins by defining basic thermodynamic
concepts and the Zeroth law which introduces the concept of thermal equilibrium and
temperature. Lecture 2 deals with the 1ST Law of thermodynamics which introduces
the concept of internal energy. Lecture 3 deals with the 2ND Law of thermodynamics
which gives the direction of natural thermodynamic processes and defines the thermal
efficiency of devices that convert heat into work and vice versa.
Lecture 4 deals with Entropy and the concept of reversibility. Lecture 5 introduces
thermodynamic potentials and Maxwell/s equations. Lastly, Lecture 6 deals with
phase changes and phase equilibrium in thermodynamic systems.
Each lecture begins with clear statement of study objectives, pertinent definitions and
simple qualitative explanations of principles together with illustrations. This is followed
by worked examples drawn from everyday life to help clarify the principles and
indicate possible applications.
To develop your creativity and understanding of the ideas, we have included a
number of problems at the end of each lecture. The questions are both Qualitative
and Quantitative (Problem sets) with the former intended to develop creative thinking
and to provoke discussions. Step-by-step solutions to selected problems are given in
the Appendix.
F. W. Nyongesa
2
SPH 302: Thermodynamics
COURSE UNIT OUTLINE
Introduction
……………………..…………........
Page
4
Lecture 1: Thermodynamic Concepts and the Zeroth Law
1.1
1.2
1.3
1.4
1.5
1.6
1.7
Introduction
…………………………………………………………
Lecture Objectives
…………………………………………………………
Introduction to Thermodynamics
…………………………………….
Thermodynamic Concepts
…………………………………….
Behaviour of Gases
…………………………………….
Zeroth Law of Thermodynamics
…………………………………….
Temperature and Temperature scales
…………………………………….
Problem set 1
………………………………………………………..
6
6
6
7
13
17
18
22
Lecture 2: First Law of Thermodynamics
2.1
2.2
2.3
2.4
2.5
2.6
Introduction
……………………………………………………….
Lecture Objectives
……………………………………………………….
First Law of Thermodynamics …………………………………………..…
Application of 1ST law of Thermodynamic ……………………………………
Heat capacities of Ideal gas
……………………………………
The Adiabatic Process of ideal gas
……………………………………
Problem set 2
……………………….………………………………
24
24
24
27
33
36
45
Lecture 3: Heat Engines and Second Law of Thermodynamics
3.1
3.2
3.3
3.4
3.5
3.5
Introduction
………………………………………………………
Lecture Objectives
………………………………………………………
Heat Engines and second law of thermodynamic
..……………..
Heat Pumps
………………………………………………………
The Carnot Engine
………………………………………………………
Internal Combustion engine
…………………………………………….
Problem set 3
……………………….……………………………..
..........52
………52
………53
………53
………60
………66
………74
Lecture 4: Entropy and Second Law of Thermodynamics
4.1
4.2
4.3
4.4
4.4
4.5
Introduction
……………………………………………………..
Lecture Objectives
……………………………………………………..
Entropy: One way process
…………………………………
Entropy and Second Law of Thermodynamics
………………………..
Entropy and Performance of Heat Engines
………………………..
Third Law of Thermodynamics
………………………………….
Problem set 4
……………………….……………………………..
………79
………79
………79
………82
………85
………88
………92
Lecture 5: Thermodynamic potentials and Maxwell’s Equations
5.1
5.2
5.3
5.4
5.5
Introduction
……………………………………………………..
Lecture Objectives
……………………………………………………..
Maxwell’s Relations
……………………………………………………..
Maxwell’s Relations using Cyclic rule
…………………………………
Applications of Maxwell’s Relations
…………………………………
Problem set 5
……………………….…………………………....
………89
………90
………95
…….100
…….102
…….104
3
SPH 302: Thermodynamics
Lecture 6: Phase Changes and phase Equilibrium
6.1
6.2
6.3
6.4
6.5
6.6
Introduction
…………………………………………………….
Lecture Objectives
…………………………………………………….
Systems, Phases and Components
…………………………………
Phase Transitions
…………………………………………...
First order phase changes
…………………………………………...
Second order phase changes …………………………………………...
Problem set 6 ……………………………………………………………….
Appendix A: Solutions to Problem sets
Appendix C: Useful Constants
……..105
……..106
……..120
……..124
……..126
……..130
……..133
…………………………….…….. …...134
…………………………………… ……138
4
SPH 302: Thermodynamics
INTRODUCTION TO THE COURSE
This course unit introduces you to Thermodynamics which is a basic science that underlines the
principles of all energy conversion devices and in particular, the transformation of Thermal Energy
into Mechanical Work and vice versa in materials and machines.
Thermodynamics has innumerable applications not only in physics but also in chemistry, biology
and engineering and as such, it is an important course in all fields of science.
Aims of the course
At the end of this course unit, you should be able to explain and apply the laws
of thermodynamics to solve problems related to energy conversion processes in
mechanical systems.
Prerequisites
To follow this course unit, you will require knowledge of partial differential equations and Structure
and Properties of Matter.
Mode of delivery and Study Skills required
This course is conducted as a mix of several teaching methods; Distance and e-learning with limited
face-to-face. This study module is also published on the web.
You are advised to set aside at least 5 hours for each lecture. At the end of every the lecture, you
should attempt all the problem sets. You will be called upon to submit solutions to selected
questions from the problem sets for assessment.
Mode of Assessment


Test and Assignments - 30%
Written Examinations - 70%
Reference materials
1.
2.
3.
4.
Finn C. B. J., Thermal Physics, 1986.
Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics.,
S. Chand & Company Ltd, New Delhi.
F.O. Akuffo, A. Brew-Hammond, F. Makau Luti and J.G.M. Massaquoi. ANST, UNESCO (1997),
Nairobi, Kenya.
Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons, Inc., NY., Pg 479489, 517-561.
5
SPH 302: Thermodynamics
Lecture
1
THERMODYNAMIC CONCEPTS AND
ZEROTH LAW OF THERMODYNAMICS
Lecture Outline
1.1 Introduction
1.2 Lecture Objectives
1.3 Introduction to Thermodynamic
1.4 Thermodynamic Concepts
1.5 Behaviour of Gases
1.6 The Zeroth Law of Thermodynamics
1.7 Temperature and Temperature Scales
1.8 Summary
A cup of hot coffee on a table at room temperature gets colder with time as it
losses heat to the environment (room). Eventually, it attains thermal equilibrium
with its environment i.e., when its temperature does not change over time. This is
the concept of the Zeroth Law of Thermodynamics.
6
SPH 302: Thermodynamics
1.1 Introduction
This lecture introduces you to the study of thermodynamics and what it entails.
We shall start by familiarizing ourselves with terminologies that are often used
when dealing with the subject of thermodynamics and it is therefore very
important that these terminologies become part and parcel of your vocabulary.
We shall also introduce the ideal gas laws and the equation of state. Thereafter,
the Zeroth law of Thermodynamics is discussed which introduces the concept of
temperature and thermometric properties.
1.2 Objectives
At the end of this lecture, you should be able to:
1. Explain what thermodynamics entails
2. Define the basic concepts relating to a thermodynamic system.
3. Define thermodynamic equilibrium and distinguish between the various
types of equilibrium
4. Explain an Ideal gas and the ideal gas Laws
5. Explain the Zeroth Law of Thermodynamics and its importance in
temperature measurements
6. Explain the concept of temperature and the various temperature scales
1.3 Introduction to Thermodynamics
Every time you drive a car, turn on an air conditioner, use a refrigerator or any
electric appliance, you reap the practical benefits of thermodynamics. The term
Thermodynamics is a combination of two words:
Thermo  associated with thermal/heat energy
Dynamics  motion or mechanical work
Literally, Thermodynamics is a science which is based upon the general laws of
nature which govern the conversion of heat into mechanical work and vice-versa.
In terms of classification, Thermodynamics is classified as part of classical
mechanics, which is a branch of physics where we study the behaviour of a
system by considering only the large scale response (macroscopic properties) or
the bulk properties (such as density, volume, pressure etc.) of the system that we
can observe and measure in experiments. This is unlike modern physics which
attempts to explain the behaviour of matter from a microscopic or atomic
viewpoint (using quantum and statistical mechanics) by taking into account the
atomic constitution of matter i.e., structure etc.
The study of thermodynamics was inaugurated by 19 th -century engineers, who
wanted to know the ultimate limitations the laws of physics impose on the
operation of steam engines and other machines that generate mechanical energy
7
SPH 302: Thermodynamics
or heat energy. Today the scope of thermodynamics has immensely increased
with innumerable applications in chemistry and engineering.
Definition
Thermodynamics is the study of relationships involving heat, mechanical work
and other aspects of energy transfer that takes place in thermodynamic systems
such as refrigerators, heat pumps, internal combustion engines etc.
These relationships are governed by the laws of thermodynamics which have
become some of the most important fundamental laws not only in physics and
other sciences but also nature in general. The four principles (referred to as
"Laws of thermodynamics") are:
Zeroth Law of Thermodynamics, which introduces the concept of
temperature and thermodynamic equilibrium where two objects attain the
same temperature when brought in thermal contact.
First Law of Thermodynamics, which mandates conservation of energy,
and represents the relationship between heat and mechanical work.
Second Law of Thermodynamics, which depicts the manner in which
energy changes take place by stating natural processes take place in a
direction such as to increase the disorder (Entropy) of the system OR
equivalently, the Entropy of an isolated macroscopic system never
decreases.
Third Law of Thermodynamics, which explains the nature of bodies in the
neighbourhood of absolute zero temperature. The entropy of a perfect
crystal tends to zero at absolute zero temperature, implying that it is
impossible to cool a system all the way to absolute zero.
1.4 Thermodynamic Concepts
To define terminologies related to thermodynamics, we shall consider a
thermodynamic device such as a refrigerator or a gasoline engine where energy
transfer takes place in the form of heat or mechanical work into or out of the
device. In a refrigerator for example, we do work (to the refrigerator) by extracting
heat from the refrigerator and expelling it to the outside (the room). The
refrigerator in this case is our working system and the room is the surrounding.
The fridge will only operate efficiently if the door is closed i.e., the system has to
be isolated from the surrounding so that we can account for any form of energy
input into or out of the system.
By considering our refrigerator as a thermodynamic system let us now define
basic thermodynamic terminologies/concepts.
Thermodynamic System:– Any portion of the material universe which can be
isolated completely and arbitrarily from the rest for consideration of the
changes which may occur within it under varying conditions. Often a system
8
SPH 302: Thermodynamics
may be considered as composed of smaller systems, which together make up
the larger system.
Take Note
A system is like a proverbial rat. To nail it, you must bring it to the open i.e.,
isolate it from the its surrounding (hiding places).
An example of a system is biological organism (Fig. 1-1) or a mechanical
device (Fig. 1-2), or a specialized material e.g., steam expanding in a turbine.
A thermodynamic system can interact (and exchange heat or matter) with its
surroundings through the boundary wall. Heat and work are the two means of
transferring energy into or out of a system.
There are three classes of systems namely closed, open and isolated
systems.
SURROUNDING
Heat
Fig. 1-1 Human digestive
system is an open system
which exchanges food (matter)
with the surrounding (body)
Boundary
Gas
system
Fig. 1-2. A gas in cylinder (closed) system can
only exchange heat with the surrounding
Open systems:- A system where the exchange of both heat and matter can
occur through the boundary. Example is a human digestive system.
Closed systems:- It is a system where only the exchange of heat between it and
the surrounding is possible through the boundary. There is no matter
exchange. Example is a refrigerator or a gas enclosed in a cylinder (Fig. 1.2).
Throughout the course, we will be interested with the exchange of heat energy
only, restricting ourselves to closed systems.
9
SPH 302: Thermodynamics
Isolated System:- This is a system which is thermally insulated and has no
communication of heat with the surrounding though work may be done on it.
Such changes are called adiabatic changes. Example is a thermos flask.
Boundary:– This is a “wall” that separates the system from the surrounding. The
boundary may allow for exchange of heat and matter between the system and
the surrounding depending on its nature. If the boundary inhibits the system
from changing its volume or shape so that no mechanical work is performed
on it, the wall is said to be rigid.
The surroundings:- Refers to the rest of the universe outside the system.
Diathermal wall:- A wall that allows for exchange of heat between the system
and the surrounding. Good examples are metallic walls. Two systems
separated by a diathermal wall are said to be in contact. In other words, two
systems are in thermal contact if heating one of them results in macroscopic
changes in the other.
Adiabatic or Adiathermal wall:- A wall that does not allow for exchange of heat
between the system and the surrounding. A good example is the walls of a
vacuum flask. A system enclosed by an adiabatic wall is called an isolated
system.
Thermal equilibrium:- This implies the system is in a steady state condition i.e.
the temperature is uniform throughout the system and remains constant in
time such that there is no flow of heat through the system.
Mechanical equilibrium:- This means that there are no unbalanced forces acting
within the system.
Chemical equilibrium:- This means that there are no chemical reactions
occurring within the system.
Thermodynamic (complete) equilibrium:- This represents a condition in which
the system experiences thermal, mechanical and chemical equilibrium. In this
case, the bulk physical properties or thermodynamic variables (Pressure,
Volume and Temperature) of the system are uniform and do not change with
time.
Questions
Think of and name examples of systems hat fall in the categories of
equilibrium defined above.
State variables:- These are directly measurable variables which are sufficient to
describe the bulk behaviour of the system. In the case of a gas system, these
include pressure (P), temperature (T), volume (V), internal energy (U), Entropy
10
SPH 302: Thermodynamics
(S) and composition (). In a magnetic solid, these are the magnetic field (H),
the magnetization (M) and temperature (T). For a homogeneous system,
consisting of a single substance, the composition () is fixed.
Usually, we require at least two state variables to specify the state of a
system. For example, of the 3 measurable variables P, V and T, in a gas
system, only two are independent and any one may be expressed in terms of
the other two e.g.,
P is a function of V and T or P = P(V, T)
V is a function of P and T or V = V(P, T)
Hence, the state of a gas is equally well specified by quoting (P,V); (P,T) or
(V,T). State variables are also known as thermodynamic variables or
coordinates.
Intensive & Extensive Variable
State variables can be extensive or intensive. Extensive variables are
macroscopic parameters that are proportional to the mass or size of the
substance present in the system and as such they correspond to some measure
of the system as a whole. Examples include internal energy (U), entropy (S),
mass (m), volume (V), length, heat capacity (C) etc. as shown in Table 1 below.
Table 1.1 Intensive and extensive variable in some systems
System
Gas or Fluid
Film
Cell
Intensive variable
Pressure, P
Surface Tension, ()
E.m.f, ()
Extensive variable
Volume, V
Area , A
Charge, Z
An intensive variable is discrete (local) in nature i.e., it is independent of mass
or size of the system but it is characteristic of the substance present in the
system. Examples of such variables are pressure, temperature, viscosity,
density, magnetic induction etc.
State functions:- These are functions that are dependent only on the value of
state variables P, V, and T at a given equilibrium state and not on the process
taking place in the system. Examples include the internal energy (U), the
enthalpy (H), and the entropy (S). On the other hand, work and heat are
dependent on nature of the process (path) and not on the final value of the
state variable and are thus not state functions.
Heat & Work:Heat is the transfer of thermal energy between a system and its environment
due to the temperature difference. Work is the transfer of mechanical energy
Heat is equivalent to work in that both represent ways of transferring energy.
Neither heat nor work is an intrinsic property of a system: that is, we cannot
say that a system contains a certain amount of heat or work unlike properties
11
SPH 302: Thermodynamics
such as pressure, temperature, and the internal energy. Thus, Heat and work
are not properties of the state of the system; they are not state functions.
Instead we say that a certain amount of energy can be transferred, either into
or out of the system, as heat or as work. Both heat and work are thus
associated with the thermodynamic process.
Activity 1.1
1. Explain the difference between heat and thermal energy
2. Compare and contrast between heat and work
3. Explain why heat and work are not state functions
Process:- It is a mechanism of bringing about a change from one equilibrium
state of a system to another. A good example is when pushing a piston and
compressing a gas in a cylinder from one equilibrium state (P1,V1) to a new
equilibrium state (P2,V2). A process may be reversible or irreversible.
Reversible process:- Reversible implies that in any such change, the system
must be capable of being returned to its original state with the surrounding
unchanged. This requires two conditions
1.
2.
The processes must be quasi-static i.e., it should take place very slowly such
that it is always in a succession of equilibrium states at any given time
There must be no hysteresis i.e., no dissipative forces such as friction are
present to dissipate energy.
P (Mpa)
a
P2 , V2
Reversible
Irreversible
b
P1 , V1
V m-3
Figure 1-3. Reversible and irreversible processes. Irreversible process cannot be plotted on
an indicator diagram.
Working substance:- This is a gas or fluid enclosed in the system that either
receives energy transfer in the form of heat or work from the surrounding or
transfers energy in the same form to the surrounding.
12
SPH 302: Thermodynamics
Usually, the working substance should remain unchanged at the end of the
process i.e., it should not introduce new parameters during the process.
Because of this, we usually consider a working substance that is ideal or
otherwise called an ideal gas and not a real gas. In the next section we look at
the behaviour of ideal gases.
Activity 1.2
1. Explain the difference between the following thermodynamic terms:
(i) Open and closed system
(ii) Diathermal and adiathermal wall
(iii) State variable and state function
(iv) Reversible and irreversible processes
(v) Intensive and extensive variables
2. Explain what is thermodynamic equilibrium
1.5 Behaviour of Gases
In a gas, the molecules are well separated and they fly about randomly in the
container. We have no way of ascertaining the initial positions and velocities of
each individual molecule. In lack of a microscopic description involving the
individual positions and velocities of the molecules of the gas, we must be
satisfied with a macroscopic description involving just a few measurable variables
(such as mass of the gas, number of moles, volume, density, pressure and
temperature of the gas) that characterize the average conditions in the volume of
a gas. We will deal with the study of macroscopic properties of gases and the
relationship between these properties with the average measurable variables of
the gas system.
In order not to introduce other macroscopic variables in our gas system that may
arise due to interactions between the gas molecules, we shall consider an “ideal”
situation or otherwise called an ideal gas.
1.5.1 What is an Ideal gas?
An ideal gas is an abstraction and it is a gas whose properties represent the
limiting behaviour of real gases at sufficiently low densities. In an ideal gas, it is
assumed that
(i)
(ii)
There are no intermolecular attractions. This implies that the internal energy
of the gas is entirely kinetic and would be dependent on temperature
The molecules themselves occupy negligible volume.
13
SPH 302: Thermodynamics
The behaviour of an ideal gas is dependent on three factors namely its
temperature, pressure and volume. These parameters obey some simple ideal
gas laws namely:
Boyle’s law
 At constant temperature, the volume of a fixed mass of a gas is inversely
proportional to its pressure i.e. P  1/V. This simply means applying pressure
compresses a gas as shown in Fig 1-2 below. Thus:
PV = constant
………….……………..(1.1)
Boyle’s Law
Or
P1V1 = P2V2
[For any two states]
P
0
V m 3C
Figure 1-5. Boyle’s law on a P-V diagram
Take Note
Boyle’s law shows that the PV curve is a hyperbola. The locus of points
( P1V1 ), ( P2V2 ), ( P3V3 ),….., ( PnVn ) in thermal equilibrium is called an
isotherm.
Pressure law
 At constant volume, the pressure (P) of a fixed mass of a gas is  absolute
temperature (T) as shown in Fig. 1-6. Thus:
P
 cons tan t
T
………………………….………..(1.2)
Pressure Law
14
SPH 302: Thermodynamics
P
oC
K
-273oC
0K
Figure 1-6. Pressure law on a P-V diagram
Take Note
Under ideal situations, the pressure goes to zero at absolute zero of
temperature (0 K or –273oC)
Charles’ Law
 At constant pressure, the volume (V) of a fixed mass of a gas (i.e. fixed
number of moles, n) is proportional to its absolute temperature (T) i.e. VT
(Fig. 1-7).
V
 cons tan t
T
…………………………….…….…(1.3)
Charles Law
V
-273oC
0K
oC
K
Figure 1-6. Charles Law on a P-V diagram
Take Note
Charles law implies that an ideal gas can be compressed to zero volume at
absolute zero temperature (0 K or –273oC). This cannot happen with real
gases
15
SPH 302: Thermodynamics

Additionally, at low pressures, V  Number of moles (n) of the gas i.e.
V n

……………………………………………....(1.4)
Combining equations (1.1) to (1.4) gives
PV = nRT
[Equation of state]
…….………....(1.5)
Equation of State
Where R is the universal gas constant = 8.314 J mol-1K-1.
A gas that obeys equation (1.15) at all temperatures and pressure is called an
ideal gas. Eqn. (1.15) is usually referred to as the equation of state of an ideal
gas.
Activity 1.3
1. List the properties of an ideal gas
2. State the three gas laws
3. Starting from first principles, derive the equation of state
Worked Examples
1. What is the pressure of 7 kg of nitrogen gas confined to a volume of 0.4 m3 at 20C ?
Solution
nRT
7  10 3 g
M
= 250 moles
From PV = nRT  P =
where n =
=
28 g/mol
V
m

P = 15.2  105 Nm-2.
2. Early in the morning, the tires of an automobile are cold (280K) and their air is at a pressure of
3.0 atm. Later in the day, after a long trip, the tires are hot (330K). What is the pressure?
Assume volume of the tires remain constant.
Solution
P P
From 1  2
T1 T2

T 
P2  P1  2   3.5 atm.
 T1 
Take Note
Pressure gauges for automobile tires are usually calibrated to read
overpressure i.e., excess above atmospheric pressure. Thus the
gauge would read 2.0 atm in the morning and 2.5 atms later in the
day.
16
SPH 302: Thermodynamics
1.5.2 Non Ideal gas
A real gas (e.g. oxygen) only approximate the ideal gas behaviour or equation
(1.4) at low enough densities (i.e., at high temperatures and low pressures) but
deviates from the law at low temperatures and high pressures. This is because a
real gas is characterized by:
(i)
(ii)
there are intermolecular attractions
The gas molecules themselves occupy a finite volume and as such, the gas
cannot be compressed to zero volume as stipulated by Charles law. Real gases
liquefy at low pressures.

