Heat and Thermodynamics
Lecture Notes 2018
Darren T Grasso
darren.grasso@uwa.edu.au
Contents
1 Introductory remarks
1
2 Brief overview
2
3 Temperature, thermometers and the zeroth law
4
4 Thermal expansion
9
4.1
Solids and linear expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.2
Volume expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5 Gases
15
5.1
Real and ideal gases. State variables . . . . . . . . . . . . . . . . . . . . . 15
5.2
Thermal equilibrium, quasi-static processes and pV diagrams . . . . . . . . 19
5.3
The kinetic theory of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5.4
Total energy in an ideal monatomic gas . . . . . . . . . . . . . . . . . . . . 27
6 Heat
29
6.1
The system and the environment . . . . . . . . . . . . . . . . . . . . . . . 29
6.2
What is heat? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
6.3
The heating of solids and liquids . . . . . . . . . . . . . . . . . . . . . . . . 32
6.4
Heating and phase transformations . . . . . . . . . . . . . . . . . . . . . . 34
6.5
Phase diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.6
Heat transfer mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
6.6.1
Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
6.6.2
Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.6.3
Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
ii
7 The first law of thermodynamics
47
7.1
Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
7.2
The first law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 48
7.3
Internal energy of an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . 50
7.4
Work done on a gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
7.5
Work done on a gas as an ‘area’ under a pV curve . . . . . . . . . . . . . . 55
7.6
Different paths – same change of internal energy . . . . . . . . . . . . . . . 58
8 Gases and special cases of the first law
8.1
Constant volume process and specific heat . . . . . . . . . . . . . . . . . . 60
8.1.1
8.2
60
Constant volume processes and ideal gases . . . . . . . . . . . . . . 61
Constant pressure process and specific heat . . . . . . . . . . . . . . . . . . 63
8.2.1
Constant pressure processes and ideal gases . . . . . . . . . . . . . 64
8.3
Molecular specific heats and degrees of freedom . . . . . . . . . . . . . . . 65
8.4
Isothermal process for an ideal gas . . . . . . . . . . . . . . . . . . . . . . 69
8.5
Adiabatic processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
8.5.1
Adiabatic process for an ideal gas . . . . . . . . . . . . . . . . . . . 70
8.6
Free expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
8.7
Final comments on problem solving and summary . . . . . . . . . . . . . . 75
9 The second law of thermodynamics, entropy and heat engines
77
9.1
Reversible and irreversible processes . . . . . . . . . . . . . . . . . . . . . . 77
9.2
Isolated systems and entropy . . . . . . . . . . . . . . . . . . . . . . . . . . 79
9.3
Definition of entropy change . . . . . . . . . . . . . . . . . . . . . . . . . . 80
9.4
The second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . 82
9.5
Microscopic and statistical interpretation of entropy . . . . . . . . . . . . . 84
iii
9.6
9.7
Heat engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
9.6.1
The idea of a heat engine
. . . . . . . . . . . . . . . . . . . . . . . 86
9.6.2
The efficiency of a heat engine . . . . . . . . . . . . . . . . . . . . . 88
9.6.3
Heat engines and the second law of thermodynamics . . . . . . . . 89
9.6.4
The Carnot engine – the most efficient heat engine . . . . . . . . . 90
Refrigerators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
iv
1
Introductory remarks
Some time ago in 1707, when physics was still called natural philosophy, Newton remarked:
“In learning the sciences examples are more useful than the precepts”. He was right (he
wasn’t always!). The most useful advice anyone can give with respect to learning physics
is that you try as many problems as you can. To understand a general concept you must
have worked through, thought about, and understood several specific examples. In this
spirit I will punctuate my lectures with as many worked problems and examples as time
permits. I will also try to give you as much practice as you can stand!
These notes will summarize almost all of what is said during the lectures. The material
which will not be found here will appear as hand written notes in the ‘Supplementary
Notes’ column in the ‘Heat and Thermodynamics Lecture Material’ file in this course’s
LMS unit (www.lms.uwa.edu.au). These handwritten notes include: (i) important derivations of general results, the details of which will not be examined; and (ii) worked examples. These notes are essentially just the ‘rules of the game’, whilst the worked examples
will show you exactly how to apply these rules.
To be absolutely clear on what is expected of you, you are only required to be able to
solve the types of problems you’ll find in the practice question sets and in the assignments.
The examination will consist of problems only of this kind.
Throughout these notes, any terms appearing in bold are being defined. You should pay
special attention to these terms.
1
2
Brief overview
Thermodynamics is one of the oldest and broadest branches of physics and engineering. It
is the study of macroscopic1 amounts of matter and its thermal properties. Central to this
branch of physics are concepts such as: heat; temperature; work; pressure; energy; and
entropy. We will carefully define all of these concepts as we proceed. It is important to
recognize that all of these terms and concepts can be defined, measured and used without
any reference to the detailed inner construction or workings of the matter. In other words,
we can understand quite a lot without having an atomic or microscopic picture of matter.
In fact, many of the details of thermodynamics were worked out long before humans had
been fully convinced that matter was made from atoms.
A far younger discipline in physics, which is very closely related to thermodynamics, is
called statistical mechanics. Broadly speaking this branch of physics is concerned with
understanding thermodynamics in terms of very large collections of atoms and molecules.
For example, you are probably aware that the temperature of a substance is a kind of
‘thermal’ energy related to the average kinetic energy of its atoms or molecules. Similarly,
you may also known that the pressure exerted on the walls of a container by a gas is related
to the average force exerted by the gas particles as a result of their continuous collisions
with the walls. Statistical mechanics is a far more fundamental approach to understanding
the world for it seeks to explain the macroscopic behavior of matter in terms of statistical
properties of large collections of molecules and atoms.
In learning about heat and the properties of matter the line between thermodynamics and
statistical mechanics becomes quite blurred. We will make no attempt to separate these
two disciplines for it serves no useful purpose. The important point to understand is that
many of the ideas we will meet in this short course can be understood and approached
from either perspective.
Just like mechanics – which is based on Newton’s three laws of motion – the subject of
thermodynamics is approached and understood by introducing three basic laws: the laws
of thermodynamics. To get a broad overview of what these laws actually mean and how
the whole picture fits together let’s take a quick look at them. They are as follows:
1
Roughly speaking the term macroscopic means large enough to be seen with the unaided eye.
2
The zeroth law of thermodynamics: This law is related to the existence and definition of temperature. This law does little more than draw our attention to some
fundamental facts about pieces of matter and their thermal interactions. As we will
see, these facts enable us to define something called ‘temperature’. At first sight
this law will seem quite obvious and perhaps even unnecessary. In being so ‘obvious’
this law was originally overlooked and its importance was not recognized until after
the other two laws had been clearly formulated. Since this law relates to temperate
and its definition, logically it should be placed before the other two laws – hence its
name.
The first law of thermodynamics: This law is nothing more than a statement of the
conservation of energy. Basically it says that energy may be transferred into or
out of a substance as either work or heat, but the total amount of energy remains
constant.
The second law of thermodynamics: At a practical level this law places a restriction
on the kinds of processes that can occur in the universe. To be more specific, some
processes occur naturally all the time, whilst some processes are never observed
to occur even though they do not violate energy conservation. Take for example
making a hot cup of coffee. Some processes which naturally occur when making
coffee include: the milk mixing with the water; the sugar dissolving; the water
cooling down over time; the cup smashing into pieces if dropped. In contrast, the
time reversed processes – the milk and water un-mixing, the sugar un-dissolving, the
water getting hotter as time passes, the cup reassembling itself – are never seen to
occur in nature, although they do not technically violate energy conservation. Why
do some processes occur and others do not? It is the second law of thermodynamics
which tells us which processes will naturally or spontaneously occur and which
processes will not. The second law has often been described as the arrow of time,
since it is one of the only laws of physics which makes a distinction between past
and future, and in which order processes can unfold. You can think of the second
law as telling you which direction in time a process will occur. At a technical level
the second law is a statement about something called entropy. From a microscopic
point of view entropy is related to randomness, disorder and probability.
3
3
Temperature, thermometers and the zeroth law
At the heart of thermodynamics is the concept of temperature. We already have a pretty
good feel for what temperature is, but what precisely is it? We will address this question
directly in a moment, but first we will recall a few facts which I assume are familiar.
Roughly speaking the temperature of a substance is a physical property relating to how
hot or cold the substance is. Physicists typically measure temperature using the Kelvin
scale – the kelvin being the SI unit of temperature – which is closely related to the Celsius
scale. If the temperature of a substance is represented by TC in the Celsius scale, and as
T in the Kelvin scale, then the connection between the two is simply:
TC = T − 273.15
(3.1)
As you know, absolute zero is the coldest possible temperature. At absolute zero we
have:
Absolute zero:
T = 0K
and
TC = −273.15 ◦ C
(3.2)
As a warning, you should always use the Kelvin scale when carrying out calculations in
thermodynamics. The symbol T which appears in our equations will always mean the
temperature as measured in the Kelvin scale.
It is very important to recognise that the increments in the Kelvin and Celsius scales are
exactly same size. This is an important connection between the two scales, since it means
that the change in temperature of a substance has the same numerical value whether
measured in degrees Celsius or kelvin. For example, if you increase the temperature of a
substance by 3 ◦ C, then this increase in temperature in the Kelvin scale is 3 K. It is for
this reason that many of the parameters that we will come across in thermodynamics are
given in either of the two scales. A common example is the specific heat of water which
can be given as either 4190 J kg−1 K−1 or 4190 J kg−1 ◦ C−1 , since specific heat is related to
temperature change (i.e. the first number tells you that it will take 4190 joules of energy
to raise one kilogram of water by 1 degree Celsius, and the second numbers says exactly
the same thing: it will take 4190 joules of energy to raise one kilogram of water by 1
kelvin).
4
An instrument which provides you with information about the change in temperature of a
substance without actually giving you an accurate reading of the temperature is known as
a thermoscope. Just like a true thermometer – which is an instrument which measures
the actual temperature of a substance – a thermoscope can be constructed by exploiting
any macroscopic physical property known to change with temperature, including: length;
volume; electrical resistance; the pressure of a confined gas; the colour of a substance;
and so on.
Defined in this way, almost anything can be a thermoscope. A crude example would be a
glass of water. If you leave a glass of water alone for long enough and it ends up frozen,
then you know that the environment is at least colder than 0 ◦ C. If the frozen water later
melts, then you know that the temperature of the environment has changed, and it must
be somewhere between 0 ◦ C and 100 ◦ C.
That the concept of temperature can exist at all hinges on a few basic facts about our
universe. Before we can see what these are we need a few definitions. Firstly, in this
course we will always assume that whenever a body is heated by some means2 , then
it will undergo some kind of macroscopically measurable change (perhaps a change of
length, volume, pressure, density, whatever). Then, two bodies are said to be in thermal
contact with one another if heating just one of the bodies results in some macroscopic
change in the other. Two bodies are said to be thermally isolated from one another if
heating just one of the bodies results in no macroscopic change in the other. Two bodies
are said to be in thermal equilibrium with one another if they are in thermal contact
and no macroscopic changes occur in either body as time passes. Note that two bodies
do not actually need to be physically touching one another to be in thermal contact.
Now imagine the following very simple experiment. Suppose you have an object – let’s
call it object Z – which somehow macroscopically changes in a very obvious way with
even the smallest amounts of heating. Object Z could achieve this by perhaps undergoing
a continuous change in length, or electrical resistance, or colour, or whatever, as it is
heated. However, just for simplicity, we will imagine that Z has an accurate numerical
digital display attached to it, where the displayed numbers grow larger as Z is heated.
The digital display will clearly indicate – to a few decimal places – any changes in Z due
2
More precisely, when a body has energy added or removed from it in some way.
5
to heating. We now place Z in thermal contact with a body A and thermally isolate them
both from the rest of the universe in a perfectly insulated box (see (a) in Figure 1 below).
We now wait for a long time. Eventually we find that the numbers on Z’s display settle
down and no longer change – the reading stays fixed. If we now study the situation very
carefully, we find that no other measurable properties of either Z or A change over time.
Any property we choose to measure of either A or Z – including volume, pressure, area,
length, electrical resistance, density, or any other macroscopic measurement – will remain
fixed and unchanging. In short A and Z are in thermal equilibrium with one another.
Z
Z
A
B
HbL
(a)
Insulating box
A
B
HcL
Figure 1: (a) A and Z in thermal contact and isolated in a box. (b) B and
Z in thermal contact and isolated in a box. (c) A and B in thermal contact
and isolated in a box.
Now we move Z into another perfectly insulated box (see (b) in Figure 1) and place it
into thermal contact with a different body B. We again wait for a long time until the
numbers on Z’s display settle down and then no longer change. Again we find that if we
study the situation very carefully, all measurable properties of Z and B stay fixed over
time. Here we have B and Z in thermal equilibrium with one another.
Now, just for the sake of this thought experiment, imagine that the number on Z’s display
when it was in thermal equilibrium with A just so happens to be the same number
6
displayed when it was in thermal equilibrium with B. (This needn’t have been the case,
but we have arranged the experiment carfeully so that the numbers are the same in both
cases.) Now suppose that we carefully place A and B into thermal contact with one
another and isolate them in a box (see (c) in Figure 1). We can now ask the key question,
as soon as they are in thermal contact with each other, will A and B be in thermal
equilibrium? Or will we be able to measure changes in some of the properties of A or B
just after we place them in thermal contact? Experiment shows that the answer to this
question is that indeed A and B will be in thermal equilibrium with each other if they
were separately in thermal equilibrium with Z. This important observation about our
world leads to the zeroth law of thermodynamics:
The zeroth law of thermodynamics: If bodies A and B are each in thermal
equilibrium with a body Z, then A and B are in thermal equilibrium with
each other.
The reason this observation is important enough to be considered a law is this: if it were
not true then there would not be something common to all three bodies and we would
not be able to speak about the idea of temperature. But since this observation is true,
the thing which is common to all three bodies can be called ‘temperature’. Thus we can
say that two bodies have the same temperature if they are in thermal equilibrium with
each other.
Said in another way, imagine a hypothetical universe in which the zeroth law were not
true. Imagine that in such a universe A and B are in thermal equilibrium with one
another, and Z is in thermal equilibrium with A. Since the zeroth law doesn’t apply here,
Z will not necessarily need to be in thermal equilibrium with B. Thus if Z is then used
as a thermoscope to measure the ‘temperatures’ of A and B, the reading on the digital
display will not necessarily be the same from each. In this hypothetical universe we could
not even consistently define the concept of ‘temperature’, and so it would be practically
useless!
In short, thermal equilibrium and the zeroth law are used to define temperature. Objects
7
in thermal equilibrium have the same temperature, and objects which have the same
temperature are in thermal equilibrium. All that is left is to agree on some kind of scale
so that we can assign numbers to different temperatures. We choose the Kelvin scale.
8
4
Thermal expansion
A well known property of almost all substances is their tendency to expand or grow in
size as their temperature increases. This phenomenon is known as thermal expansion.
The amount that a body grows or shrinks when subject to a temperature change will
depend on a number of things, including the material from which the body is made –
a fact which is easily exploited to make thermoscopes and thermometers. A standard
mercury thermometer, for example, relies on the glass expanding less than the mercury
it contains as temperature increases.
To prevent buckling or breaking under the huge forces that can occur due to thermal
expansion, engineering tasks often need to be mindful of this phenomena. Examples
include expansion joints on bridges (Figure 2 (a)), gaps in train tracks (Figure 2 (b)),
cracks in footpaths, and the strange looking loop expansion joints in pipelines (Figure 3)
(a)
(b)
Figure 2: (a) Expansion joints on bridges; (b) Gaps in train tracks. [Source:
www.wikipedia.org]
9
Figure 3: An expansion loop in a low pressure steam pipe. [Source: Museum
Victoria.]
When performing calculations relating to the thermal expansion of solids and liquids there
are just a few very simple equations. These equations are sufficient for most practical
purposes.
4.1
Solids and linear expansion
When a solid changes its temperature and undergoes thermal expansion, its length, surface
area and volume all change. We often choose the length of the solid object as a measure
of how much it grows. The first of our equations relating to thermal expansion is referred
to as the equation of linear expansion. This equation describes how measured lengths
change as the temperature changes. The term ‘linear’ in the equation’s name refers to its
mathematical form.
