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An Introduction to Thermodynamics
Classical thermodynamics deals with the flow of energy under conditions of equilibrium or
nearequilibrium and with the associated properties of the equilibrium states of matter. It is a
macroscopic theory, ignoring completely the details of atomic and molecular structure, though not
the existence of atoms and molecules to the extent required for writing chemical reactions. Time
is not recognized as a variable and cannot appear in thermodynamic equations. For students who
have become familiar with atoms and molecules, it may be surprising to find how far one can go
toward treating chemical and physical equilibria without employing any simplified models or
delving into theories of molecular structure.
The detachment of thermodynamics from molecular theory is an important asset. The
fundamental principles of thermodynamics were developed during the 19 century on the
th
foundation of two principal axioms, supplemented by a small number of definitions, long before
atomic structure was understood. Because of this lack of dependence of theory on models, even
today we need not worry about our vast ignorance at the molecular level, especially in the area s of
liquids and ionic solutions, in applying thermodynamics to real systems. It has been said, with
some justification, that if you can prove something by thermodynamics you need not do the
experiment. Such a strong statement must be handled with care, but it should become clear in t he
following pages that common practice is quite consistent with this assumption.
Two developments associated primarily with the 20 century introduced substantial new
th
insights into thermodynamics. Statistical thermodynamics, or statistical physic s, originated with
the efforts of Maxwell and of Boltzmann in the late 19 century and grew with additions by
th
Gibbs, Planck, Einstein, and many others into a companion science to thermodynamics. Because
statistical thermodynamics relies on specific models of atomic and molecular structure and
interactions, it provides important tests of those models, at the expense of substantial ly greater
mathematical complexity than classical thermodynamics. More important for pres ent purposes,
statistical mechanics provides much greater insight into the quantities that appea r in
thermodynamic equations, and thus a clearer view of why things happen. Thus we will not
hesitate to introduce some basic principles of statistical mechanics (without the extensive
mathematics) when necessary to explain what is going on.
The other new development, largely responsible for the change in physics from what is
generally considered purely Newtonian to relativistic and quantum physics, arose from the
introduction of operational definitions at the end of the 19 century. This viewpoint requires that
th
any definition (of energy, position, or time, for example) must include a statement of how we ca n
measure the quantity. Application of this criterion demands clarification of some quant ities that
were introduced casually, without a solid foundation, in the early days of thermodynamics. We
will try to be more careful in explaining what is meant by our symbols, and what can or cannot be
measured, than has been customary in thermodynamic textbooks.
One of the characteristics of thermodynamics is that most of the terms are fam iliar. Everyone
has heard of energy, of heat, and of work. The difficulty is that we must sharpen our definitions
to distinguish between loosely associated ideas. We will therefore be particularl y careful to define
these familiar quantities carefully, often emphasizing what our technical meani ngs do not include
as much as specifying the intended meanings.
These are also called vacuum flasks, because the space between the silvered double wa lls
1
has been evacuated, a design developed by Sir James Dewar. Another common name for these
and for containers of different design but for the same purpose is Thermos bottle, which is the
trade name of the American Thermos Products Co.
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One of the guidelines of early thermodynamics was that all energy transfers (under
equilibrium conditions) can be classified as either heat transfer of energy beca use of
temperature differences (thermo) or as work transfer of energy because of forc es and
motions (dynamics). It seems appropriate, therefore, to begin with definitions of energy, heat,
and work.
ENERGY. Energy has been a difficult quantity to define because it has so many faces, or forms in
which it may appear. Initially, energy was defined to be the energy of motion, or kinetic energy,
which for most objects under usual conditions is half the mass times the square of the speed,
E = ½ m
v
(1)
2
The most convenient, and generally reliable, definition of energy is that it is kinetic ener gy or
any of the other forms of energy which can be changed into kinetic energy or obtained from
kinetic energy. These other forms of energy include rotational energy (a spinning ball or
weathervane), vibrational energy (a mass oscillating up and down on a spring), and potential
energy (a skier at the top of a slope), as well as energy within an object, called internal energy.
HEAT. Most of the internal energy is associated with the nuclei or with the chemical state of the
object. We will generally ignore the nuclear energy. For any given sample of matter, the nuclear
energy typically remains unchanged. Changes of chemical energy will be considered when there
are chemical reactions. For now, we are more concerned with the relatively small portion of the
internal energy that changes when the temperature changes; it is most often called he at, or more
narrowly defined as thermal energy.
The meaning of the thermodynamic term heat can best be explored by consideration of a
few qualitative or semiquantitative experiments. For each of these we will develop a working
hypothesis, select a crucial test, and revise the hypothesis as necessary.
Our understanding of heat is based upon common experiences. When we stand before a fire,
or when we place a pan of water over a gas fire or in contact with an electrically heated coil, our
senses and the change in character of the water tell us that something passes from the fire or hot
coil to nearby objects (specifically to us or to the pan of water). The effect is to hea t the
objects, by which we mean that there is a sensation of warmth that can be verified by a
thermometer. The thermometer, in some way, measures this heat. We seek to find the
relationship between temperature, heating, and heat.
Temperature balance? As an initial hypothesis, assume that a thermometer measures the
amount of heat. If so, we should find that a loss of temperature by one body is compensated by a
gain of temperature by another. To test this we put 200 g of hot water, at 90 C, into each of two
o
Dewar flasks (Figure 1). To the first flask we add 50 g of water initially at 20 C, and st ir until
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the temperature becomes steady. The new temperature is found
to be about 76.0 C. To the second flask we add 25 g of water, at
o
20 C, and find the final temperature to be about 82.2 C. We must
o o
ask now whether the experiment shows the initial hypothesis to be
fully satisfactory or not.