The Equation of state of a real gas that takes into account the above factors
is called Van der Waals gas equation and is given by
a

 P  2 V  nb   nRT
 

……………………..…(1.6)
Where a & b are constants with a being dependent on molecular interactions,
 = V/n is the molecular volume
Activity 1.4
1. Explain why a real gas deviates from the ideal gas behaviour
17
SPH 302: Thermodynamics
1.6 The Zeroth law of thermodynamics
It is a common observation that if you
place a cup of hot coffee or glass of ice
water on a table at room temperature,
the coffee will get colder and the ice
water will get warmer, the temperature of
each approaching that of the room. In
each case, the object will tend towards
thermal equilibrium with its environment
i.e., when its temperature does not
change over time.
Fig 1-8. A steaming cup of coffee loosing heat to the
surrounding as it approaches thermal equilibrium
Such approach to thermal equilibrium must involve energy exchange (in this case
heat) between the system and the environment. This is the concept of the Zeroth
law of thermodynamics which is stated as follows:
Zeroth Law: "If body A and body C are each in thermal equilibrium with a third
body B, then, A is also in thermal equilibrium with C." such arrangement is
shown in Fig. 1-9.
A
B
C
Figure 1-9. Bodies in thermal equilibrium
It follows from the Zeroth law, a whole series of systems could be found that
would be in thermal equilibrium with each other, say, A, B, C, D, etc. All these
systems possess a common property called temperature, which determines the
direction of heat flow.
Take Note: Significance of Zeroth Law
The Zeroth Law introduces the concept of temperature and provides a
means of determining temperature –Through thermal equilibrium.
18
SPH 302: Thermodynamics
1.7 Temperature and Temperature scales
The temperature of a system is a property that determines whether or not, that
system is in thermal equilibrium with other systems. We can also define
Temperature as a measure of the degree of hotness.
The temperature of a substance is determined by measuring the change in the
thermometric property of a substance that varies linearly with change in
temperature. For example, a substance like mercury expands linearly in
proportion to the increase in temperature of the surrounding or substance whose
temperature is to be measured. Most thermometers are based on this principle.
Common types of thermometers include:
(i)
Liquid thermometers:- Based on principle of change in volume of a liquid
with change in temperature e.g., mercury thermometer.
(ii) Resistance thermometer:- Based on the principle of the change in the
resistance of a conductor is linearly dependent on temperature e.g., platinum
resistance thermometers
(iii) Thermoelectric thermometers or thermocouples:(iv) Bimetallic thermometers:1.7.1 Scales of Temperature
If X is the thermometric property that varies linearly with change in temperature
then, the temperature, TX of a substance on an X scale can be given by a linear
relationship of the form
TX  aX
Where a is a constant whose value is fixed as the reference point.
The customarily chosen reference point is the temperature at which ice, water
and vapour coexist in equilibrium, known as the triple point, TP of water and is
assigned a value of 273.16. We may thus write
 X 

TX  273.16
X
 TP 
………………..…….……. (1.7)
Which implies a zero of temperature i.e., T x= 0 at X = 0.
We can now apply Eqn (1.7) to several thermometers depending on the
thermometric property chosen. For a liquid-in-glass thermometer, X is the length
(L) of the liquid column, so Eqn (1.7) gives
 L 

TX  273.16
L
 TP 
………………..…….…...…. (1.8a)
19
SPH 302: Thermodynamics
For a gas at constant pressure, X is the volume (V) of the gas, and Eqn (1.7)
gives
V 

TV  273.16
V
 TP 
………………..…….…..…. (1.8b)
Likewise, for a platinum resistance thermometer, X is the electrical resistance (R)
of the platinum wire such that Eqn (1.7) gives
 R 

TR  273.16
R
 TP 
………………..……..….…. (1.8c)
The perfect gas scale
The perfect gas scale does not depend on the particular properties of a particular
gas but uses pressure and volume of a gas to indicate temperature. In this case,
quite a wide range of temperature can be covered. In this scale
 P 
 K
Tgas  273.16
P
 TP 
………………..…….….…. (1.8d)
Where K, the Kelvin, is the unit of temperature on the ideal gas scale.
1.7.2 Commonly used Temperature Scales
There are three commonly used temperature scales namely:
(i)
The Kelvin scale
The Kelvin, (K) is the unit of temperature on the ideal gas scale. On this
scale, the temperature between ice point and steam point is 100K
(ii) The Celsius Scale
This is a convenient scale whose range is within the range of commonly
encountered temperatures. On this scale, the temperature between ice point
and steam point is 100 oC.
20
SPH 302: Thermodynamics
(ii) Fahrenheit scale
This scale has a range of 180 equal
parts with the ice point at 32 oF and
Steam point is 212 oF.
Relationship between temperature scales
A simple Relationship between the three
temperature scales i.e., Celsius, Fahrenheit
and Kelvin is given by
C
F  32 K  273


100
180
100
Worked Example 1.1
A platinum resistance thermometer has a resistance R = 90.35  when its bulb is placed in a triple
point cell. Determine its temperature when the bulb is placed in an environment where its resistance is
96.28 .
Solution
 R 

R
 TP 
From the equation, TR  273.16

 96.28 
TR  273.16
 K  280.6 K
 90.35 
Worked Example 1.2
The temperature of the surface of the sun is about 6500 oC. What is this temperature on the Kelvin
scale?
Solution
From the equation,

C
F  32

100
180
K  6,500  273  6,773 K
21
SPH 302: Thermodynamics
1.8 Lecture Summary
In this lecture, we learned the following
1. Thermodynamics is the study of relationships involving heat,
mechanical work and other aspects of energy transfer that takes place
in thermodynamic systems such as refrigerators, heat pumps, internal
combustion engines etc.
2. The ideal gas obeys the gas law PV = nRT exactly. No such gas
exists.
3. The internal energy of an ideal gas is entirely kinetic and depends only
on its temperature
4. The Equation of state of a real gas is given by
a 

 P  2 V  nb   nRT
 

4. Statement of the Zeroth Law : "If body A and body C are each in
thermal equilibrium with a third body B, then, A is also in thermal
equilibrium with C."
5. The temperature of a system is a property that determines whether or
not, that system is in thermal equilibrium with other systems.
The Zeroth law tells us that if we introduce heat into a system, this heat has to
flow to the surrounding in order to maintain thermal equilibrium. A closed
system (such as gas enclosed in a cylinder) on the other hand can only
transfer this heat as work to the surrounding i.e., by expanding and doing work
to the environment in order to attain thermal equilibrium. The 1st Law of
Thermodynamics explores this possibility in which heat input is converted to
mechanical work. In the next lecture, we shall discuss the relationship involving
heat, mechanical work and internal energy, given by the 1ST law of
thermodynamics.
22
SPH 302: Thermodynamics
Questions for Discussion
1. Explain why a real gas behaves like an ideal gas at low densities but not at high densities
2. At the airport of La Paz, Bolivia, one of the highest in the world, pilots find it preferable to take
off early in the morning or late at night, when the air is very cold. Explain Why?
Activity
PROBLEM SET 1
The normal boiling point of liquid oxygen is -183oC. What is this temperature
on the Kelvin scale?
1.9 Further Reading
1. Finn C. B. J., Thermal Physics, 1986.
2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics
and Statistical Physics., S. Chand & Company Ltd, New Delhi.
3. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and
Sons, Inc., NY., Pg 479-489.
4. Aduda B. O., (2004). Structure and Properties of Matter, Distance
Learning study module, University of Nairobi, 2004.
Memorable quotes
Imagination is more important than knowledge. Knowledge is
limited, imagination encircles the whole world.
–Albert Einstein.
23
SPH 302: Thermodynamics
Lecture
2
FIRST LAW OF THERMODYNAMICS
Outline
2.1 Introduction
2.2 Objectives
2.3 1ST Law of Thermodynamics
2.4 Applications of 1ST Law of Thermodynamics
2.5 Heat Capacities of an Ideal Gas
2.6 The Adiabatic Process
2.7 Summary
Fig 2.1 (a) This Donkey is an engine: It requires fuel (food) which it ‘burns’ in order to work. This is the
concept of 1ST law of thermodynamics. In this picture, the work required of the donkey is more than it
can provide.
24
SPH 302: Thermodynamics
2.1 Introduction
In this lecture we shall study the first law of thermodynamics and explore possible
thermodynamic processes through which heat can be converted into work. We
shall also discuss the adiabatic processes, and derive the equation of state for the
adiabatic process.
2.2 Objectives
At the end of this lecture, you should be able to:
1. State and explain first law of thermodynamics and recognize perpetual
motion machines of the first kind
2. State the significance and limitations of the 1ST law of thermodynamics
3. Apply the 1ST law to determine work done in isothermal, isobaric,
isochoric and adiabatic processes.
4. Derive the equation of state for the adiabatic process
2.3 First Law of Thermodynamics
The 1st law of thermodynamics is an expression of the principle of conservation of
energy which states that: “Energy can be transformed (changed from one form to
another), but cannot be created or destroyed”. The 1ST law thus denies the
possibility of creating or destroying energy and broadens this principle to include
energy exchange by both heat transfer and mechanical work between the system
and the surrounding and introduces the concept of the internal energy of a
system.
Fig. 2.2. This warden at KWSTI escaping from the
wrath of the zebra is a model of 1ST law of
thermodynamics. His stored energy is converted
into mechanical work (running).
25
SPH 302: Thermodynamics
To state energy relationships (i.e., heat, work and internal energy) or the 1ST law
precisely, we shall consider a closed system consisting of a mechanical device
with a fixed mass of a compressed (ideal) gas enclosed in a cylinder with movable
piston (Fig. 2-1).
Now, If heat dQ is supplied to the system, this heat goes to increase the internal
energy (dU) of the gas as well as enable the gas do some mechanical work (dW)
on the surrounding i.e. push on the piston. If the process is reversible then:
Boundary
SURROUNDING
dQ in
1ST Law
…………………..……….. (2.1)
dQ = dU + dW
Movable piston
Gas
system
Area A
dx
Fig. 2-3. A gas in cylinder system
But, the Work done by the gas can also be given by
dW = Force  distance = PA dx
Where P = Pressure the gas applies to the piston; A = surface area of the piston
and F = PA

dW = PA dx = PdV
since Adx = volume (dV)
Hence, 1ST law (Eqn. 2.1) can also be written as
dQ = dU + PdV
1ST Law
…….…..…….…….…. (2.2)
Eqn. (2.1) is the 1ST law of thermodynamics. It states that:“Whenever we employ some process involving heat and work to change the
state of a system, the change in the internal energy of the system is given by
dU = dQ - dW and it does not depend on the path followed by the process”.
26
SPH 302: Thermodynamics
Take Note
Sign convention of dQ and dW
If the surrounding delivers heat to the system, then dQ is +ve. If the
system delivers heat to its surroundings then dQ is -ve. Likewise, if the
surrounding performs work on the system, then dW is -ve, and +ve
otherwise.
2.3.1 Internal Energy (U)
The internal energy of the system (assume an ideal gas) is the sum of the
following forms of energy
a) Kinetic energy due to translational, rotational and vibrational motion of the
molecules, all of which depend only on temperature.
b) potential energy due to intermolecular forces, which depends on the
separation between the molecules, and
c) The energy of electrons and nuclei.
Take Note: Significance of First Law
The first law provides a method for determining the internal
energy (dU) of a system.
Take Note: Limitations of The First Law
The first law does not say how much of the heat energy can be converted
into work. These limitations are addressed in the 2 ND law of
thermodynamics.
2.3.2 Work as a path dependent function
The internal energy (dU) depends only of temperature (T) of the system and it is
independent on the path taken and is termed a state function. On the other
hand, the heat (dQ) supplied to the system as well as the work (dW) done by the
gas system, both, not only depend on the initial and final states of the system but
also on the path taken. As such, dQ and dW are not state functions.
27
SPH 302: Thermodynamics
Proving that dW is path dependent
To prove that work done is dependent on the path taken in the process, let us
consider an ideal gas being taken from state A to D as shown in figure 2.2. We
see that there are infinite number of ways of reaching point D i.e., either through
path ABD or path ACD or through the curved path A  D.
Along path ABD, the work done is
dWABD = WAB + WBD = 0 + Pf (Vf – Vi)
= Pf (Vf – Vi)
……………………..…….(a)
P
Pf
Pi
B
D
A
C
Vi
Vf
V
Fig.. 2-2. Work as a path dependent function
Along path ACD, the work done is
dWACD = WAC + WCD
= Pi (Vf – Vi)
= Pi (Vf – Vi) + 0
……….…… ……..…….(b)
We see that dWABD  dWACD implying that (dW) is a path dependent
function. Thus Work done depends on the actual path taken, that is, the
way P varies with V.
Take Note:
A state function (such as internal energy, U) is a quantity which
in equilibrium depends only on the thermodynamic variables
rather than on the history of the system
28
SPH 302: Thermodynamics
2.3.4 Thermodynamic Processes
There are FOUR main thermodynamic processes namely: Isothermal,
Isobaric, Isochoric and Adiabatic.
(i) Isothermal process:- Processes that occur at constant temperature e.g., gas
expanding in a cylinder from state P1V1 to state P2V2 (along path a  b) as
shown in Fig. 1-4 below.
P (Mpa)
a
P1
Isochoric
(dV = 0)
P2
Adiabatic (dQ=0)
b
d
Isothermal (dT=0)
Th
c
P3
TC
Isobaric
(dP = 0)
V1
V2
V m-3
Figure 1-4. Isothermal, Isobaric, isochoric and adiabatic process on a P-V diagram
(ii) Isobaric process:- Processes that occur at constant pressure e.g., gas
expanding in a cylinder from state P2V1 to state P2V2 (along path d  b) as
shown in Fig. 1-4.
(iii) Isochoric process:- Processes that occur at constant volume (dV = 0) e.g.,
from state P1V1 to state P2V1 (along path a  d) as shown in Fig. 1-4.
(iv) Adiabatic process:- Processes where no heat enters or leaves the system
(dQ = 0) such as along path a  c. Such processes take place rapidly and
include rapid expansion or rapid cooling.
29
SPH 302: Thermodynamics
2.4 Applications of First Law of Thermodynamics
Since work done is path dependent, it would be of interest to explore possible
thermodynamic processes through which we can convert heat energy into
mechanical work by applying the 1ST Law of thermodynamics and find out which
process gives maximum work output. This is important in a thermodynamic
system as we shall see later in lecture 3.
(a) Work done in isothermal process (dT = 0)
If an ideal gas is taken through the isothermal process (dT = 0), then, from the
ideal gas law (PV = nRT); we have that
PV = constant
 PV diagram is hyperbola as shown in Fig 2.3. The curved path 1  2 is
the isothermal process
P
PV = c
isothermal
1
2
W
V1
V2
V
Figure 2-3. Work done in isothermal process
The work done in this process is the shaded area and is given by
V2
W   PdV
V1
But P 
nRT
.
V
Substituting for P we have
V2
dV
V V
V
W  nRT ln 2
V1
W  nRT 
1

……………………......… (2.3)
W in isothermal
process
30
SPH 302: Thermodynamics
Also from Boyle’s law, we have that P1V1  P2V2 

W  nRT ln
P1
P2
V2 P1

V1 P2
……………………..……(2.4)
W in isothermal
process
Take Note: Significance of Isothermal process
No
Since dU depends on temperature T, then in isothermal process, dU = 0.
The form of 1ST law becomes dQ  PdV .
 In isothermal process, the heat supplied = work output by the ideal gas.
This represents maximum work output from the system
(b) Work done in isochoric process (dV = 0)
Along the isochoric process (path 1  2 in Fig. 2-4), the volume remains constant
(dV = 0) and no work is done in the process. Thus, the heat supplied to the gas
only raises its internal energy and accordingly, 1st law becomes
………………………..……..(2.5)
dQ = dU
P
2
PV  T2
1
PV  T1
V1,2
V
Figure 2-4. Work done in isochoric process
(c) Work done in isobaric process (dP = 0)
In isobaric process, the pressure remains constant (dP = 0) and the work done
along path 1  2 in Fig 2.5 is given by
dW = P(V2 – V1)
……………………….(2.6)
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SPH 302: Thermodynamics
P
2
PV  T2
1
PV  T1
V1
V2
V
Figure 2-5. Work done in isobaric process
Take Note: Significance of Isobaric process
For the isobaric process, the form of 1st Law can be given by
dQ = dU + PdV = dH
No
where H is called the Enthalpy and is the heat evolved in a chemical
reaction (that occur at constant pressure). Enthalpy is a state function.
(d) Work done in Adiabatic process (dQ = 0)
In this case, no heat leaves or enters the system such that the form of first law
becomes
dU = - dW
……………………..………….(2.7)
We note however that, pressure and temperature change and we need to
evaluate the work done in terms of changes in (P) and (T). We shall discuss more
of the adiabatic process in section 2.6.
(e) Other kinds of work
(i) Extension of a wire:
For a wire extended by a torsional force F through a distance dx, the work
done is
dW = - Fdx
…………………………………..…..(2.8)
(ii) Electrolytic cell
Work done by external circuit in charging a cell (increasing its charge by dZ) is
given by
dW = - dz
……………………………..…… (2.9)
where  = emf of the cell
32
SPH 302: Thermodynamics
Worked Example 2.1
10 moles of an ideal gas are compressed isothermally and reversibly from a pressure of 1 atm. to 10
atm. at 300K. How much work is done?
Solution
W=

V2
V2
V1
V1
dV
 PdV  nRT  V
 nRT ln
V2
.
V1
But P1V1 = P2V2
P 
W  nRT ln 1  = 57.4 KJ
 P2 
Worked Example 2.2
An ideal gas is taken through the cyclic process ABCA as shown in Fig 2-6 below. Determine
(i)
The net heat transferred to the system during one complete cycle
(ii)
The net heat input for the reversed cycle ACBA.
P (kPa)
B
8
6
4
2
A
6
C
8
10
V m3
Solution
(i) From first law, we have that
dQ = dU + dW
But dW = PdV = area under the P-V curve and dU = 0 in a cyclic process.

dQ = PdV = 12 8  22  12 KJ
(ii) dQ is the same as in (i)
Worked Example 2.3
In Figure 2-7 below, an Ideal gas system can be taken from state a to c either through abc or
adc. In processes ab and bc, 600 J and 200J are given to the system respectively.
Calculate
(i)
dU along ab
33
SPH 302: Thermodynamics
(ii)
(iii)
dU along abc
Total heat added in adc
P104Pa
8
b
c
200J
600J
3
a
d
2
5
V10-3 m3
Fig.2-7. P-V diagram of ideal gas
Solution
From 1ST law (dU = dQ – dW) we have
(i)
Along ab, dW = 0:  dU = dQ = 600 J
(ii)
Total dQ along a b c = (600 + 200) J = 800 J
dW along ab = 0 since dV = 0
dW along bc = PdV = P(5 - 2) 10-3m3 = (8 104Pa)(3 10-3m3) = 240 J.
 Total dW along a b c = 240 J
 dU along a b c = dQ - dW = (800 - 240) J = 560 J
(iii)
Along a d c, dU = 560 J
(NB. Since dU is a state function, it is independent of the process such that the total dU
along a b c is the same as dU along a d c.
 dW along a d c = P1(V2 - V1 ) = 90 J since along d c, dW = 0
 dQ = dU + dW = (560 + 90)J = 650 J.
Activity 2.1
Calculate the work done by a Van der Waals gas whose equation of


state is given by  P 
a 
V  nb   nRT in expanding from a
V2 
volume V1 to a volume V2.
34
SPH 302: Thermodynamics
Activity 2.2
A sample of an ideal gas is in a vertical cylinder fitted with a piston. 5.79 kJ
of heat energy is transferred to the gas so that the state of the gas changes
from point A to point B along the semicircle shown in Figure 2-8 below.
Determine the change in internal energy of the gas.
P (KPa)
500
400
300
A
B
200
f
100
0
4
1.2
3.6
6
V m-3
Fig 2-8
35
SPH 302: Thermodynamics
2.5 Heat Capacities of An Ideal Gas
We can change the state of a system or body by transferring energy to or from it
in the form of heat or by doing work on the system. In either case, the
temperature of the system changes. The amount of heat transferred to the
system is measured by the heat capacity.
The heat capacity (symbol C) is the amount of heat required to cause a unit rise
in the temperature of a substance. It is given by
C
dQ
dT
……………………..……... (2.10)
On the other hand, the specific heat capacity (c) is the heat capacity per unit
mass or the amount of heat per unit mass required to cause a unit rise in the
temperature of a substance. It is given by
c
C
dQ
dQ


m mdT
nMdT
…………… (2.11)
Where m = mass of the substance = nM with M being the molecular mass; n =
number of moles.
The change in temperature (dT) corresponding to the transfer of a particular
quantity of heat energy dQ will depend on the circumstances under which the
heat was transferred i.e., either at constant volume or at constant pressure etc.
Hence, we speak of Molar heat capacity (Heat capacity per unit mole) at const
Volume (CV) or Molar heat capacity at const pressure (CP).
The Molar heat capacity at constant Volume (CV) is define by
 dQ 
CV  

 ndT V
…………………………….(2.12)
Where n = number of moles in the gas
Likewise, the Molar heat capacity at constant pressure (CP) is given by
 dQ 
CP  

 ndT  P
……………….……… …(2.13)
Take Note:
Usually the heat capacity increases with temperature and is given by the
expression: CP  a  bT  cT
constants
2
 dT 3 where a, b and c are
No
36
SPH 302: Thermodynamics
2.5.1 Relationship between Heat Capacities Cp and CV
We have seen from above that if the heat is introduced into the system at
 dQ 
 S.T dQ = nCPdT. In this case therefore,
 ndT  P
constant pressure then, C P  
the form of 1ST law of thermodynamics can be written as
…………….. …..…(2.14)
nCpdT = dU + nRdT.
While for a constant volume (isochoric) process (dV = 0), the form of 1 ST law
becomes
dQ = nCvdT = dU
……………..……….... …..(2.15)
Substituting for dU in Eqn (2.14) we have
nCpdT = nCvdT + nRdT