The equation of linear expansion works as follows. Imagine that I give you a solid body,
a thermometer and a normal straight ruler. I ask you to measure the initial temperature
10
Ti of the body, and any initial length Li on the body using the ruler. It is important to
understand that the measurement Li could be any length at all: perhaps the length of
the body; or the width of the body; the diameter of a hole in the body (if it has one); or
simply the distance between two pencil marks you have arbitrary placed somewhere on it.
Now imagine I take the object and change its temperature to a new (final) temperature
Tf . The change3 in temperature is usually denoted by
∆T = Tf − Ti = temperature change.
(4.1)
Now I give you back the object and ask you to measure its new temperature Tf and, using
the ruler, again measure the length you measured before. You will find that the length
has changed due to the thermal expansion and it now has a new ‘final’ length Lf , which
is given by the equation
Lf − Li = Li α∆T
linear expansion
(4.2)
where α is a constant called the coefficient of linear expansion which has the units of
K−1 or ◦ C−1 . This equation is often written as
∆L = Li α∆T
linear expansion
(4.3)
where, ∆L = Lf − Li , is the change in the measured length. The coefficient of linear
expansion, α, depends on what material the solid body is made from. For example, steel
has a coefficient of linear expansion of 11 × 10−6 K−1 , while fused quartz has a value of
0.5 × 10−6 K−1 . Although α can actually vary with the temperature of the object, we will
not consider this complication in this course. For almost all materials α is positive, but for
some materials it may be zero or even negative. The change in length ∆L is always very
small, which is reflected by the fact that α is a tiny number. A schematic representation
of thermal expansion is shown in Figure 4.
There are a few important things to note about the equation of linear expansion (4.3).
It should be stressed that the length measured can be any chosen linear length on the
body. This means, for example, if I increase the temperature of a ring or a washer, the
3
In physics the change in the value of something – which is usually indicated by the presence of the
delta symbol ∆ – always means its final value take its initial value.
11
Li
Ti
DL
Tf
Lf
Figure 4: An exaggerated representation of the thermal expansion of a metal
bar. Tf > Ti . Notice that the width and length have both increased as the
temperature has increased.
hole gets bigger (not smaller as you might have guessed). The sign of ∆T will be negative
if the body is cooled down, and so for most materials this will mean ∆L is also negative
(meaning that the Lf < Li and the body has contracted). Perhaps the simplest way of
thinking about all this is to imagine that the process of thermal expansion is a ‘re-scaling’
of the object just like the process of re-scaling a picture on your computer – absolutely
everything gets larger or smaller by the same scaling factor.
At a microscopic level thermal expansion can be understood as the increased average
movement and separation of the constituent molecules as their kinetic energy is increased
(i.e. as the body’s temperature is increased). This helps us understand why a hole drilled
in a body will grow rather than shrink as the temperature is increased – the atoms forming
the circumference of the hole must also increase their average separation.
See ‘Worked Examples 1’ on LMS for a typical example of the use of the equation of linear
expansion.
12
4.2
Volume expansion
What if we are interested in the change in the volume of a solid body due to thermal
expansion, rather than some measured length? What equation can we use for liquids
when they undergo thermal expansion? (Only a volume change can be meaningful for a
liquid.)
The answer is, when computing volume changes of either liquids or solid bodies, we use
the following formula called the equation of volume expansion:
∆V = Vi β∆T
volume expansion, liquid or solid
(4.4)
where ∆V = Vf − Vi is the change in volume of the body, ∆T = Tf − Ti is the change
temperature of the body, and β is a constant called the coefficient of volume expansion. Vi is the body’s initial volume and Vf is the body’s final volume. This equation
is clearly very similar to that of the equation of linear expansion, and is used in a very
similar way.
In the case of a solid object we can actually relate the coefficient of linear expansion α and
the coefficient of volume expansion β. In the lectures I will demonstrate – by applying
the equation of linear expansion (4.3) to each of the dimensions of a solid object, and by
using the fact that α is small – that the following relationship between β and α holds:
β = 3α
for a solid
(4.5)
See ‘Derivation 1’ on LMS.
An important example of a substance with a negative coefficient of volume expansion is
water at 1 ◦ C. Liquid water has the curious property that under standard pressures it’s
most dense at 4 ◦ C.
It’s pretty easy to exploit thermal expansion to create a thermometer. A standard mercury
or alcohol thermometer uses it: the liquid inside the glass tube expands with increasing
temperature, whereas the glass tube itself expands very little. Many thermostats also
rely on this phenomenon through a very simple device called a bimetal strip. A bimetal
strip is just two strips of different metals (with different coefficients of linear expansion)
welded together. See Figure 5 below. As the strip heats or cools it bends toward or away
from an electrical contact allowing it to open or close an electrical circuit.
13
Copper
HaL
Ti
Steel
Copper
HbL
T f > Ti
Steel
Figure 5: (a) A bimetal strip is made from copper and steel by fixing them
together at a temperature Ti . (b) At a temperature Tf > Ti the copper has
expanded more than the steel resulting in a bend in the strip. The strip would
bend in the other direction if Tf < Ti .
14
5
Gases
5.1
Real and ideal gases. State variables
In the previous section we discussed some of the thermal properties of solids and liquids.
What about gases?
Gases are somewhat simpler to deal with since, unlike solids or liquids, they are made
from molecules or atoms which are far apart and only rarely interact.
The macroscopic state of a gas in thermal equilibrium is fully specified by its temperature, pressure and volume. In other words, if you have a sample of gas whose macroscopic
properties are not changing over time, you can give me full information about the macroscopic state of the gas by telling me just those three things – there is nothing else I need
to know.
In this course we will always use the following:
• Temperature is denoted by the symbol T and has SI units of kelvin (K)
• Pressure is denoted by the symbol p and has SI units of pascals or newtons per
square metre (Pa or N m−2 )
• Volume is is denoted by the symbol V and has SI units of cubic metres (m3 )
As a general warning, always use SI units unless you have a very good reason to do
otherwise. Many equations involving temperature require that you use kelvin – failure to
do this will result in error. Remember: if all the numbers you insert into an equation are
given in SI units, then the output will be also be in SI units.
In thermodynamics the three quantities pressure, volume and temperature (or just p,
V and T for short) are examples of what are called state variables. In general a state
variable is one of a set of macroscopic quantities which together fully specify the thermodynamic ‘state’ of the substance. Two samples of a substance which are in the same
‘state’ are identical in all respects. Basically, knowledge of the values of all state variables
will allow you to exactly reproduce the state of the substance. It will not matter how you
15
actually go about reproducing the state, once the state is reproduced the substance will
be identical to the original in all respects.
For example, say I have a sample of nitrogen gas and it has a pressure of 1000 Pa, a volume
of 0.5 m3 and a temperature of 300 K. I have just told you the thermodynamic state of
my nitrogen gas: I have given you the value of the nitrogen’s three state variables. If you
now want to reproduce an exact copy of my sample, all you have to do is make sure your
sample of nitrogen has a pressure of 1000 Pa, a volume of 0.5 m3 and a temperature of
300 K. It will not matter one bit how we each made our samples – we could have used
completely different methods – once our samples have the same values of p, V and T ,
they are said to be in the same state, and will be identical in every possible way.
In thermodynamics we say that state variables are path independent to emphasize the
fact that it doesn’t matter how or by which ‘path’ we make the state variable take on a
particular value – once the value is obtained, how we got there is irrelevant.
State variables are also known as state parameters or thermodynamic variables.
Extensive experimentation with many different gases reveals that, to a very good approximation, most gases exhibit the same simple relationship between the values of p, V and
T . This relationship is called the ideal gas law or the ideal gas equation and comes
in two equivalent forms. The first version of the ideal gas law is
pV = nRT
ideal gas law
(5.1)
where n is the number of moles of gas particles4 , and R = 8.314 J K−1 mol−1 is called the
universal gas constant (since it is the same for all gases).
The second version of the ideal gas law is
pV = N kT
ideal gas law
(5.2)
where N is the number of gas particles (molecules or atoms) and k = 1.381 × 10−23 J K−1
is called Boltzmann’s constant.
4
The term ‘gas particles’ will always refer to the individual units from which the gas is made. They
may be molecules (such as CO2 molecules in carbon dioxide gas) or atoms (such as He atoms in helium
gas).
16
Any gas for which the ideal gas law holds is said to be an ideal gas.
Notice the difference between equations (5.1) and (5.2): one uses n the number of moles
of gas particles, while the other uses N the actual number of gas particles. If you put
the two equations together you find an expression relating the universal gas constant R,
Boltzmann’s constant k and Avogadro’s number NA :
NA =
N
R
=
= 6.02 × 1023 mol−1
n
k
(5.3)
If you have a sample of an ideal gas and the number of gas particles doesn’t change
throughout a particular process – which is almost always the case – then the ideal gas
law (5.1) or (5.2) is often written as
pV
= constant
T
ideal gas law, number of particles fixed
(5.4)
This is occasionally written in the more useable form
p1 V1
p2 V 2
=
T1
T2
ideal gas law, number of particles fixed
(5.5)
where p1 , V1 and T1 are respectively the pressure, volume and temperature prior to some
process, and p2 , V2 and T2 are respectively the pressure, volume and temperature after
the process. By ‘process’ we mean that something is done to the gas, like heating and
compressing it.
There are three special cases of the ideal gas law which immediately follow from (5.5).
Case 1: In the special case where the temperature and the number of particles of the gas
doesn’t change throughout a process (i.e. T1 = T2 ) we find
p1 V1 = p2 V2
Boyle’s law
(5.6)
which is called Boyle’s law.
Case 2: In the special case where the pressure and the number of particles of the gas
doesn’t change throughout a process (i.e. p1 = p2 ) we find
V1
V2
=
T1
T2
Charles’s law
which is called Charles’s law.
17
(5.7)
Case 3: In the special case where the volume and the number of particles of the gas
doesn’t change throughout a process (i.e. V1 = V2 ) we find
p1
p2
=
T1
T2
Gay-Lussac’s law
(5.8)
which is called Gay-Lussac’s law.
The ideal gas law is often called the ideal gas equation of state since it relates the
state variables p, V and T of the gas. It shows us that sometimes state variables can be
related to one another, and that you do not always need to know all of the state variables
to be able fully specify the thermodynamic state of a substance.
The word ‘ideal’ appears in the name ‘ideal gas law’ because, strictly speaking, the law
really only applies to fictitious or theoretical ‘idealized gases’. All real gases behave
slightly differently. For real gases, depending on the gas and the particular conditions,
you often need more complicated equations relating p, V and T .
An ideal gas is a theoretical gas which consists of particles which exhibit ‘ideal’ behaviour.
In particular, you can actually derive the ideal gas law (5.1) or (5.2) from a microscopic
point of view by making a few simplifying assumptions about the gas particles, and
by using some calculus and statistics. An ideal gas is one which satisfies the following
assumptions:
1. Newton’s laws apply to the gas particles.
2. The gas particles themselves take up no volume (or are very tiny compared to the
average distance between them) and are all identical.
3. The gas particles are constantly moving in random directions with a distribution of
speeds that are independent of the direction of motion.
4. There are no attractive or repulsive forces acting between the gas particles or the
surroundings. Except for collisions between other gas particles and the container
walls, the particles are completely free. They have no potential energy, just kinetic
energy.
5. The gas particles undergo elastic collisions with each other and the container walls.
No kinetic energy is lost in collisions.
18
And so the ideal gas law works for any gas that satisfies the above assumptions. In section
5.3 below we will take a detailed look at how the ideal gas law can be derived from – or
at least be shown to be consistent with – the microscopic point of view by using just the
assumptions listed above.
See ‘Worked Examples 2’ on LMS for a typical example of how to use the ideal gas law.
The ideal gas law is only approximately correct, an approximation which works extremely
well for all real gases at very low pressures. As the pressure in a real gas is increased the
gas particles begin to interact more strongly with one another (violating some of the above
assumptions) and their behaviour diverges from ideal.
5.2
Thermal equilibrium, quasi-static processes and pV diagrams
It is extremely important to understand that a gas must be in thermal equilibrium with
its immediate surroundings for us to be able to talk about its pressure and temperature
(and often volume too). This point was not stressed in the last section, but you need to
be aware that p and T (and often V ) are not defined and cannot be measured if the gas is
not in thermal equilibrium. Thus any equation of state relating such variables – whether
the gas be real or ideal – will not be valid unless the gas is in thermal equilibrium with
its immediate surroundings. To emphasize this point, consider the following example of
an ideal gas confined to a piston and cylinder system as indicated in Figure 6.
To start with, the piston is locked in an initial position and the whole device is left alone
long enough for thermal equilibrium with the room to be established. The surrounding
room has a fixed temperature Tr . In this configuration of the piston-cylinder-gas setup
we will say that the gas is in an initial state i. We now measure the initial pressure pi ,
volume Vi and temperature Ti of the gas when in this state i. Since the gas is in thermal
equilibrium with the room, the gas will have a temperature of Ti = Tr . We now very
quickly pull the piston outward to a new final position and then lock it in place. Again
we wait until the gas reaches thermal equilibrium with the room. In this new configuration
of the piston-cylinder-gas setup we say that the gas is in a final state f . We then measure
the final pressure pf , volume Vf and temperature Tf of the gas in the state f . The final
temperature is, of course, the room temperature Tf = Tr . Using the ideal gas law, for the
19
Room temp Tr
piston

gas
cylinder
Figure 6: A piston and cylinder containing a gas
gas in state i we have
pi Vi = nRTr ,
(5.9)
pf Vf = nRTr .
(5.10)
and for the gas in state f we have
The right hand sides of both of these equations are identical and so we conclude that the
initial pressure and volume is related to the final pressure and volume via
pi Vi = pf Vf = nRTr = constant .
(5.11)
The important point to note here is that although the pressure, temperature and volume
of the gas in states i and f are all easily measured, the pressure and temperature of all
in-between configurations of the piston-cylinder-gas setup (i.e. as the piston is pulled
outward) are not even defined and cannot be measured. We can plot the gas states i and
f on a so-called pV (pressure-volume) diagram, but such a plot will just have two isolated
points as shown in Figure 7.
20
p
i
f
V
Figure 7: A pV diagram for the gas in the piston-cylinder-gas setup. The
plot displays the gas in the states i and f only because the gas was not at
equilibrium at any stage in between.
Now imagine we repeat this experiment, but this time we proceed differently. Again we
will start with the gas in the same initial state i and end with the gas in the same final
state f , but this time we pull the piston outward very slowly. In fact, we don’t even
pull on the piston continuously as we move the gas from state i to f . Instead we pull
the piston outward in many tiny incremental movements or steps, and between each step
we wait long enough for the gas to establish thermal equilibrium with the surrounding
room. In thermodynamics any process like this – i.e. a process which is carried out so
slowly that equilibrium is established at all in-between points – is called quasi-static or
quasi-equilibrium. In such a process we can measure and record – so that we can plot
them later – the pressure, volume and temperature of the gas at all states between i and
f . Clearly, at each point in-between i and f we will find the temperature of the gas to be
room temperature Tr . Using the ideal gas law, the pressure p and volume V of the gas in
all in-between states will be related by
pV = nRTr = constant
(5.12)
pV = pi Vi = pf Vf = nRTr = constant .
(5.13)
or more explicitly
Note that throughout this entire quasi-static process the temperature of the gas is kept at
21
a constant value, the room temperature Tr . A quasi-static process carried out at constant
temperature is called isothermal. We can visualize this process by plotting the volume
and pressure of the gas on a pV diagram for all states that the gas passes through as it
goes from i to f . The result is given in Figure 8 and is just a constant temperature curve
of pV = constant. If you connect together all points on a pV diagram which have the
same temperature you will form a smooth curve called an isotherm, so in this example
the pV plot is part of an isotherm.
p
i
f
V
Figure 8: A pV diagram for the gas in the piston-cylinder-gas setup. The plot
displays all states of the gas as it moves from i to f as it’s taken through a
quasi-static process at constant temperature. The curve is part of an isotherm
(a line which joins all points of a given temperature).