There has indeed been a loss of temperature by the water in
the flask and a gain in temperature by the water added. But there
is clearly no temperature balance. The water in the flasks
changed temperature only slightly, whereas the water added
increased in temperature several times as much. Also, the water
in the second flask dropped in temperature less than that in the
first flask, but the water added to the second flask increased in
temperature more than that added to the first flask. Examination
of the results (Table 1) shows that the drop in temperature of the
water originally in the flasks is roughly doubled when twice as
much cool water is added.
The experiment just described suggests how the original
hypothesis might be revised. It appears that a larger amount of
water can absorb more heat for a given temperature increase. The
temperature, therefore, is more nearly a concentration of heat.
From this revised hypothesis we predict that the temperature
change times the amount of the substance should be the same for both the added and the original
water (see Table 1).
Table 1 Temperature Measurements*
Sample T
final
ΔT ΔT ΔT/ΔT ΔT M/ΔT M
s w w s w w s s
H O,50g 76.0 56.0 14.0 0.25 1.0
2
H O,25g 82.2 66.2 7.8 0.125 1.0
2
Al, 50g 86.35 66.35 3.65 0.0050 0.22
Al, 25g 88.12 68.12 1.88 0.0276 0.22
* Temperatures in C. ΔT is the temperature change of the sample and
o
s
ΔT is the temperature change of the water originally in the flask.
w
Initial temperature of the water is 90 C and of the sample, 20 C.
o o
M is the mass of the sample added and M the mass of water in the flask.
s w
It is necessary to find out whether the same relationship will hold if we exchange heat
between two different substances. To do this we again prepare two flasks, each containing 200 g
of water at 90 C, then add to one a block of aluminum, at 20 C, with a mass of 50 g and to the
o o
second a block of aluminum, at 20 C, with a mass of 25 g (Figure 2). After a few seconds we
o
may assume that the temperatures of the aluminum blocks are equal to the temperature s of the
surrounding water about 86.35 C for the larger block and 88.12 C for the smaller block.
o o
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Multiplying temperature change by mass and comparing the result for the aluminum block and the
water shows that the ratio is the same for both parts of the experiment with aluminum, but
appreciably different from the results of the earlier experiment. We conclude, ther efore, that
aluminum and water have a different heat capacity, so that a given amount of heat added to a
certain mass of one produced a different temperature change than equal heat added to the sam e
mass of the other.
Volume or mass? We have left unanswered the question whether
the heat capacity depends upon the volume or the mass of the
substance that is absorbing the heat. The choice can be easily made by
means of an experiment employing a substance, such as air, that can
readily change volume without changing mass. We fill one flask with
air, evacuate a second identical flask, and immerse both in water, with
a connection provided between the flasks, as shown in Figure 3. The
temperature of the water is measured; then the stopcock is opened,
allowing the air to expand to twice its initial volume, and the
temperature is remeasured. The temperature is found to be
unchanged. From this we conclude that the temperature of the air did
not change with the change in volume, and therefore that it is better to
define the heat capacity in terms of mass rather than of volume.
(The result is confirmed by more sensitive tests.)
Is heat gained or lost? In each of the measurements described
thus far it has been possible to follow heat as it flows from one body to
another; the amount lost by one substance has been equal to the
amount gained by the other. It is necessary to determine whether this is always true. (If it is, we
would say that heat is conserved, or that the amount of heat is constant.) Taking a hint from
the famous observations of Count Rumford, who noted the great quantities of heat evolved during
the boring of cannons, we design our next experiment to include mechanical motion, in which
energy will be added from motion ( i.e., by doing work). Instead of expanding the air from one
flask into an evacuated flask, we can let it expand against a piston, as shown in Figure 4. This
time the temperature of
the gas drops (about
50 C) during the
o
expansion, even though
we add insulation
around the cylinder to
prevent the flow of
heat outward from the
gas. The change of
temperature cannot be
solely because of the
volume change; the
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previous experiment showed that the change of volume did not cause any change of
A doubling of volume causes a temperature drop
from 25 C to about  25 C.
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temperature. The fact that the gas pushes on the piston, causing it to move, must be the
important difference.
A few additional experiments will provide more information on the relationship between
expansion, with work being done, and temperature effects on gases. (For brevity, only the results
of these experiments will be discussed.) Compression of a gas causes an increase in t he
temperature just equal to the decrease of temperature during expansion, if both expansion and
compression processes are slow. It is therefore possible, by repeated expansion and compre ssion,
to cycle the temperature between two values. Any other property of the gas that we might
measure, such as density, volume, or viscosity, will be found to depend only on the temperature as
measured by a thermometer, and not on how that temperature was achieved (for any specified
pressure). In other words, the heating effect of a compression seems to be exactly the same as
the heating effect of a flame or other source of heat. Thus it is possible to compress a ga s,
thereby raising its temperature; then extract heat from it by removing the insulation unt il the gas
has returned to room temperature; expand it into an evacuated space without change of
temperature; compress it to again increase its temperature; extract heat; and so forth, as many
times as we wish.
Clearly, heat is not a quantity that retains its identity after it is absorbed by a subst ance, for
we can add any amount of heat without changing the properties of a gas in any way (provided
only that the proper amount of work is done by the gas). There is no property that will enable us
to determine the amount of heat added to any substance, or the amount of heat removed. The
description of temperature as the concentration of heat is therefore untenable, and m ust be
abandoned.
If not heat, then what? Temperature is related to a concentration of something more
fundamental, which can give rise to heat or can cause a gas to do work and which is increased
when the substance absorbs heat or when work is done on the substance. This quantity so
directly related to temperature is called energy.