Cp - Cv = R
…………….…….…(2.16)
Mayer’s relation
This relation is known as Mayer’s relation
Take Note
dU is a state function dependent on T always. As such, total internal
energy gained either when heat is supplied at constant volume or at
constant pressure is the same. Thus, for simplicity, we can always write
dU as
dU = nCvdT
This expression holds for any ideal gas and for every kind of
process.
Why Cp is greater than Cv
Notably, heat input at constant pressure CP is greater than heat input at constant
volume CV. This is because, when heat is supplied at constant pressure, this heat
goes in doing two things
a) It raises the temperature of the gas (i.e., increase in internal energy) and
b) It does work in expanding the gas against the external pressure i.e., nCPdT
= dU + dW
On the other hand, when the gas is heated at constant volume, no work is done
(dW = 0) and all the heat supplied only raises its internal energy. Hence CP > CV.
37
SPH 302: Thermodynamics
o
An exception where CP < CV is water. When water is heated between 0 C and 4
o
C, the volume decreases such that water has its greatest density (least volume)
at 4oC as shown in Fig 2.9. The reverse occurs on cooling below 4oC which is why
ice floats on water.
This anomalous behaviour of water has an important effect on animal and plant
life in lakes. A lake cools from the surface down and at 4 oC, the cooled water at
the surface flows to the bottom (due to its greater density). As the surface cools
below 4 oC (freezes), the surface water or ice (now less dense) floats on the
warmer water below. The water below thus remains at 4 oC thereby sustaining
plant and animal life below the surface during the cold/winter seasons.
V (cm3)
ice
0
water
2
4oC
To C
Fig. 2.9. The anomalous expansion of water
Questions
Explain why the specific heat capacity at constant pressure is greater
than the specific heat capacity at constant volume for an ideal gas
2.5.2 The Gamma () Relation of Heat Capacities
The Gamma relation is the ratio of heat capacities, Cp to Cv. This relationship can
be obtained from the kinetic theory of gases as follows:
From the kinetic theory of gases, the internal energy (U) of a monatomic gas (n
=1) operating at constant volume (dV = 0) is given by (refer to SPH 203 for
complete derivation),
U  32 RT
or
dU  32 RdT
But we have seen that for isochoric process,
dU  CV dT always
38
SPH 302: Thermodynamics


3
RdT  CV dT
2
CV  32 R
……………………………. (2.17)
Also, from Equation (2.16), we have that CP - Cv = R. Combining (2.16) and
(2.17) we have

CP  CV  R  32 R  R

CP  52 R
……………………………………..(2.18)
Further, taking the rations of CP to CV we have

CP
   1.67
CV
……………………..…(2.19)
The Gamma
Relation
This is the gamma relation of heat capacities
Take Note
Molecules which contain more than 1 atom posses rotational K.E in
addition to translational K.E. When this is taken into account, it can be
shown that:
For a diatomic gas (n = 2),
CP
7
    1.4
CV
5
For a polyatomic gas,
CP
4
    1.33
CV
3
Activity 2.3
Derive the Mayer’s Relation and the Gamma relation of heat
capacities from first principles
2.6 The Adiabatic Process
In an adiabatic process there is no heat input into the system (dQ = 0) but
instead, a gas does work (-dW) to the environment by expanding such that the
39
SPH 302: Thermodynamics
internal energy (dU) decreases as the temperature drops as shown in the Figure
2.10 below. Processes that take place suddenly or quickly are adiabatic.
P (Mpa)
a
P1
Adiabatic, PV = C
Isothermal, PV = C
Th
b
TC
P2
V1
V2
V m-3
Fig. 2-10. The adiabatic ab process of ideal gas
Further, from fig 2.10, we see that in adiabatic process, all the 3 state variables P,
V and T change during the process unlike in other processes like isothermal,
isobaric or isochoric where one parameter remains constant.
Just like in isothermal process of ideal gas where we derived the relationships
between state variables P and V i.e. PV = const, it would be interesting to see
how the state variables are related in the adiabatic process. We shall look at the
following relationships: T-V, P-V and T-P for adiabatic process as well as the work
done in this process.
(a) The T-V relation
The form of 1ST law for adiabatic process can be given by (NB dU = nCVdT for
any process of an ideal gas)
………………..……..(2.20)
nCVdT = - PdV
we note that change in temperature (dT) is accompanied by change in volume
(dV). Thus, we can obtain the T-V relation by eliminating P (using P = nRT/V)
from the above equation (2.20). Doing so we obtain
nCV dT  
nRT
dV
V
40
SPH 302: Thermodynamics

dT
R dV

0
T
CV V
But
C  CV
C
R
 P
 P 1   1
CV
CV
CV

dT
dV
   1
0
T
V
for n = 1
Integrating we have
ln T    1ln V  C

ln T  ln V  1  C

ln TV  1  C

TV  1  C


 1
T1V1
OR
…….………(2.21a)
 T2V2
T-V Relation
 1
(b) The P-V relation
The above equation (2.21) can be converted into a relation between P and V by
eliminating T (using T = PV/nR) i.e.,
PV  1
V
c
nR

Or
PV   c

……………………..(2.22)
Poisson’s Law

P1V1  P1V1
where C = nR = constant
Equation (2.22) is known as Poisson’s Law. It tells us that if we allow the gas to
change its volume with no other constants, the path followed is adiabatic and can
be represented on a PV diagram by a parabola-like curve given by
PV   const with slope = .
41
SPH 302: Thermodynamics

Significance of PV  c
Since  > 1, it follows that the adiabatic curve (PV = c) has a steeper slope (
times) than the isothermal curve (PV = c) at any point at which the two curves
intersect on a P-V indicator diagram as shown in the figure 2-11 below.
P (Mpa)
Adiabatic, PV = c
Pi
Isothermal, PV = c
Pf
Vi
V m-3
Vf
Fig. 2-11. Isothermal and adiabatic curves on a P-V diagram. The
adiabatic curve is  times steeper than the isothermal curve.
The slopes of the isothermal and adiabatic can be obtained by differentiating the
Equations: PV = C and PV

 c respectively. Doing so we obtain
pdV  VdP  0
and
or
PV  1dV  V  dP  0 


 slope  
dP
P

dV
V
for isothermal
dP
P
 
dV
V
for adiabatic
dP 
 of adiabatic is  steeper than the isothermal.
dV 
Activity 2.4
 1
 C , derive the P-V
Starting from the T-V relation: TV
relationship for the adiabatic process and show that the adiabatic curve
is  times steeper than the isothermal curve on a P-V diagram.
42
SPH 302: Thermodynamics
(c) T-P relation
We can obtain a T-P relation by eliminating V in Equation (2.22) using the ideal
gas relation (PV = nRT). From Equation (2.22) we see that
But V 
PV   c


 nRT 
P
 c
P



T
c
P  1


nRT
P
or
T 
P   c
P

since nR = c
P  1
c

T
OR
TP
OR
 1 


  
c
……………………..(2.23)
T-P Relation
(d) Work done in adiabatic Process
The work done in adiabatic process when the gas expands from P1V1 to P2V2 as
shown in Fig 2-12 can be evaluated using the equations of the adiabatic process,
PV   c as follows
P (Mpa)
A
P1
Adiabatic
isothermal
T1
B
P2
a
V1
T2
b
V2
V m-3
Fig 2-9. Work done in Adiabatic process
43
SPH 302: Thermodynamics
V2
V2
dV

V V
W   PdV  C 
V1

W
i.e., by replacing for P
1
C  1
1 

1   V2 1 V1 1 
Where c is a constant that needs to be replaced. But we note that
P1V1  P2V2  c . Replacing for the constant C we get
1 C
C 
1  P2V2 P1V1 
W    1   1  

 V2
V1  1    V2 1 V1 1 

 PV  P2V2 
W  1 1



1


………………….….….[2.24]
W in Adiabatic
Process
This work is the Area (ABba) under the adiabatic curve.
Alternative Derivation
The above equation (2.24) can also be obtained from the 1ST law of
thermodynamics, noting than in this case, dQ = 0. Thus
V2
W   PdV  dU  nCV dT
V1

W  nCV T2  T1 
i.e., TH > TC (dT is –ve)
 CV T1  T2  for n = 1
………..….…………..(a)
NB. In this case, the change in volume (dV) is manifested in change in
temperature (dT) since temperature drops in the process, T 1 < T2 such that
dT is –ve.
Substituting for CV using the relationships of the specific heat (CP – CV = R
and
CP
  ) we get
CV
C
R  CV
CV  P 

R
 CV 
 1

………………..……..(b)
44
SPH 302: Thermodynamics
Replacing for CV in Equation (a) we get
W 
R
T  T2 
 1 1
………….…. (2.25)
Work done in
Adiabatic process
Further, the above equation can also be expressed in terms of P and V by
substituting for T using the ideal gas Equation (PV = nRT)
Thus: from T 
PV
R
we get T2 
P2V2
R
 PV  P2V2 
W  1 1



1



and T1 
P1V1
R
……………...(2.24)
Work done in
Adiabatic process
2.6.1 Examples of adiabatic process
(i) Air compressor
when air is let out from an air compressor like those used in gasoline (petrol)
stations or in paint-spraying equipments, it feels colder than the outside air. This
is because the air expands rapidly (nearly adiabatic) leading to temperature drop.
(ii) Opening a bottle of carbonated beverage
Similar process occurs when you open a bottle of your favourite carbonated
beverage such as coke cola. In this case, the gas enclosed in the beverage bottle
expands very rapidly as it cools and in the process cooling the surrounding air
such that the water vapour condenses forming a miniature cloud near the mouth
of the bottle.
Worked Example
A quantity of nitrogen gas at an initial pressure P 1 = 2  10-5 Nm-2 and an initial volume V1 = 4.0 m3
is compressed rapidly to a volume V2 = 2.5 m3.
(a) what is the final pressure ?
(b) if the nitrogen is now cooled to the original temperature, find the final pressure ( for nitrogen
= 1.4)
Solution
(a) A rapid compression is an adiabatic process. Thus from


P1 V 1 = P 2 V2

V 
 P2 = P1  1  = 3.86  105 Nm-2.
 V2 
(b) If final temperature = initial temperature then dT = 0 (isothermal)
Thus using P1V1 = P2V2 
P2 =
V1
P1 = 3.2  105 Nm-2.
V2
45
SPH 302: Thermodynamics
2.7 Summary
In this lecture, we have learned that
1. The formulation of 1St law of thermodynamics is given by dQ = dU + dW
where dQ = Heat input, dU = Internal energy and dW = work done
2. The work done in various thermodynamic processes is given by
V
(i) Isothermal process (dT = 0): dW  nRT ln 2
V1
(ii) Isochoric process (dV = 0):
dW = 0
T
(iIi) Isobaric process (dP = 0):
dW  nRT ln 1
T2
1
PV  P V  or
 1 1 1 2 2
R
T  T 
W
 1 1 2
(iv) Adiabatic process W 
3. Internal Energy of an ideal gas is given by dU = nCvdT
4. The heat capacities at constant pressure and volume are given by
 dQ 
 dQ 
and
CP  
Cv  


 dT  P
 dT v
5. The relationship between heat capacities is given by
[Mayer’s Relation]
CP  CV  R
CP
   1.67
CV
[The gamma Relation]
6. The relationships in the adiabatic process are given by
T-V Relation:
P-V Relation:
T-P Relation:
TV  1  C
PV   C
TP
 1 
  


c
In the next lecture, we shall study the second law of thermodynamics which provides
specific guidelines for energy transfer and conversion processes by defining the thermal
efficiency of processes that convert heat into work.
46
SPH 302: Thermodynamics
Questions for Discussion
1. Suppose that U is the balance of a bank account, Q the money paid in as cash, and W the
money drawn out by Cheque.
(a) Write an equation which relates an initial balance UI, a final balance UF, Q and W.
(b) If the bank manager could keep records of only one of these three variables, which would
he choose, and why?
(c) To what extend does the analogy between this situation and the 1st law of thermodynamics
hold well?
2. When ice melts at 0o C, its volume decreases. Is the internal energy change greater than, less
than, or equal to the heat added? How can you tell.
3. Household refrigerators always have tubing on the outside, usually at the back or bottom. When
the refrigerator is running, the tubing becomes quite hot. Where does the heat come from?
4. In constant-volume process, dU = nCVdT. But in a constant-pressure process, it is not true that
dU = nCPdT. Explain why
5. When you blow on the back of your hand with your mouth wide open, your breath feels warm.
But if you partially close your mouth to form an “O” and then blow on your hand, your breath
feels cool. Explain why.
6. Air escaping from an air hose at a gas station always feels cold. Explain
7. How can you best use a spoon to cool a cup of coffee? Stirring-which involves doing workwould seem to heat the coffee rather than cool it
8. How does a layer of snow protect plants during cold weather? During freezing spells, citrus
growers in Nairobi often spray their fruit with water, hoping it will freeze. How does that help?
9.
It is difficult to “boil” eggs in water at the top of a high mountain because water boils there at a
relatively low temperature. What is a simple, practical way of overcoming this difficulty?
10.
Will a 3-minute egg cook any faster if the water is boiling furiously than if it is simmering
quietly?
11.
Explain why the specific heat at constant pressure is greater than the specific heat at
constant volume. Can CP ever be less than CV? If so, give an example.
12.
Why is the difference between CP and CV often neglected for solids?
13.
Real gases always cool when making a free expansion, whereas an ideal gas does not.
Explain.
14.
Discuss the similarities and especially the distinctions between heat, work, and internal
energy.
47
SPH 302: Thermodynamics
PROBLEM SET 2
1. Ice at 0o C and at a pressure of 1 atm has a density of 916.23 Kgm-3 while water under these
conditions has a density of 999.84 Kgm-3. How much work is done against the atmosphere when
10Kg of ice melt into water.
2. Two moles of a monatomic ideal gas at a temperature of 300K expand reversibly and
isothermally to twice the original volume. Determine
(i)
the work done by the gas,
(ii)
the change in the internal energy.
(iii)
the heat supplied and
3. Three moles of an ideal gas have an initial temperature of 127 o C. While the temperature is kept
constant, the volume is increased until the pressure drops to 40% of its original value.
(a) Draw a P-V diagram for this process
(b) Calculate the work done by the gas
4. A gas is contained in a cylinder fitted with a frictionless piston and is taken from state a to state b
along the path acb as shown in Fig. 1. 80J of heat flow into the system and the system does 30J
of work.
P
P
c
b
a
d
P1
P2
V
Fig. 1
(i)
(ii)
(iii)
a
b
d
c
V1
V2
V
Fig. 2
How much heat flows into the system along the path adb if the work done by the gas
system is 10J.
When the system is returned from state b to a along the curved path, the work done on the
system is
20J. Find the heat transfer.
If Ua = 0 and Ud = 40J, find the heat absorbed in the process ad and db.
5. An ideal gas is taken through the cycle a  b  c d  a as shown in Fig. 2. If P1 = 3 atm,
P2 = 1 atm, V1 = 1 litre and V2 = 2 litres, find the work done in this cycle.
6. A system is taken from state a to b along the three paths shown in the figure 3 below.
(a) Along which path is the work done by the system greatest? The least
(b) If Ub > Ua, along which path is the value of the heat transfer Q the greatest? For which
path is the heat absorbed or librated by the system?
48
SPH 302: Thermodynamics
P
b
1
2
a
3
0
V
Fig.3
7. An ideal gas is taken through the cycle ABCDA as shown in Fig. 4 below. Determine
(i) The work done on the gas per cycle
(ii) The net heat energy added to the system per cycle
(iv)
The work done per cycle for 1.00 mol of the gas at 0 oC.
P
B
3Pi
C
A
Pi
D
Vi
V m-3
3Vi
8. An ideal gas expand from state i to f as shown in Fig 5 below
(i) Determine the work done on the gas
(ii) the gas is now compressed from state f back to state i along the same path. Determine
the work done.
P (Pa)
i
6  106
4  106
f
2  106
0
1
2
3
4
V m-3
49
SPH 302: Thermodynamics
9. A sample of an ideal gas is taken through the process ABCDA as shown in Fig 6 below.
Process AB is adiabatic while process BC is isobaric with 100 kJ of heat entering the system.
From C to D, the process is isothermal while from D to A, the process is isobaric with 150 kJ of
energy leaving the system. Determine the difference in the internal energy between points A
and B i.e., UB – UA.
P (atm)
dQ = 100 KJ
B
3.0
C
A
1.0
D
dQ = 150 KJ
0
0.09
0.2
0.4
1.2
V m3
10.
An ideal monatomic gas ( = 5/3) expands reversibly from a state V1, P1 to a volume V2.
Calculate the work done by the gas if the change takes place
(i)
Isothermally
(ii)
adiabatically (iii)
such that PV is a constant.
11.
For an ideal gas and starting from the same initial point, show the following processes both
on a P-V and T-S diagrams
(i)
PV = constant
(ii)
PV = constant
(iii) P = constant, and
(ii) V = constant.
12.
Show from 1ST principles that the following relationships hold for a reversible adiabatic
expansion of an ideal gas
(i)
TV  1  const (ii) PV   const (iii)
 T 
 1- 1   cont
P  
13.
Find the change in the internal energy of one mole of a monatomic ideal gas in an isobaric
expansion at 1 atm from a volume of 5 m3 to 10 m3 ( = 5/3).
14.
Show that the adiabatic curve for an ideal gas is steeper by a factor of  than the isothermal
curve at any point at which the two curves intersect on the P-V indicator diagram.
15.
During a reversible adiabatic expansion of an ideal gas, the pressure and volume at any
moment are related by PV   const where c and  are constants. Show from first principles
that the work done by the gas in expanding from state (P 1, V1) to a state (P2, V2) can be given
 P1V1  P2V2 

  1 
by W  
16.
 L  Lo  2 
The equation of state of a rubber band if given by F  aT      where Lo is the
 Lo  L  
original length and a is constant = 1.310-2 NK-1. How much work is done when the band is
50
SPH 302: Thermodynamics
stretched isothermally and reversibly from its original length of 10cm to 20cm, the
temperature being 20o C?
17. Calculate the work done by a Van der Waals gas whose equation of state is given by
a 

 P  2 V  nb   nRT in expanding from a volume V1 to a volume V2.
V 

(a) at constant pressure and
(b) at constant temperature
18. Fig. 7 shows an energy cycle with three reversible processes to which 16g of oxygen gas (M r =
32) are subjected. Calculate
(a)
the heat taken from the gas during the isovolumetric cooling,
(b)
the internal energy gained by the gas during the isobaric heating (process (2)),
(c)
the work done by the gas during process (2),
(d)
the heat supplied during process (2) (two possible methods), and
(e)
the work done on the gas while it is compressed isothermally. (C v,m for oxygen is 21
mol-1 K-1 ).
P (Mpa)
isovolumetric
(1)
0.10
(3) isothermal
(2)
isobaric
Th = 500K
TC = 300K
V m3
Figure 7
2.8 References
1. Finn C. B. J., Thermal Physics, 1986.
2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and
Statistical Physics., S. Chand & Company Ltd, New Delhi.
3. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons,
Inc., NY., Pg. 517- 536.
Memorable quotes
Without leaps of imagination or dreaming, we loose the excitement
of possibilities. Dreaming, after all, is a form of planning.
–Gloria Steinem.
51
SPH 302: Thermodynamics
Lecture
3
HEAT ENGINES AND SECOND LAW OF
THERMODYNAMICS
Outline
3.1 Introduction
3.2 Objectives
3.3 Heat Engines and the Second law
3.4 Heat Pumps
3.5 The Carnot Engine
3.6 Internal Combustion Engines
3.7 Summary
This motorcycle is a heat engine that takes in heat from the burning of petrol at a
high temperature, does work and ejects the remainder of heat as waste to the
atmosphere (low temperature reservoir) through the exhaust.
52
SPH 302: Thermodynamics
3.1 Introduction
To operate a machanical device such as a vehicle, we need to convert heat into
mechanical energy/work. Although it is easier to convert mechanical energy
completely into heat (e.g., when you step on the brakes of a car), it is impossible
to convert heat (internal energy) completely into mechanical energy. Heat engines
(steam engines, automobile engines, jet engines etc.) are only partly successful at
converting internal energy (heat) into mechanical energy (work). A closely related
process occurs in the animal kingdom where food is ‘burned’ and partially
converted into mechanical energy when muscles are at work. The rest of the
unused food is excreted. On the other hand, refrigerators are partially successful
in transporting heat from a cooler to hotter bodies (environment). The reasons to
nd
these questions lie in the directions of thermodynamic processes given by the 2
law of thermodynamics,
The second law of thermodynamics deals with the direction in which natural
processes occur. It is often said that the second law gives a preferred direction to
the “arrow of time”, teaching us that systems naturally evolve with time in one
direction but not in the other i.e., heat can only flow from high temperature
reserviour and not vice versa. The second law of thermodynamics also provides
specific guidelines for energy transfer and conversion processes by defining the
thermal efficiency of heat engines thereby imposing limitations on the efficiency of
processes that convert heat energy into mechanical energy and vice versa.
In this lecture we shall discuss the principle of operation of heat engines and
analyze their efficiency in accordance with second law. The Carnot engine is
discussed as the most ideal heat engine whose efficiency is a criteria of
perfection for all thermal power cycles.
You will also be introduced to ideal cycles upon which internal combustion
engines are based. Description of the ideal processes in refrigerators and air
conditioners and their cycle efficiencies are also treated. Lastly, we shall look at
why real engines differ from the ideal cycles.
3.2 Objectives
At the end of this lecture, you should be able to:
1. Describe the working principle of a heat engine and recognize
perpetual motion machines of the 2ND kind
2. Explain the 2nd Law of thermodynamics and state its significance.
3. Describe a Carnot engine, determine its thermal efficiency and
appreciate the importance of the Carnot efficiency as a criteria of
perfection for all thermal power cycles.
4. Explain the working principle of a refrigerator and internal combustion
engines and determine the coefficient of performance.
53
SPH 302: Thermodynamics
3.3 Heat Engines and 2nd Law of Thermodynamics
A heat engine is any device that converts heat into work by means of a cyclic
process. Example of heat engines includes steam engines, automobile engines,
jet engines etc. In any heat engine, the working substance undergoes a cyclic
process i.e., it must able to be returned to its initial configuration at the end of
each cycle, so that the process can be repeated all over again. A good example is
the steam in a steam engine.
During each cyclic process, a heat engine absorbs heat (Qh) from a high
temperature heat reservoir (at Th oC) such as a boiler in the case of a steam
engine or from the combustion of gasoline in the case of an automobile engine.
The engine then converts this heat partially into work (W) and ejects the
remainder as waste heat (Q2) into a low temperature reservoir (at T c oC) such as a
condenser in the case of steam engine or the exhaust of the automobile engine
as depicted in Fig. 3-1.
High temperature
reservoir at Th
Qh
E
W
Qc
Low temperature
reservoir at TC
Fig. 3-1: Operation of a heat Engine
Take Note
A heat reservoir is simply a body that remains at constant temperature,
even when heat is removed from or added to it.
The second law asserts that a heat engine cannot convert all the heat it receives
from a heat source into work. Some of the heat received must be rejected into a
heat sink. This is the law of nature: “If you must eat, then you must shit”. This
assertion is clearly stated in the “Kelvin-Plank’s statement” of the second law.
The Kelvin-Plank or “Engine” Statement of 2ND Law
This statement places emphasis on the operation of heat engines and states that
54
SPH 302: Thermodynamics
“it is impossible to construct a device which operating in a cycle produces
no other effect than extract heat from a high temperature reservoir and
convert the heat completely into mechanical work”.
According to this statement, it is impossible to construct a heat engine which
produces work continuously by taking heat from a single reservoir and ejecting
nothing into the low temperature sink as depicted in Fig. 3.2a. If the second law
were not true, we could power an automobile by simply cooling the surrounding
air. Therefore, a perpetual motion machine of the second kind does not exist.
Example of a perpetual machine would be a car without an exhaust or a cooling
system as shown in Fig 3-2b or an animal that does not shit. Such machine
contradicts the Kelvin-Plank’s statement of the 2ND law.
High temperature
reservoir at Th
Qh
Engine
E
Fuel Tank
W
(b) A perpetual motion machine showing a
car without exhaust or cooling system
which is naturally unrealistic.
Fig. 3-2: (a) A hypothetical heat
Engine: NOT ALLOWED by the
Kelvin-Plank statement
The 2ND law also provides a basis for us to define a criterion of performance for all
heat engines – Thermal efficiency.
Efficiency of heat engines
The “efficiency” or “effectiveness” of a heat engine is a measure of the extent to
which a heat engine is able to convert the heat supplied to it into work. In other
words, it is the ratio of the work output (i.e., what has been accomplished) to the
heat input (i.e., what you pay for to get it done). i.e.,