It is important to note here that if we chose to, we could easily take the gas from state
f back to state i quasi-statically – taking one small step at a time. That is, this overall
process is reversible since every small incremental step in the process could just as easily
be reversed. This is an important point and we will have much more to say about
reversible and irreversible processes later when we discuss entropy and the second law
of thermodynamics.
A very important concept and way of thinking has been introduced in this section of
the notes, that of the pV diagram. Plotting the progress of a gas as it quasi-statically
changes states is very common in thermodynamics. It is a convenient way to visualize
22
the thermodynamic path that the gas has taken in getting from an initial state i to a
final state f . In the quasi-static example above our gas went from i to f by following
a specific thermodynamic path, namely we forced it to expand isothermally (at constant
temperature). Generally speaking, there are an infinite number of thermodynamic paths
linking i and f , and we could have made our gas take one any of these paths by simply
altering what we did with the gas and its environment. Much more will be said about
this idea later.
With respect to pV diagrams, the concept of the isotherm is extremely useful when dealing
ideal gases. Figure 8 displays four different isotherms for an ideal gas.
p
T1 < T2 < T3 < T4
T4
T3
T2
T1
V
Figure 9: A pV diagram displaying four different isotherms of increasing temperature for an ideal gas.
5.3
The kinetic theory of gases
The ideal gas law, pV = nRT , was discovered first through extensive experimentation.
However, as mentioned above, we can make a few basic assumptions about the microscopic
behaviour of a collection of identical particles and show how the ideal gas law emerges.
We will not present the full microscopic derivation here as it would take too long and it is
a little too involved. Rather, we just want to get the flavour of how it all works without
getting lost in the mathematical details. The notion that certain macroscopic properties
of a gas – like temperature and pressure – can be derived from a microscopic point of
23
view using only a few assumptions about the gas as a collection of particles is called the
kinetic theory of gases.
In section 5.1 we listed all of the assumptions we needed to make about our gas particles
so that we can derive the ideal gas law. They were:
1. Newton’s laws apply to the gas particles.
2. The gas particles take up no volume (or are very tiny compared to the average
distance between them) and are all identical.
3. The gas particles are constantly moving in random directions with a distribution of
speeds that are independent of the direction of motion.
4. There are no attractive or repulsive forces acting between the gas particles or the
surroundings. Except for collisions between other particles and the container walls
the particles are free. They have no potential energy, just kinetic energy.
5. The gas particles undergo elastic collisions with each other and the container walls.
No kinetic energy is lost in collisions.
Although we don’t actually need to, we will make one additional simplifying assumption
so that we can avoid getting lost in the maths. We will assume that the particles do not
collide with each other, they only collide with the sides of the container.
Now imagine that we have N gas particles which satisfy all of the above assumptions and
are confined inside a cubic box of side length L. Setting up a Cartesian coordinate system
as shown in Figure 10, we will first consider the right hand wall – the wall shaded in the
Figure.
If a single gas particle of mass m is moving toward the right hand wall with a velocity with an x-component of vx , it will eventually collide and rebound with a velocity
with x-component −vx (since we have assumed all collisions are elastic). The change in
momentum of this particle in the x-direction must be
∆px = −mvx − mvx = −2mvx .
24
(5.14)
y
m
x
L
L
L
z
Figure 10: A cube of side length L containing an ideal gas. One of the N gas
particles is shown along with its direction of motion.
This particle must strike the shaded wall repeatedly forever, traveling a distance in the
x-direction of 2L between each collision. The time ∆t between collisions with the wall is
therefore
∆t =
2L
.
vx
(5.15)
Since the rate of change of momentum is force – Newton’s second law5 – this molecule
will experience an average6 force in the x-direction, F̄x , from the wall of
F̄x =
2mvx
∆px
mv 2
=−
=− x.
∆t
2L/vx
L
(5.16)
Thus the magnitude of the average force supplied to the wall by all N molecules will be
F̄x,total
2
2
2
mvx1
mvx2
mvxN
=
+
+ ... +
L
L
L
m 2
2
2
vx1 + vx2 + . . . + vxN
=
L
(5.17)
where vxi is the x-component of the velocity of the ith particle.
The pressure p exerted on the shaded wall is the total average force divided its area L2 :
p=
5
F̄x,total
m 2
2
2
=
v
+
v
+
.
.
.
+
v
.
xN
x1
x2
L2
L3
(5.18)
Newton’s second law for a particle with momentum p~ is F~ = d~
p/dt. Only if the mass of the particle
doesn’t change do we have the more familiar form of the law: F~ = d(m~v )/dt = md~v /dt = m~a.
6
The line over the top of a symbol means that an average of some kind has been taken.
25
This value of p will also be the same pressure exerted on each of the walls of the container
– it is the pressure of the gas inside.
Now, writing the average value of the square of the x-component of the speeds of the
particles as
vx2 =
2
2
2
vx1
+ vx2
+ . . . + vxN
N
(5.19)
and noting that V = L3 is the volume of the gas, we find
p=
mN 2
v
V x
(5.20)
For any gas particle, the square of its speed v 2 is given by
v 2 = vx2 + vy2 + vz2 .
(5.21)
One of our original assumptions was that the gas particles are constantly moving in
random directions with a distribution of speeds that are independent of the direction of
motion. Since all of the particles are moving in random directions, the average values of
the squares of their velocity components must be equal, and so we have vx2 = vy2 = vz2 .
The square of the average speed for all the gas particles in the container, v 2 = vx2 +vy2 +vz2 ,
will therefore be
1
vx2 = v 2
3
(5.22)
Inserting this last expression into (5.20) gives
p=
mN 2
v .
3V
(5.23)
The square root of the quantity v 2 is a measure of the average speed of the particles in
the gas and is often called the root-mean-squared speed or thermal speed and is
given the symbol vth :
vth =
p
v2 .
(5.24)
Rearranging the expression (5.23) gives
2
pV = N
3
26
1 2
mv
2
(5.25)
where we have placed the average translational kinetic energy of the particles in brackets.
This looks very much like the ideal gas law and we have arrived at the result from just
a few assumptions and mechanics. The only real difference between our last expression
above and the ideal gas law is that there is no mention of temperature anywhere in our
expression. If we now compare our result with the ideal gas law, pV = N kT , we find that
they will be the same if the average translational kinetic energy per particle Kave,trans is
3
1
Kave,trans = mv 2 = kT .
2
2
(5.26)
Thus, a collection a particles obeying Newton’s laws and a few other assumptions will
satisfy the ideal gas law and the temperature of the gas is to be identified with the average
kinetic energy associated with the random translational motion of the particles. Quite
remarkable is the fact that it doesn’t matter what the mass m of the particles is – it’s
the average translational kinetic energy per particle Kave,trans which is to be related to the
temperature.
Rearranging the last equation and taking the square root yields
r
3kT
vth =
,
m
(5.27)
an expression for the thermal speed of the particle in terms of the gas temperature and
particle mass.
5.4
Total energy in an ideal monatomic gas
For later developments it will be very important to consider how much energy in total is
to be found within a gas. In the simplest case of an ideal gas consisting of many identical
point-like particles, the total energy within the gas is just all of the translational kinetic
energy carried by the particles. An ideal gas consisting of point-like particles is said to
be an ideal monatomic gas since the gas particles are just free atoms (like He atoms),
rather than something more complicated like diatomic molecules (like H2 molecules) or
polyatomic molecules (like H2 O molecules). The reason this is important will be discussed
later, where it will be necessary that we distinguish between monatomic, diatomic and
polyatomic ideal gases.
27
In the simplest case of a monatomic ideal gas, the only energy available within the gas –
all we could ever extract from it – is to be found within the translational kinetic energy of
each of the particles. By our original assumptions the particles do not have any potential
energy. Thus, in this case, the total energy within the gas is the average kinetic energy
per particle multiplied by the number of particles N :
3
3
total energy = N × Kave,trans = N kT = nRT
2
2
(5.28)
with n the number of moles and R the universal gas constant (having used equation (5.3)).
28
6
6.1
Heat
The system and the environment
It is extremely important that we now clearly define the terms ‘system’ and ‘environment’,
for they will play a major role in all that follows.
In the most general situation in thermodynamics we restrict our attention to just a specific
part of the universe. This ‘part of the universe’ might be anything at all – it is just the
thing that we are interested in. It may be a gas inside a container, or the water in a
cup, or a chicken – it doesn’t matter. The part of the universe that we are interested in
is called the system. Everything else around the system which may interact with it in
some way is called the environment. The two dimensional surface which separates the
system and the environment – whether it be real or imagined – is called the boundary.
As a simple example consider a freshly made cup of coffee. If we are interested in the
thermodynamics of a cup of coffee cooling down over time, we consider the cup of coffee
to be the system, and the surrounding air in the room to be the environment.
As a second example consider the cylinder-piston-gas setup discussed in the last section of
these notes. In that example the gas inside the cylinder was the system and together the
piston, cylinder and room formed the environment. The system (the gas) went from an
initial state i to a final state f . Generally speaking, the procedure by which you change
a system from one state to another state is called a thermodynamic process.
This very simple idea of a system and environment is shown schematically in Figure 11.
The system is the thing of interest and the environment is everything else.
Broadly speaking, systems will be either isolated, closed or open, which are defined below:
Open system: A system without any restrictions. It can exchange energy and matter
with its environment.
Closed system: A system which can exchange energy with its environment, but not
matter.
Isolated system: A system which cannot interact in any way with its environment. No
energy or matter can be exchanged with its environment.
29
system
environment
Figure 11: The system and its surrounding environment.
A few words to clarify these definitions. Firstly, later we will learn that the energy which
can be exchanged between the environment and a system comes in two forms: heat and
work. Secondly, by ‘matter exchange’ we mean mass of some kind flowing across the
boundary between the system and its environment. The total mass of closed and isolated
systems are therefore constant.
Open systems are usually very complicated and will not be considered in this course. The
human body is an example of an open system.
The cooling cup of coffee is a closed system (ignoring evaporation).
Strictly speaking, isolated systems are not found anywhere in nature, but very good
approximations are found in many situations. Hot coffee in a tightly sealed thermos flask
sitting on a desk, for example, is approximately an isolated system since it is thermally
isolated from its environment and no exchange of energy and matter can occur.
6.2
What is heat?
In physics ‘heat’ and ‘temperature’ are technical terms. They are completely different
concepts and should not be confused. We will now define heat.
Consider a general situation of a non-isolated system and its surrounding environment,
and imagine that at some specific instant we measure their temperatures. Let us denote
the temperature of the system as Ts and the temperature of the environment as Te . We
know from experience that, if left alone, there are three possibilities:
30
Case I: If Ts = Te then nothing happens. The system and environment are in thermal
equilibrium.
Case II: If Ts > Te then the system is hotter than the environment and energy will be
transferred from the system to the environment.
Case III: If Ts < Te then the system is colder than the environment and energy will be
transferred from the environment to the system.
In short, energy is transferred (or flows) from a hotter thing to a colder thing. The
energy transferred across the boundary between a system and its environment due to a
temperature difference between them is called heat. (Note that it is extremely common
in the physics literature to say ‘the heat transferred’ rather than just ‘heat’.)
Since heat is a transfer of energy it is measured in SI units of joules (J). The amount of
heat transferred into a system from the environment is denoted by Q. Thus if Ts < Te
then heat energy flows into the system and Q > 0. If Ts > Te then heat energy leaves
the system and Q < 0. If no heat energy transferred then Q = 0. All this is shown
schematically in Figure 12:
Ts = Te
Q=0
system
Ts
environment
(a)
Ts > Te
Q<0
system
Ts
Te
environment
(b)
Q
Te
Ts < Te
Q>0
system
Ts
environment
Q
Te
(c)
Figure 12: Heat transfer. (a) Thermal equilibrium and no heat is transferred.
(b) The system is hotter than the environment and heat energy (indicated by
the arrow) is transferred to the environment. (c) The system is colder than
the environment and heat energy (indicated by the arrow) is transferred to the
system.
31
It is important to understand that in many situations the temperature of the system
and the environment will change as heat energy is transferred, but this is not necessarily
always the case. There are many examples of physical situations where adding heat to a
system has no effect on the system’s temperature. For instance you can continuously add
heat to water when it’s boiling without changing its temperature.
And of course heat is not the only kind of energy transfer that is possible, since we can
also transfer energy to a system by doing work on it. For example you can increase the
temperature of your hands by rubbing them together. As you may already know, work
is the energy transferred between a system and its environment due to a force acting over
some displacement. The work done on a system is denoted by W and is measured in
joules. In the cylinder-piston-gas example we met earlier, when the piston was slowly
pulled upward the gas performed work on the piston, since the gas placed a force on the
piston (due to the pressure of the gas) as the piston moved. We’ll have much more to say
about work later.
To conclude this subsection we put some of the above technical terms together: During a
thermodynamic process which takes a system from an initial state to a final state, energy
is transferred between the system and the environment as either work W or heat Q or
both.
6.3
The heating of solids and liquids
Experimentally we know that heating a solid or liquid usually results in a temperature
change. If you add an amount of heat Q to a given sample of a solid or liquid (the system)
then the temperature change ∆T of that sample is given by the very simple formula
Q = C∆T = C(Tf − Ti )
heat capacity
(6.1)
where Tf and Ti are the initial and final temperatures of the sample and C is a constant
which depends on the sample. Since only the temperature difference appears in this
formula, Tf and Ti can be measured in either ◦ C or K. The constant C of proportionality
between the heat added and the temperature change is called the heat capacity of the
sample and has units of J/K or J/ ◦ C. The constant C can be determined experimentally.
32
The heat capacity is a positive constant and so the sign of Q and ∆T is the same. This
means that if we remove heat energy from a sample (Q < 0 by definition), then ∆T < 0
and so the sample’s temperature will decrease.
Equation (6.1) is of little use to us in practice. However, the following observation makes
life simpler: something which is made from the same material but has twice the mass will
require twice the heat Q to bring about the same temperature change. In other words,
the heat capacity is proportional to the mass of the sample. So instead of (6.1) we use
Q = mc∆T
specific heat
(6.2)
where: Q is the heat added to the sample; ∆T = Tf − Ti is the temperature change of the
sample; m is the mass of the sample; and c is called the specific heat of the material from
which the sample is made. The specific heat c depends on what material we are dealing
with and is measured in J kg−1 K−1 or J kg−1 ◦ C−1 . For obvious reasons the specific heat
of a material is also known as the heat capacity per unit mass.
The specific heat of a material can be determined experimentally. Liquid water, for example, has a specific heat of 4184 J kg−1 K−1 , whereas ice has a specific heat of 2050 J kg−1 K−1 .
You can look up the specific heat of other common materials in the textbook.
Often it can be more convenient to use the molar specific heat of a substance (otherwise
known as the heat capacity per mole), which we will denote by C. Molar specific heat has
SI units of J mol−1 K−1 or J mol−1 ◦ C−1 . When using the molar specific heat we use
Q = nC∆T
molar specific heat
(6.3)
to relate the heat Q transferred and temperature change of the substance, where n is the
number of moles of the substance.
See ‘Worked Examples 3’ on LMS for a typical example of how to use the specific heat
equations.
It is of some importance to realize that the specific heat can depend on the conditions
under which the energy is transferred as heat. Firstly, the specific heat may vary slightly
with temperature and so it is not truly a constant. Secondly, the specific heat also
depends on whether the substance being heated changes its volume or internal pressure.
33
We actually find that a substance will have two distinct specific heats: a specific heat
when the substance is heated at constant pressure, and a different specific heat when the
substance is heated at constant volume.
We almost always heat solids and liquids at constant pressure (usually atmospheric), but
it is possible to heat them under constant volume conditions by increasing the external
pressure in such a way that the solid or liquid cannot thermally expand. Although it is
experimentally very difficult to arrange heating without volume changes for solids and
liquids, theoretical calculations reveal that the difference between the specific heats at
constant pressure and specific heats at constant volume is negligible for solids and liquids.
Gases however, are a completely different story. It makes a big difference as to whether
you heat a gas at constant pressure or volume – the specific heats for these two processes
are not the same, and you need to know which one to use. We will return to this issue
later.