In classical physics, any measurement of energy is necessarily an energy difference. We often
Be particularly careful to distinguish between Q, which is an amount of thermal energy
2
transferred, and the change in a property, such as thermal energy or E. To write Q displays a
confusion to all who may read your notes.
Be aware that older thermodynamic literature quite generally defined work with opposite
3
sign. Because of an emphasis on steam engines, W was taken to be work done by an object,
rather than work done on an object. Some textbooks have chosen both definitions, changing from
one chapter to another. Again, the amount of energy transferred is W, not W.
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write E for energy, where we mean E, the difference between the current energy value and some
implied reference level of energy. There are several meanings for energy. Total energy includes
information on where the sample is located and how it is moving. Measurements made on a
sample at rest with respect to our apparatus would measure the internal energy. A small portion
of that internal energy is the thermal energy, which is a collection of several different forms of
energy, or modes of storage of energy, in matter, that can be changed by a change of temperature.
It will be sufficient to let E represent total energy, keeping in mind that the changes we are
primarily concerned with will be changes of internal energy and, most often, changes of thermal
energy.
Transfer of heat. Furthermore, in thermodynamics we are primarily concerned with the
transfer of energy to or from a system, including the transfer of thermal energy. This quantity is
nearly always represented by the symbol Q. Thus Q represents a change of energy of the system;
2
Q = E (2)
To ensure that all the energy transfer occurs as transfer of thermal energy, only, we sharpen
the definition of Q:
Q (thermal energy transfer) is the transfer of energy between two objects in
physical contact as a consequence of a difference in temperature between the
objects.
The importance of this refinement will become apparent later.
WORK. We distinguish work from play; we say that machinery works or doesnt work;
and we read that if a force acts on a body and the body moves through some distance, work is
done on the body. But in thermodynamics, work has a special meaning.
First of all, work is a transfer of energy from one object to another by the action of a force. If
work is done on an object, that increases the energy of the object. Thus, if the only energy
3
transfer is as work,
W = E (3)
Second, and equally important, all investigations thus far of transfer of energy as work
∑
∑
∫
×==
i i
iii
WW (4) dxf
∫
∑
×
dsf
i
i
The product of force and distance is an integral of a scalar product, or dot product.
4
That will be automatically taken care of in most of our applications. In mechanics, we often
measure work done on a particle (or, better, a physical particle), which can change only its
kinetic energy. For such a particle it is easily shown that work is If·ds. We are concerned here
with more general bodies, subject to rotation and deformation, for which the definition of work
must be more explicit.
The imposter equation, (p)/2m =
I
f ·ds, where p = m
v,
has been called the second
5 2
net
law equation, SLI. It is a valid statement of Newtons second law, net force is equal to mass
times acceleration, but except for special circumstances it has no connection to W.
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indicate that the amount of work done is equal to a product of the force exerted (on the object)
4
times the displacement of the point of application of the force.
If more than one force is acting, each force must be multiplied by the distance through which
it acts. Then the work terms may be added.
What is the imposter equation? The expression for work done is often abbreviated to work is
equal to force times distance, but this is often interpreted to mean a net force tim es the distance
traveled by the object. Is this imposter equation ever wrong?
Imposter equation W =?If ·ds = (5)
net
Yes, if it is interpreted as giving the work done. We give here just two examples. First, if
you jump, you move upward because the floor exerts a force on you (which is the reaction force
to the force you exert on the floor). You move, and momentum p = ( m
v
) is transferred between
you and the floor, but the force exerted on you times the distance you move (while in contact
with the floor) cannot be equal to work done on you because the floor does not transfer any
energy to you. The point of application of the force is at the floor, and that point doesnt move.
Second, if you drag a box across the floor, it is easy to find the force you exert on the cord
and the distance the cord moves, so work done by you on the cord is known. Similarly, the force
exerted by the cord on the box can be measured, as well as the distance the box moves, so work
done by the cord on the box is known. (Energy transferred, as work, to the cord is equal to
energy transferred, as work, by the cord to the box) But the forces of friction between the box
and the floor are molecularlevel forces, not known in detail. And the distances over which the se
molecularlevel forces act, individually, cannot be known. A simple force times dista nce
calculation gives an answer, but it is demonstrably wrong!
5
Analogues. Fortunately, we can usually avoid ambiguities in the definition of work in our
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analysis of equilibrium thermodynamics. We must, however, be sensitive to the differences
between what is transferred and what is stored.
The description of energy transfers as heat (thermal energy transfer) and as work may be
compared with deposits and withdrawals from a savings bank. The deposit slip may ask for a
separate listing of bills and coins, and a withdrawal may be in the form of bills or coins. Ye t the
account balance itself is neither bills nor coins. In the same manner, energy may be put into a
substance, or withdrawn from it, either as Q, a thermal energy transfer (heat) or as W, work,
but it exists within the substance only as energy not as Q or W. (Remember that money may
also be deposited by check or electronic transfer. We must be alert to the possibility of
transferring energy by means other than Q and W.)
If we are being very casual, it may be sufficient to describe thermal energy and therma l
energy transfers as heat, without additional labels. That, however, is very much like an
accountant choosing a single label (perhaps money) to indicate income and net worth, or trying
to balance your check book without distinguishing between net deposits for the month and
balance at the end of the month. Learning to distinguish between thermal energy and Q, the
transfer of thermal energy, will go a long way toward helping you understand discussions of
thermodynamics.