W
Qh
……………………………………….(3.1)
Where W = work performed and Qh = heat obtained from the hot reservoir
The work (W) done by a heat engine can be obtained from the 1 St law of
thermodynamics (dW = dQ – dU). Since in such a cyclic process, there is no
change in the internal energy of the system (i.e., dU = 0), it follows that dW = dQ
55
SPH 302: Thermodynamics
i.e., the net heat input is converted to work output [Principle of heat engine].
Therefore
………………….….(3.2)
W = Qh - Qc
W Qh  Qc

Qh
Qh



  1
Qc
Qh
………….……….….(3.3)
If the working substance is an ideal gas, then it can also be shown that
Qh QC

Th TC

  1
Th
TC
……..……………..…….….(3.3)
Experimental evidence suggests that even under ideal conditions, QC  0 (never)
and as such, it is impossible to construct a heat engine with 100% thermal
efficiency as stipulated in the 2ND law of thermodynamics. This is because a heat
engine always ejects substantial amount of heat (QC) into the environment.
Activity 3.1
Show from first principles that
Qh QC

Th TC
3.4 Heat Pumps
A heat pump is a heat engine operated in reverse i.e., it is an engine which
requires work input in order to transfer heat from a colder reservoir to a hot
reservoir as shown in Fig 3.3. This is the principle involved in refrigerators, heat
pumps and air conditioners.
Heat pumps are widely used to heat buildings during the cold weather. They
operate by transferring heat from the outside (which is at lower temperature QC)
to the insider which is at a higher temperature Q h. In comparison, a heat pump is
merely a refrigerator turned inside out so that its cold end is outdoors and its
warm end indoors.
56
SPH 302: Thermodynamics
HOT RESERVOIR
(Outside) at Th
HOT RESERVOIR
(Outside) at Th
Qh
Qh
R
R
W
QC
QC
COLD RESEVIOR
(Inside Refrigerator) at TC
COLD RESEVIOR
(Inside Refrigerator) at TC
Fig. 3-3: (a) Operation of Refrigerator
(b) Hypothetical Refrigerator:
Clausius -NOT ALLOWED
The second law also tells us that it is impossible to construct a heat pump which
can transfer heat from a cold reservoir to a hotter region with no input of
mechanical work”. This is stated in the “Refrigerator” or “Clausius Statement” of
the second law.
The Clausius Statement of 2ND Law
“No cyclic process can transfer heat from a cold reservoir to a hotter region with
no input of mechanical work”.
This simply means: “For you to work, you must eat”
Thus the second law denies the possibility of reversing the natural tendency for
heat to flow from a hotter to a colder body without external interference (in the
form of work). The second law also allows us to define a criterion of performance
of heat pumps. This criterion or the “effectiveness” of a heat pump is called the
HP
coefficient of performance (COP ) and is defined by
COP HP 
Heat transferre d to high temp reservoir
Work input
COP HP 
Qh
Qh

W Qh  QC
………………(3.4)
For heat pumps, COPHP >1 and it depends on the outside temperature. If the
outside temperature is say 4oC or higher, COPHP  4 i.e., amount of energy
transferred to the building is about 4 times greater than the work done by the
motor in the heat pump. However, as the outside temperature decreases (TC 
0), the heat delivered becomes equal to the work required and the COPHP can fall
below unit. Under such circumstances, it becomes difficult for the heat pump to
57
SPH 302: Thermodynamics
extract sufficient energy from the outside air and the heat pump has no economic
advantage over electric heaters. Under such circumstances, it would be cheaper to
burn the original fuel directly for the heat, rather than generate electricity to
operate a heat pump.
Thus, the use of heat pumps that extract energy from the air is only satisfactory in
moderate climate but it is not appropriate in areas where cold (winter)
temperatures are very low. An alternative is to bury the external coils deep in the
ground so that the energy is extracted from the ground, which is warmer than the
air above.
3.4.1 The Refrigerator
A refrigerator is a heat engine which requires work input to transfer heat from a
colder reservoir to a hot reservoir as shown in Fig 3.4. Practical refrigerators use
gas and liquid Freon (dichlorodifluoromethane) as working fluid (refrigerants), and
their cycle differs from the Carnot cycle. Refrigerants (Freon etc.) have a boiling
point near room temperature when at high pressure, but a boiling point below at
0o C when at low pressure.
In each cycle, Liquid Freon at low pressure and low temperature enters the
cooling coils in the refrigerator box and absorbs heat while evaporating into Freon
gas (see Fig 3.4a). This gas flows to the compressor (outside the refrigerator
box), where its pressure, temperature and density are increased by the push of a
piston. The high-pressure gas then circulates through the condenser coils, which
are exposed to the atmospheric air as the Freon gas loses its heat and
condenses into a liquid (see Fig 3.4b and c). This high-pressure liquid then
passes through an expansion valve (a small orifice) where its pressure is reduced
to match the low pressure in the cooling coils. This return of the fluid to the
cooling coils completes the cycle.
Figure 3-4(a) and (b) Principle of a mechanical refrigeration cycle
58
SPH 302: Thermodynamics
Figure 3.4(c) The coils on the
back of a refrigerator transfer energy
by heat to the air outside
(d) Inside a refrigerator
The effectiveness of a refrigerator is sometimes called the coefficient of
performance (COPR). For an ideal Carnot refrigerator, (COPR) is defined by
Heat transferre d from cold reserviour
Work input
Q
QC
TC
 C 

..................(3.5)
W Qh  QC Th  TC
COP R 
Carnot
Refrigerator
It can also be shown that
COP HP 1  COP R
……………………(3.6)
Activity 3.2
Show from first principles that COP
HP
1  COP R
59
SPH 302: Thermodynamics
From an economic point of view, the best refrigerator is one that removes the
greatest amount of heat QC from the inside of refrigerator for the least amount of
work (W). The value of COPR is  5. However, if (Th – TC) or Th (the room
temperature) is very high, then COPR  0 and more work will be required to
transfer a given quantity of heat from the refrigerator. That is why it becomes
difficult to obtain cooling at very low temperatures in the refrigerator, since more
work is needed.
Note also that the refrigerator supplies heat to the heat pump at the back of the
refrigerator. The function of such a heat pump is to deliver heat to some reservoir
which is at a higher temperature than its surrounding.
Take Note
Work is always needed to operate a refrigerators. If no work were
needed, COPR would be infinite and such device is a workless
refrigerator; it is a mythical beast, like the unicorn and the free lunch.
Worked Example 3.1
Suppose that QC = 200J, W = 100J. Determine the COP of the refrigerator
Solution:
In this case, Qh = 200 + 100 = 300J

COP R 
QC
200

Qh  QC 100
2
or COP
R
 200%
Take Note
In the case of a heat engine, efficiency cannot be more than 100% but in
the case of a refrigerator, COPR can be higher than 100%.
Worked example 3.2
In a refrigerator, the cooling chamber is maintained at 290K while the outside temperature is 305K.
The motor (located outside) has compression cylinders operating at 320K and the expansion coils
inside the chamber operating at 280K. If the motor operates reversibly
(i)
(ii)
(iii)
(iv)
Give a schematic diagram illustrating the sequence of events
Calculate the efficiency of the motor
Calculate the COP of the refrigerator
How much work must be done for each transfer of 5,000J of heat from the chamber?
60
SPH 302: Thermodynamics
Solution
In this case, the refrigerator is powered by the motor which supplies work to it as shown below.
(i)
Room at
Th = 305K
HOT RESERVOIR
at ThE = 320K
QhE
Qh
R
W
E
QCE
QC
Inside Fridge
at TC = 290K
(ii)
COLD RESEVIOR
at TCE = 280K
The refrigerator is powered by the motor (E). Hence the efficiency of the motor is

(iii)
COP R 
(iv)
From
Q  QCE 320  280
W
 hE

 12.5%
QhE
QhE
320
QC
QC
TC


W Qh  QC Th  TC
 19.3
QhE ThE
T

 QhE  hE 5000  5,258.6 J
QCE TCE
TCE

W  QhE  QCE  258.6 J
61
SPH 302: Thermodynamics
3.4.2 The Air Conditioner
The air conditioner employs a similar refrigeration cycle but in this case, the
refrigerator box becomes a room or an entire building. The evaporator coils are
inside, the condenser is outside, and fans circulate air through these as shown in
the figure 3.6 below.
Cold air to the room
Hot air to outside
Fig 3-5: An air conditioner working on the same principle as a refrigerator
The parameters of interest are the rate of heat removal (heat current H from the
room) and the power input P ( =W/t) to the compressor. H is the heat removed per
sec = (QC/t). Hence coefficient of performance is
COP AC 
QC Ht H
TC

 
W
Pt P Th  TC
……………..…(3.7)
A typical value of COPAC is about 2.5 and is dependent on the inside and
outside temperatures.
62
SPH 302: Thermodynamics
3.5 The Carnot Engine
Since the conversion of work to heat is an irreversible process, a heat engine only
partly reverses this process. For maximum efficiency, the thermodynamic process
within the engine must be reversible i.e., the engine can, in principle be operated
in reverse converting work into heat at the same rate as in the forward direction.
Thus, when the engine takes heat from the hot reservoir at Th, The working
substance should also be at Th i.e., no heat should be lost. Similarly, when the
engine discards heat to the cold reservoir at TC, the engine itself must be at TC.
Thus, every process involving heat transfer must be isothermal at either Th or TC.
Conversely, for any process in which the working substance is intermediate
between Th and TC, there must be no heat transfer (adiabatic process) between
the engine and either reservoir otherwise, such a process could not be reversible.
Additionally, thermal and mechanical equilibrium must be maintained at all times.
The only engine with maximum possible efficiency is a hypothetical, idealized
reversible heat engine called the Carnot Engine.
A Carnot engine is a hypothetical engine which is assumed to be reversible with
no dissipative effects such as turbulence and friction. The Carnot engine operates
between two reservoirs with an ideal gas as the working fluid enclosed in a
cylinder with a piston.
Take Note
It is important to emphasize that reversibility is a standard of perfection. In
real life, no process is totally reversible; reversibility is approached but it is
never achieved in practice. A reversible process is thus an abstraction
because all natural processes result in energy loss in terms of turbulence,
friction or other dissipative effects.
A reversible engine therefore represents the perfect engine in the sense
that it converts the highest fraction of heat into work.
3.5.1 The Carnot Cycle
The Carnot cycle consists of a sequence of 4 steps (with two isothermal and two
adiabatic processes) as depicted in the P-V diagram by cycle abcd in Fig. 3.6.
These are namely:
Path a  b: Heat Qh enters the system and the gas expands isothermally (dT = 0)
from volume V1 to V2 doing work on the piston
Path b  c: Gas expands adiabatically (dQ =0) as temperature decreases from
Th to TC.
63
SPH 302: Thermodynamics
SADI CARNOT, French
Engineer (1796–1832)
Carnot is considered to be
the founder of the science of
thermodynamics. Some of
his notes found after his
death indicate that he was
the first to recognize the
relationship between work
and heat.
Qh
P
W
V1, P1
a
Qh Isothermal expansion, Th
V2, P2
b
Adiabatic
compression
V4, P4
Adiabatic
expansion
d
V3, P3
Isothermal
compression
Qc
c
W
Temp cools with
little work done
 Exhaust
Qc
Fig. 3-6. The Carnot Cycle on a P-V diagram
V
Path c  d: Gas is compressed isothermally from volume V 3 to V1 thereby
ejecting heat (QC) into the low temperature reservoir at TC.
Path d  a: Gas is compressed adiabatically from (V4, P4) to (P1,V1) as
temperature rises from TC to Th. The total done in the cycle is the area
abcd.
64
SPH 302: Thermodynamics
The efficiency of a Carnot engine can be obtained using equation (3.1) i.e.,

Qh
W

Qh Qh  QC
………..……………..(3.1)
In this case we note that, for the isothermal expansion a  b, dUab = 0 (since dT
= 0). Hence we have that.
V 
Qh  dWab  nRTh ln  2 
 V1 
Similarly,
V 
V 
QC  dWcd  nRTh ln  4   nRTC ln  3 
 V4 
 V3 


 V3  
TC ln   
Q

 V4  
 1 C 1 

Qh


V
 T ln  2  
 h  V1  
……………..………(3.8)
But for the two adiabatic processes b  c and d  a, we have
ThV2 1  TCV3 1
and
ThV1 1  TCV4 1
From which
V2 1 V3 1

V1 1 V4 1
Or
V2 V3

V1 V4
Substituting these in equation (3.8) gives the efficiency as
Q
T
 1 C 1 C
Qh
Th
…….………..……(3.9)
Carnot
Efficiency
This is the Carnot efficiency for a heat engine and it is the maximum possible
efficiency attained by any engine operating between 2 reservoirs.
We note that the efficiency () can only be = 1 (100%) iff TC = 0 K, which is not
possible. No real engine can have a thermal efficiency greater than the Carnot
efficiency. This assertion is stated in the Carnot’s theorem.
65
SPH 302: Thermodynamics
Carnot's Theorem
Carnot’s theorem is a result of the second law of thermodynamic. It is a principle
that specifies limits on the maximum efficiency any heat engine can obtain, which
solely depends on the difference between the hot and cold temperature
reservoirs.
Carnot's theorem states that:

All real (irreversible) heat engines operating between any two heat
reservoirs are less efficient than a Carnot’s engine operating between the
same reservoirs.

All reversible heat engines operating between the same two heat
reservoirs have the same efficiency (as a Carnot’s engine operating
between the same reservoirs).
Activity 3.3
1. Show that the efficiency of a heat engine can be given as
 1
QC
T
1 C
Qh
Th
2. State the Carnot’s Theorem
Take Note: Why the Carnot Efficiency is the best possible:
It assumes that there is no temperature difference between the heat
reservoir and the working fluid, so that the entropy gained by one exactly
matches the entropy lost by the other, with no net change in entropy for the
system as a whole. This condition is of course an ideal one, and cannot be
met in practice by any real machine. Even theoretically; all real machines will
be strictly worse than this.
66
SPH 302: Thermodynamics
Worked Example 3.3
A man uses a Carnot engine operating in reverse as a heat pump to extract heat
from the outside air (at -10o C) and inject it into his house (at 20o C). What is the
amount of work that must be supplied to pump 4.2 KJ of heat from the outside to
the inside?
Solution
Th = (20 + 273) K, TC = (-10 + 273) K

QC TC

 0.9
Qh Th
Or Qh = (Qc /0.9) = 4.67 KJ
 Work supplied = Qh - QC = 0.47 KJ
This is more economical heating method than the expenditure of 4.2 KJ of fuel in
a conventional furnace or electric heater.
3.5.3 The steam engine
The operation of a steam engine resembles the Carnot cycle but uses steam and
water as working fluid instead of an ideal gas. It consists of a cylinder fitted with a
piston that performs cyclic motions of expansion and compression (Fig. 3.7).
condenser
boiler
cylinder
water
heat
Fig. 3-7. A steam engine (schematic)
Figure 3.7(b). This steam-driven locomotive
that used to run from mombasa to kisumu obtains
its energy by burning wood or coal. The generated
energy vaporizes water into steam, which powers
the locomotive. (This locomotive must take on
water from tanks located along the route to
replace steam lost through the funnel.)
67
SPH 302: Thermodynamics
In each cycle of operation, the high-pressure steam enters the cylinder and
pushes against the piston doing work. The low-pressure, spent steam is then
exhausted from the cylinder and sent to condenser where an external coolant (air
or flowing water) condenses the steam into liquid water which is pumped back to
the boiler and the cycle is repeated.
The Efficiency of a steam engine is defined by
 Steam  1 
QC
T
 1 C
Qh
Th
………………(3.10)
Such simple steam engine has efficiencies of about 5-18%. To maximize the
efficiency, the designer must make the intake temperature Th as high as possible
and TC as low as possible (= room temperature of water). Th is limited by the
mechanical strength of the boiler (since pressure increases with temperature) and
maximum value of Th possible is 500 oC (pressure = 235 atm in present day
steam boilers). Most modern steam engines employ a turbine wheel instead of
the cylinder and piston and can achieve efficiencies of up to 40%.
Worked Example 3.3
A steam engine has a boiler that operates at 500 K. The cold reservoir’s
temperature is that of the outside air, approximately 300K. Determine the
maximum thermal efficiency of this steam engine.
Solution
 Steam  1 
1
QC
T
 1 C
Qh
Th
300 K
 0.4 or
500 K
40%
This is the max possible (theoretical efficiency).
68
SPH 302: Thermodynamics
3.6 Internal Combustion Engines
Internal combustion engines are where the release of energy by combustion
takes place within the confines of the engine. Examples of such engines include
cars, trucks etc and they employ a fluid which remains gaseous throughout.
These engines can broadly be divided into two categories
(i)
Spark ignition or compression ignition types with either 2 or 4 stroke
(ii)
Gas turbines
We shall briefly describe the working principle of the spark ignition (petrol)
engines and the compression ignition (diesel) engines.
(a) The Gasoline Engine (Otto Cycle)
In real engines, the cycles proceed rapidly with friction and turbulence with no
quasistatic and the cycles are irreversible. To discuss a real engine cycle, we
replace it by an idealized model (called the Otto cycle) in which the working
substance is assumed to be an ideal gas and all processes are assumed to be
reversible.
Fig 3.8a below shows the idealized Otto cycle representing a gasoline engine
operating in 4 steps (strokes) viz:
Intake Stroke, e  a: The intake valve opens and the mixture of air and gasoline
vapour flows into the cylinder as the piston descends as shown in Fig 3-8a
below. This is a quasistatic isobaric process (dP = 0) where the volume
increases from V1 to V2. with V2 = rV1 where r is the compression ratio.
Compression Stroke, a  b: Intake valve closes and the mixture is compressed,
approximately adiabatically from volume V2 to volume V1 as the
temperature rises from Ta to Tb as shown in Fig 3-8b. This processes can
be represented by the adiabatic expression;
TaV2 1  TbV1 1
…………………….(a)
Ignition and Power Stroke, b  c  d: Mixture is ignited by a spark plug (path b
 c) ejecting heat Qh into the system. This process occurs in a very short
period of time and is not one of the strokes of the cycle. During this time,
the temperature and pressure increases rapidly. However, the volume
remains approximately constant because of the short time interval and as
a result, approximately no work is done on the gas.
The heat input along path b  c therefore occurs at constant volume
(isochoric) with
Qh = CV(Tc – Tb)
…………………….(b)
During the power stroke (path c  d), the mixture expands approximately
adiabatically back to volume V2 pushing on the piston and doing some
69
SPH 302: Thermodynamics
work. In this case, the temperature drops from T c to Td according to the
adiabatic expression;
TdV2 1  TcV2 1
………………………(c)
The work done in this process is the area under the curve cd.
P
P2
P
c
Qh
b
Qh
P1
Adiabatics
c
Adiabatic
W out
W out
b
W in
d
QC
d
P0
QC
e
a
V1
V2
e
V
Fig. 3-8(a). P-V diagram of Otto Cycle
Representing the Gasoline Engine Cycle
(a) Intake
(b) Compression
a
V
rV
V
Fig. 3-9(b). P-V diagram for a Diesel Cycle
(c) Ignition &
Power Stroke
(d) Exhaust
Figure 3-8b: The real cycles of a 4-stroke gasoline (petrol) engine
70
SPH 302: Thermodynamics
Exhaust Stroke d  a e: Exhaust valve opens allowing the gas to cool at nearly
constant volume down to temperature T a as the pressure drops. During
this interval, the volume remains approximately constant (isochoric) as
heat Qc is ejected into the environment where
Qc = CV(Td – Ta)
………………………(d)
Along path a e, the combustion products are pushed out (isobarically)
as the volume decreases from V2 to V1 leaving the cylinder for the next
intake stroke.
The thermal efficiency of this idealized engine is thus given by
  1
QC
C T  Ta 
 1 V d
Qh
CV TC  Tb 
But from equations (a) and (c) we get
Td  Ta V2 1  Tc  Tb V1 1
Or
 V2 
 
