CHAPTER TWELVE
THERMODYNAMICS
12.1 INTRODUCTION
In previous chapter we have studied thermal properties of
matter. In this chapter we shall study laws that govern
thermal energy. We shall study the processes where work is
converted into heat and vice versa. In winter, when we rub
our palms together, we feel warmer; here work done in rubbing
produces the ‘heat’. Conversely, in a steam engine, the ‘heat’
of the steam is used to do useful work in moving the pistons,
which in turn rotate the wheels of the train.
In physics, we need to define the notions of heat,
temperature, work, etc. more carefully. Historically, it took a
long time to arrive at the proper concept of ‘heat’. Before the
modern picture, heat was regarded as a fine invisible fluid
filling in the pores of a substance. On contact between a hot
body and a cold body, the fluid (called caloric) flowed from
the colder to the hotter body! This is similar to what happens
when a horizontal pipe connects two tanks containing water
up to different heights. The flow continues until the levels of
water in the two tanks are the same. Likewise, in the ‘caloric’
picture of heat, heat flows until the ‘caloric levels’ (i.e., the
temperatures) equalise.
In time, the picture of heat as a fluid was discarded in
favour of the modern concept of heat as a form of energy. An
important experiment in this connection was due to Benjamin
Thomson (also known as Count Rumford) in 1798. He
observed that boring of a brass cannon generated a lot of
heat, indeed enough to boil water. More significantly, the
amount of heat produced depended on the work done (by the
horses employed for turning the drill) but not on the
sharpness of the drill. In the caloric picture, a sharper drill
would scoop out more heat fluid from the pores; but this
was not observed. A most natural explanation of the
observations was that heat was a form of energy and the
experiment demonstrated conversion of energy from one form
to another–from work to heat.
12.1 Introduction
12.2 Thermal equilibrium
12.3 Zeroth law of
Thermodynamics
12.4 Heat, internal energy and
work
12.5 First law of
thermodynamics
12.6 Specific heat capacity
12.7 Thermodynamic state
variables and equation of
state
12.8 Thermodynamic processes
12.9 Heat engines
12.10 Refrigerators and heat
pumps
12.11 Second law of
thermodynamics
12.12 Reversible and irreversible
processes
12.13 Carnot engine
Summary
Points to ponder
Exercises
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Thermodynamics is the branch of physics that
deals with the concepts of heat and temperature
and the inter-conversion of heat and other forms
of energy. Thermodynamics is a macroscopic
science. It deals with bulk systems and does not
go into the molecular constitution of matter. In
fact, its concepts and laws were formulated in the
nineteenth century before the molecular picture
of matter was firmly established. Thermodynamic
description involves relatively few macroscopic
variables of the system, which are suggested by
common sense and can be usually measured
directly. A microscopic description of a gas, for
example, would involve specifying the co-ordinates
and velocities of the huge number of molecules
constituting the gas. The description in kinetic
theory of gases is not so detailed but it does involve
molecular distribution of velocities.
Thermodynamic description of a gas, on the other
hand, avoids the molecular description altogether.
Instead, the state of a gas in thermodynamics is
specified by macroscopic variables such as
pressure, volume, temperature, mass and
composition that are felt by our sense perceptions
and are measurable*.
The distinction between mechanics and
thermodynamics is worth bearing in mind. In
mechanics, our interest is in the motion of particles
or bodies under the action of forces and torques.
Thermodynamics is not concerned with the
motion of the system as a whole. It is concerned
with the internal macroscopic state of the body.
When a bullet is fired from a gun, what changes
is the mechanical state of the bullet (its kinetic
energy, in particular), not its temperature. When
the bullet pierces a wood and stops, the kinetic
energy of the bullet gets converted into heat,
changing the temperature of the bullet and the
surrounding layers of wood. Temperature is
related to the energy of the internal (disordered)
motion of the bullet, not to the motion of the bullet
as a whole.
12.2 THERMAL EQUILIBRIUM
Equilibrium in mechanics means that the net
external force and torque on a system are zero.
The term ‘equilibrium’ in thermodynamics appears
in a different context : we say the state of a system
is an equilibrium state if the macroscopic
variables that characterise the system do not
change in time. For example, a gas inside a closed
rigid container, completely insulated from its
surroundings, with fixed values of pressure,
volume, temperature, mass and composition that
do not change with time, is in a state of
thermodynamic equilibrium.
Fig. 12.1 (a) Systems A and B (two gases) separated
by an adiabatic wall – an insulating wall
that does not allow flow of heat. (b) The
same systems A and B separated by a
diathermic wall – a conducting wall that
allows heat to flow from one to another. In
this case, thermal equilibrium is attained
in due course.
In general, whether or not a system is in a state
of equilibrium depends on the surroundings and
the nature of the wall that separates the system
from the surroundings. Consider two gases A and
B occupying two different containers. We know
experimentally that pressure and volume of a
given mass of gas can be chosen to be its two
independent variables. Let the pressure and
volume of the gases be (P
A
, V
A
) and (P
B
, V
B
)
respectively. Suppose first that the two systems
are put in proximity but are separated by an
(a)
(b)
* Thermodynamics may also involve other variables that are not so obvious to our senses e.g. entropy, enthalpy,
etc., and they are all macroscopic variables. However, a thermodynamic state is specified by five state
variables viz., pressure, volume, temperature, internal energy and entropy. Entropy is a measure of disorderness
in the system. Enthalpy is a measure of total heat content of the system.
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adiabatic wall – an insulating wall (can be
movable) that does not allow flow of energy (heat)
from one to another. The systems are insulated
from the rest of the surroundings also by similar
adiabatic walls. The situation is shown
schematically in Fig. 12.1 (a). In this case, it is
found that any possible pair of values (P
A
, V
A
) will
be in equilibrium with any possible pair of values
(P
B
, V
B
). Next, suppose that the adiabatic wall is
replaced by a diathermic wall – a conducting wall
that allows energy flow (heat) from one to another.
It is then found that the macroscopic variables of
the systems A and B change spontaneously until
both the systems attain equilibrium states. After
that there is no change in their states. The
situation is shown in Fig. 12.1(b). The pressure
and volume variables of the two gases change to
(P
B
′, V
B
′) and (P
A
′, V
A
′) such that the new states
of A and B are in equilibrium with each other*.
There is no more energy flow from one to another.
We then say that the system A is in thermal
equilibrium with the system B.
What characterises the situation of thermal
equilibrium between two systems ? You can guess
the answer from your experience. In thermal
equilibrium, the temperatures of the two systems
are equal. We shall see how does one arrive at the
concept of temperature in thermodynamics? The
Zeroth law of thermodynamics provides the clue.
12.3 ZEROTH LAW OF THERMODYNAMICS
Imagine two systems A and B, separated by an
adiabatic wall, while each is in contact with a third
system C, via a conducting wall [Fig. 12.2(a)]. The
states of the systems (i.e., their macroscopic
variables) will change until both A and B come to
thermal equilibrium with C. After this is achieved,
suppose that the adiabatic wall between A and B
is replaced by a conducting wall and C is insulated
from A and B by an adiabatic wall [Fig.12.2(b)]. It
is found that the states of A and B change no
further i.e. they are found to be in thermal
equilibrium with each other. This observation
forms the basis of the Zeroth Law of
Thermodynamics, which states that ‘two
systems in thermal equilibrium with a third
system separately are in thermal equilibrium
with each other’. R.H. Fowler formulated this
law in 1931 long after the first and second Laws
of thermodynamics were stated and so numbered.