By going beyond thermodynamics, into statistical mechanics, the internal energy can be
described in terms of thermal energy and other forms of energy (such as chemical), and the
thermal energy may be further broken down into kinetic energy and potential energy of individual
atoms and molecules. But for purposes of thermodynamics we need know nothing more about
energy than that it includes a component related to the temperature. This component, called
thermal energy, can be internally converted to increase or decrease the temperature without
changing the internal energy. (If a mixture of oxygen and hydrogen is ignited, the gases become
substantially hotter, without any transfer of energy to or from the surroundings.) Or the therma l
energy can be transferred to or from the surroundings either as thermal energy transfers or as
work. We can find by experiments, such as the compressioncooling experiment or a variety of
others performed by Joule, how much thermal energy transfer is equivalent to how much work
and, for a given substance, how much energy (by either transfer method) must be put in for a
given temperature rise. The relationships between temperature, energy, thermal ene rgy transfer,
and work will be considered quantitatively in the discussion of the first law of thermodynami cs.
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1 The First Law of Thermodynamics
Thermodynamics is based on a small number of postulates, or assumptions. These are called
the laws of thermodynamics because they are suggested by a great amount of accumulated
experimental evidence. In fact it is extremely important to keep in mind that thermodynamics is
important just because there is total agreement between the results of thermodynamics (properly
applied) and all careful experimental results available to us. Because it is not possible to prove
the fundamental assumptions of thermodynamics, both the postulates and the derived results of
thermodynamics have often been challenged. In every showdown thus far, thermodynamics has
been shown to be correct.
Energy
The first step in understanding thermodynamics, and making it serve your purposes, is to
learn how to evaluate changes in energy in any object. As you probably recognize already, that
means understanding how to evaluate Q and W, the transfer of thermal energy to (or from) the
object and the work done on (or by) the object. To do so, we will write what is known as the
first law equation, which directly relates Q and W to the change of energy. But before we can
move ahead, we must look at how we define the object or quantity that is to give up or receive
energy; we must consider what is meant by a property; and we must examine the meaning of a
conservation law.
System and Surroundings. When we consider a ball projected through the air, or a block
sliding down a plane, it is quite satisfactory to talk about the ball, or the block, or simply the
object. If we wish to consider a gas or a solution, object is no longer a very appropriate
description, but we could still refer to the gas or the solution or simply call it i t. A better
method is to call whatever is of primary interest (ball, or block, or gas, or solution, or whate ver)
the system. One advantage is that we neednt change descriptions when we substitute one gas for
another, or a solution for a gas, and so forth. Another advantage is that it allows for a smooth
transition to problems in which we choose an open system, that is, where we allow material, as
well as energy, to pass back and forth, into or out of the system. Unless specifically indicat ed
otherwise, however, we will assume our systems to be closed.
Whatever else is around, we simply call the surroundings. Then, ideally, we need consider
only system and surroundings. To avoid difficulties that we might encounter in trying to describe
the far reaches of the universe, it is quite sufficient to limit the term surroundings to include all of
the universe that might be affected by whatever change, or process, we are considering.
It is important that we clearly define what we mean by our system. When we have done so,
the surroundings is usually adequately defined.
There will be times when you will find it more convenient to deal with the interactions of two
bodies, or two systems, with each other, ignoring anything outside those systems. That is
perfectly legal. Terminology such as system and surroundings is meant to be an aid, not a
liability. On the other hand, it has been shown many times that disdaining the definitions of
7/10/07 1 10
system and surroundings provides sufficient ambiguity to allow erroneous conclusions to be
"proved". Also note that system and surroundings are essentially equivalent and arbitrary labels,
so we are free to interchange them if we wish.
Properties, or State Functions. Examine a block of copper and you will find that it has a
certain shape, a certain volume, a certain temperature, a certain density, and various other
properties that characterize the sample. Each of these may be changed, by changing the
temperature of the copper, or to some extent by changing the pressure exerted on the copper.
Neglecting the shape, which depends on the specific sample but not on the nature of the copper
itself, we call each of these measurable quantities a property of the copper. The state of the
copper, or state of the system, is defined by these properties, so they are often called state
functions.
A gas, such as a gram of oxygen, O, is adequately described if we know the temperature, the
2
pressure, and the volume, but two of these are sufficient to determine the third, so only two state
functions are sufficient to define the state of the system. Other substances, such as iron or plastic,
may have additional properties (such as magnetic state or strain) that must be specified.
Conservation Laws. In mechanics we pay special attention to quantities that do not change
during a process we are studying. A free particle (not acted on by any net force) will have a
constant velocity no change in speed or in direction. Such a constant of the motion may be
said to be preserved under the special conditions of the process. In a frictionless syst em, energy
will be preserved, or unchanged. When a liquid is poured from one container to another, the
volume of the liquid is preserved. Unfortunately, it has become common practice to refer, in s uch
instances, to the unchanging quality as being conserved, although the term conservation has
quite a different, very important meaning.
Volume is something we can see and often measure with a meter stick, whether the sam ple is
a copper block or the gram of oxygen gas (in a rigid container). For many processes, the volume
of the system (the liquid) is "preserved", but we also are aware that the volume of such a system
can be changed. With a gas, we need only move a piston, or change the pressure or the
temperature. With the block of copper or a free liquid, a change of temperature will change t he
volume.
What we cannot do is change the volume of our selected system without causing some
change elsewhere. If the volume of the system increases, then less volume is availabl e outside the
system, so the volume of the surroundings must decrease, by the same amount. Thus we have the
seemingly trivial, but very important, conclusion that the volume of the system plus the volume of
the surroundings is constant. That, of course, is because of the assumed properties of local spa ce.
We state this result as a conservation law. Volume is a conserved quantity. The volume of a
gas, or liquid, or a solid may be changed, but the volume of the system plus the volume of the
surroundings remains unchanged for all changes we make. Let V = V  V represent the change
2 1
in V, the volume. Then the law of conservation of volume would be written
(V ) = 0 (1)
system + surroundings
The bestknown of the three thermodynamic postulates is known as the first law of
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thermodynamics. It is just the principle of conservation of energy. It may be stated in the form:
First Law The energy of the universe is constant.