 V1 
The ratio

 1

Tc  Tb 
Td  Ta 
V2
is called the compression ratio (rc) so
V1
V 
  1   2 
 V1 
 1
 1
1
rc 1
………………………………………….(3.11)
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SPH 302: Thermodynamics
Take Note
We can also express this efficiency in terms of temperature by noting from
Equations (a) and (b) that.
 V2 
 
 V1 
 1

Ta Td

Tb Tc
Therefore, Equation (3.11) becomes
T
Tb
T
……………..……(3.12)
Tc
  1 a  1 d
NB. During the Otto cycle, the lowest temperature is Ta and the highest is
Tc such that the efficiency of the Carnot engine operating between these
two temperatures will be given by   1 
Ta
, which is greater than the
TC
efficiency of the Otto cycle given by Equation (3.12) as expected..
Equation (3.11) shows that efficiency increases as the compression ratio (rC)
increases. For typical compression ratios of 8 (for most engines), and with  =1.4
(for air), we predict the theoretical efficiency of 1 
1
 56% for an engine
80.4
operating in the idealized Otto cycle.
Although it is important to have a high compression as possible, it is noted that
increasing the compression ratio also increases the temperature during the
adiabatic compression of the air-fuel mixture causing the mixture to explode
spontaneously during compression before ignition. This is pre-ignition or
detonation and causes “knocking”, damaging the engine. Pre-ignition lower the
value of rC to about 7 particularly for unleaded low octane petrol thereby lowering
its efficiency.
Activity 3.5
Explain why lead is usually added to petrol used in car engines
72
SPH 302: Thermodynamics
Take Note
In real engines, the cycles proceeds rapidly with friction, turbulence and no
quasistatic (hence irreversible), loss of heat to cylinder walls and incomplete
combustion of air-fuel mixtures leading to low efficiency values (  20 30%). Incomplete combustion occurs if the mixture has less gasoline with
the resulting products of CO and unburned hydrocarbons (air pollution)
instead of CO2 and H2O.
(b) The Diesel Engine
It is similar to the petrol engine only that there is no fuel in the cylinder at the
beginning of the compression stroke. Instead, air is adiabatically compressed (Fig
3-9) and just before the power stroke, the injectors eject fuel directly into the
cylinder. Due to the high temperature, this fuel ignites spontaneously. No spark
plugs are needed.
Take Note
Since no fuel is compressed, pre-ignition does not occur in diesel engines
resulting in high compression ratios and high efficiency values (r  15 – 20, 
 65 – 70%). Diesel engines are often heavier and harder to start but they
need no carburetor or ignition system. However, the fuel-injection system
requires expensive high-precision machining.
(c) Turbochargers and Intercoolers
The power output of an engine is proportional to the mass of air forced into the
engine’s cylinder. In turbocharged systems, air is compressed (nearly
adiabatically with rise in temperature) before entering the engine. This gives a
greater mass of air per unit volume. In intercoolers. The compressed air is cooled
by passing through an internal cooling system where it loses heat to the
surrounding (at constant pressure), and then compressed further. Due to high T
and P values in these systems, turbochargers and intercoolers are more powerful
than ordinary engines.
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SPH 302: Thermodynamics
Worked Example 3.4
The compression ratio of a diesel engine is 15 to 1 i.e., air is compressed to 1/15
of its initial volume. If the initial pressure and temperatures are 1.0  105 Pa and
27 oC (300K), Determine
(a) The final pressure and temperature after compression (assume ideal gas
conditions,  = 1.4)
(b) The work done by the gas if V1 = 1.0 litres (Cv = 20.8 J mol-1K-1).
Solution
V1
(a) Since
 15
V2
And
V 
P2  P1  1 
 V2 
V 
 T2  T1  1 
 V2 
 1
 613o C
 1
 44.8  10 5 Pa  44 atm
 P1V1 
  0.0405 mol
RT
 1
1
PV  P V   494 J
 W  nCV T2  T1  
 1 1 1 2 2
(b) Using PV = nRT  n  
NB. –ve sign since compression
Take Note: Significance of the Second Law of Thermodynamics
In this lecture, we have seen that the second law is important for the following
reasons
1. The second law gives a preferred direction in which thermodynamic
processes occur and It imposes limitations on the efficiency of processes
that convert heat energy into mechanical energy and vice versa by
defining the thermal efficiency of heat engines. In particular it stipulates
that it is impossible to construct a heat engine with 100% thermal
efficiency.
2. The Second law also establishes Entropy and absolute temperature
as thermodynamic properties and provides us with objective criteria to
determine reversibility
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SPH 302: Thermodynamics
3.7 Summary
So far, we have seen that where as the 1ST law denies the possibility of
creating or destroying energy, the 2ND law limits the availability of energy
and the way in which it can be used and converted i.e., it is the law of nature
describing the direction of natural thermodynamic processes.

A heat engine is a device that takes in energy by heat and, operating in
a cyclic process, expels a fraction of that energy by means of work.

The second law of thermodynamics can be stated in the following two
ways:
 “it is impossible to construct a device which operating in a cycle
produces no other effect than extract heat from a high
temperature reservoir and convert the heat completely into
mechanical work” [The Kelvin-Planck Statement].

“No cyclic process can transfer heat from a cold reservoir to a
hotter region with no input of mechanical work” [The Clausius
Statement].

Carnot’s theorem states that no real heat engine operating (irreversibly)
between temperature TC and Th can be more efficient than a Carnot
engine operating reversibly between the same two temperatures.

The Thermal efficiency of a heat engine operating in the Carnot cycle is
given by
 1

The Coefficient of performance of refrigerator is given by
COP R 

QC
T
1 C
Qh
Th
QC
QC
TC


W Qh  QC Th  TC
The efficiency of internal combustion engine is given by
1
  1   1
rc
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SPH 302: Thermodynamics
Question for Discussion
1.
The lady in the figure below is trying to cool herself by keeping the door of a refrigerator open.
Will she achieve her objective? Explain.
NB. The purpose of a refrigerator is to keep its contents cool. Beyond the attendant increase in
your electricity bill, there is another good reason you should not try to cool the kitchen on a hot
day by leaving the refrigerator door open. What might this reason be?
2.
Does a refrigerator full of food consume more power if the room temperature is 28 oC than if it
is 20 oC. Explain.
3.
Why are heat pumps used for heating houses in mild climates but not in very cold climates?
4.
Hot air rises. Why then does the temperature decrease when one climbs a mountain?
5.
A room can be warmed by opening the door of an oven but it cannot be cooled by opening
the door of a self-contained refrigerator. Discus and explain.
6.
Briefly explain whether it is a violation of the 2ND law of thermodynamics to
(a) Convert mechanical energy completely into heat
(b) Convert heat completely into work
7.
An electric motor has its shaft coupled to that of an electric generator. The motor drives the
generator, and some current from the generator is used to run the motor. The excess current
is used to light a home. What is wrong with this scheme.
8.
Explain the difference in the operation of a gasoline engine and a diesel engine.
9.
Why is the actual efficiency of a gasoline engine less than the theoretical efficiency? Explain.
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SPH 302: Thermodynamics
PROBLEM SET 3
1. A steam engine uses pressurized steam at 470 K, and exhausts it at 373K. What would be the
efficiency of the machine if it were ideal? Explain why the actual efficiency will be lower than
this.
2. A reversible Carnot cycle engine operates between temperatures of 1000K and 250K. If 1.5 kJ
of heat are transferred to the engine at 1000K in one cycle, find
(i)
the efficiency of the engine
(ii) the heat transferred from the engine at 250K
3. A Carnot engine takes 2000 J of heat from a reservoir at 500K, does some work, and discards
some heat to a reservoir at 350K. Determine
(i)
The work done and the heat discarded by the engine
(ii) The thermal efficiency
4. A gasoline engine takes in 16,100 J of heat and delivers 3,700 J of work per cycle. The heat is
obtained by burning gasoline with a heat of combustion of 4.6  104 Jg-1. Determine
(i)
The thermal efficiency
(ii) The amount of heat discarded and the mass of fuel burned in each cycle
(iii) The power of the engine (in HP) if the engine goes through 60.0 cycles per second.
5. A Carnot engine operates between a heat source at a high temperature Th, and a heat sink at
a low temperature Tc. Which will have the greater effect on the efficiency, raising the value of
Th by T, or lowering that of TC by the same value?
6. An inventor claims to have developed an engine which takes in 11  107 J at 400K, rejects 5 
107 J at 200K and delivers 16.67 kW hours of work. Would you advice investing money in this
project? Explain
7. An inventor claims to have developed a heat pump that draws heat from a lake at 3.0oC and
delivers heat at a rate of 20 kW to a building at 35oC, while using only 1.9 kW of electrical
power. How would you judge the claim?.[Resnick]
8. A Carnot engine working as a refrigerator between 260 K and 300 K receives 2100 J of heat
from the reservoir at the lower temperature. Calculate
(i)
The amount of heat ejected to the high temperature reservoir
(ii) The work done in each cycle to operate the refrigerator
(iii) Its efficiency
9. 50 Kg of water at 0 oC has to be frozen into ice in a refrigerator. The room temperature is 20 oC.
What is the minimum work input to the refrigerator to achieve this. [Latent heat of fusion of
water is = 3.33  105 JKg-1].
10. 0.02 moles of an ideal diatomic gas ( = 1.4) undergoes a Carnot cycle as shown in Figure 1
with temperatures of 227 oC and 27 oC. The initial pressure, Pa = 10.0  105 Pa, and during the
isothermal expansion at the high temperature, the volume doubles. Determine
(i) The pressure and volume at each of the points a, b, c and d
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SPH 302: Thermodynamics
(iii)
(iv)
Q, W and dU for each process in the cycle and for the entire cycle
The efficiency of the cycle using the results of part (ii) above and compare it with the
efficiency of the Carnot engine.
P
Q1 Isothermal at TC
Adiabatic
compression
Adiabatic
expansion
Isothermal, TH Q2
Fig 1
V
11.
A refrigerator does 153 J of work to transfer 568 J of heat from its cold compartment.
(a) Calculate the COP of the refrigerator
(b) How much heat is exhausted to the kitchen? [Resnick]
12.
Two moles of a monatomic ideal are taken through the cycle shown in the Figure 2 below.
Calculate.
(i) The heat added to the gas
(ii) The heat leaving the gas
(iii) The net work done by the gas, and
(iv) The efficiency of the cycle. [Ref: Resnick]
P (atm)
10.4
B
Adiabatic
A
1.22
C
9.13 M3
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SPH 302: Thermodynamics
13.
A hypothetical engine, with an ideal gas as the working substance, operates in the cycle
shown in the Fig. 3 below. Show that the efficiency of the engine can be given by
1 1  P P 
3
1

  1  
  1  V1 V3 
Q1
P
1
P1
2
Q2
Adiabatic
P3
3
V1
14.
V3
V
Figure 4 below shows an operation cycle for an idealized diesel engine where fuel is sprayed
into the cylinder at pt B and the combustion occurs in the isobaric process B  C. Show that
the efficiency of the engine can be given by
e  1
1  TD  TA 

 where symbols have their usual meaning
  TC  TB 
P
Q1
B
C
Adiabatics
D
Q2
E
VB
A
VA
Fig. 4. P-V diagram for an ideal diesel engine
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SPH 302: Thermodynamics
15.
The motor in a refrigerator has a power output of 210 W. The freezing compartment is at –
3.0oC and the outside air is at 26oC. Assuming that the efficiency is 85% of the ideal, calculate
the amount of heat that can be extracted from the freezing compartment in 15 min.
16.
In a refrigerator, the cooling chamber is maintained at 290K while the outside temperature is
305K. The motor (located outside) has compression cylinders operating at 320K and the
expansion coils inside the chamber operating at 280K. If the motor operates reversibly
(i) Give a schematic diagram illustrating the sequence of events
(iii) Calculate the COP of the refrigerator
(iv) How much work must be done for each transfer of 5000J of heat from the chamber?
(v) Determine the entropy change inside and outside the chamber for this amount of
refrigeration?
17. The T-S diagram of a reversible engine is shown in Figure 5 below. Calculate its efficiency.
T (oC)
B
400
C
300
A
500
1000
(J/K)
References
4. Finn C. B. J., Thermal Physics, 1986.
5. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and
Statistical Physics., S. Chand & Company Ltd, New Delhi.
6. F.O. Akuffo, A. Brew-Hammond, F. Makau Luti and J.G.M. Massaquoi. ANST,
UNESCO (1997), Nairobi, Kenya.
7. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons,
Inc., NY., Pg. 545- 561.
Memorable quotes
“Wisdom is knowing when you can’t be wise”
–Anonymous.
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SPH 302: Thermodynamics
Lecture
4
ENTROPY AND SECOND LAW OF
THERMODYNAMICS
Outline
Introduction
Objectives
Entropy: One way process
Entropy and 2ND Law of Thermodynamics
4.5 Entropy and Performance of Heat Engines: T - S diagrams
4.6 Third Law of Thermodynamics
4.7 Summary
This professor’s room is untidy and disorderly. We say it has high entropy compared to his
assistant’s room which is orderly representing low entropy.
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SPH 302: Thermodynamics
4.1 Introduction
We saw in lecture 1 that the Zeroth law of thermodynamics leads to the concept
of temperature and in lecture 2 that the first law leads to the concept of internal
energy. In lecture 3 we saw that the second law of thermodynamics determines
the direction in which natural processes occur i.e., it gives a preferred direction to
the “arrow of time”, teaching us that systems naturally evolve with time in one
direction but not in the other. We also saw that the second law in conjunction with
the fundamental thermodynamic relation places limits on a system's ability to do
useful work.
In this lecture, we shall see that the second law establishes still another concept;
Entropy, a quantity in terms of which the second law of thermodynamics is
expressed. Entropy is a thermodynamic property that can be used to determine
the energy not available for useful work in a thermodynamic process, such as in
energy conversion devices – engines and machines. Such devices have a
theoretical maximum efficiency when converting energy to work. During this
processes, entropy (or disorder) accumulates in the system, but has to be
removed by dissipation in the form of waste heat. Entropy is a quantitative
measure of disorder. The importance of Entropy is that it provides us with a
criterion of reversibility and thereby enable us determine the direction of naturally
occurring processes.
We shall study Entropy and also discuss T-S diagrams and calculate efficiency of
Carnot cycle from T-S diagrams. Lastly we shall discuss the Third law of
thermodynamics and explain some of the consequences associated with the third
law.
4.2 Objectives
At the end of this lecture, you should be able to:
1.
2.
3.
4.
Define entropy and state its significance
State the 2ND law in terms of Entropy
State the central Equation of Thermodynamics
Explain T-S diagrams and obtain efficiency of Carnot engine from T-S
diagrams
5. Explain the 3RD Law of thermodynamics and state some of the
consequences of the 3Rd law
4.3 Entropy: One-Way Process
There is a property of things that happen naturally in the world around us that is
strange beyond belief. Yet we are so used to it that we hardly ever think about it.
It is this:
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SPH 302: Thermodynamics
All naturally occurring processes proceed in one direction only: They never,
of their own accord, proceed in the opposite direction.
Consider the following example:
Example 1:
If you drop a stone, it falls to the ground. A stone resting on the ground never,
of its own accord, leaps up into the air.
Example 2:
If you put a drop of ink in a glass of water, the ink molecules eventually spread
out uniformly throughout the volume of water. They never, of their own accord,
regroup into a drop-shaped clump.
Such spontaneous one-way processes are irreversible. You cannot make them go
backward by making any small change in their environment. Essentially, all
naturally occurring processes are irreversible.
Notably, it is not the energy that controls the direction of irreversible processes; it
is a property called Entropy (S) of the system which is a measure of randomness
or disorder of a system.
One feature that distinguishes entropy from other such concepts as energy,
momentum and angular momentum in that it does not obey a conservation law
and there is no principle of conservation of entropy. In fact, since it is not a
conserved quantity, entropy can be created at will. No matter what changes occur
within a closed system, the energy of that system remains constant. Its entropy,
however, always increases for irreversible processes. Thus, entropy has its
central property called the Entropy principle or the principle of increasing
Entropy which is stated as:
“The Entropy of an irreversible processes always increases; It never
decreases”.
There are two equivalent ways to define the change in entropy of a system:
(1) A macroscopic approach, involving heat transfer and the temperature at
which the transfer occurs, and
(2) a microscopic approach, involving counting the ways in which the atoms or
molecules that make up the system can be arranged. We use the first
approach in this lecture.
Take Note
1.
Entropy is a state function just internal energy (U), temperature (T),
volume (V) and Pressure (P).
2. Entropy is an extensive property since it depends on the mass of the
working substance.
3. It is difficult to form a physical concept of entropy since there is
nothing physical to represent it. Moreover, it cannot be felt like
temperature or pressure etc.
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SPH 302: Thermodynamics
4.3.1 Defining Entropy change
The concept of entropy is defined by the second law of thermodynamics, which
states that the entropy of an isolated system always increases or remains
constant. Thus, entropy is a measure of the tendency of a process, such as a
chemical reaction, to be entropically favored, or to proceed in a particular
direction. It determines that thermal energy always flows spontaneously from
regions of higher temperature to regions of lower temperature, in the form of heat
(see Fig. 4-1). These processes reduce the state of order of the initial systems,
and therefore entropy is an expression of disorder or randomness.
Fig. 4-1. Ice melting in a warm room is a common example of
increasing entropy since it represents an increase in the
desegregation of ice molecules. Whereas the entropy of ice
increases, that of the room reduces
For example, when mixing ink with water, the system starts from low entropy state
(each fluid is separate and distinct hence orderly with dS = 0) to a disordered
state of high entropy (when particles are thoroughly mixed). No spontaneous unmixing (net decrease of entropy) is ever observed. Likewise, a perfect crystal, free
of impurities and defects, would have zero entropy at 0K. As the temperature of
the crystal is increased, its atoms absorb energy by thermal motion, some
disordering occurs, and as a consequence, the entropy of the crystal increases to
a value characteristic of that temperature and degree of randomness.
We can now define Entropy using a macroscopic approach by considering heat
transfer in a system. Using he above example, the entropy change (dS) of a
system such as the above crystal at some temperature (T) that has occurred from
the initial state (i) to final state (f), assuming a reversible process, is given
mathematically by
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SPH 302: Thermodynamics
f
dQ
Heat added / Re moved
Change in Entropy (dS ) 
 R
Absolute Temperatur e i T
i.e.
dS  
dQR
T
………………..…….(4.1)
Where QR = the heat absorbed reversibly.
Entropy as measure of disorder
We have seen that entropy determines the direction of flow of thermal energy and
such a direction should always be such as to reduce the state of order of the
initial systems. Therefore entropy is an expression of disorder or randomness.
For example, when heat is added to any system, there is always a change in
volume. A gas system expands when heated and will be in a more disorderly
state since the gas molecules now move in a larger volume and have more
randomness of position. From 1ST law of thermodynamics, we have that
dQ  PdV  nRT
dV
V
We see that the fractional increase in the volume is given by
dV
dQ

V
nRT

dQ
dV
C
 dS
T
V
dV is a measure of the increase in disorder and is  dQ
T
V
or the Entropy (S). We can therefore see that entropy is a qualitative measure of
disorder or randomness of a system since any heat input into a system creates
disorder and increases the entropy.
The fractional increase
Definition of Entropy
We can therefore define Entropy as a measure of randomness or disorder and it
is a thermal property of a system that increases in every irreversible process but
remains constant in a reversible (adiabatic) process (just as temperature remains
constant in an isothermal process).
4.4 Entropy and 2ND Law of Thermodynamics
The principle of increasing Entropy states categorically that “The Entropy of an
irreversible processes always increases; It never decreases”, since in such
processes, there is an increase in disorder. This statement can be written as
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SPH 302: Thermodynamics
dSirrev >0.
On the other hand, for a reversible (adiabatic process, dQ = 0), the Total entropy
change is always zero i.e.,
dSRev = 0.
From these expressions, we can obtain the general statement for the definition of
entropy as:
Irreversible
dS  0
[for any cycle]
Reversible
dQ
T
But since
dS 