The Zeroth Law clearly suggests that when two
systems A and B, are in thermal equilibrium,
there must be a physical quantity that has the
same value for both. This thermodynamic
variable whose value is equal for two systems in
thermal equilibrium is called temperature (T ).
Thus, if A and B are separately in equilibrium
with C, T
A
= T
C
and T
B
= T
C
. This implies that
T
A
= T
B
i.e. the systems A and B are also in
thermal equilibrium.
We have arrived at the concept of temperature
formally via the Zeroth Law. The next question
is :how to assign numerical values to
temperatures of different bodies ? In other words,
how do we construct a scale of temperature ?
Thermometry deals with this basic question to
which we turn in the next section.
Fig. 12.2 (a) Systems A and B are separated by an
adiabatic wall, while each is in contact
with a third system C via a conducting
wall. (b) The adiabatic wall between A
and B is replaced by a conducting wall,
while C is insulated from A and B by an
adiabatic wall.
* Both the variables need not change. It depends on the constraints. For instance, if the gases are in containers
of fixed volume, only the pressures of the gases would change to achieve thermal equilibrium.
(a)
(b)
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12.4 HEAT, INTERNAL ENERGY AND WORK
The Zeroth Law of Thermodynamics led us to
the concept of temperature that agrees with our
commonsense notion. Temperature is a marker
of the ‘hotness’ of a body. It determines the
direction of flow of heat when two bodies are
placed in thermal contact. Heat flows from the
body at a higher temperature to the one at lower
temperature. The flow stops when the
temperatures equalise; the two bodies are then
in thermal equilibrium. We saw in some detail
how to construct temperature scales to assign
temperatures to different bodies. We now
describe the concepts of heat and other relevant
quantities like internal energy and work.
The concept of internal energy of a system is
not difficult to understand. We know that every
bulk system consists of a large number of
molecules. Internal energy is simply the sum of
the kinetic energies and potential energies of
these molecules. We remarked earlier that in
thermodynamics, the kinetic energy of the
system, as a whole, is not relevant. Internal
energy is thus, the sum of molecular kinetic and
potential energies in the frame of reference
relative to which the centre of mass of the system
is at rest. Thus, it includes only the (disordered)
energy associated with the random motion of
molecules of the system. We denote the internal
energy of a system by U.
Though we have invoked the molecular
picture to understand the meaning of internal
energy, as far as thermodynamics is concerned,
U is simply a macroscopic variable of the system.
The important thing about internal energy is
that it depends only on the state of the system,
not on how that state was achieved. Internal
energy U of a system is an example of a
thermodynamic ‘state variable’ – its value
depends only on the given state of the system,
not on history i.e. not on the ‘path’ taken to arrive
at that state. Thus, the internal energy of a given
mass of gas depends on its state described by
specific values of pressure, volume and
temperature. It does not depend on how this
state of the gas came about. Pressure, volume,
temperature, and internal energy are
thermodynamic state variables of the system
(gas) (see section 12.7). If we neglect the small
intermolecular forces in a gas, the internal
energy of a gas is just the sum of kinetic energies
associated with various random motions of its
molecules. We will see in the next chapter that
in a gas this motion is not only translational
(i.e. motion from one point to another in the
volume of the container); it also includes
rotational and vibrational motion of the
molecules (Fig. 12.3).
Fig. 12.3 (a) Internal energy U of a gas is the sum
of the kinetic and potential energies of its
molecules when the box is at rest. Kinetic
energy due to various types of motion
(translational, rotational, vibrational) is to
be included in U. (b) If the same box is
moving as a whole with some velocity,
the kinetic energy of the box is not to be
included in U.
Fig. 12.4 Heat and work are two distinct modes of
energy transfer to a system that results in
change in its internal energy. (a) Heat is
energy transfer due to temperature
difference between the system and the
surroundings. (b) Work is energy transfer
brought about by means (e.g. moving the
piston by raising or lowering some weight
connected to it) that do not involve such a
temperature difference.
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What are the ways of changing internal
energy of a system ? Consider again, for
simplicity, the system to be a certain mass of
gas contained in a cylinder with a movable
piston as shown in Fig. 12.4. Experience shows
there are two ways of changing the state of the
gas (and hence its internal energy). One way is
to put the cylinder in contact with a body at a
higher temperature than that of the gas. The
temperature difference will cause a flow of
energy (heat) from the hotter body to the gas,
thus increasing the internal energy of the gas.
The other way is to push the piston down i.e. to
do work on the system, which again results in
increasing the internal energy of the gas. Of
course, both these things could happen in the
reverse direction. With surroundings at a lower
temperature, heat would flow from the gas to
the surroundings. Likewise, the gas could push
the piston up and do work on the surroundings.
In short, heat and work are two different modes
of altering the state of a thermodynamic system
and changing its internal energy.
The notion of heat should be carefully
distinguished from the notion of internal energy.
Heat is certainly energy, but it is the energy in
transit. This is not just a play of words. The
distinction is of basic significance. The state of
a thermodynamic system is characterised by its
internal energy, not heat. A statement like ‘a
gas in a given state has a certain amount of
heat’ is as meaningless as the statement that
‘a gas in a given state has a certain amount
of work’. In contrast, ‘a gas in a given state
has a certain amount of internal energy’ is a
perfectly meaningful statement. Similarly, the
statements ‘a certain amount of heat is
supplied to the system’ or ‘a certain amount
of work was done by the system’ are perfectly
meaningful.
To summarise, heat and work in
thermodynamics are not state variables. They
are modes of energy transfer to a system
resulting in change in its internal energy,
which, as already mentioned, is a state variable.
In ordinary language, we often confuse heat
with internal energy. The distinction between
them is sometimes ignored in elementary
physics books. For proper understanding of
thermodynamics, however, the distinction is
crucial.
12.5 FIRST LAW OF THERMODYNAMICS
We have seen that the internal energy U of a
system can change through two modes of energy
transfer : heat and work. Let
∆Q = Heat supplied to the system by the
surroundings
∆W = Work done by the system on the
surroundings
∆U = Change in internal energy of the system
The general principle of conservation of
energy then implies that
∆Q = ∆U + ∆W (12.1)
i.e. the energy (∆Q) supplied to the system goes
in partly to increase the internal energy of the
system (∆U) and the rest in work on the
environment (∆W). Equation (12.1) is known as
the First Law of Thermodynamics. It is simply
the general law of conservation of energy applied
to any system in which the energy transfer from
or to the surroundings is taken into account.