A better form for our purposes is to write
First Law ( E) = 0 (2)
system + surroundings
Remember that (Greek capital delta) indicates a change in the quantity that follows, in this
instance a change in the energy, E. Expressed in symbols,
E / E  E (3)
final initial
Because we are primarily interested in changes within that part of the universe tha t we
choose to call the system, which generally is a small, fixed quantity of substance, a symbol such as
E (unless specifically labeled otherwise) will always refer to the system unde r discussion. When
the energy of the system increases, E is positive; when the energy of the system decreases, E is
negative. In any given problem, the system to be considered must be carefully defined. The
surroundings are then adequately defined by what is outside the system. It should be clear that
equation 2 is, indeed, a statement of a conservation law, equivalent to the previous statement of
the first law.
It is often convenient to think in terms of a system that is totally isolated from its
surroundings. Unfortunately, such arrangements are not possible. We may prevent energy from
moving to or from a system, or hold volume of the system constant, or keep the pressure on the
system (exerted by the surroundings) constant, or make other special provisions, but there is no
practical way to define a truly isolated system and, if we had one, we could not make
measurements on the system. We must therefore carefully specify in what sense we want our
system to be isolated. It may be isolated in the sense that no energy is transferred to or f rom the
system, or it may be isolated in a more narrow sense. If the system is insulated, there is no
thermal energy transfer, Q. If the volume of the system is being held constant, there can be no
transfer of energy as work of expansion or compression. To simply call the system isolate d
leaves the question indeterminate as to what is being prevented or what is maintained cons tant.
Heat, Work, and the FirstLaw Equation. Thus far we have considered energy transfers, to
or from a system, only as Q or as W. These will suffice for the present. Then we can write the
total energy change for the system in the form
E = ( E) + ( E) (4)
heat work
but this is awkward. It is better, and conventional, to introduce new symbols for the terms on
the righthand side, as we have already indicated. We let Q be the amount of thermal energy
transferred to the system from the surroundings, and let W be the amount of work done on the
system. Then, allowing for both modes of energy transfer, at the same time,
FirstLaw Equation E = Q + W (4a)
The notation ( V) at the left of the equation indicates that the equation is valid when the
1
volume of the system is held constant. We are also assuming, for the present, that there is no
other form of work, such as electrical work.
Other choices are δQ and ñQ, each of which is often misinterpreted, including confusing
2
the small magnitude with an inexact differential (i.e., change).
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Note that this firstlaw equation, by itself, tells us nothing about whether the energy f or the
entire universe changes or not.
Roughly, the division of energy transfers into heat (thermal energy transfer) and work
corresponds to a division into random motion and directed motion (or collective motion ). At this
point we have no good criterion for deciding whether light, for example, should be called heat
or work, but for the present it is arbitrary. Later we will find that the second law of
thermodynamics deals with these distinctions.
Insofar as all energy transfers may be measured either as heat (thermal energy transfer) or as
work, or a sum of these, this equation accounts for all possible changes in the energy of the
system. This equation is called the firstlaw equation. If we combine this equation with the first
law (conservation of energy) we may conclude that Q =  Q and that W = 
system surroundings system
W. It is necessary to add additional terms to the firstlaw equation when there are other
surroundings
modes of energy transfer (including especially the transfer of matter to or from the system). A
few simple examples of energy changes, in the following sections, will show the meaning of the
terms in equation 4.
TEMPERATURE CHANGES AND HEAT CAPACITY, C. Consider first the process of heating a gas
V
confined in a container of constant volume. Because there is no directed motion, no work is
done.
( V ) E = Q (5)
1
We are often concerned with very small (infinitesimal) changes in energy, which would imply
an infinitesimal value for Q. The temptation is to write dQ, but that would be misunderstood.
2
Because Q = E, it follows that Q = (E) and dQ = d(E), which is not at all what we
intend. We therefore choose the notation
q = dE (5a)
intended to serve as a reminder that we want an infinitesimal amount of thermal energy transfer.
The amount of thermal energy absorbed by the system, Q, divided by the temperature rise is
the quantity commonly called heat capacity, C. C = Q/T. However, because C may itself vary
with temperature, it is better to take the limiting value of the ratio for small changes of
(6) LimLim
00
dT
q
T
Q
T
E
C
TT
V
=
=
=
®®
V
T
E
(6a)
V
V
T
E
C
=
( )
( )7 dTCdT
T
E
dT
dT
dE
dEV
V
V
=
==
( ) ( )
( )
∫
∫
==
==
2
1
2
1
12
12
8
T
T
VV
T
T
TTCdTC
dEEEEV
See Appendix.
3
For convenience, numbers specified for illustrative purposes, such as amounts of
4
material, temperatures, and pressures, will be treated as integers, known exactly. Also, for
present purposes we approximate C for hydrogen as 5/2 R (see Table I), although the actual
V
value is somewhat less and increasing with temperature.
7/10/07 1 13
temperature. This is just a derivative.
3
It is common practice to represent a derivative, such as dE/dT, which is evaluated with some
variable (in this case V ) held constant, by the special symbol
These are called partial derivatives. Employing this notation we arrive at the us ual form of
equation 6.
Consider now the change in energy for 2 mol of H warmed at constant volume from 25 C to
2
o
50 C. For a diatomic gas near room temperature, C is constant and is about 5/2 R = 21
o 4
V
J/mol·K. The gas can do no work at constant volume so the total energy change is equal to the
thermal energy transfer (heat absorbed).
The total energy change is obtained by integration.