TdS  dQ  0
………………………(4.2)
Equation (4.2) gives the direction of transformation from one state to another and
we see that any such direction should be such as to increase the Entropy. This is
the general form of the Second Law of Thermodynamics.
Thus the second law can be stated quantitatively in terms of entropy as follows:
“No natural process is possible in which the total entropy decreases when all
systems taking part in the process are included”. This is equivalent to the “Engine”
and/or the “Refrigerator” statement of the 2 ND law.
The above statement implies that heat will not flow from a colder body to a hotter
body without the application of work (the imposition of order) to the colder body.
Secondly, it is impossible for any device operating on a cycle to produce net work
from a single temperature reservoir without ejecting any heat to the colder
reservoir. As a result, there is no possibility of a perpetual motion system.
Finally, it follows from the second law of thermodynamics that the entropy of a
system that is not isolated may decrease. An air conditioner, for example, may
cool the air in a room, thus reducing the entropy of the air of that system. The
heat expelled from the room (the system), involved in the operation of the air
conditioner, will always make a bigger contribution to the entropy of the
environment than will the decrease of the entropy of the air of that system. Thus,
the total of entropy of the room plus the entropy of the environment increases, in
agreement with the second law of thermodynamics.
Take Note; Significance of Entropy
Since Entropy is a measure of randomness or disorder in a system, it thus
provides us with a criterion of reversibility and thereby enable us determine
the direction of naturally occurring thermodynamic processes.
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SPH 302: Thermodynamics
Take Note; Significance of 2nd Law
We see from Equation (4.2) that the 2 ND law limits the availability of energy
and the way in which it can be used and converted. It is the law of nature
describing the direction of natural thermodynamic processes.
Take Note: Entropy of the universe
Since all natural processes taking place in the universe are irreversible i.e.,
processes involving friction, free expansion and even life itself is irreversible,
this essentially means that the entropy of the universe is increasing i.e.,
dS[Universe] > 0. The universe is therefore running down and we are approaching
the “heat death” of the universe. Is this true?
4.4.1 The central Equation of Thermodynamics.
From the first law of thermodynamics, we have that
dQ = dU + PdV
Combining the 1ST and 2ND law of thermodynamics, we get
TdS  dU + PdV
[For any process]
……………..…(4.3)
Equation (4.3) holds for all processes (both reversible and irreversible) since all
the functions in the equation are state function. This is called the central
Equation of Thermodynamics.
4.4.2 Entropy and degradation of energy
From the Central Equation of thermodynamics (TdS  dU + PdV), we have that,
dW  TdS  dU
This expression implies that the system does less work in an irreversible process
(since TdS > 0). Hence, of the available energy, some must go into increasing the
Entropy (S) and hence the energy is degraded in that it is less useful for doing
work. The energy involved in the entropy change is tied up in the random
arrangement and thermal motion of the atoms, it is often referred to as
unavailable energy. For a reversible process, dSsystem = -dS surrounding i.e., total
entropy change in system + surrounding is zero. More work is done in this case.
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SPH 302: Thermodynamics
Worked Examples
1. One kilogram of ice at 0 oC is melted and converted to water at 0 oC.
Compute the change in Entropy. Take the specific latent heat of fusion of ice
(SLHF) as = 333,624.2 JKg-1.
Solution
Since the temperature remains constant,
S  S 2  S1 
1
Q 333.624.2
dQ  
 1222.1JK 1

T
T
273
2. 1 kg of water at 0oC is heated to 100oC. Compute the change in entropy
Solution
From dQ  mcdT
T2
T2
1
1
T
dS
dT
dS  
  mc
 mc ln 2  1.3110 3 JK 1
T
T1
T T
T

3. Compute the entropy change that occurs when 1kg of water at 100oC is
mixed with 1 kg of water at 0oC.
Solution
We shall assume the entropy of water is zero when it is in the liquid state at
0oC.
 From the last example, the entropy of water at 100 oC is 1310.85
JK-1 and the entropy of 1kg of water at 0oC is zero.

Entropy of the system before mixing = 1310.85JK-1

After the hot and cold water have been mixed, we have 2 Kg of
water at 50oC. The entropy of the system is now:
mc p ln

323
323
 2 x4200 x ln
 1412.72 JK 1
273
273
Total increase in entropy = 1412.72  1310.85  101.87 JK 1
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SPH 302: Thermodynamics
4.5 Entropy and the performance of heat Engines: T-S
Diagrams
We saw in lecture 1 that thermodynamic changes in the state of a substance can
be represented by plotting P-V diagrams. Such changes can also be represented
on Temperature – Entropy or (T – S) diagrams.
The T – S diagram can be obtained from P – V diagram as shown below. For
example, Figure 4.1 shows a Carnot’s reversible cycle abcd on a P-V diagram
while Fig 4.2 shows the same cycles on a T-S diagram whereby the resulting T-S
diagram is a rectangle abcd.
The isothermal process ab in fig 4-1 is represented by a straight line on a T-S
diagram in Fig 4-2. For the adiabatic process (dQ = 0) such that dS = 0. These
processes are represented by isentropic lines (dS = 0) as shown in Figure 4-2
below.
P
V1, P1
a
Isothermal expansion, Th
Qh
b
Adiabatic
compression
V4, P4
V2, P2
Adiabatic
expansion
d
c
QC
Fig. 4-1. The Carnot Cycle on a P-V diagram
V
Isothermal expansion, Th
T
Tc
V3 , P 3
Qh
a
b
Isentropic
dS2 = 0
Isentropic
dS1 = 0
Th
d
QC
S1
c
S2 S
Fig. 4-2. The T-S Diagram of a Carnot cycle
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SPH 302: Thermodynamics
Along the isothermal expansion ab, the gain in entropy of a working
substance is
dS = S 2  S1 
Qh
Th
………………………….(a)
Along bc and along da, there in no change in Entropy, dS = 0
Along cd, loss in the entropy of the working substance is
dS = S 2  S1 
QC
TC
…………………………(b)
Combining Equations (a) and (b) we get
Qh  QC  Th  TC S 2  S1 
Where Qh – QC = external work done in the reversible cycle and
Th  TC S 2  S1 = Area of rectangle on T-S diagram.
Thus, the net heat absorbed is area adcd = external work done in a
reversible Carnot’s cycle.
4.5.1 Efficiency of Carnot’s Engine on T-S diagram
The efficiency of a Carnot engine can be obtained using a T-S diagram as follows
Efficiency,  
External work in each cycle
Heat absorbed from source


Qh  QC Th  TC S 2  S1 

Qh
Th S 2  S1 


Or
 1 C
T  T 
h
C
Th
T
Th
…..………………………………..(4.5)
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SPH 302: Thermodynamics
Worked Examples
1. A fridge working on the reversed Carnot cycle has a power requirement of 5
KW. If the maximum and minimum temperatures in the cycle are 40 oC and 10 oC respectively. Draw a T-S diagram and determine
(i)
The coefficient of performance
(ii)
The rate of heat extracted from the cold space
(iii)
The change in entropy for each of the processes in the cycle, given
a refrigerant mass flow rate of 0.32 Kg/s
Solution
Given W = - 5 Kw, TC = 273.15 + 40 = 313.15 K, TC = 273.15 – 10 =
263.15K
T
Q1
4
3
T2
1
Q2
T1
2
S
Reversed Carnot Cycle
T2
 5.263
T1  T2
Q
(b) From COP 
 Q1  COP WNET   26.315KW
WNET
(a) COP 
(c) From 2ND law, dS 
dQ
T
Along Process 1 2
2
Q
Q
dQ 1 1
dS  
  dQ  1  1  0.3125 KJ Kg  K
T1 1
T1 mT1
1 T
Along Process 2 3 and 41, it is isentropic S.T dS = 0
Along Process 32

dS 
Q2
mT2
dS 
Q2
 0.3125 KJKg 1 K 1
mT2
but Q2  dW  Q1  31.315kW
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SPH 302: Thermodynamics
4.6 Third Law of Thermodynamics
The third law of thermodynamics is concerned with the limiting behaviour of
systems (or Entropy) as temperatures approach absolute zero.
All Statements of the 3RD law of thermodynamics (i.e., Plank’s statement,
Simon’s statement and The Nernst statement) point out that the entropy of a
system tends to zero at absolute zero of temperature.
The Nernst Statement
It states that :
“The Entropy change in a process associated with a change in the external
parameters (e.g., P and T) tends to zero as temperatures approach
absolute zero” i.e.,dS  0 as T  0.
This means that as T 0, a strange behaviour is expected in materials i.e., an
orderly arrangement of molecules should occurs in materials. For example, ice is
more orderly than water.
The reason for this orderly arrangement can be explained from a microscopic
viewpoint (Quantum mechanics) where is assumed that particles of a system can
exist in different quantum states with different discrete energies. However, as T
0, all particles pack in the lowest energy levels (ground state) i.e., assuming they
are bosons. Since the ground state in nondegenerate, g = 1, and the number of
ways the particles can be arranged is 1. Hence, there is complete order and S =
0.
In real cases however, the decrease in S depends infact on the number of states
per unit energy range (density of states) with energy rather than just on the
occupancy of the ground level as our simple theorem suggest. In this case, S may
not fall to zero as T 0 for both fermions and bosons (refer to your statistical
mechanics for details).
4.6.1 Some Consequences of the 3RD Law
As a consequence of dS  0 as T 0, certain measurable parameters vanish at
the absolute zero of temperature. Typical examples are:
(a) The Heat Capacity (CV) vanishes as T 0
Recall the heat capacity at constant volume is given by
 dQ 
 S   S 
CV  
  T   

 dT V
 T V   ln T V
Thus, as T 0, dS  0 but d(lnT)  0 since lnT  . Hence, CV should 
0 as T 0. Similar arguments hold for other heat capacities.
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SPH 302: Thermodynamics
(b) The Thermal expansion Coefficient  vanishes as T 0
The volume cubic expansivity (or isobaric expansivity),  is defined as the
fractional increases in the volume per unit temperature rise i.e.,

1  dV 


V  dT  P
By suing Maxwell’s relation, this becomes
1  S 

V  P T
  

Thus, as T 0, dS  0 and   0.
(c) Slope of phase boundary in 2ND order phase transition vanishes as T 0
Recall: The Slope of phase boundary in 2ND order phase transition is given by:
 P   S 
 

 T   V 
 P 
 or the slope  0. This is observed in
 T 
Thus, as T 0, dS  0 and 
4
He as shown in the figure below.
P
Solid
Helium
Slope = 0
Liquid
Helium II
Liquid
Helium I
2.2K
K
Fig. 4-6. The phase boundary of liquid He as T 0.
Questions
Can you now list some of the consequences of the 3rd law of
thermodynamics
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SPH 302: Thermodynamics
4.6.2 The Unattainability of Absolute Zero
The other statements of 3RD law says that
“It is impossible to reach absolute zero even on using a finite number of
processes”.
Although much progress has been made in recent years to reach extremely very
low temperatures e.g., the super fluid of 3He, the attainability of absolute zero is
still elusive
4.7 Summary
In this lecture, we have learned that:
1. Entropy is a measure of randomness or disorder and is a thermal property of
a system that increases in every irreversible process
2. The change in Entropy (dS) of a system is given by dS 
dQR
dT
3. The Principle of Increasing Entropy states that:- “The entropy of a thermally
isolated system increases in every irreversible process and is unaltered in a
reversible process” i.e., dS  0
4. The significance of Entropy is that it provides us with criteria for reversibility
and enables us determine the direction of naturally occurring processes
5. The 2nd law describes the direction of natural thermodynamic processes and
states that “No natural process is possible in which the total entropy of the
system decreases”
6. The central Equation of Thermodynamics is given by
TdS = dU + PdV
7. The Third Law of thermodynamics says that “The Entropy change of a system
tends to zero as temperatures approach absolute zero” i.e.,dS  0 as T 
0.
8. Some of the consequences of the 3RD law are
(a). The Heat Capacity (CV) vanishes as T 0
(b) The Thermal expansion Coefficient  vanishes as T 0
(c) Slope of phase boundary in 2ND order phase transition vanishes as T 0
9. Attainability of absolute zero is still elusive
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SPH 302: Thermodynamics
Questions for Discussion
1.
Discuss the change in entropy of a gas that expands
(a) at constant temperature and
(b) adiabatically
2.
Discuss the entropy change that occurs when you
(a) bake a loaf of bread and
(b) consume the bread.
3.
Discuss three common examples of natural processes that involve an increase in entropy. Be
sure to account for all parts of each system under consideration.
4.
If a supersaturated sugar solution is allowed to evaporate slowly, sugar crystals form in the
container. Hence, sugar molecules go from a disordered form (in solution) to a highly ordered
crystalline form. Does this process violet the second law of thermodynamics?. Explain.
5.
Suppose your roommate is “Mr. Clean” and tidies up your messy room after a big party.
Because your roommate is creating more order, does this represent a violation of the second
law of thermodynamics?
6.
“Energy is the mistress of the universe and Entropy is her shadow”. Writing for an audience of
general readers, argue for this statement with examples. Alternatively, argue for the view that
“Entropy is like a decisive hands-on executive instantly determining what will happen, while
energy is like a wretched back-office bookkeeper telling us how little we can afford”.
7.
When we put cards together in a deck or put bricks together to build a house, for example, we
increase the order in the physical world. Does this violet the second law of thermodynamics?
Explain.
8.
“The first law of thermodynamics says that you can’t really win, and the second law says you
can’t even break even.” Explain how this statement applies to a particular device or process;
alternatively, argue against the statement.
Memorable Quotes
ABILITY is what we are capable of doing, MOTIVATION determines what we do, and
ATTITUDE determines how well we do it.
-Lou Holtz
95
SPH 302: Thermodynamics
PROBLEM SET 4
1.
2.0 Kg liquid water at 363 K is mixed adiabatically and at constant pressure with 3.0 Kg liquid
water at 283 K. Determine the total entropy change resulting from this process.
2.
1 mole of an ideal gas for which CV = 25.12 and CP = 33.44 J mol-1 K, expands adiabatically
from an initial state at 340 K and 500 KPa to a final state where its volume has doubled. Find
the final temperature of the gas, the work done and the entropy change of the gas for
(ii)
a reversible expansion
(iii)
a free expansion of the gas into an evacuated space (Joule's expansion).
1.
The specific heat capacity of a solid at low temperature is given by
C  aT  bT 2
where a and b are constants
Determine the Entropy of the solid as a function of temperature if the entropy is zero at T =
0oC.
2.
10 Kg of water at 20 oC is converted to superheated steam at 250 oC at constant atmospheric
pressure. Determine the Entropy change of water. [C P (liquid) = 4180 J/Kg oC; CP (vapour) =
1670 + 0.4904 T + 1.86  106 T-2; Latent heat of vapourization (100oC) = 22.6  105 J/Kg].
3.
One gram of water when converted to steam at atmospheric pressure occupies a volume of
1671 cm3. The latent heat of vapourization at this temperature is 539 cal/gm.
(i)
Compare the volume of steam with the volume that would be occupied at this
temperature and pressure if the water vapour were an ideal gas
(ii)
Compute the increase in the internal energy U, and the Entropy S when one gram of
water is evaporated at this temperature and pressure.
4.
In a refrigerator, the cooling chamber is maintained at 290 K while the outside temperature is
305 K. The motor (located outside) has compression cylinders operating at 320 K and the
expansion coils inside the chamber operating at 280 K. If the motor operates reversibly
(i)
Give a schematic diagram illustrating the sequence of events
(ii)
Calculate the efficiency of the refrigerator
(iii)
How much work must be done for each transfer of 500 J of heat from the chamber?
(iv)
What entropy change occurs inside and outside the chamber for this amount of
refrigeration?
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SPH 302: Thermodynamics
4.8 References
1. Finn C. B. J., Thermal Physics, 1986.
2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and
Statistical Physics., S. Chand & Company Ltd, New Delhi.
3. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons,
Inc., NY., Pg. 517-561.
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SPH 302: Thermodynamics
Lecture
5
THERMODYNAMIC POTENTIALS AND
MAXWELL’S RELATIONS
Outline
5.1 Introduction
5.2 Objectives
5.3 Maxwell’s Relations
5.4 Maxwell’s Relations using Cyclic rule
5.5 Applications of Maxwell’s Relations
5.6 Summary
5.1 Introduction
A mechanical system is said to be in a stale equilibrium when the potential
energy of the system is minimum. This means that the system must proceed in
such a direction so as to acquire minimum potential energy. This is what we
observe in nature, viz. water flows from a higher to lower level, electric current
flows from higher to lower potential, heat flows from higher to lower temperature,
a body falls from higher to lower potential due to gravitational field and so on.
Likewise, in thermodynamics, the behaviour of the state functions namely; internal
energy (U), Helmholtz Free Energy (F), Enthalpy (H), and Gibbs Free Energy (G)
is similar to potential energy in mechanics. To achieve thermodynamic
equilibrium, the direction of thermodynamical processes must be in such a
direction as to minimize the respective thermodynamical function (potential).
Since the four functions U, F, H and G play in thermodynamics the same role as
played by potential energy in mechanics, they are called thermodynamic
potentials and all have the dimensions of energy.
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SPH 302: Thermodynamics
The four thermodynamic potentials are all functions of their natural thermal state
variables (temperature T or entropy S ) and their mechanical state variable
(pressure P or volume V ). For example, the internal energy (U) is given by
U
= TS
Thermal
Variable
–
PV
Mechanical
Variable
Therefore, the thermodynamic variables associated with a state function can be
derived from the thermodynamic functions by their differentiations with respect to
the independent variables associated with them. The resulting set of equations
are called Maxwell’s Equations
In this lecture, we shall derive some (four) useful general thermodynamic relations
between the four thermodynamic variables viz: P, V, T, and S referred to as the
four Maxwell’s equations and discuss the significance of each of the
thermodynamic potentials.
5.2 Objectives
At the end of this lecture, you should be able to:
1. Define the four thermodynamic potential (U, F, H and G) and state their
significance
2. Derive Maxwell’s equations from the thermodynamic potentials and also
by using the cyclic rule
3. Apply Maxwell’s equations to solve problems related the mechanical
behaviour of materials
5.3 Maxwell’s Relations
In lecture 2, we introduced the thermodynamic potential (U) i.e., internal energy of
a system and in lecture 4, we introduced the concept of Entropy (S). By
combining the first and the second laws of thermodynamics, we came up with the
central equation of thermodynamics given by
TdS  dU  PdV
Though this equation is quite powerful, it is not well suited in analysis of some
certain thermodynamic processes. It is therefore convenient to introduce three
additional thermodynamic potentials (also called state functions), all closely
related to (U) and all having dimensions of energy. These are, the Enthalpy (H),
the Helmholtz Free Energy (F), and the Gibb’s function (G). There is also a fifth
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SPH 302: Thermodynamics
function, the chemical potential (  ), which is useful in discussing the
thermodynamics of open systems where the mass of the system is not constant.
The four thermodynamic potentials are defined in terms of their natural thermal
state variables (T and S) and mechanical state variables (P and T). (for a system
with TWO degrees of freedom) as follows:
Internal Energy;
U = TS – PV
Enthalpy;
H = U + PV
Helmholtz Free Energy;
F = U – TS
Gibbs Free Energy;
G = H – TS = U – TS + PV
We see that the four potentials U(S,V), F(T,V) H(S,P) and G(P,T) are functions of
thermodynamic variables S, T, P and V. Thus, these thermodynamic variables
can be derived from the thermodynamical functions by their differentiations with
respect to the independent variables associated with them. The resulting is the
equalities of each of the four thermodynamic potentials w.r.t their natural variables
(T,S,P and V) and is called Maxwell’s Equations.
Let us now derive the Maxwell’s equation from the thermodynamical potentials:
Take Note
Since the state of a system can be specified by any pair of the four
quantities, viz, pressure (P), volume (V), temperature (T), and entropy (S).
Taking two of the four state variables at a time, we see that there are six
possible pairs, i.e., (P,V), (P,T), (P,S), (V,T), (V,S) and (T,S).
However, the pair (P,V) is connected with composite and is exact differential
quantity dW as dW = PdV and the pair (T,S) with dQ as dQ = TdS. Hence
these two pairs can be eliminated. Thus, we are left with four pairs of the
thermodynamical variables (P,T), (P,S), (V,T), and (V,S) and corresponding
to each pair, we can write a thermodynamical relations known as Maxwell’s
Thermodynamic relations
Questions
1. What is a thermodynamic potential
2. Write down the four Thermodynamical potentials in terms of their
thermodynamic variables
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SPH 302: Thermodynamics
5.3.1 Internal Energy and The First Maxwell’s Equation
Consider the central Equation of Thermodynamics given by
dU  TdS  PdV
(5.1)
This suggests that U is a function of Entropy (S) and volume (V) and we write U in
terms of its independent pair of variables (its natural variables) S and V i.e.,
U  U ( S ,V )
Thus, the change in U is thus given by
 U 
 U 
dU  
 dS  
 dV
 S V
 V  S
(5.2)
By comparing the two expressions (5.1) and (5.2) we see that;
 U 
T 

 S V
and
 U 
P  

 V  S
(5.3)
Thus, if U is known in terms of its natural function S and V, then T and P can be
found from the natural functions.
Now, since U is a state function, dU is an exact differential. By using the condition
that the differential to be exact (see condition below) we get
 T 
 P 

   
 V  S
 S V
(5.4)
This is the First Maxwell’s Equation.
Notice that the natural variables of U, are S and V which are the qualities
appearing outside the partial differentials.
Activity 5.1
Beginning from the central equation of thermodynamics, derive the first
Maxwell’s relation
101
SPH 302: Thermodynamics
Take Note: Condition for a differential to be Exact.
Consider a mathematical function G x, y  of x and y that takes unique
values for each pair of x and y. when x and y change by dx and dy, the
infinitesimal change in G is
 G 
 G 
 dy
dG  
 dx  

x

y

y

x
 Xdx  Ydy
where X and Y are the general functions of both x and y
The condition for  to be exact is that the differential of the coefficient of dx
with respect to y while holding x constant should be equal to the differential
of the coefficient of dy when differentiated with respect to x while holding y
constant i.e.,
 X   Y    2G 

     


y

x

x

y


y

x


We see that this two partial differentials are equal, the order of the
differential being in material. Thus the necessary condition for dG to be
exact differential is that
 X 
 Y 