Let us put Eq. (12.1) in the alternative form
∆Q – ∆W = ∆U (12.2)
Now, the system may go from an initial state
to the final state in a number of ways. For
example, to change the state of a gas from
(P
1
, V
1
) to (P
2
, V
2
), we can first change the
volume of the gas from V
1
to V
2
, keeping its
pressure constant i.e. we can first go the state
(P
1
, V
2
) and then change the pressure of the
gas from P
1
to P
2
, keeping volume constant, to
take the gas to (P
2
, V
2
). Alternatively, we can
first keep the volume constant and then keep
the pressure constant. Since U is a state
variable, ∆U depends only on the initial and
final states and not on the path taken by the
gas to go from one to the other. However, ∆Q
and ∆W will, in general, depend on the path
taken to go from the initial to final states. From
the First Law of Thermodynamics, Eq. (12.2),
it is clear that the combination ∆Q – ∆W, is
however, path independent. This shows that
if a system is taken through a process in which
∆U = 0 (for example, isothermal expansion of
an ideal gas, see section 12.8),
∆Q = ∆W
i.e., heat supplied to the system is used up
entirely by the system in doing work on the
environment.
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If the system is a gas in a cylinder with a
movable piston, the gas in moving the piston does
work. Since force is pressure times area, and
area times displacement is volume, work done
by the system against a constant pressure P is
∆W = P ∆V
where ∆V is the change in volume of the gas.
Thus, for this case, Eq. (12.1) gives
∆Q = ∆U + P ∆V (12.3)
As an application of Eq. (12.3), consider the
change in internal energy for 1 g of water when
we go from its liquid to vapour phase. The
measured latent heat of water is 2256 J/g. i.e.,
for 1 g of water ∆Q = 2256 J. At atmospheric
pressure, 1 g of water has a volume 1 cm
3
in
liquid phase and 1671 cm
3
in vapour phase.
Therefore,
∆W =P (V
g
–V
l
) = 1.013 ×10
5
×(1671×10
–6
) =169.2 J
Equation (12.3) then gives
∆U = 2256 – 169.2 = 2086.8 J
We see that most of the heat goes to increase
the internal energy of water in transition from
the liquid to the vapour phase.
12.6 SPECIFIC HEAT CAPACITY
Suppose an amount of heat ∆Q supplied to a
substance changes its temperature from T to
T + ∆T. We define heat capacity of a substance
(see Chapter 11) to be
T
Q
S
∆
∆
=
(12.4)
We expect ∆Q and, therefore, heat capacity S
to be proportional to the mass of the substance.
Further, it could also depend on the
temperature, i.e., a different amount of heat may
be needed for a unit rise in temperature at
different temperatures. To define a constant
characteristic of the substance and
independent of its amount, we divide S by the
mass of the substance m in kg :
s
S
m m
Q
T
= =
1 ∆
∆
(12.5)
s is known as the specific heat capacity of the
substance. It depends on the nature of the
substance and its temperature. The unit of
specific heat capacity is J kg
–1
K
–1
.
If the amount of substance is specified in
terms of moles µ (instead of mass m in kg ), we
can define heat capacity per mole of the
substance by
1
S Q
C
T
µ µ
∆
= =
∆
(12.6)
C is known as molar specific heat capacity of
the substance. Like s, C is independent of the
amount of substance. C depends on the nature
of the substance, its temperature and the
conditions under which heat is supplied. The
unit of C is J mo1
–1
K
–1
. As we shall see later (in
connection with specific heat capacity of gases),
additional conditions may be needed to define
C or s. The idea in defining C is that simple
predictions can be made in regard to molar
specific heat capacities.
Table 12.1 lists measured specific and molar
heat capacities of solids at atmospheric pressure
and ordinary room temperature.
We will see in Chapter 13 that predictions of
specific heats of gases generally agree with
experiment. We can use the same law of
equipartition of energy that we use there to
predict molar specific heat capacities of solids
(See Section 13.5 and 13.6). Consider a solid of
N atoms, each vibrating about its mean
position. An oscillator in one dimension has
average energy of 2 × ½ k
B
T = k
B
T. In three
dimensions, the average energy is 3 k
B
T.
For a mole of a solid, the total energy is
U = 3 k
B
T × N
A
= 3 RT (
∵
k
B
T × N
A
= R)
Now, at constant pressure, ∆Q = ∆U + P ∆V ≅
∆U, since for a solid ∆V is negligible. Therefore,
C
Q
T
U
T
R= = =
∆
∆
∆
∆
3
(12.7)
Table 12.1 Specific and molar heat capacities
of some solids at room
temperature and atmospheric
pressure
As Table 12.1 shows, the experimentally
measured values which generally agrees with
Substance
Speci"c heat
–v
(J kg K )
–1 –1
Molar speci"c
heat (J mol K )
–1 –1
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predicted value 3R at ordinary temperatures.
(Carbon is an exception.) The agreement is
known to break down at low temperatures.
Specific heat capacity of water
The old unit of heat was calorie. One calorie
was earlier defined to be the amount of heat
required to raise the temperature of 1g of water
by 1°C. With more precise measurements, it was
found that the specific heat of water varies
slightly with temperature. Figure 12.5 shows
this variation in the temperature range 0 to
100 °C.
Fig. 12.5 Variation of specific heat capacity of water
with temperature.
For a precise definition of calorie, it was,
therefore, necessary to specify the unit
temperature interval. One calorie is defined
to be the amount of heat required to raise the
temperature of 1g of water from 14.5 °C to
15.5°C. Since heat is just a form of energy,
it is preferable to use the unit joule, J.
In SI units, the specific heat capacity of water
is 4186 J kg
–1
K
–1
i.e. 4.186 J g
–1
K
–1
. The so
called mechanical equivalent of heat defined
as the amount of work needed to produce
1 cal of heat is in fact just a conversion factor
between two different units of energy : calorie
to joule. Since in SI units, we use the unit joule
for heat, work or any other form of energy, the
term mechanical equivalent is now
superfluous and need not be used.
As already remarked, the specific heat
capacity depends on the process or the
conditions under which heat capacity transfer
takes place. For gases, for example, we can
define two specific heats : specific heat
capacity at constant volume and specific
heat capacity at constant pressure. For an
ideal gas, we have a simple relation.
C
p
– C
v
= R (12.8)
where C
p
and C
v
are molar specific heat
capacities of an ideal gas at constant pressure
and volume respectively and R is the universal
gas constant. To prove the relation, we begin
with Eq. (12.3) for 1 mole of the gas :
∆Q = ∆U + P ∆V
If ∆Q is absorbed at constant volume, ∆V = 0
C
Q
T
U
T
U
T
v
v v
=
=
=
∆
∆
∆
∆
∆
∆
(12.9)
where the subscript v is dropped in the last
step, since U of an ideal gas depends only on
temperature. (The subscript denotes the
quantity kept fixed.) If, on the other hand, ∆Q
is absorbed at constant pressure,
C
Q
T
U
T
P
V
T
p
p p p
=
=
+
∆
∆
∆
∆
∆
∆
(12.10)
The subscript p can be dropped from the
first term since U of an ideal gas depends only
on T. Now, for a mole of an ideal gas
PV = RT
which gives
P
V
T
R
p
∆
∆
=
(12.11)
Equations (12.9) to (12.11) give the desired
relation, Eq. (12.8).