Thermodynamic temperature is expressed on the Kelvin scale, by adding 273.15 to the
This scale was formerly called centigrade in Englishspeaking countries, a name that
5
can be confused with 1/100 degree in other languages.
7/10/07 1 14
temperature on the Celsius scale. Temperature differences are the same for the two scales. With
5
the appropriate numerical substitutions,
E = 2 mol x 21 J/mol·K x (323  298) K
= 1.05 kJ/mol
This is the amount of (thermal) energy that must be added to the gas to raise its temperature
25 C.
o
IDEALGAS EXPANSIONS. Now let 2 mol of hydrogen, at 3 atm pressure, expand at a constant
temperature of 50 C through a pinhole (to maintain a slow expansion) and against a piston.
o
Assume the external pressure acting on the piston is 1 atm, as in Figure 1. The work done on the
gas is the product of the force exerted on the gas (by the piston) and the distance through which
the piston moves, with a negative sign because the pressure (of the gas) acts opposite to the
pressure acting on the gas. The force, on the gas, times the displacement, is f dx =  (PA) dx = 
P (Adx) =  P dV. Because of the unusual arrangement (effusion controlled), the pressure acting
on the gas at the point of displacement is the external pressure, P. Therefore the work done on
ext
the gas is
(Effusion controlled) W = I f dx =  I P dV (9)
ext
Because P is maintained constant, it may be removed from the integral.
ext
W =  P I dV =  P V (9a)
ext ext
3
5
1
m 0.0530
Pa
10
x
013
.
1
K 323K x J/mol .3148 x mol 2
=
×
=V
3
5
2
m 0.01767
Pa
10
x
1.013
x
3
K 323K x J/mol .3148 x mol 2
=
×
=V
Kmol
atmml 06.82
Kmol
cal 987.1
Kmol
J 314.8
×
×
=
×
=
×
=R
The term ideal means simply that the substance obeys a certain equation. An ideal gas
6
obeys the equation PV = nRT; in later chapters we will encounter the ideal solution, which
obeys an equation known as Raoults law. The idealgas equation combines Boyles law and
Charles, or GayLussacs, law into a single, more convenient expression. The temperat ure must
be on an absolute scale, which we will always take as the Kelvin scale; n is the number of moles of
gas; and R is a universal constant, whose value depends on the units chosen for pressure and
volume. It should be noted that the product of pressure and volume has the dimensions and units
of energy (although PV is not a measure of energy). Such gases as He, H, O, and N closely
2 2 2
follow the idealgas equation at room temperature; such easily condensable gases as CO or H O
2 2
vapor follow the equation less closely.
7/10/07 1 15
The initial and final volumes can be calculated from the idealgas equation, PV = nRT.
6
Thus the work is
W =  P V =  P (V  V )
ext ext f i
=  1.013 x 10 Pa (0.0530  0.01767) m =  3.58 kJ
5 3
The necessary constants are given in Table 1.
Table 1 GAS CONSTANT
AND CONVERSION FACTORS
1 joule = 1 newton·meter = 0.239 cal = 1 pascal·meter
3
1 cal (thermochemical) = 4.1840 joule = 41.3 ml·atm
*
1 ml·atm = 0.1013 joule = 0.0242 cal
1 atm = 1.01325 x 10 N/m
5 2
* Various experimentally defined values of the calorie appear in
the literature (including the dietician's Calorie = 1 kcal). Only the
thermochemical calorie is defined exactly.
In order that the temperature may remain constant it is necessary to supply thermal energy to
the gas to compensate for the energy expended in doing work. From the firstlaw equation,
7/10/07 1 16
Q = E  W
For the special case of an ideal gas, the energy depends only on the temperature, not on the
pressure or volume. At constant temperature, therefore, E = 0 and the thermal energy that must
be supplied is
Q =  W = 3.58 kJ
A more important example than the expansion against a constant external pressure is the
reversible and isothermal (constant temperature) expansion. A process is said t o be
thermodynamically reversible if it can be reversed at any stage by an infinitesimal increase in the
opposing force or an infinitesimal decrease in the driving force. It should be clear that such a
reversible process is an example of a limiting case, which is only approximately achieved in
practice. Not only does it require a frictionless mechanism, but if the unbalance of forces is
infinitesimal, the rate will be infinitesimal and the time required will be infinite. The question of
reversibility will be considered in more detail later.
From the assumption that the operation is isothermal and the gas is ideal,
E = 0
and
Q =  W = I P dV
Because the expansion is to be reversible, the pressure on the piston differs only infinitesimally
from the equilibrium pressure of the gas, P = nRT/V. The work done on the gas is then
(T, I.G., rev.) W =  I nRT dV/V =  nRT ln(V/V ) (10)
f i
A completely general proof that the reversible work is a minimum for the isothermal
7
reversible process cannot be given here, but this important result can be illustrated by considering
the expansion or compression of a gas. When the piston moves away from the gas molecules, the
molecules do not strike the piston as hard as when the piston is stationary. The pressure exerted
by the gas on the piston, and hence by the piston on the gas, is therefore less than the equilibrium
pressure of the gas. Hence IP dV < I P dV; the work done by the gas is a maximum for a
eff
reversible expansion. Changing the sign, to obtain the work done on the gas, changes the
direction of the inequality; the work done on the gas, W, is a minimum when the effective pressure
is equal to the equilibrium pressure, for expansion or compression.
One may see elsewhere the expression W =  I P dV applied when P, the external
ext ext
pressure, differs from the pressure of the system (= the gas). Except in unusual circumstances
(e.g., an effusioncontrolled expansion), this is not even a good approximation. See Am. J. Phys.
37, 675679 (1969).