  

 y  x  x  y
This condition will be used in derivation of Maxwell’s Relations
eeeeeeeeeeeeeeeeeeee
Activity 5.2
Consider a differential given by dG  2 xy dx  4 x y dy . Show
that dG is an exact differential
4
Solution
X  2 xy 4
and
3
y  4x 2 y 3
 X 
 Y 
  8 xy 3   
 x  y
 y  x
 

2
Hence dG is an exact differential
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SPH 302: Thermodynamics
Questions
Check whether the following differential is exact?
dF  xy 4 dx  4 x 2 y 3 dy
Heat capacity at constant volume, CV
Let’s now write down an expression for CV in terms of S and T. Now from the
central equation of Thermodynamics, we have that for an isochoric reversible
process (dV =O)
TdS  dU  CV dT
(5.5)
Where Cv is the molar heat capacity
It follows that
 dU 
 dS 
CV  
  T

 dT V
 dT V
(5.6)
5.3.2 Enthalpy and The Second Maxwell’s Equation
We have defined enthalpy (H) as;
  U  PV
(5.7)
Differentiating we obtain;
dH  dU  PdV  VdP
Or
dH  TdS  VdP
(5.8)
This suggests that H is a function of Entropy (S) and pressure (P) and we write H
in terms of its independent pair of variables (its natural variables) S and P i.e.,
H  H ( S , P)
Thus, the change in H is thus given by
  
  
d  
 dS  
 dP
 S  P
 P  S
(5.9)
By comparing the two expressions (5.8) and (5.9) we see that;
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SPH 302: Thermodynamics
  
T 

 S  P
and
 H 
V 

 P  S
(5.10)
Which means that if we know H in terms of its natural variables, S and P we can
find both the temperature T and the volume V. Using the condition that dH is an
exact differential we have?
 T   V 

  


p

 S  S  P
(5.11)
This is The Second Maxwell’s Equation.
Take Note
Significance of Enthalpy
From dH = TdS + VdP
We see that for a reversible isobaric process (dP = 0), then

dH = dQP
The Enthalpy change is the heat evolved in a reversible isobaric process
such as a chemical reaction.
If dH is +ve, heat is absorbed during the reaction and the reaction is called
endothermic. If dH is –VE, heat is evolved during the reaction and the
reaction is called exothermic.
Heat capacity at constant pressure, Cp
Likewise, we can also write the expression for CP in terms of S and T. Now from
the central equation of Thermodynamics, we have that for an isochoric reversible
process (dP =O)
TdS  dH  CP dT
(5.12)
Where CP is the molar heat capacity
Thus, if the temperature dependency of the heat capacity is known, the change in
heat content which occurs during the heating of a given material can be
calculated from the following expression
T2
dH   C P dT
T1
It follows that
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SPH 302: Thermodynamics
 dH 
 dS 
CP  
  T

 dT  P
 dT  P
(5.13)
5.3.3 Helmholtz Free Energy and The Third Maxwell’s Equation
For processes where temperature and volume are the important variables, we
introduce the state function: The Helmholtz Free Energy (F) defined by
F  U  TS
(5.14)
For an infinitesimal change (differentiating), we have;
dF  dU  TdS  SdT
But
dU  TdS  PdV

dF  PdV  SdT
(5.15)
From which we see that the natural variables of F are V and T. Which suggests
that?
F = F(V,T)
(5.16)
Thus, the change in F is thus given by
 F 
 F 
dF  
 dV  
 dT
 V T
 T V
(5.17)
Comparing the coefficients of the two expressions for dF we see that;
 F 
P  

 v T
and
 F 
S   
 T V
(5.18)
Hence if we know F in terms of V and T we can calculate P and S.
As F is a function of state, dF is an exact differential and using the conditions for
an exact differential, we obtain;
 P   S 
  

 T V  V T
(5.19)
This is the Third Maxwell’s Equation, with the natural variables V and T
appearing outside the differentials.
Take Note
Significance of Helmholtz Free Energy
From dF = -PdV - SdT
We see that for a reversible isothermal process (dT = 0), then
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SPH 302: Thermodynamics
Activity 5.3
The Helmholtz Free Energy (F) defined by . F  U  TS . Derive the
related Maxwell’s Equation and state the significance of Helmholtz Free
Energy (H)
5.3.4 Gibbs function (G) and The Fourth Maxwell’s Equation
This state function is designed for use in problems where pressure and
temperature are the important variables. It finds enormous application in the study
of equilibrium conditions in systems that are mixtures of two phases such as the
ice and water mixture.
The Gibbs function is defined as;
G  H  TS
(5.20)
For an infinitesimal change;
dG  dH  TdS  SdT
But
dH  dU  PdV  VdP

dG  dU  PdV  VdP  TdS  SdT
But also

TdS  dU  PdV
dG  VdP  SdT
(5.21)
From which we see that the natural variables of G are thus P and T.
Thus, we can write G in terms of its independent pair of variables (its natural
variables) P and T as;
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SPH 302: Thermodynamics
G = G(P,T)
(5.22)
From which we obtain the differential;
 G 
 G 
dG  
 dP  
 dT

P

T

T

P
(5.23)
Comparing the two equations (5.21 and 5.23) for dG it shows that
 G 
V 

 P  T
and
 G 
S  

 T  P
(5.24)
Hence if we know G as a function of P and T we can find the volume and the
entropy. Since G is a state function, dG is an exact differential, thus;
 V 
 S 

   
 T  P
 P T
(5.25)
This is the Fourth Maxwell’s Equation with the natural variables P and T
appearing outside the partial differentials.
Take Note: Significance of Gibbs Function (G)
From dG  dH  TdS  SdT
We see that in a chemical reaction, P and T remain constant such that
dG  dH –TdS
or dG = dH - dQ
Now, a chemical reaction requires heat (dQ) in order to take place. Hence:
If dQ > dH, then dG < 0 and the reaction is spontaneous.
If dQ = dH, then dG = 0, and the reaction is at equilibrium
If dQ < dH, then dG > 0, and the is no reaction
Thus, the Gibbs free energy, G is a measure of the available energy. It
represents the driving force for a reaction. For a reaction to occur
spontaneously, the free energy change must be negative (the free energy of
the system decreases). At equilibrium, the Gibbs function is at its minimum
(dG = 0).
“The condition for thermodynamic equilibrium in a system in thermal and
mechanical contact is that the Gibbs function is a minimum”.
Example 5.1
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SPH 302: Thermodynamics
Consider the reaction: ice  water . For ice to melt, dS has to increase (dS >0) as the system
moves from order to disorder. Also, dH is +ve since ice requires latent heat in order to melt. dH
goes to increase the potential energy of the molecules as they break free from their bound positions
in solid. Now

When heat is supplied to ice at very low temperatures ( < 0 oC), then, dH > TdS, hence dG >
0 and ice does not melt.

At high temperatures, ( > 0 oC), TdS > dH and dG < 0. Hence, ice melts into water.

At ( = 0 oC), dH =TdS, and dG = 0. At this temperature, water and ice coexist.
Activity 5.4
1. Show that the enthalpy change in an isobaric chemical reaction is equal
to the heat of reaction.
2. State the significance of G
The above four are by no means the only Maxwell relationships. When other work
terms involving other natural variables besides the volume work are considered or
when the number of particles is included as a natural variable, other Maxwell
relations become apparent.
Take Note: Significance of Thermodynamic Potentials
As mentioned earlier, a mechanical system is said to be in state of
equilibrium when the potential energy of the system is minimum. This means
that for thermodynamic equilibrium, the behaviour of the internal energy (U),
Helmholtz free energy (F), enthalpy (H), and Gibbs free energy (G) is similar
should proceed in a such a direction such as to minimize these potentials.
As we have seen, the direction of isothermal-isochoric process is to make
Helmholtz free energy (F) minimum. In isothermal-isobaric process, Gibbs
free energy (G) tends to be minimum while in an isobaric-adiabatic process,
the Enthalpy (H) tends to be minimum.
Questions
State the significance of thermodynamic potential. Why are they important in
thermodynamics
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SPH 302: Thermodynamics
5.4 Derivation of Maxwell’s Relationships using cyclic
rule
Recall the four thermodynamic potentials as functions of their natural thermal and
mechanical variables (for a system with TWO degrees of freedom) are given as
follows:
Internal Energy;
U = TS – PV
Enthalpy;
H = U + PV
Helmholtz Free Energy;
F = U – TS
Gibbs Free Energy;
G = H – TS = U – TS + PV
Their differential forms are
dU = TdS – PdV
dH = TdS + VdP
dF = -SdT – PdV
dG = -SdT + VdP
where the two terms on the right hand side correspond to the 2 degrees of
freedom
The corresponding Maxwell’s equations are
From the internal energy, U;
 2U
 T 
 P 

    

V

S
 S V VS
From enthalpy, H;
 T   V 
  

 P  S  S  P
From the Helmholtz function, F;
 P   S 
  

 T V  V T
From the Gibb’s function, G;
 V 
 S 

   
 T  P
 P  T
These equations relating the partial differentials of the thermodynamic variables
P, V, T and S can be easily derived using the Mnemonic as follows
STEP 1
Write P, V, T and S as a mnemonic as follows
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SPH 302: Thermodynamics
S
P
V
T
The phrase for remembering the sequence is “Let’s join the Society for the
Prevention of Teaching of Vectors”.
STEP 2
To derive the Maxwell’s equation, all one has to do is to go round this array
cyclically in opposite directions. For example the First Maxwell’s equation would
follow from starting at T as;
S
P
S
V
P
T
V
T

 T 
 P 

   
 V  S
 S V
STEP 3
The sign is positive if T appears with P in the differential e.g.,
S
P
S
V
P
T
V
T

 T   V 
  

 P  S  S  P
Activity 5.5
Using the cyclic rule, derive the four Maxwell’s equations
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SPH 302: Thermodynamics
5.5 Application of Maxwell’s Relationships to simple
thermodynamic systems
The state of a system can be specified by any pair of quantities, viz, pressure (P),
volume (V), temperature (T), and entropy (S). In solving any thermodynamical
problem, the most suitable pair of variables as given by Maxwell’s relations is
chosen.
Worked Example 5.2
The central equation of thermodynamics is given by TdS  dU  PdV . If the Entropy S is a
function of T and V i.e., S = S(T,V), using Maxwell’s relationships, show that the above equation
can be written as
 dP 
TdS  CV dT  T 
dV
 dT 
where symbols have their usual meanings.
Solution
If S  S T ,V 
 S 
 S 
dS  
 dT  
 dV

T

V

V

T

By using the mnemonic
S
S
P
P
V
T
V
 S   P 
  
 V T  T V

T
 S 
 P 
 dT  T   dV
 T V
 T V
 TdS  T 
 S 
  CV
 T V
But T 

 dP 
TdS  CV dT  T 
dV
 dT 
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SPH 302: Thermodynamics
5.6 Summary

The four thermodynamic potential are
Internal Energy;
U = TS – PV
Enthalpy;
H = U + PV
Helmholtz Free Energy; F = U – TS
Gibbs Free Energy;

G = H – TS = U – TS + PV
The respective Maxwell’s equations are
 2U
 T 
 P 




 
 V  S
 S V VS
From U; 
 T   V 
 

 P  S  S  P
From H; 
From F;
 P   S 
  

 T V  V T
 V 
 S 
   
 T  P
 P  T
From G; 




The Enthalpy change is the heat evolved in a reversible isobaric process
such as a chemical reaction.
The dF represents the maximum amount of work done by the system in an
isothermal process and is equal to the decrease in the Helmholtz function
The condition for thermodynamic equilibrium in a system in thermal and
mechanical contact with a heat and pressure reservoir, is that the Gibbs
function is a minimum.
Maxwell’s equations can be applied to solve problems in any mechanical
system such as magnetic systems, elastic rods etc.
5.7 References
8. Finn C. B. J., Thermal Physics, 1986.
9. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and
Statistical Physics., S. Chand & Company Ltd, New Delhi.
10. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons,
Inc., NY., Pg. 545-561.
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SPH 302: Thermodynamics
PROBLEM SET 5
1.
The central equation of thermodynamics is given by TdS  dU  PdV . If the Entropy S
is a function of T and V i.e., S = S(T,V), using Maxwell’s relationships, show that the above
equation can be written as
 dV 
TdS  C P dT  T 
 dP
 dT  P
where symbols have their usual meanings.
2.
One gram of water when converted to steam at atmospheric pressure occupies a volume of
1671 cm3. The latent heat of vapourization at this temperature is 539 cal/gm.
(iii) Compare the volume of steam with the volume that would be occupied at this
temperature and pressure if the water vapour were an ideal gas
(iv) Compute the increase in the internal energy U, the Entropy S, the Enthalpy H, and
Gibbs function G when one gram of water is evaporated at this temperature and
pressure.
(v) Using your answer for G above, what can you say about the nature of this process
3.
The Gibbs function of 1 mole of a certain gas is given by
G  RT ln P  A  BP  C
P2
P3
D .
2
3
where A, B, C and D are constants. Find the equation of state of the gas.
4.
Show that the difference between the isothermal and the adiabatic compressibility can be
given by
 V 2 

K T  K S  T 
 CP 
5.
A wire undergoes an infinitesimal change from one equilibrium state to another. Obtain
expressions for the infinitesimal changes in length and in tension in terms of the Young’s
modulus and the thermal expansivity.
Memorable Quotes
“Ignorant people make life so difficult for some of us”
-Albert Einstein
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SPH 302: Thermodynamics
Lecture
6
PHASE CHANGES AND PHASE
EQUILIBRIA
Outline
6.1 Introduction
6.2 Objectives
6.3 Systems, phases and components
6.4 Phase transitions
6.5 First order phase changes
6.6 Second order phase changes
6.7 Summary
(a) a glass of water showing the three
phases of water: ice, water and vapour
(b) ice, water and steam coexisting naturally
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SPH 302: Thermodynamics
6.1 Introduction
The chemical and physical properties of most materials are related to the
number, composition, and distribution of the phases present. Temperature,
pressure, and composition are the principle variables which determine the kinds
and amounts of the phases present under equilibrium conditions. For example,
water changes to vapour when heated to a temperature of 100 oC at a pressure of
760 mmHg. On the contrary, at low temperatures of 0 oC, water turns to ice. Ice,
water and steam are the three phase of a single component called water.
An understanding of phase equilibrium in material systems is central to the
utilization and development of materials particularly in refractories, glass and
other high-temperature technologies. Phase equilibrium address significant
questions related to the flexibility and constraints dictated by forces of nature, on
the evolution of phase assemblages in materials. Phase boundaries also assist in
the evaluation of the service stability of materials specially ceramics, both in the
long and short time frames. Thus, knowledge of the stability of a material (e.g.,
ceramic or glass) component in high-temperature or high-pressure environments
can often be obtained from an appropriate stable or metastable phase diagram.
The phase rule developed by J. Willard Gibbs was derived from the first and
second laws of thermodynamics and phase relationships can be presented in
temperature-composition-pressure diagrams. In a system with more than one
phase, each phase may be considered as a separate system within the whole.
Thermodynamics parameters of the whole system may then be applied to the
component phases to determine the equilibrium conditions for the two phases.
In this lecture, we shall study the change of phase that occur in a substance as a
result of change in temperature and pressure as well as the effect of Pressure on
the melting and boiling point of a substance. We shall study the equilibrium
conditions for existence of two phases and the characteristics of first order and
second order phase changes respectively. We shall restrict ourselves to a single
component system such as water since 2 or 3 component systems are more
complex to analyze their phase changes.
6.2 Objectives
At the end of this lecture, you should be able to
1. Explain what is a phase and what causes phase changes
2.
3.
Explain the importance of phases changes in thermodynamic systems
State the equilibrium conditions for existence of two phases
4.
Explain the effect of pressure on the melting and boiling point of a
substance
5.
State the characteristics of first order and second order phase transitions.
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SPH 302: Thermodynamics
6.3 Systems, Phases, and components
Before proceeding with the development of the concept of equilibrium between
phases and the derivation of the phase rule, it will be useful to define certain
terms commonly used in a treatment of phase equilibrium.
6.3.1 Definitions
Phase:- Any portion including the whole of a system which is physically
homogeneous within itself and bounded by a surface so that it is mechanically
separable from any other potions. Example, Ice, water and steam are the
three phases of water.
Components:- Are the smallest number of independently variable chemical
constituents necessary and sufficient to express the composition of each
phase present in any state of equilibrium. For example, in the water system,
H2O is the component of the system while in the alumina-silica system, Al2O3
and SiO2 are the components of the system since all phases and reactions
can be described by using only these two materials.
Equilibrium:- In lecturer 1, we defined thermodynamic equilibrium as a state in
which a system experiences thermal, mechanical and chemical equilibrium. A
more comprehensive thermodynamic definition of equilibrium states that a
system at equilibrium has a minimum free energy as defined by the Gibbs free
energy (G).
6.3.2 Phase diagrams of One-Component System
Since the change in phase of a substance is dependent mainly on P and T,
phase diagrams are best illustrated on P-T diagrams. For pressure and
temperature changes, typical P- T projections are as shown in Fig. 6-1.
P (Pa)
B
C
Melting
freezing
SOLID
LIQUID
Vapourization
condensation
Sublimation
TP
VAPOUR
deposition
TTP
TC T (K)
Fig. 6-1. Hypothetical P-T projections of one-component system
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SPH 302: Thermodynamics



TP is the triple point i.e., the temperature (TTP) at which all the three phases (liquid, solid
and vapour) coexist.
Point C is the critical point i.e., where liquid and vapour are indistinguishable.
Below pt C, the system exists in two phases (liquid and vapour), while beyond C, (or
temperature TC), the system is gaseous.
At the triple point (TP), the solid and liquid are in equilibrium. If the volume were
held constant as heat is added to the system, the solid would melt. The increased
volume of the melt increases the pressure and temperature and as such, the
equilibrium between the solid and liquid shifts to a position of higher pressure and
temperature i.e., towards pt B on curve IB (Fig. 6-1). Similar projections can be
worked out for solid-vapour and for liquid-vapour systems at equilibrium.
6.4 Phase Transitions
6.4.1 Equilibrium Conditions for Two Phases
In the example above, we saw that a single component such as water can exist in
3 phases (solid, liquid and vapour) depending on the conditions of Pressure and
temperature. We need to determine the conditions under which any 2 phases e.g.
solid and liquid can exist in equilibrium.
The conditions for thermodynamic equilibrium of any two phases of a system, is
that the Gibbs function for both phases (phases 1 and phase 2) should be
minimum. Thus by Applying Gibbs potential (G), it is possible to investigate the
equilibrium between a liquid and its vapor or between any two phases of a
substance.
For example, consider a closed system containing a liquid in equilibrium with its
saturated vapour. For equilibrium conditions, the temperature and pressure are
equal in both the phases and must remain constant throughout the phases. The
thermodynamic coordinate V, S, U and G will be equal to the product of the
specific value and the mass of the substance in that phases.
Suppose the phases have masses M1 and M2, where M1 + M2 = M (total
mass) and Gl and G2 are Gibbs functions for the two phases respectively. Then,
for the whole system,
G  m1G1  m2G2
…………….…….(6.1)
(NB: G is extensive variable i.e., dependent on system size. If a small quantity of
the liquid changes into vapour, then the change in G is obtained by differentiating
equation (6.1) giving
dG  m1G1  m2G2
………….……….(6.2)
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SPH 302: Thermodynamics
But for equilibrium condition, dG = 0.

m1G1  m2G2  0
But since the system is closed, the total mass is constant i.e., m1 +
m2 = M (total mass) or m1  m2

G1  G2
…………………..……..(6.3)
This shows that the equilibrium condition for existence of 2 phases is that their
specific Gibbs functions (G) must be equal. Equation (6.3) is applicable to the
processes of evaporation, fusion and sublimation. This equality of (G) leads to
Clausius-Clapeyron Heat Equation which characterizes phase transition
between two phases of matter.
Activity 6.1
1. Explain what is a phase
1. Show that the condition for thermodynamic equilibrium of 2 phases in a
system in thermal and mechanical contact with a heat and pressure
reservoir, is that their Gibb’s function must be equal.
6.4.2 The Clausius-Clapeyron Equation
The Clausius–Clapeyron relation is a way of characterizing a discontinuous phase
transition between two phases of matter. On a pressure-temperature (P–T)
diagram, the line separating the two phases is known as the coexistence curve.
The Clausius–Clapeyron relation gives the slope of this curve and as such, this
relation can be used to (numerically) find the relationships between pressure and
temperature for the phase change boundaries. Entropy and volume changes (due
to phase change) are orthogonal to the plane of the P-T diagram.
To derive the Clausius–Clapeyron relation, let us consider a system in equilibrium
such as a liquid and its saturated vapour, such that G1  G2 . If p and T are
changed (by dT and dP respectively) as shown by a portion of a phase boundary
in Fig. 6-2 below, but in such a way that the system is still kept in equilibrium, then
dG1  dG2
as the equilibrium condition
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SPH 302: Thermodynamics
P
B
PHASE 1
(T+dT, P+dP)
PHASE 2
A
(T,P)
T (K)
Fig. 6.2. Portion of a phase boundary between two phases
Remembering that (i.e., from Gibbs Relation)
dG  d  VdP  SdT
Then 
But
V1dP  S1dT  V2 dP  S 2 dT

V  V dP  S  S dT

dP S

dT V
S  
dQ Q L
 
T
T T
1
2
2
1
………………………..………(a)
……..…………..…….(b)
where L = the latent heat of vapourization
Substituting (b) into (a) we get
dP
L

dT T (V2  V1 )
where
………………………….………(6.4)
dP
is the slope of the coexistence curve, and ΔV is the volume change
dT
of the phase transition.
Equation (6.4) is the Clausius-Clapeyron equation for the slope of the phase
boundary and is used to characterize first order phase transitions.
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SPH 302: Thermodynamics
Activity 6.2
Can you now deduce and state the first latent heat equation,
dP
L

.
dT T (V2  V1 )
6.4.3 Applications of Clausius-Clapeyron Equation
(a) Effect of Pressure on the melting point of a substance
When a solid is converted into a liquid, there is a change in volume. Usually, the
substance may expand on melting in which case the volume of the liquid (V L =
V2) > the volume of the solid (VS = V1) or it may contract on melting such that VL
< VS. We wish to look at the effect of pressure on these two cases.
(i)
if VL > VS i.e., for substances that expands on melting
Then,
dP
> 0 i.e., the slope is +ve  the melting point of the substance
dT
will increase with increase in pressure and vice versa as shown in Fig 6-3
below by the graph IB.
P (Pa)
D
Under
Normal P
B
dP
0
dT
dP
0
dT
CP
LIQUID
SOLID
I
VAPOUR
TP
TTP
TC T (K)
Fig. 6-3. Substances that expand and contract on melting
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SPH 302: Thermodynamics
(ii)
if VL < VS i.e., for substances that contract on melting such as water and
antimony, the volume of water formed is less than the volume of ice
taken:
In this case,
dP
< 0 or the slope is –ve  the melting point of the
dT
substance will decrease with increase in pressure and vice versa as
shown by graph ID.
Thus, at high pressure, ice melts at a temperature lower than 0oC. NB.
Ice melts at 0 oC only at a normal pressure of 76 cm of Hg. Hence,
increase in pressure lowers the melting point of ice.
(b) Effect of Pressure on the boiling point
Let us now look at the effect of pressure on the boiling point. When a liquid is
converted into a gaseous state, the volume V g (Vg = V2) of the gas is always
greater than the corresponding volume VL (VL = V1) of the liquid i.e., Vg > VL.
Therefore,
dP
is a +ve quantity.
dT
Thus, increase in pressure increases the boiling point of a substance and vice
versa. For example, water boils at 100 oC only at 76 cm of Hg pressure. In the
labs, while preparing steam, the boiling point is less than 100 oC because the
atmospheric pressure is less than 76 cm of Hg. In pressure cookers, the liquid
boils at a higher temperature because the pressure inside is more than the
atmospheric pressure and this makes food cook faster.
Questions
Can you now explain the following
1. Will water on top of mount Kenya boil at a lower or higher temperature
compared to the same water at the coast
2. Meat cocks faster in a pressure cooker than in a sufuria
Worked Example 6.1
Calculate the change in the boiling point of water when the pressure is increased by 1 atmosphere.
Boiling point of water is 373 K. Specific volume of steam = 1.671 m3 kg-1 and latent heat of steam is
2.268  106 J Kg-1.
Solution
Given V2 = 1.671 m3Kg-1, V1 = 110-3m3 Kg-1, T = 373K, dP = 1atm = 105 Nm-2.