6.4
Heating and phase transformations
As mentioned above, when we heat a substance the temperature doesn’t necessarily need
to change. Instead the energy can go into changing the phase of the substance rather than
changing its temperature. By a phase change or phase transition we mean the state
of the substance (i.e. solid, liquid or gas) is altered. More specifically, when undergoing
a change of phase a substance can melt, freeze, vaporize, condense, sublimate, or deposit
(desublimate), and when doing so will take up or release energy.
Very roughly, whenever we heat something, the energy transferred can either go into: (i)
increased kinetic energy of the consistent atoms or molecules, which registers as an increase
in temperature; (ii) overcoming attractive forces between the molecules (i.e. increasing
the potential energy shared between molecules), which registers as a phase transformation;
or simultaneously both (i) and (ii).
For pure substances (i.e. substances made from only one type of molecule or atom), (i)
or (ii) can occur but never both at the same time. That is, for pure substances the
temperature remains constant as it changes phase. For substances which are mixtures
(i.e. not pure) the behaviour can be far more complicated and will not be considered here.
34
For pure substances, a typical heat added vs temperature plot – known as a heating curve
– is shown in Figure 13.
Temperature H°CL
gas phase
BP
liquid and gas HboilingL
MP
solid phase
solid and liquid HmeltingL
liquid phase
Heat supplied HJL
Figure 13: A typical heating curve for a pure substance. The plot shows that
the temperature of the substance remains constant at its melting point (MP)
and boiling point (BP).
As shown in Figure 13, when a pure substance is heated, its temperature will rise until it
reaches a phase transition temperature (i.e. a melting, boiling or sublimation point). At
this stage further added heat energy will melt, boil or sublimate the substance whilst the
temperature remains constant. For a pure substance at a phase transition temperature,
the amount of heat Q added to the sample (the system) is related to the amount of the
substance m (in kilograms) which transforms (melts, boils or sublimates) via:
Q = mL
heat of transformation
(6.4)
Here L is a constant depending on what the substance is made from and which phase
transformation is occurring. The constant L is generally called the heat of transformation. It has SI units of J/kg.
If the phase transformation is from solid to liquid, the constant L is denoted by Lf
and is called the heat of fusion or heat of melting. If the phase transformation is
from liquid to gas, L is denoted by Lv and is called the heat of vaporization. If the
35
phase transformation is from solid to gas, L is denoted by Ls and is called the heat of
sublimation.
The equation Q = mL can actually be used to give either the heat absorbed by the
substance in bringing about melting, boiling or sublimating, or the heat energy released in
the reverse process (i.e when the substance freezes, condenses or desublimates). However,
it is important to keep in mind that since the symbol Q is defined as the heat transferred
into a system from the environment, we must have Q > 0 if heat is going into the system
and Q < 0 if heat is being removed. Thus we must be careful to add in a negative sign
by hand when using Q = mL for a system which freezes, condenses or goes straight from
the gas to the solid phase.
See ‘Worked Examples 4’ on LMS for a typical example of how to use these ideas.
6.5
Phase diagrams
In general the occurrence of a phase transformation of a substance will depend on temperature and pressure. For instance, at standard pressure (1 atmosphere or 101.3 kPa)
water boils at 100 ◦ C, whereas at the top of Mount Everest where atmospheric pressure
is much lower (about 26 kPa), water will boil at about 69 ◦ C.
A phase diagram is a simple means of showing under which conditions a substance
undergoes phases transitions, and under which conditions the substance is a solid, liquid
or gas. Typically a phase diagram is a plot – on a graph of pressure vs temperature – of
the conditions under which liquid/gas, liquid/solid and gas/solid are in equilibrium. A
typical phase diagram for a substance is shown in Figure 14. The so-called triple point
is a point on the plot under which all three phases are in thermal equilibrium7 . The
so-called critical point is the point at which the boundary between the gas and liquid
phases terminates. At temperatures and pressures above the critical point there ceases to
exists a distinction between gas and liquid8 . A fascinating demonstration in one of the
lectures will clearly show what happens at the critical point for a substance called diethyl
ether.
7
8
For water the triple point occurs at 273.16 K and 611.73 Pa.
For water the critical point occurs at 374 ◦ C and 22 MPa.
36
Pressure
Melting
Solid
Critical point
Liquid
Gas
Triple point
Sublimation
Boiling
Temperature
Figure 14: A phase diagram showing solid, liquid and gas phases. The curves
represent conditions under which phases are in equilibrium.
Figure 15 shows the following typical phase transitions:
constant pressure melting:
solid −→ liquid
(f −→ g)
constant pressure freezing:
liquid −→ solid
(g −→ f )
constant pressure sublimation:
solid −→ gas
(e −→ d)
constant pressure deposition:
gas −→ solid
(d −→ e)
constant temperature boiling:
liquid −→ gas
(a −→ c)
constant pressure boiling:
liquid −→ gas
(a −→ b)
constant temperature condensing:
gas −→ liquid
(c −→ a)
constant pressure condensing:
gas −→ liquid
(b −→ a)
A video (see LMS) will be shown in the lectures which demonstrates the remarkable
properties of cornstarch. This substance is a liquid at room temperature and atmospheric
pressure, but is readily seen to become solid under greater pressures at room temperature.
The phase transition occurring in this substance is shown in Figure 16.
37
Pressure
Liquid
Solid
g
f
a
b
Gas
c
e
d
Temperature
Figure 15: A phase diagram showing various phase transitions.
Pressure
Liquid
Solid
b
a
Gas
Temperature
Figure 16: A phase diagram for cornstarch. The liquid/solid phase transition
induced by a pressure change at room temperature is shown as a ←→ b
38
The phase diagram of water is shown in Figure 17. The Figure shows that water is an
unusual substance as its solid/liquid equilibrium curve ‘bends’ in the opposite direction
compared to most other substances. The liquid/solid phase transition which allows ice
skates to melt ice and thus slide on a thin layer of liquid water is shown in Figure 17.
The increased pressure on the ice due to the weight of the person through the blade of
ice skate temporarily melts the ice.
Pressure
Liquid water
b
Ice
a
Water vapour
Temperature
Figure 17: A phase diagram for water. The solid/liquid phase transition induced by a pressure change at constant temperature is shown as a ←→ b. Such
a phase transition occurs under the blade of an ice skate.
6.6
Heat transfer mechanisms
We now conclude this section of the notes on heat by briefly discussing the different
mechanisms by which energy is transferred as heat.
Recall our earlier definition of heat: the transferred energy between a system and its
environment due to a temperature difference between them. Here we are concerned with
exactly how this energy is transferred.
We will consider three ways in which energy is transported from a hotter body to a colder
body: conduction, convection and radiation. In simple situations we need only consider
39
one of these, whilst more complicated situations my require that we take two or all of
them into account.
6.6.1
Conduction
Generally speaking, thermal conduction is heat transfer by direct physical contact.
More specifically, at the microscopic level, conduction is an energy transferred from more
energetic particles to less energetic particles via continuous collisions between them. Two
simple examples will help make this idea more concrete.
Conduction Example 1: If you place two objects with different temperatures in physical contact with one another and thermally isolate them from the rest of the universe,
they will eventually reach thermal equilibrium. This will be achieved by energy being transferred from the hotter body to the colder body by conduction.
Conduction Example 2: If you hold one end of a metal spoon and place the other end
over a flame, before long the handle of the spoon will become far too hot to hold.
Energy is transferred from the flame to the handle of the spoon by conduction along
the length of the spoon.
The underlying microscopic reason for why conduction occurs is easy to understand if
you recall that at the microscopic level the temperature of a body is nothing more than
the average kinetic energy of the particles which make up the body. So, in Conduction
Example 1 above the hotter body consists of particles (i.e. atoms and/or molecules) which
have on average more kinetic energy than those which make up the colder body. When
the two bodies are placed in direct physical contact the particles of each of the bodies
are then able to collide with one another. On average, over time, the collisions between
the particles of the hotter body will transfer kinetic energy to the particles of the colder
body. Over time the average kinetic energies of the particles of both bodies will be the
same, thermal equilibrium will be established, and their temperatures will be equal.
In Conduction Example 2 we first note that the end of the spoon which is placed in the
flame becomes hot due to conduction (i.e. this end of the spoon is a cold body in direct
contact with the flame which is a hot body). Once the end of the spoon in the flame gets
40
hotter – which means that the atoms and electrons in this part of the spoon have more
kinetic energy than the atoms and electrons in other parts of the spoon – the continuous
collisions of the particles within the spoon will pass energy along the length of the spoon.
Let’s now consider a very simplistic but useful example of this process. Consider a slab9
of some material which has a face area A and thickness L. The two faces are maintained
at constant temperatures Tc and Th by placing the faces it in direct contact with thermal
reservoirs10 with these temperatures, as indicated in Figure 18.
L
k
Q
Hot reservoir
Cold reservoir
Th
Tc
Figure 18: A cross-sectional view of a slab in thermal contact with hot and cold
thermal reservoirs at temperatures Th and Tc with Th > Tc . Heat Q flows at a
steady rate through the slab from the hot face to the cold face via conduction.
This setup is in a steady-state which means that, as a consequence of conduction, heat
energy will steadily flow through the slab from the hot face to the cold face at a constant
rate. Experimentation shows that if Q is the heat energy exchanged between the faces in
a time t, then the rate of heat flow or conduction rate H from the hot face to the cold
9
10
The term ‘slab’ here is taken to mean a rectangular prism.
In thermodynamics a thermal reservoir or heat reservoir or heat bath is an idealized body which
will maintain a constant temperature when placed in thermal contact with some system of interest. In
practice a thermal reservoir could be any body which has such a large heat capacity that its temperature
is effectively constant when in thermal contact with the system of interest. For example, a huge body
of water at say 300 K would effectively stay at 300 K if we placed a hot metal spoon in contact with it.
Alternatively we might think of a thermal reservoir as a kind of ‘hot plate’ whose temperature we can
fix at any value of our choosing.
41
face is given by
H=
Th − Tc
Q
= kA
t
L
heat conduction of a slab
(6.5)
The quantity k here is a constant called the thermal conductivity, has the SI units of
W m−1 K−1 , and depends on the material from which the slab is made from. You can look
up values of k for various substances in the textbook. The higher the value of k, the faster
energy is transferred from the hot face to the cold face of the slab. Materials with higher
values of k are said to be better thermal conductors than materials with lower values of
k.
Often one defines the thermal resistance R of a slab of material to be
R=
L
kA
(6.6)
which has SI units of K/W. The thermal resistance is not a property of the material
from which the slab is made, but rather depends upon the specific details of the slab in
question. The higher the thermal resistance of a slab, the better an insulator it is. In
terms of thermal resistance, equation (6.5) becomes
H=
Th − Tc
.
R
(6.7)
We can now extend this picture to two slabs, both having the same face area A, but being
made from different materials and having different lengths, as shown in Figure 19.
This setup is also in a steady-state which means that the temperature of any fixed point
somewhere within one of the slabs will remain constant as time passes. It also means that
the rate of heat flow H1 through slab 1 must be equal to the rate of heat flow H2 through
slab 2: H1 = H2 . (If this was not the case energy would accumulate somewhere between
the slabs and we would not have a steady-state.) So, the overall rate of heat flow H from
the hot face to the cold face is the same as the rate of heat flow through either of the two
slabs H = H2 = H1 . If we let the temperature at the interface of the two slabs be Tx , we
find that
H = k1 A
Th − Tx
Tx − Tc
= k2 A
L1
L2
(6.8)
k1 L2 Th + k2 L1 Tc
,
k2 L1 + k1 L2
(6.9)
which allows us to find Tx
Tx =
42
L1 L2
k1
k2
Q
Cold reservoir
Hot reservoir
Th
Tx
Tc
Figure 19: A cross-sectional view of two slabs made of different materials
in thermal contact with one another, and each in contact with a thermal
reservoir. Heat Q flows at a steady rate from the hot reservoir to the cold
reservoir through both slabs via conduction. The faces of the two slabs which
are in contact both have a constant temperature Tx , where Th > Tx > Tc .
which in turn, can be eliminated from (6.8) to give
H=
A(Th − Tc )
.
(L1 /k1 ) + (L2 /k2 )
(6.10)
This result can be further extended to a set up with n slabs all with the same face area
A, sitting between two thermal reservoirs as in Figure 20.
The rate of heat conduction between the hot and cold face is:
A(Th − Tc )
H = Pn
i=1 (Li /ki )
(6.11)
where Li and ki is the thickness and thermal conductivity of the ith slab.
6.6.2
Convection
In physics the term fluid is a technical term which means liquid or gas. (You are immersed
in a fluid right now!) The heat transfer mechanism of convection only occurs in fluids.
Generally speaking, when a fluid comes into thermal contact with a substance whose
temperature is hotter than the fluid’s, convection will transfer the energy away from the
43
L2
Li
Ln
k1 k2
ki
kn
L1
Hot
Q
reservoir
Cold
reservoir
Th
Tc
Figure 20: n slabs made of different materials in thermal contact with one
another between two thermal reservoirs. Heat Q flows at a steady rate from
the hot reservoir to the cold reservoir via conduction.
substance and around the fluid in the following steps (think of heating water on a hot
plate if you want a specific example):
1. The temperature of the part of the fluid in thermal contact with the hot substance
will increase.
2. In getting hotter, the part of the fluid which is in thermal contact with the hot
substance will thermally expand and decrease in density.
3. Having a lower density than the surrounding fluid, the hotter part of the fluid will
rise due to buoyant forces.
4. Some of the surrounding cooler and more dense fluid will move in to replace the
upward moving hotter fluid.
5. Some cooler fluid is now in thermal contact with the heat source and the process
repeats.
In this way, a ‘pattern of movement’ in the fluid is established: hot fluid rises and is
replaced with cooler fluid and so on. A macroscopic fluid flow will be established by
convection which is called a convection current. You witness convection currents all
44
the time when you watch smoke rise above a fire: the smoke particles are drawn upward
with the hotter fluid.
In the above steps we have described convection where energy is taken away from a hot
substance when in thermal contact with a cooler fluid. For completeness it should be
mentioned that the reverse also readily occurs. More specifically, when a fluid comes into
thermal contact with a substance whose temperature is colder than the fluid’s, energy can
be added to the substance and taken from the fluid via conduction. You should be able
to describe the steps by which this occurs (for example, it relies on the fact that cooler
fluid sinks because it has a higher density than the surrounding warmer fluid.)
The convection heat transfer mechanism is extremely important in may natural processes,
including climate and weather. The main mechanism for the cooling of a hot cup of coffee,
for example, is through convection. In fact, the cooling of hot cup of coffee involves two
fluids: the water and the air. The air directly above the hot water warms and rises,
drawing in cool air etc, and all the while removing energy from the coffee. The water at
the surface of the coffee is simultaneously cooled and sinks, allowing hotter water to rise
and come into contact with the air. If you wish to keep your coffee warmer for longer,
simply interrupt the convection cooling process by placing a lid on the cup.
The process of heat transfer by convection requires that buoyant forces be present. You
may want to read more about the origin of these forces in the textbook. Interestingly,
buoyant forces do not exist in the absence of a gravitational fields or if the fluid is in
free-fall. Thus, heating or cooling utilising convection is not possible on the international
space station!
6.6.3
Radiation
The final mechanism by which a system and its environment can exchange energy as heat
is through electromagnetic radiation or light. Often the radiation exchanged between
two bodies – or a system and its environment – due to a temperature difference is called
thermal radiation.
Unlike the other methods of heat transfer, the radiation mechanism allows bodies to
exchange energy without physical contact and even when separated by a vacuum: no
45
medium is required to transmit this energy. This is because electromagnetic waves (light)
require no medium to propagate.
The most obvious example of this is the sun heating the earth. More personally, when
you feel the heat of the sun on your face, you are experiencing heat transfer by radiation.