12.7 THERMODYNAMIC STATE VARIABLES
AND EQUATION OF STATE
Every equilibrium state of a thermodynamic
system is completely described by specific
values of some macroscopic variables, also
called state variables. For example, an
equilibrium state of a gas is completely
specified by the values of pressure, volume,
temperature, and mass (and composition if
there is a mixture of gases). A thermodynamic
system is not always in equilibrium. For example,
a gas allowed to expand freely against vacuum
is not an equilibrium state [Fig. 12.6(a)]. During
the rapid expansion, pressure of the gas may
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not be uniform throughout. Similarly, a mixture
of gases undergoing an explosive chemical
reaction (e.g. a mixture of petrol vapour and
air when ignited by a spark) is not an
equilibrium state; again its temperature and
pressure are not uniform [Fig. 12.6(b)].
Eventually, the gas attains a uniform
temperature and pressure and comes to
thermal and mechanical equilibrium with its
surroundings.
Fig. 12.6 (a) The partition in the box is suddenly
removed leading to free expansion of the
gas. (b) A mixture of gases undergoing an
explosive chemical reaction. In both
situations, the gas is not in equilibrium and
cannot be described by state variables.
In short, thermodynamic state variables
describe equilibrium states of systems. The
various state variables are not necessarily
independent. The connection between the state
variables is called the equation of state. For
example, for an ideal gas, the equation of state
is the ideal gas relation
P V =
µ
R T
For a fixed amount of the gas i.e. given µ, there
are thus, only two independent variables, say P
and V or T and V. The pressure-volume curve
for a fixed temperature is called an isotherm.
Real gases may have more complicated
equations of state.
The thermodynamic state variables are of two
kinds: extensive and intensive. Extensive
variables indicate the ‘size’ of the system.
Intensive variables such as pressure and
temperature do not. To decide which variable is
extensive and which intensive, think of a
relevant system in equilibrium, and imagine that
it is divided into two equal parts. The variables
that remain unchanged for each part are
intensive. The variables whose values get halved
in each part are extensive. It is easily seen, for
example, that internal energy U, volume V, total
mass M are extensive variables. Pressure P,
temperature T, and density ρ are intensive
variables. It is a good practice to check the
consistency of thermodynamic equations using
this classification of variables. For example, in
the equation
∆Q = ∆U + P ∆V
quantities on both sides are extensive*. (The
product of an intensive variable like P and an
extensive quantity ∆V is extensive.)
12.8 THERMODYNAMIC PROCESSES
12.8.1 Quasi-static process
Consider a gas in thermal and mechanical
equilibrium with its surroundings. The pressure
of the gas in that case equals the external
pressure and its temperature is the same as
that of its surroundings. Suppose that the
external pressure is suddenly reduced (say by
lifting the weight on the movable piston in the
container). The piston will accelerate outward.
During the process, the gas passes through
states that are not equilibrium states. The non-
equilibrium states do not have well-defined
pressure and temperature. In the same way, if
a finite temperature difference exists between
the gas and its surroundings, there will be a
rapid exchange of heat during which the gas
will pass through non-equilibrium states. In
due course, the gas will settle to an equilibrium
state with well-defined temperature and
pressure equal to those of the surroundings. The
free expansion of a gas in vacuum and a mixture
of gases undergoing an explosive chemical
reaction, mentioned in section 12.7 are also
examples where the system goes through non-
equilibrium states.
Non-equilibrium states of a system are difficult
to deal with. It is, therefore, convenient to
imagine an idealised process in which at every
stage the system is an equilibrium state. Such a
* As emphasised earlier, Q is not a state variable. However,
∆
Q is clearly proportional to the total mass of
system and hence is extensive.
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process is, in principle, infinitely slow, hence the
name quasi-static (meaning nearly static). The
system changes its variables (P, T, V ) so slowly
that it remains in thermal and mechanical
equilibrium with its surroundings throughout.
In a quasi-static process, at every stage, the
difference in the pressure of the system and the
external pressure is infinitesimally small. The
same is true of the temperature difference
between the system and its surroundings
(Fig.12.7). To take a gas from the state (P, T ) to
another state (P ′, T ′ ) via a quasi-static process,
we change the external pressure by a very small
amount, allow the system to equalise its pressure
with that of the surroundings and continue the
process infinitely slowly until the system
achieves the pressure P ′. Similarly, to change
the temperature, we introduce an infinitesimal
temperature difference between the system and
the surrounding reservoirs and by choosing
reservoirs of progressively different temperatures
T to T ′, the system achieves the temperature T ′.
Fig. 12.7 In a quasi-static process, the temperature
of the surrounding reservoir and the
external pressure differ only infinitesimally
from the temperature and pressure of the
system.
A quasi-static process is obviously a
hypothetical construct. In practice, processes
that are sufficiently slow and do not involve
accelerated motion of the piston, large
temperature gradient, etc., are reasonably
approximation to an ideal quasi-static process.
We shall from now on deal with quasi-static
processes only, except when stated otherwise.
A process in which the temperature of the
system is kept fixed throughout is called an
isothermal process. The expansion of a gas in
a metallic cylinder placed in a large reservoir of
fixed temperature is an example of an isothermal
process. (Heat transferred from the reservoir to
the system does not materially affect the
temperature of the reservoir, because of its very
large heat capacity.) In isobaric processes the
pressure is constant while in isochoric
processes the volume is constant. Finally, if
the system is insulated from the surroundings
and no heat flows between the system and the
surroundings, the process is adiabatic. The
definitions of these special processes are
summarised in Table. 12.2
Table 12.2 Some special thermodynamic
processes
We now consider these processes in some detail:
12.8.2 Isothermal process
For an isothermal process (T fixed), the ideal gas
equation gives
PV = constant
i.e., pressure of a given mass of gas varies inversely
as its volume. This is nothing but Boyle’s Law.
Suppose an ideal gas goes isothermally (at
temperature T ) from its initial state (P
1
, V
1
) to
the final state (P
2
, V
2
). At any intermediate stage
with pressure P and volume change from V to
V + ∆V (∆V small)
∆W = P ∆ V
Taking (∆V → 0) and summing the quantity
∆W over the entire process,
W = P V
V
V
1
2
d
∫
= RT
V
V
RT
V
V
V
V
1
2
2
1
µ µ
d
∫
= In
(12.12)
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PHYSICS312
where in the second step we have made use of
the ideal gas equation PV = µ RT and taken the
constants out of the integral. For an ideal gas,
internal energy depends only on temperature.
Thus, there is no change in the internal energy
of an ideal gas in an isothermal process. The
First Law of Thermodynamics then implies that
heat supplied to the gas equals the work done
by the gas : Q = W. Note from Eq. (12.12) that
for V
2
> V
1
, W > 0; and for V
2
< V
1
, W < 0. That
is, in an isothermal expansion, the gas absorbs
heat and does work while in an isothermal
compression, work is done on the gas by the
environment and heat is released.
12.8.3 Adiabatic process
In an adiabatic process, the system is insulated
from the surroundings and heat absorbed or
released is zero. From Eq. (12.1), we see that
work done by the gas results in decrease in its
internal energy (and hence its temperature for
an ideal gas). We quote without proof (the result
that you will learn in higher courses) that for
an adiabatic process of an ideal gas.
P V
γ
= const (12.13)
where
γ
is the ratio of specific heats (ordinary
or molar) at constant pressure and at constant
volume.