V2 – V1 = 1.670 m3Kg-1, Steam = 2.268 J/Kg.
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SPH 302: Thermodynamics
From

dP
L

dT T (V2  V1 )
dT 
T (V2  V1 )dP
 27.47 K
L
Worked Example 6.2
Find the increase in the boiling point of water at 100 oC when the pressure is increased by one
atmosphere. Latent heat of vaporization of steam is 540 cal/gram and 1 gram of steam occupies a
volume of 1677 cm3.
Solution
Given V1 = 1.00 cm3, V2 = 1,677 cm3, T = 373K, dP = 7613.6980 dynes cm-2.
From
dP
L

dT T (V2  V1 )
T (V2  V1 )dP
 27.92 K
L

dT 

Increase in boiling point = 27.92K or 27.97 oC
6.5 First Order Phase Changes
In the modern classification scheme, phase transitions are divided into two broad
categories, namely First-order and Second order phase transitions.
First-order phase transitions are those that take place at constant temperature
and pressure and involve a latent heat. During such a transition, a system either
absorbs or releases a fixed (and typically large) amount of energy as the
temperature remains constant. The system is in a "mixed-phase regime" in which
some parts of the system have completed the transition and others have not.
Familiar examples are the melting of ice or the boiling of water (the water does
not instantly turn into vapour, but forms a turbulent mixture of water and vapor
bubbles).
Definition
Apart from the transference of heat, a more concise classification of phases
transitions is based on Ehrenfest classification. According to Ehrenfest
classification, phase transitions are classified based on the behavior of the
thermodynamic free energy (G) as a function of other thermodynamic variables.
Under this scheme, A first order phase transition is defined as one in which the
Gibbs function (G) remains the same in both the phases at equilibrium. However,
the first order derivative of the Gibbs function with respect to pressure and
temperature change discontinuously at the transition point.
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SPH 302: Thermodynamics
The various solid/liquid/gas transitions are classified as first-order transitions
because they involve a discontinuous change in density, which is the first
derivative of the free energy (G) with respect to chemical potential.
6.5.1 Characteristics of 1St order phase transitions
First order phase transition is characterized by
(i)
The Gibbs function remains the same in both the phases at equilibrium i.e.,
dG = 0 (since GS = GL).
(ii)
A transference of heat and a subsequent change in specific volume (V) and
entropy (S) leading to discontinuity in the first order derivative of the Gibbs
function with respect to pressure and temperature respectively at the
 G 
 G 
 and V  
 are discontinuous are

T

P

T

P
transition point i.e., S  
shown in the G-T and G-P curves of Figs 6-5b and 6-7b respectively.
Activity 6.3
1. List the characteristics of a First order phase change
1. Give an example of a First order phase change
6.5.2 Variation of G, S and V in First Order Phase Transitions
Since a first order phase is characterized by a transfer of latent heat, we thus
expect a change in volume (V) and entropy (S). These changes are characterized
by the Clausius-Clapeyron heat equation.
(a) Change in S
The entropy S can be obtained from the expression dG = VdP – SdT where we
see that
 G 
S  
 = slope of G vs. T curve
 T  P
The change in entropy (S) can be determined by considering the variation of the
Gibb’s function (G), along a constant pressure section (XY) on the P-T diagram
shown in Figure 6-4a below
From the graph, the corresponding G-T plots for both the liquid and solid phases
i.e., GL and GS are shown in Fig. 6-4b. Note that the G vs. T plots has a –ve slope
as S is always +ve. At the phase boundary, (at temperature To), the Gibbs
functions for the solid and liquid phases are equal (i.e., Gs = GL). For T < To, the
solid phase is more stable, and the Gs curve is much lower in this section in order
to minimize G. For T > To, the liquid phase is more stable, and the GL curve is
much lower in this section compared to the Gs curve.
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SPH 302: Thermodynamics
G
P
X
Y
L
S
TP
GL
GS
CP
V
To
S
L
T (K)
To
Fig 6-4 (a) P-T curve of water
T (K)
(b) G-T curve along X-Y in Fig (a)
The G-T curves of Figure 6-4(b) can be combined to give a general variation of G
(= Gs + GL) shown in Fig. 6-5a. Note the discontinuity in the slope of this graph at
To. The corresponding variation in Entropy (S) vs. T curve is shown in Fig 6-5b.
Higher temperature phase has a higher entropy (i.e., SL> Ss). Note the
discontinuity in S at To.
S
Discontinuity
in Slope
FIRST
ORDER
G

S
G
T
S
L
L
To
T (K)
6-5(a) General G-T curve
To
T (K)
(b) S-T curve from Fig (a)
(b) Change in V
The volume of the phases (V) can be obtained from the expression dG = VdP –
SdT
 G 
  0 = slope of G vs. P curve
 P T
 V 
Thus, the change in volume (V) can be determined from the variation of the
Gibb’s function (G), along a constant temperature section (X’Y’) on the P-T
diagram shown in Figure 6-6a below
124
SPH 302: Thermodynamics
Figure 6.5(b) shows the G-T plots for both the liquid and solid phases i.e., GL and
GS as function of P as obtained from Fig 6(a). Note that the G vs. P plots have a
+ve slope since the volume of a substance should increase upon decrease in
pressure.
For P < Po, the liquid phase is more stable, and the GL curve is much lower in this
section in order to minimize G. For P > Po, the solid phase is more stable, and
hence GL curve is much lower in this section as compared to the G s curve.
X’
P
G
Po
L
S
CP
TP
GS
V
GL
Y’
S
L
T (K)
Po
Fig 6-6 (a) P-T curve of water
P(Mpa)
(b) G-T curve along X’-Y’ in Fig (a)
The general variation of the above G-T curves (G = Gs + GL) is shown in Fig. 67a. Note the discontinuity in the slope of this graph at Po. The slope of this graph
gives the variation in specific volume (V) which is plotted as a function of P in Fig
6-7b. Note that higher pressure phase (solid phase) has a smaller specific volume
(i.e., VL> Vs) since substances suffer a decreases in volume upon increase in
pressure (obvious). The discontinuity in V at Po is a result of the discontinuity in
Fig 6-7a.
V
L
Discontinuity
in V at Po
G
dP
L
S
Po
VL > V S
V
Discontinuity
in Slope
FIRST
ORDER
G
P(Mpa)
Fig. 6-7(a) General G-P curve along X’Y’
S
Po
P(Mpa)
(b) V-P curve
125
SPH 302: Thermodynamics
St
Example of 1 order phase transitions
When water is heated at 100 oC and 1 atmosphere, it changes from liquid to
vapour. The density of water is 1000 Kg cm -3, while that of vapour is 0.6 kg m-3
representing an increase in volume and consequently an increase in the Entropy
(S). Therefore, the transformation of water into vapour at constant temperature
and pressure is a first order phase transition. Similarly, the transformation of ice
into water at 0 oC and at 1 atmospheric pressure is a first order phase transition
since the volume of water is less than the volume of ice due to increase in
density.
6.6 Second Order Phase Changes
Recent investigations have revealed that during second order phase transitions,
there is no transfer of heat and hence, there is no change in volume. Second
order phase transitions are defined as the phenomena that take place with no
change in volume and entropy at constant pressure and temperature respectively.
As such, second-order phase transitions are also called continuous phase
transitions
6.6.1 Characteristics of 2ND order phase transitions
Second order phase transition are characterized by
(i)
No transfer of heat and hence no change in specific volume (V) and entropy
(S)
(ii)
No discontinuity in first order derivatives of the Gibbs function with respect to
 G 
 = 0
 T  P
pressure and temperature at the transition point i.e., S  
 G 
 = 0  S and V are continuous at the
 P T
and V  
phase boundary as shown in Fig 6-8(b) and 6-9(b) respectively.
(iii)
  2G 
C
 S 
    P 0
However, second order derivatives of G i.e.,  
2 
 T  P  T  P T
  2G   V 
 
and 

2 

P

T  P T
change discontinuously at the phase boundary as shown in Fig 6-8(c).
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SPH 302: Thermodynamics
S 
 S 
CP   
 T  P
Discontinuity in slope
No Discontinuity
in Slope
SECOND
ORDER
G
G
T
S
S
L
To
T (K)
Superfluid
Liquid 4He II
Normal
Liquid 4He I
L
To
T (K)
Fig 6-8(a) General G-T curve
in 2ND order phase transition
 G 
V 

 P T
No Discontinuity
in Slope
L
L
K
(c) The  anomaly in CP of 4He.
(b) S-T curve from Fig (a)
showing continuity in S
SECOND
ORDER
G
2.2K
S
S
Po
P(Mpa)
Fig 6-9(a) General G-P curve
Po
P(Mpa)
 G 
 vs. P curve showing
 P T
(b) 
continuity in slope
6.6.2 Implications of second order phase changes
The discontinuity in the second order derivatives of G implies that certain material
properties such as CP, β and K that are derivatives of Entropy (S) and Volume (V)
are discontinuous. For example, the heat capacity is defined by
 S 
CP  T  
 T  P
is discontinuous for liquid helium at 2.2K
Likewise The volume cubic expansivity () and the bulk modulus defined by
127
SPH 302: Thermodynamics

1  dV 
1  S 

   
V  dT  P
V  P T
K 
1  S 
 
V  P  T
is discontinuous and
is discontinuous
This classification of phase changes (Ehrenfest Classification) can also be
extended to higher order phase changes.
Examples of 2ND Order Phase Changes
1. Transition in helium from normal liquid helium I to superfluid (liquid helium II)
at very low temperature (λ = 2.2K). In a superfluid, there is complete
absence of internal friction (viscosity) and a rotating fluid would go on
rotating forever. 4He has two phases and its heat capacity has a discontinuity
at 2.2K (see Fig. 6-10)
2.
Transition of ferromagnetic material to a paramagnetic material at the curie
point
3.
Transition of superconducting metal into an ordinary conductor in the
absence of a magnetic field
 S 
CP   
 T  P
Superfluid
Liquid Helium II
Normal Liquid
Helium I
2.2K
K
Fig. 6-10. The  anomaly in CP of 4He.
Activity 6.4
1. List the characteristics of a second order phase change
2. Give examples of second order phase transitions
128
SPH 302: Thermodynamics
Take Note
The superfluidity of 4He can be explained with aid of Quantum mechanics.
4He are bosons with zero integral spins (2P with opposing spins  and 
and 2N with opposing spins  and ). At low temperatures, all the 4He
atoms pack in lowest same quantum states and their effect on each other is
minimal-hence the superfluidity.
On the other hand 3He are fermions (i.e., with ½ integral spins:  for
protons and  for neutrons). As such, they do not pack in the same quantum
state.
1.7 Summary
In this lecture, we have learned that
1. A Phase is any portion including the whole of a system which is
physically homogeneous within itself and bounded by a surface so that it
is mechanically separable from any other potions.
2. The equilibrium condition for the existence of two phases is the
minimum in the Gibbs function G.
3. The slope of the phase boundary of a substance is defined by the
dP
L
Clausius-Clapeyron equation given by
.

dT T (V2  V1 )
4. Increase in pressure increases the melting point and the boiling point of
a substance and visa versa.
5. First order phase transition is characterized by
(i) the Gibbs function remains the same in both the phases at
equilibrium i.e., dG = 0 (since GS = GL).
(ii) A transference of heat and hence a change in specific volume (V)
and entropy (S) leading to discontinuity in the first order derivative of
the Gibbs at the transition point.
6. Second order phase transition is characterized by
(i) No transference of heat and hence no change in volume (V) and
entropy (S). Therefore no discontinuity in the first order derivatives
of the Gibbs function at the transition point.
 S   S 
 V 
(ii) Second order derivatives of G i.e., 
 ,   and 
 ,
 T  P  P  T
 T  P
 V 
and  
 change discontinuously at the transition point.
 P  T
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SPH 302: Thermodynamics
PROBLEM SET 6
1.
Calculate under what pressure ice freezes at 272K if the change in specific volume when I kg
of water freezes is 91  10-6 m3. Given latent heat of ice = 3.36  105 JKg-1.
2.
Calculate the depression of melting point of ice produced by one atmosphere increase of
pressure. Given that latent heat of ice = 80 cal/gm and specific volume of ice and water at
0oC are 1.091 cm3 and 1.0 cm3 respectively.
3.
Calculate the change in the melting point o ice when it is subjected to a pressure of 100
atmospheres.
4.
Calculate the change in temperature of boiling water when the pressure is increased by 27.12
mm Hg. The normal boiling point of water at atmospheric pressure is 100 oC. Explain why the
boiling temperature of a liquid increase with pressure?
5.
Discuss the effect of increase of pressure on ice and wax under isothermal conditions.
6.
7.
Show that for a single component system, the necessary conditions for any of its two phases
to be at thermodynamic equilibrium is the equivalence of their specific Gibbs functions.
Hence derive the Clausius – Clapeylon equation for a First – Order phase change.
dP
L
Establish and state the Clausius-Clapeyron’s equation,
and explain the

dT T (V2  V1 )
effect of pressure on
(a) boiling point of a liquid, and (b) melting point of a solid.
8.
The Figure below shows a P-T diagram of a substance which expands on melting.
P (Pa)
C.P
Liquid
Solid
T.P
Vapour
T(K)
(a)
(b)
(c)
9.
Make a clearly labeled P-T diagram of a real substance which contracts on melting.
Briefly explain the effect of increasing pressure on the melting point of the substance
whose diagram you have.
Suggest an example of such a substance.
The phase diagram for water is given in the figure below where TP is the triple point.
P
Po
X
Y
S
L
CP
130
SPH 302: Thermodynamics
(a)
(b)
10.
Discuss with the aid of G–T diagrams the variation of Gibbs function along the section
XY for both FIRST ORDER and for SECOND ORDER phase changes.
Sketch the corresponding Entropy-Temperature behaviour for both FIRST ORDER and
SECOND ORDER phase changes and compare the entropy of the solid phase with the
entropy of the liquid phase in both the two cases
The phase diagram for water is given in the figure below where TP is the triple point.
X’
P
Po
LIQUID
CP
SOLID
TP
Y’
To
(a)
(b)
VAPOUR
T (K)
Discuss with the aid of G–T diagrams the variation of Gibbs function along the section
X’Y’ for both FIRST ORDER and for SECOND ORDER phase changes.
Sketch the corresponding Volume-Pressure behaviour for both FIRST ORDER and
SECOND ORDER phase changes and compare the volume of the solid phase with the
volume of the liquid phase in both the two cases.
References
1. Finn C. B. J., Thermal Physics, 1986.
2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and
Statistical Physics., S. Chand & Company Ltd, New Delhi.
3. Bergeron C. G., et al, (1984), Introduction t Phase Equilibrium in Ceramics,
American ceramic society, Inc, Columbus, Ohio, pp 1- 19.
131
SPH 302: Thermodynamics
APPENDIX A
SOLUTIONS TO PROBLEM SETS - EVEN QUESTIONS
PROBLEM SET 2
2.
Given n = 2, T = 300K, dT = 0 and V2 = 2V1
 V2 
 = 3.46 103 J
V
 1
(i) dW  nRT ln 
(ii) From 1ST law, dQ = dU + dW. But dU = 0J for a reversible process
 dQ = dW = 3.46 103 J.
(iii) dU = 0J
4.
P
W = 30 J
c
b
dQ= 80J
10 J
a
dQ = ?
d
From 1ST law: dQ = dU + dW
Along a c b, dQ = 80J and dW = 30J
 dU = (80 – 30)J = 50J.
V
NOTE: dU depends only on temperature or the final states and not the path taken. Thus
dUacb = dUadb.
(i)
Along a d b, dW = 10J and dU = Ub – Ua = 50J
 dQ = dU + dW = (50 + 10) J
 dQ = 60 J absorbed
(ii)
For the curved path b  a, dW = -20J and dU = Ua – Ub = -50J
 dQ = dU + dW = (-50 - 20) J
 dQ = -70 liberated
(iii)
dUad = 40 J.
 dQ = dU + dW = (40 + 10) J
 dQ = 50J
For d b, dUdb = (50 – 40) J = 10 J and dW = 0J
 dQ = dU = 10 J.
132
SPH 302: Thermodynamics
10.
18.
Since pressure remains constant, then
(i)
For isothermal process, W  PV2  V1 
(vi)
  V  23 
3
For adiabatic process, W  P1V1 1   1  
2
  V2  


(iii)
For PV = const,
W
Given m = 16g, Mr =32  n 
1
PV  P V 
 1 1 1 2 2
m 16

 0.5
M r 32
(a)
For isovolumetric cooling, dV = 0
1. dQ = dU = nCvdT = 0.5 (21)(200K) = 2100 J
(b)
For process 2, dP = 0
 dU = dQ – nRdT = nCpdT – nRdT
But Cp = R + Cv = 29.314 J
1.
dU = 2100 J
(c)
dW = PdV = nRdT = 831.45 J
(d)
dQ = dW + dU = nCpdT = 2931 J
(e)
W = PdV = 831 J
PROBLEM SET 3
2. Given Th = 1,000K, Tc = 250K, Qh = 1.5 KJ
(i)
T
Th
 1 c 1
(ii) From
250
 0.75   =0.75 or 75%
1000
T 
Qh Th
 250 
 Qc  Qh  c   1,500

 J  375 J
Qc Tc
 1000 
 Th 
8. Given Qh = ?, Qc = 2100J, Th = 300K, Tc = 260K
(i) From
Qh Th

Qc Tc
T 
 Tc 
 Qh  Qc  h   2,100
 
300 
 J  2,432 J
 260 
(v) W = Qh – Qc = 323 J
133
SPH 302: Thermodynamics
PROBLEM SET 4
2.
Given dQ =0 with n = 1, Cv = 25.12, Cp = 33.44, T1= 340K, P1 = 500Kpa, V2 = 2 V1
 1
(i)
(ii)
C 
V 
From the adiabatic expression; T2  T1  1  with    P   1.33
C 
 V2 
 V
0.33
 T2  340  0.5  270.3K

By 1ST law, W = -dU = -CPdT = 1,351 J

dS = 0 since process is reversible
In a free, dW = 0, dQ = 0 (adiabatic)
 dU = 0 = CvdT
But Cv ≠ 0, therefore dT = 0  T2 = T1

Entropy must change if process is irreversible. Hence from central equation of
thermodynamics

TdS – PdV = dU = CvdT


3.
CV dT PdV R

 dV for n = 1
T
T
V
V
dS  R ln 2  5.76 JK 1
V1
dS 
Given m = 10 Kg, T1 = 20 oC = 293K, T2 = 100 oC = 373K, T3 = 250 oC = 523K, dP = 0, Cp
(Liquid) = 4180 J/Kg oC, Cp (vapour) = 1670 + 0.4904T + 1.86  106 T-2, L (100 oC) = 22.6 
105 J/Kg.
Now, for heating water to 373 K, we have
dQ
, but dQ  C P dT
T
373
T
dT
 C P ln 2  10,090 J
 dS1  C P 
T1
293 T
dS1 
When water is converted to steam, Q = mL where L = latent heat of vapourization
 dS 2 
mLCP (vapour)
 60,590 J
T
When water is converted to superheated steam,
523
dT
 1670  0.49T  1.86  10 6 T 2 
m 
 dS 3  mC P 
dt
T

T T
273
T3
2
= 1.25  1012 J
Thus dSTOTAL  dS1  dS 2  dS 3  7.7110 J
4
134
SPH 302: Thermodynamics
135
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