All bodies with a temperature will emit some radiation. The rate Pem at which an object
will emit energy via electromagnetic radiation depends on the object’s surface area A and
its temperature T (in kelvin) in accordance with the Stefan-Boltzmann law:
Pem = eσAT 4
Stefan-Boltzmann law
(6.12)
Here Pem is the total power emitted by the object in watts, σ = 5.67 × 10−8 W m−2 K−4 is
the Stefan-Boltzmann constant and e is the emissivity of the object’s surface. The
emissivity is a dimensionless number which has a value between 0 and 1 depending on
the composition of the object. A perfect emitter – a so-called blackbody radiator – has
an emissivity of e = 1. Blackbody radiator’s are idealized objects which are not actually
found in nature.
The rate Pabs at which an object will absorb energy via electromagnetic radiation from its
environment, assuming that the environment has a uniform (the same in all directions)
temperature Tenv , is given by
4
Pabs = eσATenv
(6.13)
where again A is the surface area of the object and e the emissivity (the same value used
in (6.12)). A blackbody radiator with e = 1 is also a perfect absorber and will absorb all
radiation falling upon it, rather than reflecting any.
Considering an object which simultaneously emits and absorbs electromagnetic radiation
via (6.12) and (6.13), we find the object’s net exchange of electromagnetic radiation, Pnet ,
is given by
4
Pnet = Pabs − Pem = eσA(Tenv
− T 4) .
(6.14)
Note that Pnet may be positive or negative depending on whether the object is gaining
or losing energy. When the object is in thermal equilibrium with the environment (i.e.
T = Tenv ) we find that Pnet = 0 and no energy is gained or lost as expected.
See ‘Worked Examples 5’ on LMS for examples of the use of the heat transfer mechanism
equations.
46
7
7.1
The first law of thermodynamics
Work
Before we can finally meet the first law of thermodynamics, we first need to discuss the
idea of the work done on or by a system. We briefly mentioned the idea of work earlier
when we first encountered the definition of heat. In section 6.2 we first defined heat as
an energy exchange between the system and environment due to a temperature difference
between the two. It was then briefly mentioned that this was not the only kind of energy
transfer that is possible, we can also transfer energy to a system by doing work on it.
Although both are measured in the SI units of joules, these two types of energy transfer
– work and heat – are distinct from one another and you must clearly understand the
difference. Briefly:
Heat: Energy transfer due to temperature difference.
Work: The mechanical transfer of energy due to a force acting over a displacement.
As an example of energy transfer due to work, imagine that I vigorously stir a cup of
water with a spoon – both at room temperature. After a while the water will have a
higher temperature – the water molecules have increased their average kinetic energy.
This energy transfer was not due to heating since both the spoon and the water are at
the same temperature. The energy transfer is a result of the spoon doing work on the
water. The same comment applies to rubbing your hands together – they will increase
their temperatures not through heat exchange, but through mechanical work.
In the most general case in thermodynamics where we have a system and its environment,
we denote the work done on the system by the environment as W and the amount of
heat transferred into the system from the environment by Q. Carefully note that in
being defined this way, either of the two separate quantities Q and W might be positive,
negative or zero depending on what exactly is going on.
47
7.2
The first law of thermodynamics
During a specific thermodynamic process where we change the state of the system from
an initial state i to a final state f , we have two distinct means be which we can transfer
energy from the environment to the system: heat and work. If an amount of heat Q is
transferred to the system as the state changes from i to f and an amount of work W is
done on the system during as the state changes from i to f , then the total energy received
by the system is Q + W .
Now, we could actually bring about the change of state of the system from i to f in any
number of different ways. For example we might be able to bring about this change of
state by purely heating the system and performing no work on it, in which case W = 0.
Or we might be able to bring about the same change of state by only performing work (i.e.
with no heating so Q = 0). For a specific example, think of increasing the temperature
of a glass of water by 5 ◦ C: we could just heat it with a flame, or just stir it vigorously.
In general there will be an infinite number of ways in which we could bring about the
change of state of the system from i to f , each with their own different Q and W values.
It is perhaps surprising to find out that the value of the quantity Q + W , which is the
total energy transferred to the system, is the same for all processes which take the system
from i to f .
In the jargon of thermodynamics, we describe a specific way in which the system gets
from i to f by saying the system has followed a specific thermodynamic path. In getting
from from i to f in different ways we say the system follows different paths. Thus we
refer to the observation that quantity Q + W does not depend on the actual path taken
by the system as path independence.
In other words, the quantity Q + W depends only on the initial and final states of the
system, and not the specific process by which the system gets from i to f . This is telling
us that the quantity Q+W must represent a change of some intrinsic or ‘internal’ property
of the system. This internal property must depend only on the state of the system – it
is a state variable. We call this state variable the internal energy of the system and
denote it by the symbol U . The name internal energy is very suggestive, and you should
think of the internal energy as the total energy (kinetic plus potential) stored within the
48
system.
So, if a system starts in an initial state i and has an initial internal energy Ui , and then
has its state changed via some thermodynamic process and ends up in a final state f with
a final internal energy Uf , then the change in internal energy of the system is
∆U = Uf − Ui = Q + W
(7.1)
where Q is the heat transferred to the system and W the work done on the system during
the process.
Equation (7.1) is called the first law of thermodynamics, which can be stated in words
as:
The first law of thermodynamics: The change of internal energy of a system
is the sum of the heat added to and work done on the system, and depends
only on the initial and final states of the system and not on the particular
process involved.
Clearly equation (7.1) is nothing more than the conservation of energy: if the system
starts off with a total energy Ui , and you add an amount of energy Q + W , the total final
energy Uf must be Uf = Ui + Q + W .
If the system undergoes an infinitesimally small change in state (i.e. the system is taken
from one state to an infinitesimally different state), then the first law of thermodynamics
can be written in infinitesimal form:
dU = dQ + dW
(7.2)
where dU is the infinitesimal change in internal energy of the system, dQ is the infinitesimal heat added to the system during the change and dW is the infinitesimal work done
to the system during the change. (If the idea of ‘infinitesimals’ is new to you, you are
to imagine that equation (7.2) means exactly the same as (7.1), except that it applies to
final and initial states which are very close to one another).
If we consider a process which moves a system between finitely separated11 states as being
made up of infinitely many infinitesimal steps, we can use the infinitesimal version of the
11
‘Finitely separated’ means the two states are not infinitesimally close.
49
first law (7.2) to obtain the total change of the internal energy, ∆U , for the process.
All we need to do is integrate equation (7.2) over the thermodynamic path taken in the
process.
7.3
Internal energy of an ideal gas
You will recall from section 5.4 that we derived an expression for the total energy within
a monatomic ideal gas. In that case we argued that the only energy available within the
gas – all we could ever extract from it – was to be found in the translational kinetic energy
of each of the particles. From our original assumptions the particles did not have any
potential energy. Thus we already have an expression for the internal energy for an ideal
monatomic gas, it is just the expression (5.28):
3
U = nRT
2
internal energy, monatomic ideal gas
(7.3)
with n the number of moles of the ideal monatomic gas and R the universal gas constant.
Therefore, for any process whatsoever, the change of internal energy of an ideal monatomic
gas is just
3
∆U = nR∆T
2
monatomic ideal gas, any process
(7.4)
where ∆T is the temperature change of the gas for the process.
Carefully note from the above expression that the internal energy depends only on the
temperature (and not on the pressure or volume). The above expression can be generalized to incorporate all ideal gases and later we will give the expression for diatomic and
polyatomic ideal gases.
As a consequence, it is extremely important to understand the following statement:
The internal energy of any ideal gas is a function only of its temperature and
does not depend on any other variable. The internal energy of any ideal gas will
not change if its temperature does not change.
50
The reason for this is quite simple: since the gas is ideal its molecules do not have any
potential energy. The internal energy of the ideal gas is to be found only within the
kinetic energy of the gas particles. So, if in a particular process the gas doesn’t change its
temperature the total available kinetic energy carried by the gas particles will not change,
and therefore the internal energy will not change.
A pV diagram illustrating this idea is given in Figure 21. In this Figure we show an ideal
gas starting in a state i at a temperature Ti . The gas is then taken by three separate
processes to three different final states f all with the same temperature Tf , but with
different volumes and pressures. Since the change of temperature ∆T = Tf − Ti is the
same for all three processes, the change of internal energy for each process is the same.
p
f
3
f
1
f
2
Tf
i
Ti
V
Figure 21: Two isotherms at temperatures Ti and Tf are shown by dashed lines.
The three paths shown, labeled 1, 2 and 3, illustrate three processes what take
an ideal gas from a state i with temperature Ti a state f with temperature
Tf > Ti . Although each process is different and finishes at different volumes
and pressures, the internal energy change for each of the processes is the same.
51
7.4
Work done on a gas
Most often in thermodynamics we are interested in the work done on a gas during some
thermodynamic process, so let’s discuss this situation in detail.
Consider any gas (ideal or real) that is confined to a piston-cylinder set-up as shown
in Figure 22 below. The cylinder walls are filled with a perfect insulating material so
that the gas is thermally isolated from the room12 . The bottom of the cylinder is in
thermal contact with a thermal reservoir which has an adjustable temperature which can
be set to whatever temperature we want. In terms of the terminology introduced earlier,
the gas inside the cylinder is the system, and the thermal reservoir and piston form the
environment.
insulated cylinder wall
piston
gas
thermal reservoir
with variable temperature
Figure 22: The cross section of a gas confined to a cylinder with movable
piston. The cylinder walls are perfect insulators, allowing no heat exchange
with the surrounding room. The gas is in thermal contact with the thermal
reservoir which has an adjustable temperature.
12
The piston itself is also assumed to be made from a perfect insulating material.
52
We will now take this gas through a thermodynamic process from an initial state i to
some final state f . As usual, when in the initial state i, the volume, temperature and
pressure of the gas will be denoted by Vi , Ti and pi respectively. Similarly, when in the
final state f , the volume, temperature and pressure of the gas will be denoted by Vf , Tf
and pf respectively.
The thermodynamic process we will be considering will be quasi-static so that at all stages
between states i and f the gas is in thermal equilibrium with the thermal reservoir. As
such, at all stages between the states i and f , the volume, temperature and pressure of
the gas can be measured. In practice this will require us to carry out the process very
slowly. At each step during the process we will need to: (i) move the piston only a very
small amount; (ii) adjust the temperature of the reservoir by only a very small amount;
and (iii) wait until the gas has established thermal equilibrium with the reservoir before
we can adjust the piston or temperature of the reservoir again. At any arbitrary state
between i and f we will denote the volume, temperature and pressure of the gas by the
variables V , T and p respectively.
In moving the system (the gas) from the state i to f it will absorb a total amount of heat
Q from the environment (the reservoir), and will have a total amount of work W done
on it by the piston. As given by the first law of thermodynamics the change of internal
energy ∆U of the gas during this process is
∆U = Q + W
(7.5)
where, as usual, the change of internal energy is ∆U = Uf − Ui with Uf the internal
energy of the gas in state f and Ui the internal energy of the gas in state i.
Let us now determine the work done on the gas by the piston up to some arbitrary stage
within this quasi-static process. Consider an arbitrary point during the process where the
volume of the gas is V and its pressure is p. Let the length of the cylinder confining the
gas at this stage be L, and the cross-sectional area of the piston in contact with the gas
be A, as shown below in Figure 23.
53
piston area A
L
gas
Figure 23: The cross section of a gas confined to a cylinder with movable piston
at some arbitrary point during a process. The length of the cylinder confining
the gas is L and the cross-sectional area of the piston in contact with the gas
is A.
If we now make a minute change to the volume of the gas by moving the piston outward
an infinitesimal distance dL, what infinitesimal amount of work is done by the gas? Since
work is computed by force times displacement, and the force F exerted on the piston over
the minute distance dL is
F = pA
(7.6)
then the infinitesimal amount of work done by the gas is:
infinitesimal work done by gas = F dL = pAdL .
(7.7)
Since the infinitesimal change in volume of the gas in this case is just
dV = AdL
(7.8)
infinitesimal work done by gas = pdV .
(7.9)
we can write:
54
We therefore conclude that the infinitesimal work done on the gas, dW , is
dW = −pdV
infinitesimal work done on a gas
(7.10)
Here the negative sign reflects the fact that if the volume change13 dV is negative then
the volume has decreased in which case the gas has positive work done on it. If we now
add up (i.e. integrate) all the infinitesimal amounts of work done dW at each stage during
the change of state from i to f we will compute the total work W done on the gas. Since
the pressure p may continuously change during the process we end up with:
Z
Z Vf
pdV
total work done on a gas
W = dW = −
(7.11)
Vi
It should be stressed here that integral must be carried out over the specific thermodynamic path which starts in state i and finishes in state f . Different ways of getting from i
to f will yield different amounts of work W done on the gas, and so W depends upon what
process is carried out. More specifically, we will need to know exactly how the pressure of
the gas p changes as a function of volume V (i.e. we need to know p as a function of V ,
which we write as p(V ), for the path in question). We will see exactly how to compute
W for various paths in a moment.
Finally, it should also be stressed that the expressions (7.10) and (7.11) are always true
for any gas as it undergoes any kind of quasi-static volume change (and not just for the
special case of a gas enclosed in a cylinder). We will not attempt to prove this general
result here.
7.5
Work done on a gas as an ‘area’ under a pV curve
In the most general case of a gas being taken from a state i to f via some arbitrary
quasi-static process, the work done on the gas can easily be seen to be related to the area
beneath the curve that the gas traces out on a pV diagram as it goes from i to f . If we
assume that Vf > Vi so that the gas has expanded, then the curve that a gas traces out
on a pV diagram as it goes from i to f might look something like that shown in Figure
24. In this case, since the work done on the gas is given by
Z
Z Vf
W = dW = −
pdV ,
Vi
13
Recall that change in volume means final volume take the initial volume.
55
(7.12)
it follows that the work is nothing more than the negative of the ‘area’ beneath the pV
curve traced out by the gas.
p
i
f
work = -area
V
Figure 24: The work done on a gas in going from i to f where Vf > Vi is the
negative of the shaded ‘area’ beneath the pV curve traced out by the gas.
If we assume that Vf < Vi so that the gas has contracted, the work done on the gas is
again computed by (7.12), and we find that this value is just the ‘area’ beneath the pV
curve traced out by the gas. See Figure 25.
p
i
f
work = area
V
Figure 25: The work done on a gas in going from i to f where Vi > Vf is the
shaded ‘area’ beneath the pV curve traced out by the gas.
It should now be clear that if the gas is taken from a state i to f via a quasi-static process
which is made up from a series of separate stages – an example of such a case is given in
56
Figure 26 – then the work done on the gas in going from i to f is simply the sum of the
work done on the gas in each of the of separate stages.
p
3
2
4
5
f
i
1
work = - total area
V
Figure 26: A quasi-static process taking a gas from i to f via five separate
stages labeled 1 to 5. The work done on a gas in going from i to f where
Vf > Vi is the negative of the total shaded ‘area’ beneath the pV curve.
In the special case where the gas is taken through a cycle, which means that the state i is
the same as the state f (i.e. the gas starts and ends in the same state), the work done on
the gas is related to the ‘area’ enclosed by the pV curve. If the gas traces out a clockwise
path, the work done on the gas is just the negative of the ‘area’ enclosed by the pV curve
– See Figure 27. If the gas traces out a counter-clockwise path, the work done on the gas
is just the ‘area’ enclosed by the pV curve.
p
f
i
W = - area enclosed
V
Figure 27: A quasi-static process taking a gas through a cycle starting and
ending at the same point (i.e. i and f are the same point). In this case the
path is a clockwise direction and so the work done on the gas is the negative
of the ‘area’ enclosed by the curved.
57
7.6
Different paths – same change of internal energy
As was indicated above, if we take a gas from a state i to f , the change of internal energy
is just ∆U = Uf − Ui . If we take this gas from i to f via different quasi-static paths, the
change of internal energy will always be the same value no matter what the actual process
is since internal energy is just the total energy stored in the gas. However, the work W
done on the gas and the heat Q added to the gas do depend on the actual process or path
taken. For this reason W and Q are said to be path-dependent quantities, whilst ∆U is
said to be a path-independent quantity.