7/10/07 1 17
For example, if 2 mol of hydrogen at 3 atm and 50 C is expanded isothermally and reversibly
o
to a final pressure of 1 atm,
Q =  W = 2 mol x 8.314 J/mol·K x 323 K x ln V/V
f i
Because the temperature is constant, V/V = P/P = 3.
f i i f
Q =  W = 5.37 kJ ln 3 = 5.90 kJ
Notice that the amount of thermal energy that must be supplied is greater for the revers ible
expansion than in the previous example of expansion against a constant external pressure. A
reversible expansion (or compression) must always give a maximum value for Q, and accordingly
a minimum value for W.
7
An integral may be represented by the area under a curve (see Appendix) and the three
examples considered above may be compared graphically. When the gas was warmed, at constant
volume, the pressure increased by the factor T/T = 323/297 = 1.09. In a plot of pressure against
f i
volume, this is represented by a vertical line (Figure 2a) and the area under such a line is zero.
No work is performed.
The expansion against a constant opposing force is represented by the plot in Figure 2b. The
gas, in escaping through the pinhole, changes its pressure from 3 atm to 1 atm, but there is no
force exerted on moving surroundings during this escape. When the gas is through the pinhole,
however, it has the pressure of 1 atm and the volume increases, at this pressure, while the gas
pushes against the piston (the surroundings), until all of the gas has reached the same pressure
and occupies the total volume V. The work done by the gas,  W, is the area under the vertical
f
line, which is zero, plus the area under the horizontal line, which is 1 atm times the volume
change.
When the pressure of the gas changes smoothly, as in Figure 2c, the area under the curve (
W) is a maximum for the given temperature. The pressure follows the curve PV = constant
7/10/07 1 18
(where the constant is nRT) which is part of a hyperbola.
The condition to be satisfied in order that an expansion or compression of any fluid should be
thermodynamically reversible, and thus that the work done on the fluid should be  IP dV, is
simply that the fluid have a welldefined, uniform pressure throughout. In other words, there
must be no leaks, allowing different pressures at different points, and the motion of the pi ston
confining the fluid must be slow compared to the relaxation time of the fluid, or the time f or
pressures to equilibrate in the fluid. For a gas, the expansion or compression must be very slow
compared to the speed of sound in the gas. Unless there is such a welldefined pressure, or in
certain situations (see Figure 1) two or more welldefined pressures, for the system,
thermodynamics cannot be applied. It is then necessary to apply the more difficult methods of
nonequilibrium, or irreversible, thermodynamics. That is, equilibrium thermodynamics is
sufficient for reversible processes and certain types of irreversible processes, whereas irreversible
thermodynamics deals with the timedependent equations that become necessary in other
problems involving irreversible phenomena, such as diffusion rates and shockwave propagation.
The existence throughout the process of a welldefined state (an equilibrium state, for
which pressure is defined) for the system is sufficient to ensure that the expansion or compression
will be reversible, but there may well be other irreversibilities in the total process. Some of the
work done by the gas may be against frictional forces (in the surroundings), or work done on a
massive piston may appear as kinetic energy of the piston, subsequently converted to thermal
energy by collision of the piston with mechanical stops (in the surroundings). Or there may be
conduction of thermal energy to or from the surroundings at a lower or higher temperature. Thus
the total process may consist of a sum of parts, some of which are reversible and some
irreversible. It is particularly important in such circumstances to define carefully the system and
surroundings and the exact process to be considered. Interchanging system and surroundings
labels may be helpful in analyzing the total process.
PHASE CHANGES. Let 5 g of ice melt at 0 C under 1 atm pressure. Experimental measurements
o
have shown that the amount of thermal energy that must be supplied, called the heat of fusion,
is 334 J/g. The work done is  IP dV =  1 atm x V. The volumes of ice and water are 1.09
cm/g and 1.00 cm/g at 0 C. The energy change for the ice can be calculated from the firstlaw
3 3 o
equation.
E = Q + W = 5 g x 334 J/g  1 atm(1.00  1.09)ml/g x 5 g
Conversion of the work term from ml·atm to J gives
E = 1.67 kJ + 0.046 J = 1.67 kJ
Although the total energy absorbed by the ice, E, is slightly greater than the heat of fusion, Q,
the work done on the icewater system by the atmosphere (0.05 J) is negligible for nearly all
purposes.
If 5 g of water is vaporized at 100 C and 1 atm pressure, the energy change can be calculated
o
This heat of vaporization is required to separate the molecules from each other that i s,
8
it represents an increase in the potential energy of the molecules. Because the temperature does
not change, the kinetic energy remains constant.
7/10/07 1 19
in the same manner. The heat of vaporization is 2.258 kJ/g, and for calculation of the work done
8
it is sufficient to assume that the water vapor is an ideal gas.E = Q + W
= 2.258 kJ/g x 5 g  1 atm[ 5/18 mol x (8.314 J/mol·K)/1 atm x 373.15 K  5 x 10 m ]
6 3
= 11.290 kJ  0.861 kJ = 10.43 kJ
In problems of this type, however, an important shortcut is often satisfactory. The volume of
the liquid is sufficiently small compared to that of the vapor that it may be neglected. Then the
work term becomes
W =  P(V  V ) =  PV =  PnRT/P = nRT
f i f
The value of R in J/mol·K may be inserted into this expression to avoid completely the units of
volume and the conversion from atm to Pa. The entire problem can thus be written
E = Q + W = 5 g x 2258 J/g  (5/18) mol x 8.314 J/mol·K x 373 K
= 11.290 kJ  0.861 kJ = 10.43 kJ
CHEMICAL REACTIONS. An example of a gasphase chemical reaction is the combustion of
hydrogen at 100 C and 1 atm pressure (or slightly less) to give water vapor.