γ
=
C
p
C
v
Thus if an ideal gas undergoes a change in
its state adiabatically from (P
1
, V
1
) to (P
2
, V
2
) :
P
1
V
1
γ
= P
2
V
2
γ
(12.14)
Figure12.8 shows the P-V curves of an ideal
gas for two adiabatic processes connecting two
isotherms.
Fig. 12.8 P-V curves for isothermal and adiabatic
processes of an ideal gas.
We can calculate, as before, the work done in
an adiabatic change of an ideal gas from the
state (P
1
, V
1
, T
1
) to the state (P
2
, V
2
, T
2
).
W = P V
V
V
d
1
2
∫
(12.15)
From Eq. (12.14), the constant is P
1
V
1
γ
or P
2
V
2
γ
W =
P V
V
P V
V
2 2
2
1 1
1
1
1 −
− −
−
γ
γ
γ
γ
γ
1 1
= P V P V
R(T T
1
1 −
−
−
−
[ ]
=
γ
µ
γ
2 2 1 1
1 2
)
1
(12.16)
As expected, if work is done by the gas in an
adiabatic process (W > 0), from Eq. (12.16),
T
2
< T
1
. On the other hand, if work is done on
the gas (W < 0), we get T
2
> T
1
i.e., the
temperature of the gas rises.
12.8.4 Isochoric process
In an isochoric process, V is constant. No work
is done on or by the gas. From Eq. (12.1), the
heat absorbed by the gas goes entirely to change
its internal energy and its temperature. The
change in temperature for a given amount of
heat is determined by the specific heat of the
gas at constant volume.
12.8.5 Isobaric process
In an isobaric process, P is fixed. Work done by
the gas is
W = P (V
2
– V
1
) =
µ
R (T
2
– T
1
) (12.17)
Since temperature changes, so does internal
energy. The heat absorbed goes partly to
increase internal energy and partly to do work.
The change in temperature for a given amount
of heat is determined by the specific heat of the
gas at constant pressure.
12.8.6 Cyclic process
In a cyclic process, the system returns to its
initial state. Since internal energy is a state
variable, ∆U = 0 for a cyclic process. From
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Eq. (12.1), the total heat absorbed equals the
work done by the system.
12.9 HEAT ENGINES
Heat engine is a device by which a system is
made to undergo a cyclic process that results
in conversion of heat to work.
(1) It consists of a working substance–the
system. For example, a mixture of fuel
vapour and air in a gasoline or diesel engine
or steam in a steam engine are the working
substances.
(2) The working substance goes through a cycle
consisting of several processes. In some of
these processes, it absorbs a total amount
of heat Q
1
from an external reservoir at some
high temperature T
1
.
(3) In some other processes of the cycle, the
working substance releases a total amount
of heat Q
2
to an external reservoir at some
lower temperature T
2
.
(4) The work done (W ) by the system in a cycle
is transferred to the environment via some
arrangement (e.g. the working substance
may be in a cylinder with a moving piston
that transfers mechanical energy to the
wheels of a vehicle via a shaft).
The basic features of a heat engine are
schematically represented in Fig. 12.9.
Fig. 12.9 Schematic representation of a heat engine.
The engine takes heat Q
1
from a hot
reservoir at temperature T
1
, releases heat
Q
2
to a cold reservoir at temperature T
2
and delivers work W to the surroundings.
The cycle is repeated again and again to get
useful work for some purpose. The discipline of
thermodynamics has its roots in the study of heat
engines. A basic question relates to the efficiency
of a heat engine. The efficiency (
η
) of a heat
engine is defined by
1
Q
W
=
η
(12.18)
where Q
1
is the heat input i.e., the heat
absorbed by the system in one complete cycle
and W is the work done on the environment in
a cycle. In a cycle, a certain amount of heat (Q
2
)
may also be rejected to the environment. Then,
according to the First Law of Thermodynamics,
over one complete cycle,
W = Q
1
– Q
2
(12.19)
i.e.,
1
Q
2
Q
− = 1
η
(12.20)
For Q
2
= 0,
η
= 1, i.e., the engine will have
100% efficiency in converting heat into work.
Note that the First Law of Thermodynamics i.e.,
the energy conservation law does not rule out
such an engine. But experience shows that
such an ideal engine with
η
= 1 is never possible,
even if we can eliminate various kinds of losses
associated with actual heat engines. It turns
out that there is a fundamental limit on the
efficiency of a heat engine set by an independent
principle of nature, called the Second Law of
Thermodynamics (section 12.11).
The mechanism of conversion of heat into
work varies for different heat engines. Basically,
there are two ways : the system (say a gas or a
mixture of gases) is heated by an external
furnace, as in a steam engine; or it is heated
internally by an exothermic chemical reaction
as in an internal combustion engine. The
various steps involved in a cycle also differ from
one engine to another.
12.10 REFRIGERATORS AND HEAT PUMPS
A refrigerator is the reverse of a heat engine.
Here the working substance extracts heat Q
2
from the cold reservoir at temperature T
2
, some
external work W is done on it and heat Q
1
is
released to the hot reservoir at temperature T
1
(Fig. 12.10).
Fig. 12.10 Schematic representation of a refrigerator
or a heat pump, the reverse of a heat
engine.
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PHYSICS314
A heat pump is the same as a refrigerator.
What term we use depends on the purpose of
the device. If the purpose is to cool a portion of
space, like the inside of a chamber, and higher
temperature reservoir is surrounding, we call
the device a refrigerator; if the idea is to pump
heat into a portion of space (the room in a
building when the outside environment is cold),
the device is called a heat pump.
In a refrigerator the working substance
(usually, in gaseous form) goes through the
following steps : (a) sudden expansion of the gas
from high to low pressure which cools it and
converts it into a vapour-liquid mixture, (b)
absorption by the cold fluid of heat from the
region to be cooled converting it into vapour, (c)
heating up of the vapour due to external work
done on the system, and (d) release of heat by
the vapour to the surroundings, bringing it to
the initial state and completing the cycle.
The coefficient of performance (
α
) of a
refrigerator is given by
2
Q
W
α
=
(12.21)
where Q
2
is the heat extracted from the cold
reservoir and W is the work done on the
system–the refrigerant. (
α
for heat pump is
defined as Q
1
/W) Note that while
η
by definition
can never exceed 1,
α
can be greater than 1.
By energy conservation, the heat released to the
hot reservoir is
Q
1
= W + Q
2
i.e.,
2
1 2
–
Q
Q Q
α
=
(12.22)
In a heat engine, heat cannot be fully
converted to work; likewise a refrigerator cannot
work without some external work done on the
system, i.e., the coefficient of performance in Eq.
(12.21) cannot be infinite.
12.11 SECOND LAW OF THERMODYNAMICS
The First Law of Thermodynamics is the principle
of conservation of energy. Common experience
shows that there are many conceivable
processes that are perfectly allowed by the First
Law and yet are never observed. For example,
nobody has ever seen a book lying on a table
jumping to a height by itself. But such a thing
Pioneers of Thermodynamics
Lord Kelvin (William Thomson) (1824-1907), born in Belfast, Ireland, is
among the foremost British scientists of the nineteenth century. Thomson
played a key role in the development of the law of conservation of energy
suggested by the work of James Joule (1818-1889), Julius Mayer (1814-
1878) and Hermann Helmholtz (1821-1894). He collaborated with Joule
on the so-called Joule-Thomson effect : cooling of a gas when it expands
into vacuum. He introduced the notion of the absolute zero of temperature
and proposed the absolute temperature scale, now called the Kelvin scale
in his honour. From the work of Sadi Carnot (1796-1832), Thomson arrived
at a form of the Second Law of Thermodynamics. Thomson was a versatile
physicist, with notable contributions to electromagnetic theory and
hydrodynamics.