In Figure 28 we see, on the same pV diagram, three examples of different quasi-static
processes which take a gas from i to f .
p
a
b
2
i
1
c
3
f
V
Figure 28: Three different quasi-static processes – labeled 1, 2 and 3 – which
take a gas from i to f . The work done on the gas is each process is the negative
of the ‘area’ below the curve.
In each process shown in Figure 28 the internal energy change ∆U will be the same value,
but we can easily see that the work done on the gas W , which is the negative of the ‘area’
under the curve, is different for each path. Since the first law of thermodynamics holds,
∆U = Q + W ,
(7.13)
we can therefore conclude that the heat added must also be different for each path. We
say that heat and work are path-dependent quantities, whereas the internal energy is
path-independent.
58
The processes labeled 2 and 3 shown in Figure 28 could easily be achieved using a gas confined to an insulated cylinder with a piston and removable and adjustable heat reservoir
by following the series of steps outline below. The gas starts in state i with temperature Ti , pressure pi and volume Vi , and is initially thermally isolated from the universe.
(Remember these are quasi-static processes, so we need to proceed very slowly.)
Process 2: First lock the piston in place fixing the volume at Vi . Next place the gas in
thermal contact with the reservoir which is set to a temperature Ti . Very slowly
increase the temperature of the heat reservoir until the pressure of the gas has
reached point a in the diagram. Now unlock the piston so it can move. Incrementally
draw the piston outward – adjusting the temperature at each step, making sure that
the pressure of the gas remains constant – until the volume has reached point b in
the diagram. Now lock the piston in place fixing the volume at Vf . Slowly decrease
the temperature of the heat reservoir until the pressure of the gas has reached point
f in the diagram.
Process 3: First lock the piston in place fixing the volume at Vi . Next place the gas in
thermal contact with the reservoir which is set to a temperature Ti . Very slowly
decrease the temperature of the heat reservoir until the pressure of the gas has
reached point c in the diagram. Now unlock the piston so it can move. Incrementally
draw the piston outward – adjusting the temperature at each step until, making sure
that the pressure of the gas remains constant – until the volume has reached point
f in the diagram.
See ‘Worked Examples 6’ on LMS for a simple example which brings many of these ideas
together.
In the next chapter we will closely examine some special cases of quasi-static processes
and determine the work done and heat transferred to the gas in each case.
59
8
8.1
Gases and special cases of the first law
Constant volume process and specific heat
If a gas is heated or cooled and the volume doesn’t change (for example the gas is confined
to a vessel with fixed volume), we speak of a constant volume process, also known as an
isochoric, isometric or an isovolumertic process (‘iso’ means equal). It should be clear
that a gas cannot have any work done on it if its volume does not change. Thus, for any
constant volume process whatsoever we will always have the work done on the gas being
zero:
W =0
(8.1)
constant volume process
The first law of thermodynamics therefore tells us that for a constant volume process
∆U = Q
(8.2)
constant volume process
A quasi-static constant volume process appears as a straight vertical line on a pV diagram,
an example of which is shown in Figure 29.
p
f
i
V
Figure 29: An example of a quasi-static constant volume process. No work
can be done in such a process since there is zero area under the curve.
The amount of heat added to a sample of n moles of gas during a constant volume process
is computed via the following equation
Q = nCV ∆T
constant volume process
60
(8.3)
where ∆T = Tf − Ti is the change in temperature of the gas and CV is a constant known
as the molar specific heat of the gas at constant volume. The value of CV will
depend on what gas is being considered.
So, for a constant volume process we have
∆U = nCV ∆T
8.1.1
constant volume process
(8.4)
Constant volume processes and ideal gases
For a monatomic ideal gas we earlier deduced that
3
∆U = nR∆T
2
monatomic ideal gas, any process
(8.5)
and so combining this with equation (8.4) gives
3
CV = R = 12.5 J mol−1 K−1
2
monatomic ideal gas
(8.6)
Later we will come up with expressions for CV for diatomic and polyatomic ideal gases.
It is extremely important to appreciate that for any ideal gas undergoing any process
from i to f (not necessarily a constant volume process) the change of internal energy is
always just
∆U = nCV ∆T
ideal gas, any process
(8.7)
We do not even require the process to be quasi-static because the change of internal energy
is process independent. That the above equation holds for any ideal gas for any process
is a remarkable result which should be clearly understood.
Consider an arbitrary process where an ideal gas starts in some initial state i and ends in
some final state f . This process could be anything at all. We know from the first law of
thermodynamics that the change of internal energy depends only on the initial and final
states of the system and not on the particular process involved. Thus, we can compute
the change of internal energy by imagining that we take the gas from i to f in a simple
two step process where we know exactly the change of internal energy in each step. The
steps are as follows (see Figure 30):
61
p
Example of an actual process
i
isotherms
Step 1 (isochoric)
Step 2 (isothermal)
Ti
f
Tf
V
Figure 30: To compute the change of internal energy for any ideal gas undergoing any process which starts in state i to f (an example of such a process is
shown), we imagine the gas moves from i to f via the two step process shown.
Step 1 is a constant volume process; and Step 2 is a constant temperature
process.
Step 1: At a constant volume Vi , heat or cool the gas so that its moves between the
isotherms from a temperature Ti to Tf . Since this is a constant volume process, in
this step W = 0 and Q = nCV ∆T . Therefore, in this step, the change of internal
energy is just nCV ∆T .
Step 2: At a constant temperature of Tf , take the gas from volume Vi to a state with a
volume of Vf . The gas is now in the final state f . (In other words, alter the volume
of the gas from Vi to Vf by moving it along an isotherm.) Since the gas is ideal and
the temperature is constant, the change of internal energy in this step is zero14 .
The total change of the internal energy in going from i to f in this two step process is
the sum of changes in internal energies in each step. Therefore the change of internal
energy for the imagined two step process is just ∆U = nCV ∆T . Since the change of
internal energy depends only on the initial and final states of the system and not on
14
Why? Remember the text in the box on page 50?
62
the actual process, we immediately conclude that for any process the change of internal
energy between the states i and f for the ideal gas is just (8.7).
8.2
Constant pressure process and specific heat
Let’s take a look at how to use the equation
Z
Z
W = dW = −
Vf
pdV
(8.8)
Vi
in the simplest non-trivial case where a gas expands at constant pressure. In thermodynamics a constant pressure process is generally known as an isobaric process. Such
processes are very common since any process exposed to the atmosphere will be carried
out at constant (atmospheric) pressure.
Consider a gas starting in a state i and ending in a state f as shown in Figure 31.
p
i
f
V
Figure 31: A constant pressure process.
As shown in the Figure, this gas simply expands from i to f at some constant pressure p
(the line joining all points of constant pressure is called an isobar). An example of this
would be a gas being heated at atmospheric pressure. Since the pressure p of the gas is a
constant for the entire process, the p can be taken outside the integral in equation (8.8),
and the work done on the gas is just
Z Vf
Z
W =−
pdV = −p
Vi
Vf
h i Vf
dV = −p V
= −p(Vf − Vi )
Vi
Vi
63
and so
W = −p∆V
constant pressure process
(8.9)
where ∆V = Vf − Vi is the change in volume.
The amount of heat added to a sample of n moles of gas during a constant pressure
process is computed via the following equation
Q = nCp ∆T
constant pressure process
(8.10)
where ∆T = Tf − Ti is the change in temperature of the gas and the constant Cp is the
molar specific heat of the gas at constant pressure.
Note that unlike liquids and solids, when we heat gases we need to be careful about which
specific heat to use. If the process is a constant pressure process we must use Cp , whereas
if it is a constant volume process we must use CV .
8.2.1
Constant pressure processes and ideal gases
Let us now consider an ideal gas undergoing a constant pressure process and apply all
that we have learned from the above discussions.
Firstly, an ideal gas will satisfy the ideal gas law pV = nRT . So, if it experiences a
constant pressure process at a pressure p we can write
Vi =
nRTi
,
p
Vf =
nRTf
p
(8.11)
and so the work done on the gas is just
W = −p(Vf − Vi ) = −p
nRTf
nRTi
−
p
p
= −nR∆T .
(8.12)
We also know that for any process for an ideal gas the change of internal energy is given
by
∆U = nCV ∆T .
(8.13)
So, for a constant pressure process for an ideal gas we have the following:
Q = nCp ∆T ,
W = −nR∆T ,
64
∆U = nCV ∆T .
(8.14)
Inserting these expressions into the first law of thermodynamics, ∆U = Q + W , gives
nCV ∆T = nCp ∆T − nR∆T
(8.15)
which, upon dividing through by n∆T , immediately leads to a relationship between the
molar specific heats at constant pressure and volume for ideal gases:
CV = Cp − R
(8.16)
Cp = CV + R .
(8.17)
or
This is a remarkable result that always holds for all ideal gases.
8.3
Molecular specific heats and degrees of freedom
Earlier we established that for any ideal gas the change of internal energy for any process
is
∆U = nCV ∆T
ideal gas, any process
(8.18)
In the specific case for a monatomic ideal gas we found that its molar specific heat at
constant volume was
3
CV = R
2
monatomic ideal gas
(8.19)
and using the general ideal gas result Cp = CV + R, we see that its molar specific heat at
constant pressure is
5
Cp = R
2
monatomic ideal gas
(8.20)
For real gases we can measure CV quite easily. We confine the gas to a fixed volume,
heat it, measure how much the temperature changes ∆T as a known amount of heat Q is
added, and use
Q = nCV ∆T
constant volume process
(8.21)
The value for CV for real monatomic gases, such as helium and argon, is very close to
the predicted value of 3R/2. However, when we heat real gases which consist of diatomic
65
and polyatomic molecules – such as nitrogen or water vapour respectively – at constant
volume and measure their specific heats CV we find they do not equal 3R/2.
The central reason for this is that as we add energy to an ideal gas which consists of
molecules with some kind of extended structure – rather than just a single point or sphere
– the energy is seen to not only go into moving the particles around faster (the so-called
translational kinetic energy15 ) but also into making them spin and perhaps even vibrate.
Thus, a gas can ‘store’ its internal energy in a number of different ways if the molecules
have some kind of structure.
The number of ways in which a particle of gas can store the internal energy is accounted
for by a theorem called the equipartition of energy theorem. This theorem basically says
that as we add energy to a gas, the energy will be divided equally amongst each of the
different ways in which the internal energy can be stored within the gas. Before we can
state this theorem precisely we first need a concept from classical mechanics known as
the number of degrees of freedom of an object.
Quite simply, the number of degrees of freedom of an object is the minimum number
of parameters or ‘coordinates’ that you need to give to be able to fully specify the object’s
location and orientation in space. For example, if the object is a point or a sphere you
will need only 3 parameters to fully specify its position – for example you could use the
standard choice of x, y and z coordinates. A point or sphere therefore has 3 degrees of
freedom. See Figure 32. The degrees of freedom associated with the x, y and z coordinates
are called the translational degrees of freedom.
Now imagine an idealized dumbbell or diatomic molecule like object, which consists of
two points separated by some fixed distance. How many degrees of freedom will it have?
A moments reflection will show you that you will need 5 parameters to fully specify the
location and orientation of a dumbbell. You will need 3 parameters to locate the centre of
the dumbbell (for example its x, y and z coordinates – 3 translational degrees of freedom),
and another 2 parameters (for example angles – called rotational degrees of freedom) to
indicate exactly how the dumbbell is orientated at that position in space. See Figure 33.
15
Translational kinetic energy means the energy associated with the linear speed.
66
y
x
3
2
1
z
Figure 32: Three parameters – labeled 1, 2, and 3 – are required to fully specify
the position of a point or sphere.
y
5
4
x
3
2
1
z
Figure 33: Five parameters – labeled 1 to 5 – are required to fully specify the
position and orientation of dumbbell.
67
Finally, an idealized polyatomic molecule consisting of three or more atoms (not in a line),
will have 6 degrees of freedom. You will need 3 parameters to locate the centre of the
molecule and another three angles to indicate its orientation in space.
If we include the modes of vibration that individual molecules can undergo, the number of
degrees of freedom increases further still. We will not consider such complications in this
course since experimentation shows that the vibrational modes can be ignored in most
cases.
Having defined the number of degrees of freedom in this way we can now state the equipartition of energy theorem:
Equipartition of energy: On average a molecule will have an energy of 12 kT
for each degree of freedom.
We are encouraged to think about each of the degrees of freedom f of a molecule as an
independent means of storing the internal energy. Thus the total internal energy U of a
ideal gas which consists of N molecules each with f degrees of freedom will be
1
U = number of molecules × number of degrees of freedom × kT
2
1
1
U = f N kT = f nRT
2
2
(8.22)
where
monatomic ideal gas:
f =3
diatomic ideal gas:
f =5
polyatomic ideal gas:
f =6
We immediately see that this agrees with our previous results for a monatomic ideal gas
(7.3).
Again we note that the internal energy of an ideal gas depends only on the temperature.
When undergoing some process, any ideal gas will have an associated change of internal
energy of
1
∆U = f nR∆T = nCV ∆T
2
ideal gas, any process
68
(8.23)
which gives
f
CV =
R
2
8.4
ideal gas, f degrees of freedom
(8.24)
Isothermal process for an ideal gas
An isothermal process is a process carried out at constant temperature.
Consider a sample of an ideal gas which undergoes an isothermal quasi-static expansion
or contraction from a state i to a state f . (This just means that our ideal gas expands
or contracts very slowly at a constant temperature.) Such a process could be achieved,
for example, by placing the sample of ideal gas inside an insulated cylinder-piston set-up
which sits on a thermal reservoir with a fixed temperature T , and slowly withdrawing or
inserting the piston.
As was stressed earlier, the internal energy of any ideal gas is a function only of its
temperature. Thus, when our sample of ideal gas undergoes an isothermal quasi-static
expansion or contraction from i to f at a constant temperature T , the internal energy
change must be zero:
∆U = 0
ideal gas, constant temp process
(8.25)
This fact along with the first law of thermodynamics, ∆U = Q + W , tells us
Q = −W
ideal gas, constant temp process
(8.26)
That is, if there is no change in the internal energy, the heat added to the gas must be
equal to the negative work done on the gas.
Since our gas is ideal it must always satisfy the ideal gas law pV = nRT . Using this we
can easily determine the work done on the gas by inserting it into
Z Vf
W =−
p dV .
(8.27)
Vi
More specifically, since the pressure of the ideal gas changes with volume
p=
nRT
V
69
(8.28)
we can insert it into the work integral to get
Z Vf
nRT
W =−
dV .
V
Vi
(8.29)
Since the temperature T is constant in the process being considered, as is the number of
moles of gas particles n, we can bring them outside the integral and integrate:
Z Vf
h
iVf
1
Vf
dV = −nRT ln V
= −nRT ln
.
W = −nRT
Vi
Vi
Vi V
(8.30)
We therefore arrive at a final set of expressions for an ideal gas undergoing an isothermal
expansion or contraction:
Q = −W ,
∆U = 0 ,
8.5
W = −nRT ln
Vf
Vi
.
(8.31)
Adiabatic processes
An adiabatic process is any thermodynamic process in which there is no heat exchanged
between the system and the environment, i.e.
Q=0
adiabatic process
(8.32)
Such a process can occur if the system is thermally isolated from the environment, or if
the process is so fast that there is simply no time for heat exchange. The latter is not a
quasi-static process and its progress cannot be plotted on a pV diagram.
For any adiabatic process the first law tells us that the change in internal energy for the
process is just the work done on the gas
∆U = W
8.5.1
adiabatic process
(8.33)
Adiabatic process for an ideal gas
If we are dealing with ideal gases there is a whole lot more we can say about adiabatic
processes. We already know that an ideal gas always satisfies the ideal gas law pV = nRT .
In addition to this law, if the number of particles is fixed, it turns out that for any ideal
gas undergoing an adiabatic process the following relationship between the pressure and
70
the volume of the gas also holds:
pV γ = constant
adiabatic process, ideal gas
where
γ=
Cp
CV
ratio of specific heats
is the dimensionless ratio of the molar specific heats for the gas and is sometimes called
the adiabatic constant. See ‘Derivation 2’ on LMS for a derivation of this equation. The
ratio γ = Cp /CV depends on whether the ideal gas is monatomic, diatomic or polyatomic:
γ=
5
≈ 1.67
3
γ=
7
= 1.4
5
γ=
4
≈ 1.33
3
monatomic ideal gas
diatomic ideal gas
polyatomic ideal gas
(8.34)
(8.35)
(8.36)
If the adiabatic process is carried out quasi-statically we can plot the progress of the gas
on a pV diagram. The curve will be a plot of
p=
constant
.