Rudolf Clausius (1822-1888), born in Poland, is generally regarded as
the discoverer of the Second Law of Thermodynamics. Based on the work
of Carnot and Thomson, Clausius arrived at the important notion of entropy
that led him to a fundamental version of the Second Law of
Thermodynamics that states that the entropy of an isolated system can
never decrease. Clausius also worked on the kinetic theory of gases and
obtained the first reliable estimates of molecular size, speed, mean free
path, etc.
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THERMODYNAMICS 315
would be possible if the principle of conservation
of energy were the only restriction. The table
could cool spontaneously, converting some of its
internal energy into an equal amount of
mechanical energy of the book, which would
then hop to a height with potential energy equal
to the mechanical energy it acquired. But this
never happens. Clearly, some additional basic
principle of nature forbids the above, even
though it satisfies the energy conservation
principle. This principle, which disallows many
phenomena consistent with the First Law of
Thermodynamics is known as the Second Law
of Thermodynamics.
The Second Law of Thermodynamics gives a
fundamental limitation to the efficiency of a heat
engine and the co-efficient of performance of a
refrigerator. In simple terms, it says that
efficiency of a heat engine can never be unity.
According to Eq. (12.20), this implies that heat
released to the cold reservoir can never be made
zero. For a refrigerator, the Second Law says that
the co-efficient of performance can never be
infinite. According to Eq. (12.21), this implies
that external work (W) can never be zero. The
following two statements, one due to Kelvin and
Planck denying the possibility of a perfect heat
engine, and another due to Clausius denying
the possibility of a perfect refrigerator or heat
pump, are a concise summary of these
observations.
Kelvin-Planck statement
No process is possible whose sole result is the
absorption of heat from a reservoir and the
complete conversion of the heat into work.
Clausius statement
No process is possible whose sole result is the
transfer of heat from a colder object to a hotter
object.
It can be proved that the two statements
above are completely equivalent.
12.12 REVERSIBLE AND IRREVERSIBLE
PROCESSES
Imagine some process in which a thermodynamic
system goes from an initial state i to a final
state f. During the process the system absorbs
heat Q from the surroundings and performs
work W on it. Can we reverse this process and
bring both the system and surroundings to their
initial states with no other effect anywhere ?
Experience suggests that for most processes in
nature this is not possible. The spontaneous
processes of nature are irreversible. Several
examples can be cited. The base of a vessel on
an oven is hotter than its other parts. When
the vessel is removed, heat is transferred from
the base to the other parts, bringing the vessel
to a uniform temperature (which in due course
cools to the temperature of the surroundings).
The process cannot be reversed; a part of the
vessel will not get cooler spontaneously and
warm up the base. It will violate the Second Law
of Thermodynamics, if it did. The free expansion
of a gas is irreversible. The combustion reaction
of a mixture of petrol and air ignited by a spark
cannot be reversed. Cooking gas leaking from a
gas cylinder in the kitchen diffuses to the
entire room. The diffusion process will not
spontaneously reverse and bring the gas back
to the cylinder. The stirring of a liquid in thermal
contact with a reservoir will convert the work
done into heat, increasing the internal energy
of the reservoir. The process cannot be reversed
exactly; otherwise it would amount to conversion
of heat entirely into work, violating the Second
Law of Thermodynamics. Irreversibility is a rule
rather an exception in nature.
Irreversibility arises mainly from two causes:
one, many processes (like a free expansion, or
an explosive chemical reaction) take the system
to non-equilibrium states; two, most processes
involve friction, viscosity and other dissipative
effects (e.g., a moving body coming to a stop and
losing its mechanical energy as heat to the floor
and the body; a rotating blade in a liquid coming
to a stop due to viscosity and losing its
mechanical energy with corresponding gain in
the internal energy of the liquid). Since
dissipative effects are present everywhere and
can be minimised but not fully eliminated, most
processes that we deal with are irreversible.
A thermodynamic process (state i → state f )
is reversible if the process can be turned back
such that both the system and the surroundings
return to their original states, with no other
change anywhere else in the universe. From the
preceding discussion, a reversible process is an
idealised notion. A process is reversible only if
it is quasi-static (system in equilibrium with the
surroundings at every stage) and there are no
dissipative effects. For example, a quasi-static
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isothermal expansion of an ideal gas in a
cylinder fitted with a frictionless movable piston
is a reversible process.
Why is reversibility such a basic concept in
thermodynamics ? As we have seen, one of the
concerns of thermodynamics is the efficiency
with which heat can be converted into work.
The Second Law of Thermodynamics rules out
the possibility of a perfect heat engine with 100%
efficiency. But what is the highest efficiency
possible for a heat engine working between two
reservoirs at temperatures T
1
and T
2
? It turns
out that a heat engine based on idealised
reversible processes achieves the highest
efficiency possible. All other engines involving
irreversibility in any way (as would be the case
for practical engines) have lower than this
limiting efficiency.
12.13 CARNOT ENGINE
Suppose we have a hot reservoir at temperature
T
1
and a cold reservoir at temperature T
2
. What
is the maximum efficiency possible for a heat
engine operating between the two reservoirs and
what cycle of processes should be adopted to
achieve the maximum efficiency ? Sadi Carnot,
a French engineer, first considered this question
in 1824. Interestingly, Carnot arrived at the
correct answer, even though the basic concepts
of heat and thermodynamics had yet to be firmly
established.
We expect the ideal engine operating between
two temperatures to be a reversible engine.
Irreversibility is associated with dissipative
effects, as remarked in the preceding section,
and lowers efficiency. A process is reversible if
it is quasi-static and non-dissipative. We have
seen that a process is not quasi-static if it
involves finite temperature difference between
the system and the reservoir. This implies that
in a reversible heat engine operating between
two temperatures, heat should be absorbed
(from the hot reservoir) isothermally and
released (to the cold reservoir) isothermally. We
thus have identified two steps of the reversible
heat engine : isothermal process at temperature
T
1
absorbing heat Q
1
from the hot reservoir, and
another isothermal process at temperature T
2
releasing heat Q
2
to the cold reservoir. To
complete a cycle, we need to take the system
from temperature T
1
to T
2
and then back from
temperature T
2
to T
1
. Which processes should
we employ for this purpose that are reversible?
A little reflection shows that we can only adopt
reversible adiabatic processes for these
purposes, which involve no heat flow from any
reservoir. If we employ any other process that is
not adiabatic, say an isochoric process, to take
the system from one temperature to another, we
shall need a series of reservoirs in the
temperature range T
2
to T
1
to ensure that at each
stage the process is quasi-static. (Remember
again that for a process to be quasi-static and
reversible, there should be no finite temperature
difference between the system and the reservoir.)
But we are considering a reversible engine that
operates between only two temperatures. Thus
adiabatic processes must bring about the
temperature change in the system from T
1
to T
2
and T
2
to T
1
in this engine.
Fig. 12.11 Carnot cycle for a heat engine with an
ideal gas as the working substance.
A reversible heat engine operating between
two temperatures is called a Carnot engine. We
have just argued that such an engine must have
the following sequence of steps constituting one
cycle, called the Carnot cycle, shown in Fig.
12.11. We have taken the working substance of
the Carnot engine to be an ideal gas.
(a) Step 1 → 2 Isothermal expansion of the gas
taking its state from (P
1
, V
1
, T
1
) to
(P
2
, V
2
, T
1
).
The heat absorbed by the gas (Q
1
) from the
reservoir at temperature T
1
is given by
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THERMODYNAMICS 317
Eq. (12.12). This is also the work done (W
1 → 2
)
by the gas on the environment.
W
1 → 2
= Q
1
=
µ
R T
1
ln
V
V
2
1
(12.23)
(b) Step 2 → 3 Adiabatic expansion of the gas
from (P
2
, V
2
, T
1
) to (P
3
, V
3
, T
2
)
Work done by the gas, using
Eq. (12.16), is
W
R T T
2 3→
=
−
( )
−
µ
1 2
1
γ
(12.24)
(c) Step 3 → 4 Isothermal compression of the
gas from (P
3
, V
3
, T
2
) to (P
4
, V
4
, T
2
).
Heat released (Q
2
) by the gas to the reservoir
at temperature T
2
is given by Eq. (12.12). This
is also the work done (W
3
→ 4
) on the gas by the
environment.
W RT
V
V
3 4→
= =
Q
2
µ
2
3
4
ln
(12.25)
(d) Step 4 → 1 Adiabatic compression of the
gas from (P
4
, V
4
, T
2
) to (P
1
,V
1
, T
1
).
Work done on the gas, [using Eq.(12.16), is
W R
T T
4 1→
=
−
µ
γ
1 2
-1
(12.26)
From Eqs. (12.23) to (12.26) total work done
by the gas in one complete cycle is
W = W
1 → 2
+ W
2 → 3
– W
3
→ 4
– W
4 → 1
=
µ
RT
1
ln
V
V
2
1
–
µ
RT
2
ln
V
V
3
4
(12.27)
The efficiency
η
of the Carnot engine is
1
2
1
1
W Q
Q
Q
η
= = −
= −
1
T
T
V
V
V
V
2
1
3
4
2
1
In
In
(12.28)
Now since step 2 → 3 is an adiabatic process,
T V T V
1 2 2 3
γ
γ
−
−
=
1
1
i.e.
V
V
=
T
T
2
3
2
1
−1 1/( )
γ
(12.29)
Similarly, since step 4 → 1 is an adiabatic
process
T V T V
2 4 1 1
γ γ
− −
=
1 1
i.e.
V
V
=
T
T
1
4
2
1
−1 1/
γ
(12.30)
From Eqs. (12.29) and (12.30),
V
V
=
V
V
3
4
2
1
(12.31)
Using Eq. (12.31) in Eq. (12.28), we get
η
= 1 −
T
T
2
1
(Carnot engine) (12.32)
We have already seen that a Carnot engine
is a reversible engine. Indeed it is the only
reversible engine possible that works between
two reservoirs at different temperatures. Each
step of the Carnot cycle given in Fig. 12.11 can
be reversed. This will amount to taking heat Q
2
from the cold reservoir at T
2
, doing work W on
the system, and transferring heat Q
1
to the hot
reservoir. This will be a reversible refrigerator.
We next establish the important result
(sometimes called Carnot’s theorem) that
(a) working between two given temperatures T
1
and T
2
of the hot and cold reservoirs respectively,
no engine can have efficiency more than that of
the Carnot engine and (b) the efficiency of the
Carnot engine is independent of the nature of
the working substance.
To prove the result (a), imagine a reversible
(Carnot) engine R and an irreversible engine I
working between the same source (hot reservoir)
and sink (cold reservoir). Let us couple the
engines, I and R, in such a way so that I acts
like a heat engine and R acts as a refrigerator.
Let I absorb heat Q
1
from the source, deliver
work W ′ and release the heat Q
1
- W
′
to the sink.
We arrange so that R returns the same heat Q
1
to the source, taking heat Q
2
from the sink and
requiring work W = Q
1
– Q
2
to be done on it.
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SUMMARY
1. The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a
third system separately are in thermal equilibrium with each other’. The Zeroth Law leads
to the concept of temperature.
2. Internal energy of a system is the sum of kinetic energies and potential energies of the
molecular constituents of the system. It does not include the over-all kinetic energy of
the system. Heat and work are two modes of energy transfer to the system. Heat is the
energy transfer arising due to temperature difference between the system and the
surroundings. Work is energy transfer brought about by other means, such as moving
the piston of a cylinder containing the gas, by raising or lowering some weight connected
to it.
3. The first law of thermodynamics is the general law of conservation of energy applied to
any system in which energy transfer from or to the surroundings (through heat and
work) is taken into account. It states that
∆Q = ∆U + ∆W
where ∆Q is the heat supplied to the system, ∆W is the work done by the system and ∆U
is the change in internal energy of the system.
Now suppose
η
R
<
η
I
i.e. if R were to act
as an engine it would give less work output
than that of I i.e. W < W′ for a given Q
1
. With R
acting like a refrigerator, this would mean
Q
2
= Q
1
– W > Q
1
– W ′. Thus, on the whole,
the coupled I-R system extracts heat
(Q
1
– W) – (Q
1
– W ′) = (W ′ – W ) from the cold
reservoir and delivers the same amount of work
in one cycle, without any change in the source
or anywhere else. This is clearly against the
Kelvin-Planck statement of the Second Law of
Thermodynamics. Hence the assertion
η
I
>
η
R
is wrong. No engine can have efficiency greater
than that of the Carnot engine. A similar
argument can be constructed to show that a
reversible engine with one particular substance
cannot be more efficient than the one using
another substance. The maximum efficiency of
a Carnot engine given by Eq. (12.32) is
independent of the nature of the system
performing the Carnot cycle of operations. Thus
we are justified in using an ideal gas as a system
in the calculation of efficiency η of a Carnot
engine. The ideal gas has a simple equation of
state, which allows us to readily calculate η, but
the final result for η, [Eq. (12.32)], is true for
any Carnot engine.
This final remark shows that in a Carnot
cycle,
2
1
2
1
T
T
=
Q
Q
(12.33)
is a universal relation independent of the nature
of the system. Here Q
1
and Q
2
are respectively,
the heat absorbed and released isothermally
(from the hot and to the cold reservoirs) in a
Carnot engine. Equation (12.33), can, therefore,
be used as a relation to define a truly universal
thermodynamic temperature scale that is
independent of any particular properties of the
system used in the Carnot cycle. Of course, for
an ideal gas as a working substance, this
universal temperature is the same as the ideal
gas temperature introduced in section 12.11.
I
R
W
Fig. 12.12 An irreversible engine (I) coupled to a
reversible refrigerator (R). If W
′
> W, this
would amount to extraction of heat
W
′
– W from the sink and its full
conversion to work, in contradiction with
the Second Law of Thermodynamics.