Vγ
(8.37)
A curve connecting all points on a pV diagram such that Q = 0 is said to be an adiabat.
An adiabat is steeper than an isotherm16 , and crosses through isotherms as shown in
Figure 34. Figure 35 shows three adiabats passing through the same point for monatomic,
diatomic and polyatomic gases.
In practice when an ideal gas starts in a state i and is taken to another state f via an
adiabatic process we use the equation
pV γ = constant
adiabatic process, ideal gas
(8.38)
in the form
pi Viγ = pf Vfγ
16
adiabatic process, ideal gas
Remember an isotherm is a plot of pV = constant.
71
(8.39)
p
isotherms
adiabat
V
Figure 34: Adiabat and isotherms.
p
monatomic
diatomic
polyatomic
V
Figure 35: Adiabats for monatomic, diatomic and polyatomic ideal gases.
72
If one is interested in how the final and initial temperatures of the gas are related we can
insert the ideal gas law into (8.38) and show
nRT
V γ = constant
V
adiabatic process, ideal gas
(8.40)
which leads to
T V γ−1 = constant
adiabatic process, ideal gas
(8.41)
or – if you know the initial and final values – the equivalent form
Ti Viγ−1 = Tf Vfγ−1
adiabatic process, ideal gas
(8.42)
These equations help us understand why, for example, when we open a soft-drink bottle
or spray a CO2 fire extinguisher around, a ‘fog’ is momentarily produced. As we open the
drink bottle or spray the fire extinguisher – which is sealed under high pressure – the gas
inside quickly expands, doing work on the atmosphere. The process is so fast that the
gas has little time to absorb any heat, and so we have an adiabatic process. Since Q = 0,
but work is done by the gas on the atmosphere, the first law tells us that the internal
energy of the gas must drop. The temperature must therefore also drop, and in doing so
it condenses any water vapor into tiny droplets thus forming the fog. The last equation
above lets us understand this process at a quantitative level (if Vi < Vf for a gas, then
Tf < Ti ).
If an adiabatic process is carried out quasi-statically on an ideal gas we can compute the
work done on an the gas as follows. During the process we know that
pV γ = G = pi Viγ = pf Vfγ
where G is some constant. We insert this into
Z Vf
W =−
p dV ,
(8.43)
(8.44)
Vi
which gives
W =−
=
=
Vf
h V 1−γ iVf
GV −γ dV = −G
1 − γ Vi
Vi
!
1−γ
1−γ
GVf − GVi
γ−1
Z
pf Vf − pi Vi
.
γ−1
73
(8.45)
8.6
Free expansion
The final process we will consider is called ‘the free expansion of a gas’. Unlike the other
processes described above, this process cannot be carried out quasi-statically and cannot
be plotted as a continuous curve on a pV diagram. A free expansion of a gas is both
adiabatic and one in which no work is done. Since Q = 0 and W = 0 for a free expansion,
the first law tells us that the internal energy must remain unchanged:
Q = 0,
W = 0,
∆U = 0
free expansion
(8.46)
For an example of a free expansion, consider the following. Place a balloon filled with a
gas into a sealed and thermally isolated box which contains nothing but a vacuum. If we
now pop the balloon, the gas will undergo a free expansion: Q = 0 since no heat energy
is exchanged by the gas with anything; and W = 0 as no work is done by the gas as it
expands since it meets no resistance as it does so.
Another more practical example which is easier to carry out in a lab is to connect two
insulated boxes by a tube. The tube contains a valve which can be opened or closed to
allow or prevent the flow of gas down the tube between the boxes. See Figure 36. Now
close the valve and fill only one of the boxes – the other is evacuated. On opening the
valve, the gas will freely expand to fill both boxes.
If should be clear that if the gas undergoing the free expansion is ideal, its temperature
in state i and f is the same:
Ti = Tf
free expansion, ideal gas
(8.47)
This result can be understood in two ways: (i) by thinking about the gas particles and
how they move apart but do not change the energy they carry; or (ii) by recalling that
the internal energy of an ideal gas depends only on temperature, and that internal energy
is not changed in a free expansion. Since the ideal gas law holds and the temperature
remains constant, we can relate the initial and final pressures and volumes via Boyle’s
law
pi Vi = pf Vf
free expansion, ideal gas
74
(8.48)
system (gas)
valve
vacuum
insulation
Figure 36: The free expansion of a gas. Opening the value allows the gas to
freely expand to occupy both boxes.
It should be clear from these examples that the process of free expansion is not quasistatic17 . In a free expansion the gas is only at equilibrium at the initial and final states:
the volume and pressure of the gas at any stage after the balloon is popped, but before
the gas has fully expanded to fill the box, are not defined. One therefore cannot plot
the progress of such a process on a pV diagram, only the initial and final points can be
plotted.
8.7
Final comments on problem solving and summary
It is hopefully clear at this stage that if a gas undergoes some process from i to f , where
the entire process is built from a series of steps, with each step being one of the above
special processes (isothermal, adiabatic, free expansion, etc), then you can compute the
total heat Q added to the system, the total work W done on the system or the total
change of internal energy of the system, as the system goes from i to f , by simply adding
together the heat, work and internal energy change for each of the individual steps.
17
It is possible to conceive of a highly impractical and contrived situation in which you might be able
to construct a quasi-static version of a free expansion, but we’ll ignore this possibility.
75
Below is a table which serves to summarize some of the material presented in this chapter.
Defining
Other
Process
characteristic
First law
Work
relationships
Isochoric
V = const
∆U = Q
W =0
Q = nCV ∆T
Isobaric
p = const
∆U = Q + W
W = −p∆V
Q = nCp ∆T
Isothermal
T = const
0=Q+W
Adiabatic
Q=0
∆U = W
W = −nRT ln
W =
Vf
Vi
pf Vf −pi Vi
γ−1
pV = const
pV γ = const
T V γ−1 = const
Free expansion
W =Q=0
∆U = 0
W =0
−
It should be noted that some of the equations appearing in the above table apply in the
most general settings, while others only apply to idea gases. You should make sure you
know which is which.
See ‘Worked Examples 7’ on LMS for some examples of how to use the equations of this
chapter.
76
9
The second law of thermodynamics, entropy and
heat engines
9.1
Reversible and irreversible processes
So far we have spent much time discussing quasi-static processes: thermodynamic processes which are carried out incredibly slowly by making small incremental changes to a
system’s environment. These processes are carried out in such a way that the system and
environment are effectively in equilibrium at all points in time. An important feature of a
quasi-static processes is that it is reversible18 : if a system starts in a state i and is taken
through a quasi-static processes to a state f , we can easily reverse the process and take
the system back from f to i. The reason that all quasi-static processes can be reversed is
due to the fact that they are carried out by making small incremental steps, all of which
are reversible.
The whole idea of a quasi-static or reversible process is mostly just a theoretical device
that we use to carry out calculations or draw pV diagrams etc. Such processes are rarely
if ever seen in nature. Most processes in nature are irreversible: one way processes where
a system cannot be changed back into its original state by making small incremental
changes to its environment. The simple act of cooking an egg is a common example: the
egg cannot be uncooked by reversing the steps taken to cook the egg. Once cooked, the
egg stays cooked. Mixing colours of paint is another example of an irreversible process
– we cannot unseparate the colours. A cup falling to the floor and smashing is another
example – once broken there is no fixing it. Most processes in nature only go one way.
The irreversibility of almost all processes around us is intimately linked with our psychological notion of a direction to the flow of time. Everyone agrees that there seems to be
a forward direction and a backward direction to the flow of time, and this idea is closely
associated with our continuous observation of irreversible processes. For instance, you
can almost always tell if a video is being played backwards, and when you can you are
witnessing a recording of an irreversible process. If you cannot tell if the video is being
18
We are taking the terms ‘quasi-static’ and ‘reversible’ to mean exactly the same thing, although
technically speaking they are not. The distinction is not important for us.
77
played backwards or forwards, then you are probably watching a rare reversible event.
The recording of an event using a video and playing it backwards helps drive home the
fact that we naturally take irreversible processes for granted. If an irreversible process
were to spontaneously (all by itself) go the wrong way, we would be amazed and would
notice immediately.
Place a small droplet of food colouring into a glass of water. The food colouring will
spread out and uniformly colour the water. Never do we observe the reverse process
whereby some uniformly spread out food colouring gathers itself together into a nice tiny
droplet. Clearly this is another irreversible process.
The simple process of a hot cup of coffee cooling down is also an irreversible process:
we would be astonished if a cold cup of coffee spontaneously reheated to nearly boiling
point. Note that the conservation of energy – the first law of thermodynamics – would
not prohibit the cold coffee from heating up spontaneously. The cup of coffee cooling
down to room temperature or heating up to boiling point are both allowed by energy
conservation – its just the energy will have flowed ‘in the wrong direction’ if the coffee
heats up. What laws of nature tell us that one of these directions of energy flow is wrong,
and the other is ok? The first law of thermodynamics clearly does not set the direction of
a thermodynamic process. As far as the first law of thermodynamics is concerned there
is no direction to time.
We need another law to help us understand the difference between the two types of
processes reversible and irreversible. We need another law to tell us which direction a
given process is allowed to spontaneously occur, and in which direction it is forbidden
to spontaneously occur. This law is the second law of thermodynamics. The direction
in which a thermodynamic process will spontaneously occur is determined by a quantity
called entropy.
The term spontaneous is key to understanding exactly what we mean by irreversible.
After after an irreversible process has occurred we can often return the system to its
original state by means of significant effort, but the point is that the original state will
never be seen to re-emerge spontaneously. Additionally, the ‘significant effort’ required
to re-establish the initial state will itself involve irreversible processes. For example,
78
imagine you place two identical copper cubes – one at 20 ◦ C and the other at 60 ◦ C – in
thermal contact with one another, in a vacuum, inside a thermally insulated container.
You then leave them alone. Eventually thermal equilibrium will be established and both
cubes will be at 40 ◦ C. This process – one cube heating up and the other cooling – is an
irreversible process. The original configuration of this set-up (one cube being at 20 ◦ C and
the other at 60 ◦ C) will never spontaneously arise if we leave two 40 ◦ C cubes together in
the container. We could, however, return this physical set-up to its original configuration
through significant effort: we open the container, place one of the cubes in a fridge and
the other on a hot plate, separately return the cubes to their original temperatures, and
place them back in the container. However, we can never get the cubes back to their
original temperatures without performing some kind of action. This is telling us that
something in the universe has changed when the irreversible process has occurred, and
we can never go back. This something is called entropy.
9.2
Isolated systems and entropy
The internal energy of a system is a measure of a quantity which is intrinsic to the system.
The value of the change of internal energy, ∆U = Uf − Ui , is independent of the actual
thermodynamic process which brought about the change. Or, said in other words, ∆U
depends only on the initial and final states of a system and not on the path taken during
the process. Depending only on the state of a system, internal energy is a state variable.
Similarly, entropy is a state variable. It is a quantity which is intrinsic to a system. If a
system starts in a state i it will have an entropy Si and if it ends in a state f it will have
an entropy Sf . When a system changes from i to f , the change in entropy ∆S = Sf − Si
associated with the change in state is independent of the actual thermodynamic process
which brought about the change. Simply put: ∆S = Sf − Si depends only on the initial
and final states of a system and not on the path taken during the process. The existence
of this thing called ‘entropy’ and the fact that it is a function only of the state of a system
has been established through extensive experimentation.
Before we learn how entropy is defined and computed, let’s get a feeling for how it is related
to the irreversibility of a process. First we recall a definition. An isolated system is
79
a system which cannot interact in any way with its environment – it is completely and
perfectly isolated from the rest of the universe. No energy or matter can flow between an
isolated system and its environment. We now come to the first use of the idea of entropy:
If an irreversible process occurs within an isolated system, the entropy S of the
system increases.
Entropy is clearly very different from energy. In an isolated system, energy will be conserved. During an irreversible process the entropy always grows in a isolated system. In
terms of an equation, the above statement in the box would read: if an isolated system
undergoes an irreversible process taking it from a state i to a state f , then for the system
∆S > 0
irreversible process, isloated system
(9.1)
This is quite peculiar since it is an inequality, something which appears rarely in physics.
9.3
Definition of entropy change
There are two very different looking but entirely equivalent definitions for the change
of entropy of a system. The first is a macroscopic definition and is related to the heat
exchanged by a system and the temperature at which the heat exchange occurs. The
second is a microscopic definition and is related to the number of ways in which the
constituent particles of a system can be arranged. Although the microscopic definition
provides us with a good feel for what entropy really is, it is hard to use for practical
calculations. In this course we will stick with the macroscopic definition and only briefly
mention the microscopic interpretation. Historically the macroscopic definition came first.
Consider a system which goes from a state i to a state f via some reversible or quasi-static
process. When in state i the system has an entropy Si , and when in state f the system has
an entropy Sf . We define the change of entropy for the system during the reversible
process as
Z
∆S = Sf − Si =
i
f
dQ
T
entropy change, reversible process
80
(9.2)
Here dQ is an infinitesimal amount of heat transferred to the system during an infinitesimal step during the process, and T is the temperature of the system19 at which this
infinitesimal transfer of heat occurs. The integral is just the ‘sum’ over all infinitesimal
steps between the states i and f . That is, the integral must be carried out over the path
of the process in question.
The infinitesimal version of (9.2) is simply:
dS =
dQ
T
infinitesimal reversible process
(9.3)
where dS is the infinitesimal entropy change of a system as it changes from one state to
another which are infinitesimally separated. The quantity dQ is the infinitesimal heat
transferred to the system during the infinitesimal change of state, and T the temperature
of the system at which the heat exchange occurs. The finite version, (9.2), is obtained
from the infinitesimal version by integrating both sides – which amounts to adding up a
whole pile of entropy changes for all the small steps linking the states i and f .
Note that since T is always positive, dS must have the same sign as dQ. Since T has
units of K and Q units of J, the units of entropy are J/K.
At this stage the attentive reader may have spotted a potentially serious problem. On
the one hand we have just learned that entropy is supposed to tell us something about
the irreversibility of a process. On the other hand the definition of entropy change given
above only applies to reversible processes. The definition requires that we integrate over a
continuous path between two states, and this is only possible if the process is quasi-static
or reversible. How is this definition – which can only be used on reversible processes –
ever going to provide any information about irreversible processes?
The key to resolving this problem is to recognize that the change of entropy of a system
does not depend on the process, it only depends on the initial and final states. We can
use the definition given above to compute the entropy change of an irreversible process if
we are crafty! This leads us to the following extremely important idea:
19
Warning: the temperature must be in kelvin.
81
To find the entropy change of a system which starts in a state i and undergoes
an irreversible process ending in the state f , you perform a thought experiment.
You imagine a situation in which that system undergoes a reversible process with
exactly the same initial and final states i and f . You now compute the entropy
change for this imagined process using the definition (9.2). Since entropy change
does not depend on the actual process involved – only on the initial and final
states – your computation will give the entropy change for the irreversible process.
To compute the entropy change of an isolated system which undergoes some process, it
is often convenient or necessary to imagine that the system is made up of several parts
or subsystems. One then computes the entropy change for each of the subsystems for the
given process (by using the idea in the box above), and then adds the entropy changes
for all of the subsystems together to determine the total entropy change for the entire
isolated system. Some worked examples will make this idea clearer in a moment.
9.4
The second law of thermodynamics
We are now finally ready to state the second law of thermodynamics. There exists a
few equivalent versions of the second law of thermodynamics, which at face value seem
to have nothing to do with each other. In these notes we are taking a modern and
perhaps more abstract approach. Many textbooks (yours included) do not even discuss
the concept of entropy until after stating the second law in terms of something called heat
engines and refrigerators. The heat engine/refigerator approach more closely follows the
historical development of thermodynamics and is less abstract. However, the statement
of the second law in such terms is more difficult to use in more general settings and one
is eventually forced to deal with entropy anyway!