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THERMODYNAMICS 319
4. The specific heat capacity of a substance is defined by
s
m
Q
T
=
1
∆
∆
where m is the mass of the substance and ∆Q is the heat required to change its
temperature by ∆T. The molar specific heat capacity of a substance is defined by
1
Q
C
T
µ
∆
=
∆
where µ is the number of moles of the substance. For a solid, the law of equipartition
of energy gives
C = 3 R
which generally agrees with experiment at ordinary temperatures.
Calorie is the old unit of heat. 1 calorie is the amount of heat required to raise the
temperature of 1 g of water from 14.5 °C to 15.5 °C. 1 cal = 4.186 J.
5. For an ideal gas, the molar specific heat capacities at constant pressure and volume
satisfy the relation
C
p
– C
v
= R
where R is the universal gas constant.
6. Equilibrium states of a thermodynamic system are described by state variables. The
value of a state variable depends only on the particular state, not on the path used to
arrive at that state. Examples of state variables are pressure (P ), volume (V ), temperature
(T ), and mass (m ). Heat and work are not state variables. An Equation of State (like
the ideal gas equation PV =
µ
RT ) is a relation connecting different state variables.
7. A quasi-static process is an infinitely slow process such that the system remains in
thermal and mechanical equilibrium with the surroundings throughout. In a
quasi-static process, the pressure and temperature of the environment can differ from
those of the system only infinitesimally.
8. In an isothermal expansion of an ideal gas from volume V
1
to V
2
at temperature T the
heat absorbed (Q) equals the work done (W ) by the gas, each given by
Q = W =
µ
R T ln
1
2
V
V
9. In an adiabatic process of an ideal gas
PV
γ
= constant
where
p
v
C
C
γ
=
Work done by an ideal gas in an adiabatic change of state from (P
1
, V
1
, T
1
) to (P
2
, V
2
, T
2
)
is
(
)
– 1
1 2
R T T
W
µ
γ
−
=
10. Heat engine is a device in which a system undergoes a cyclic process resulting in
conversion of heat into work. If Q
1
is the heat absorbed from the source, Q
2
is the heat
released to the sink, and the work output in one cycle is W, the efficiency
η
of the engine
is:
1
2
1 1
Q
W
Q Q
η
= = −
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PHYSICS320
11. In a refrigerator or a heat pump, the system extracts heat Q
2
from the cold reservoir and
releases Q
1
amount of heat to the hot reservoir, with work W done on the system. The
co-efficient of performance of a refrigerator is given by
21
22
QQ
Q
W
Q
−
==
α
12. The second law of thermodynamics disallows some processes consistent with the First
Law of Thermodynamics. It states
Kelvin-Planck statement
No process is possible whose sole result is the absorption of heat from a reservoir and
complete conversion of the heat into work.
Clausius statement
No process is possible whose sole result is the transfer of heat from a colder object to a
hotter object.
Put simply, the Second Law implies that no heat engine can have efficiency η equal to
1 or no refrigerator can have co-efficient of performance α equal to infinity.
13. A process is reversible if it can be reversed such that both the system and the surroundings
return to their original states, with no other change anywhere else in the universe.
Spontaneous processes of nature are irreversible. The idealised reversible process is a
quasi-static process with no dissipative factors such as friction, viscosity, etc.
14. Carnot engine is a reversible engine operating between two temperatures T
1
(source) and
T
2
(sink). The Carnot cycle consists of two isothermal processes connected by two
adiabatic processes. The efficiency of a Carnot engine is given by
1
2
T
T
1 −=
η
(Carnot engine)
No engine operating between two temperatures can have efficiency greater than that of
the Carnot engine.
15. If Q > 0, heat is added to the system
If Q < 0, heat is removed to the system
If W > 0, Work is done by the system
If W < 0, Work is done on the system
Quantity Symbol Dimensions Unit Remark
Co-efficienty of volume
α
v
[K
–1
] K
–1
α
v
= 3 α
1
expansion
Heat supplied to a system ∆Q [ML
2
T
–2
] J Q is not a state
variable
Specific heat capacity s [L
2
T
–2
K
–1
] J kg
–1
K
–1
Thermal Conductivity K [MLT
–3
K
–1
] J s
–1
K
–1
H = – KA
d
d
t
x
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THERMODYNAMICS 321
POINTS TO PONDER
1. Temperature of a body is related to its average internal energy, not to the kinetic energy
of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature
because of its high speed.
2. Equilibrium in thermodynamics refers to the situation when macroscopic variables
describing the thermodynamic state of a system do not depend on time. Equilibrium of
a system in mechanics means the net external force and torque on the system are zero.
3. In a state of thermodynamic equilibrium, the microscopic constituents of a system are
not in equilibrium (in the sense of mechanics).
4. Heat capacity, in general, depends on the process the system goes through when heat is
supplied.
5. In isothermal quasi-static processes, heat is absorbed or given out by the system even
though at every stage the gas has the same temperature as that of the surrounding
reservoir. This is possible because of the infinitesimal difference in temperature between
the system and the reservoir.
EXERCISES
12.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C.
If the geyser operates on a gas burner, what is the rate of consumption of the fuel if
its heat of combustion is 4.0 × 10
4
J/g ?
12.2 What amount of heat must be supplied to 2.0 × 10
–2
kg of nitrogen (at room
temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular
mass of N
2
= 28; R = 8.3 J mol
–1
K
–1
.)
12.3 Explain why
(a) Two bodies at different temperatures T
1
and T
2
if brought in thermal contact do
not necessarily settle to the mean temperature (T
1
+ T
2
)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent
the different parts of a plant from getting too hot) should have high specific
heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert
at the same latitude.
12.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature
and pressure. The walls of the cylinder are made of a heat insulator, and the piston
is insulated by having a pile of sand on it. By what factor does the pressure of the
gas increase if the gas is compressed to half its original volume ?
12.5 In changing the state of a gas adiabatically from an equilibrium state A to another
equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the
gas is taken from state A to B via a process in which the net heat absorbed by the
system is 9.35 cal, how much is the net work done by the system in the latter case ?
(Take 1 cal = 4.19 J)
12.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock.
A contains a gas at standard temperature and pressure. B is completely evacuated.
The entire system is thermally insulated. The stopcock is suddenly opened. Answer
the following :
(a) What is the final pressure of the gas in A and B ?
(b) What is the change in internal energy of the gas ?
(c) What is the change in the temperature of the gas ?
(d) Do the intermediate states of the system (before settling to the final equilibrium
state) lie on its P-V-T surface ?
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12.7 A steam engine delivers 5.4×10
8
J of work per minute and services 3.6 × 10
9
J of heat
per minute from its boiler. What is the efficiency of the engine? How much heat is
wasted per minute?
12.8 An electric heater supplies heat to a system at a rate of 100W. If system performs
work at a rate of 75 joules per second. At what rate is the internal energy increasing?
12.9 A thermodynamic system is taken from an original state to an intermediate state by
the linear process shown in Fig. (12.13)
Fig. 12.13
Its volume is then reduced to the original value from E to F by an isobaric process.
Calculate the total work done by the gas from D to E to F
12.10 A refrigerator is to maintain eatables kept inside at 9
0
C. If room temperature is 36
0
C,
calculate the coefficient of performance.
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