In words the second law states:
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The second law of thermodynamics: In an isolated system, the entropy
cannot decrease. More specifically, if a process occurs in an isolated system,
the entropy change of the system will be zero if the process is reversible,
and will be greater than zero if irreversible.
In equation form the second law of thermodynamics states: for an isolated system
undergoing some process the entropy change ∆S of the system is
∆S ≥ 0 .
(9.4)
Here the greater-than sign applies if the process is irreversible and the equal sign applies
if the process is reversible.
As mentioned earlier, almost all real processes – often due to the presence of friction and
turbulence – are irreversible and so are accompanied by an increase in entropy.
Considering the entire universe as an isolated system, the second law of thermodynamics
tells us that its entropy must always increase.
Entropy may decrease for a system provided it is not isolated.
A subsystem within an isolated system can have a entropy decrease for a particular
process, but in order to not violate the second law of thermodynamics, other subsystems
within the isolated system must have an increase in entropy which is equal to or greater
than the decrease.
Like anything is physics, you need a few detailed examples to make sense of all this. See
‘Worked Examples 8’ on LMS.
83
9.5
Microscopic and statistical interpretation of entropy
It is quite likely that at this stage you are left with a sense of confusion: we can compute
the entropy change but what does it actually mean? Here we want to get a feel for what
entropy is at a fundamental microscopic level.
Entropy can be understood in a number of different ways which are closely related to one
another: it can be thought of as a measure of the disorder of a system; or as a measure
of how likely it is that a system will be in its current state; or as how much useful work a
system has to offer. (For a discussion on this last point, see your textbook, section 19.4.)
The greater the entropy a system has, the more disordered the system is and the more
probable its current state is.
Understanding entropy as a measure of disorder, the second law of thermodynamics tells
us that the disorder of an isolated system (or the entire universe) increases over time.
But what do we mean by disorder? The analogy of your bedroom has been used by a few
physicists to try to describe what is meant by disorder without giving a precise definition.
As time passes your bedroom will naturally tend to get messier or more disordered, and
it will take effort to restore everything to its correct location, to return it to an ordered
state.
Let’s take a more specific physics example to illustrate all this. Imagine the process of
melting ice. As shown in one of the worked examples, a system consisting of water has
more entropy when it is in the liquid state in comparison to when it is frozen. In the
frozen state the water molecules are all locked together in a tightly organized fashion –
very few, if any, will be out of place. They are all aligned in a very regular way. When in
the liquid state, however, the water molecules are all stacked randomly together in any
old way. The frozen water is a very ordered state compared to the very disordered liquid
water. This is basically what we mean when we speak about the order and disorder of a
system.
More technically correct than the vague concept of disorder is to think about the number
of ways in which you can arrange a collection of particles in a system but still have
that system appear, at a macroscopic level, to represent the same thermodynamic state.
For example, if I was asked to organize a collection of water molecules and have them,
84
on a macroscopic scale, appear as ice, there would be very few ways of achieving this.
Relatively speaking, there are not many ways in which the water molecules could be
placed so that they formed a regular organized ice lattice. In contrast there are many
more ways in which I could arrange water molecules and have them appear as a liquid –
I could arrange them in pretty much any way I liked, they would just have to be stacked
close together. For a given macroscopic state, the larger the number of ways you can
organize a collection of particles and have them form that state, the more entropy the
system has.
Take as another example an ordinary room with a balloon in one of its corners. The
balloon is filled with a gas X. You now pop the balloon. Just after you pop the balloon
all the gas X particles will be located in one small area in the corner – this is the initial
state i of the gas X. As time passes the X particles will spread out and uniformly fill the
room – this is the final state f of the gas X. The thermodynamic process of the gas X
particles spreading out, i → f , is clearly irreversible, and in this process the entropy of
the gas X has hugely increased. If the gas X particles started in the state f you would
be astounded to suddenly see them all move towards the corner and collect together in
one small balloon sized area. It is virtually impossible that the state i will spontaneously
emerge. In terms of the concept of disorder, the X particles are in a far more organized
state when in the corner, as compared to the far more random state of being spread
throughout the room. Being collected all together in one small pocket of the room is a
highly organized arrangement for the X particles to be in. In terms of the number of
ways in which you can arrange the X particles – and macroscopically have them resemble
the state i – there are comparatively speaking very few, since the volume in which they
start is quite small. In contrast, the number of ways in which you can arrange the X
particles and macroscopically have them resemble the state f is vastly greater, since the
volume is so much larger. In terms of probability, if you were to peek into the room at
some random time and you were to find all the X particles to be located in one small
region in a corner, you would think that this is an exceedingly improbable. In contrast
if you were to peek into the room at some random time and you were to find all the X
particles to be uniformly spread throughout the room, you would think that this a very
likely situation with a high probability.
85
9.6
9.6.1
Heat engines
The idea of a heat engine
Historically speaking, the concept of entropy is intimately linked with heat engines. As we
shall see, the second law of thermodynamics can be formulated in terms of heat engines.
A heat engine is a device which operates in cycles and converts heat into useful mechanical work, and in each cycle the heat engine:
1. Absorbs some energy as heat from a hot thermal reservoir;
2. Uses some of this absorbed energy by performing some mechanical work;
3. Dumps the remainder of the absorbed energy as heat into a cold reservoir.
Since heat engines works in cycles, the above steps 1 to 3 are repeated again and again,
returning the heat engine to its starting point each time. Examples of heat engines include
steam engines (such as those used to generate electricity in power stations) and the petrol
and diesel engines inside cars.
The thermodynamic system inside a heat engine (the substance which absorbs the heat
and performs the work) – which is a fluid (a gas or liquid or both) – is called the working
substance of the heat engine. For each complete cycle the change of internal energy of
the system is ∆U = 0, since the system starts and finishes in the same state. Thus for
each cycle, the first law of thermodynamics tells us
0=Q+W
for each cycle of a heat engine
(9.5)
where Q is the heat added into the system, and W the work done on the system in each
cycle.
When dealing with heat engines it is very convenient to break with our previous conventions for W and Q. In this section of the notes, for each cycle of the heat engine we
will:
• Use Wby to denote the work done by the system in each cycle instead of using W
which is the work done on the system.
86
• Split Q, the heat absorbed by the heat engine in each cycle, into the following two
distinct parts:
– A part Qin which tells you how much heat energy has flowed into the system
in each cycle (i.e. how much heat went into increasing the internal energy in
each cycle).
– A part Qout which tells us how much heat energy has flowed out of the system
in each cycle (i.e. how much heat left the system and caused a decrease in
internal energy in each cycle).
More specifically we will write equation (9.5) as
0 = Qin − Qout − Wby
for each cycle of a heat engine
(9.6)
where Wby is the mechanical work performed by the system in each cycle; Qin is the energy
flow as heat into the system (and thus raises the internal energy) during each cycle; and
Qout is the energy flow as heat out of the system (and thus lowers the internal energy)
during each cycle. A schematic representation of a heat engine is given in Figure 37.
Hot reservoir
Qin
Wby
System
Qout
Cold reservoir
Figure 37: A schematic representation of a heat engine for one cycle. Energy
conservation requires that Qin = Qout + Wby .
87
Each of the quantities Wby , Qout and Qin are positive by definition. Thus we no longer
consider negative and positive heat flow, we only speak of heat in and heat out. The
relationship between the old convention and this convention is
Wby = −W ,
(9.7)
Q = Qin − Qout .
(9.8)
When the working substance of the heat engine is a gas we can display the cycle on a pV
diagram. An example of a cycle is shown in Figure 38.
p
f
i
Wby = area enclosed
V
Figure 38: An example of a pV diagram for one cycle for a heat engine. The
pV curve of the gas is always clockwise for a heat engine. The area enclosed
by the pV curve is Wby the work done by the gas in one cycle.
As shown in Figure 38 heat engines usually operate by moving the system through a series
of distinct steps. To compute the three quantities Wby , Qout and Qin we usually need to
determine the work and heat for each of the steps.
9.6.2
The efficiency of a heat engine
From a practical point of view the purpose of the heat engine is to perform work for us.
To recap the information above, in each cycle a heat engine the working substance:
88
1. Absorbs Qin from a hot thermal reservoir;
2. Performs an amount Wby of mechanical work;
3. Dumps Qout into into a cold reservoir;
where energy conservation (the first law of thermodynamics) requires
Qin = Qout + Wby
for each cycle of a heat engine
(9.9)
The thermal efficiency, e, of a heat engine is a measure of how well the heat engine
does its job of converting the energy it absorbs from the hot thermal reservoir into useful
work:
Wby
thermal efficiency definition
Qin
Using equation (9.9), we can express the thermal efficiency as
e=
e=
Wby
Qin − Qout
Qout
=
=1−
.
Qin
Qin
Qin
(9.10)
(9.11)
The thermal efficiency of a heat engine is a number between 0 and 1, usually written as a
percentage. The case were e = 0 is a completely useless engine since it performs no work
(i.e. Wby = 0). The case where e = 1 or 100% would give a perfect heat engine such that
all the absorbed energy is converted into work (i.e. Wby = Qin ) and no energy is wasted
(Qout = 0).
See ‘Worked Examples 9’ on LMS to get a feeling for how to compute the thermal efficiency
for a given heat engine.
9.6.3
Heat engines and the second law of thermodynamics
The link between heat engines and the second law of thermodynamics is the following
statement, which can be shown the be equivalent to the second law of thermodynamics
as was given earlier:
The second law of thermodynamics (Kelvin-Plank version): It is impossible to construct a heat engine operating in cycles that extracts heat
from a reservoir and delivers an equal amount of work.
89
In other words, it can be shown that the second law of thermodynamics is equivalent to
the statement that it is impossible to construct a perfect heat engine:
e 6= 1
Second law of thermodynamics (Kelvin-Plank version)
A schematic representation of the impossible perfect engine is shown in Figure 39.
Hot reservoir
Qin
Wby
System
Cold reservoir
Figure 39: A schematic representation of the perfect engine. Here we have
Qin = Wby . According to the second law of thermodynamics such a perfect
heat engine is impossible.
9.6.4
The Carnot engine – the most efficient heat engine
So we cannot build a perfect heat engine. But how close can we get?
An ideal heat engine is a theoretical heat engine in which all processes are reversible
and no wasteful energy transfers occur due to friction and turbulence etc. Such heat
90
engines cannot ever be built in practice, but they are very useful since they place upper
limits on the efficiencies which can achieved.
It turns out that the most efficient of all ideal heat engines is the so-called Carnot engine.
This heat engine represents the most efficient heat engine one could construct in principle:
it is the best at converting heat energy into work. The Carnot engine operates using an
ideal gas as its working substance, and passes through a four step reversible cycle called
the Carnot cycle. In the Carnot cycle the ideal gas passes through the four following
steps (a pV plot of the Carnot cycle is shown in Figure 40):
Step 1: a → b is an isothermal expansion at a temperature TH (which is the temperature
of the hot thermal reservoir). This is the only step where the gas absorbs heat, and
so the heat absorbed in this step is Qin .
Step 2: b → c is an adiabatic expansion taking the gas from TH to TC .
Step 3: c → d is an isothermal compression at a temperature TC (which is the temperature of the cold thermal reservoir). This is the only step where heat leaves the gas,
and so the heat leaving the gas in this step is Qout .
Step 4: d → a is an adiabatic compression taking the gas from TC to TH , back to the
staring point.
As an exercise you should try to compute the efficiency of the Carnot engine eCarnot
yourself. In doing so you will need to assume the ideal gas has a given adiabatic constant
γ, and you will also need to assume that TH , TC , Va (the volume at point a of the cycle)
and Vb (the volume at point b of the cycle) are all given constants. See Section 19.2 of
your textbook if you need help. You will find that the efficiency of the Carnot engine is
actually independent of most of these constants, and only depends on the temperatures
of the hot and cold reservoirs:
eCarnot = 1 −
TC
TH
Carnot engine efficiency
(9.12)
It turns out that there is a much easier way to compute the efficiency of the Carnot
engine without having to go through the usual steps of computing Wby , Qout and Qin for
91
p
a
Qin
b
Wby
TH
d
Qout
c
TC
V
Figure 40: A pV plot of an ideal gas in the Carnot cycle. The cycle has four
steps: a → b and c → d are isothermal; b → c and d → a are adiabatic.
the cycle. To see this we first note that since the gas moves in a cycle, the entropy change
for the entire cycle must be zero, since the gas starts and finishes at the same point:
∆Scycle = 0
enrtopy change for cycle
(9.13)
Since each step in the Carnot cycle is reversible, we can easily compute its entropy change
using the definition:
Z
∆S =
i
f
dQ
.
T
In Step 1 of the cycle, a → b, the entropy change of the gas is
Z b
Z b
Qin
dQ
1
∆Sab =
=
dQ =
.
TH a
TH
a T
(9.14)
(9.15)
In Steps 2 and 4, b → c and d → a, the processes are adiabatic so dQ = 0. Thus the
entropy change of the gas in these steps are
Z c
dQ
∆Sbc =
= 0,
T
b
Z a
dQ
∆Sda =
= 0.
T
d
92
(9.16)
(9.17)
In Step 3 of the cycle, c → d, the entropy change of the gas is
Z d
Z d
dQ
1
Qout
∆Scd =
=
dQ = −
,
T
TC c
TC
c
(9.18)
paying careful attention to the two conventions we have introduced.
And so we finally arrive at the total change of entropy for the entire cycle:
∆Scycle = ∆Sab + ∆Sbc + ∆Scd + ∆Sda
(9.19)
which, upon insertion of ∆Scycle = 0 and the values computed above, gives
0=
Qin Qout
−
TH
TC
(9.20)
or
Qin
Qout
=
TH
TC
(9.21)
Qout
TC
=
.
TH
Qin
(9.22)
or equivalently
Upon inserting the last expression above into the expression for the efficiency of a heat
engine (9.11), we immediate obtain the result quoted above (9.12),
eCarnot = 1 −
TC
Qout
=1−
Qin
TH
Carnot engine efficiency
(9.23)
The hypothetical Carnot engine is important because no heat engine can be more efficient
than it (when operating between the same two temperatures). All real heat engines must
be less efficient than the Carnot heat engine. Carnot’s theorem reflects these facts:
Carnot’s theorem: All Carnot engines operating between the same two temperatures TH and TC have the same efficiency eCarnot , and no other engine
operating between these two temperature can be more efficient.
It can be shown that if an engine was more efficient than the Carnot engine, then it would
violate the second law of thermodynamics.
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9.7
Refrigerators
Refrigerators are simply heat engines operating in reverse. Therefore, a refrigerator is a
device which operates in cycles, and having work performed upon the working substance,
shifts energy from a cold reservoir to a hotter reservoir.
A schematic representation of a refrigerator is given in Figure 41.
Hot reservoir
Qout
System
Won
Qin
Cold reservoir
Figure 41: A schematic representation of a refrigerator for one cycle. Energy
conservation requires that Qout = Qin + Won .
Examples of refrigerators include ordinary household fridges (surprise!), freezers and air
conditioners. In consuming mechanical work energy, they take heat energy from the cold
reservoir (the inside of the house in the case of an air conditioner) and dump it into the
hotter reservoir (the outside of the house).
The link between refrigerators and the second law of thermodynamics is the following
statement, which can be shown the be equivalent to the other two versions of the second
law of thermodynamics given earlier:
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The second law of thermodynamics (Clausius version): It is impossible
to construct a refrigerator operating in cycles that extracts heat from a
cold reservoir and delivers an equal amount of heat to a hot reservoir.
In other words, it can be shown that the second law of thermodynamics is equivalent to
the statement that it is impossible to construct a perfect refrigerator that moves heat
from a cold body to a hot body without work being performed.
A schematic representation of the impossible perfect refrigerator is shown in Figure 42.
Hot reservoir
Qout
System
Qin
Cold reservoir
Figure 42: A schematic representation of the perfect refrigerator. Here we have
Qin = Qout . According to the second law of thermodynamics such a perfect
refrigerator is impossible.
Refrigerators will not be assessed in this course.
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