o
2 H + O &6 2 H O
2 2 2
The heat of reaction (the amount of thermal energy absorbed by the system) is  242.5
kJ/mol(H O). That is, a negative amount of thermal energy is absorbed by the reacting system, or
2
a positive amount of thermal energy is given up by the reacting system to the surroundings. The
work done on the system is
W =  IP dV = P(V  V ) =  P(n RT/P  n RT/P) =  (n  n )RT
f i f i f i
=  (n)RT (11)
and the energy change of the reacting system is
E = Q + W
= 2 mol( 242.5 kJ/mol)  ( 1 mol) x 8.314 J/mol·K x 373 K
=  481.90 kJ + 3.10 J =  481.90 kJ/(2 mol H O)
2
The same reaction, at 100 C and 1 atm pressure (or slightly more) but producing liquid water
o
rather than water vapor, will have a heat of reaction that differs from that of the pre ceding
Later we will consider work done in other ways, especially by electrical fields. We are
9
explicitly excluding such work terms for the present because they are unnecessary here.
7/10/07 1 20
problem by the heat of vaporization of water.
Q =  481.90 kJ  2 mol x 18 g/mol x 2.258 kJ/g =  481.90 kJ  81.2 kJ
=  563.19 kJ/(2 mol H O)
2
The work differs, also, because V is different. Neglecting the volume of the liquid, V =
(n)RT/P = ( 3)RT/P.
W =  P V = 3 RT = 3 mol x 8.314 J/mol·K x 373K = 9.30 kJ
Then the energy change of the system is
E = Q + W =  563.19 kJ + 9.30 kJ
=  553.89 kJ/(2 mol H O)
2
Enthalpy
We have covered, in brief outline, the methods of finding the change of energy of any system,
by measuring or calculating the amount of thermal energy transfer, Q, and the amount of work
done, W. But already, we have seen that complications arise, such as having to find work done,
against the atmosphere, when that work term is itself of little interest.
The way to avoid these annoying correction terms, in general, is to define a new quality that
focuses on what we do want to know. Our second step, therefore, in understanding an applying
thermodynamics, is to learn how to define and calculate new variables that will simplify our
measurements and calculations.
Chemical reactions and phase changes are more often carried out at constant pressure than at
constant volume. At constant volume (no work done) the energy change is equal to Q, the
amount of thermal energy transferred to the system, but under constant pressure a correction
must be made for the work performed on the system by the external pressure. Consider a
completely general constantpressure process in which the only work done is because of the
volume change.
9
( P) E = Q + W = Q  P(V  V )
2 1
which can be written
E = Q  (P V  P V )
2 2 1 1
where P = P. Set E = E  E and rearrange.
1 2 2 1
Q = E  E + P V  P V
2 1 2 2 1 1
As you might guess, H was chosen as a symbol because H = Q, the heat ( i.e., the
10
amount of thermal energy transfer). One must, however, avoid any attempt to associate H with
the (total) thermal energy. Warning: PV has the dimensions, and units, of energy but, as shown
explicitly in equation 13, PV is not energy.
7/10/07 1 21
= E + P V  (E + P V )
2 2 2 1 1 1
or
(P) Q = (E + PV) (12)
This last equation will be encountered so often that it is a great convenience to introduce a new
symbol for the quantity E + PV.
Definition of Enthalpy H = E + PV (13)
Then, if pressure is constant and the only form of work is W = IP dV, equation 12 becomes
(P) H = Q (14)
The function H is called the enthalpy.
10
Note that the enthalpy is actually defined by equation 13, which contains no restrictions on
pressure, temperature, or volume. Nevertheless, the enthalpy will be found to be most convenient
for problems in which pressure is constant.
A comparison between energy and enthalpy, E and H, can be made by calculating the
enthalpy changes for the same processes for which energy changes were previously found.
TEMPERATURE CHANGES AND HEAT CAPACITY, C. We found that for 2 mol of H warmed at
P 2
constant volume from 25 C to 50 C, E = 1.05 kJ. For the same process, the enthalpy change is
o o
H = (E + PV ) = E + (PV ) = E + V P
= E + V(nRT/V  nRT/V) = E + nRT
f i
= E + 2 mol x 8.3144 J/mol·K x 25 K
= 1.05 kJ + 416 J = 1.47 kJ
The enthalpy change is greater than the energy change because the increase in temperature causes
an increase in P and thus in the product PV.
If the gas is warmed at constant pressure, rather than at constant volume, the enthalpy
change can be calculated in much the same way.
H = E + (PV) = E + P V = E + nR T
= 1.47 kJ
The enthalpy, like the energy, depends only on the temperature, for an ideal gas. It is for this
reason that we find the same H (as well as the same E) when the hydrogen gas is warmed by
25 C whether the process is at constant volume or constant pressure, or under other conditions.
o
dT
q
dT
dH
=
(15)
P
P
C
T
H
=
∫ ∫
=
= dTCdT
T
H
H
P
P
(
)
(
)
RC
dT
RTd
dT
dE
dT
PVd
dT
dE
dT
dH
C
VP
+=+=+==
7/10/07 1 22
It should be observed that in the constantpressure process the correction term to obtain the
enthalpy change from the energy change is a work term, W = P V. But in the constantvolume
process the correction term, though of equal magnitude, is not a work term, having instead the
form V P. (You will learn very quickly that I V dP … W.)
The constantpressure warming process can be treated in a somewhat different manner.
Employing equation 14, for constant pressure,
(P ) dH = q
Then, dividing by dT,
(P )
or
For a diatomic gas near room temperature, C is about 7/2 R = 29 J/mol·K. Inserting this
P
(approximate) value for H,
2
= 2 mol x 29 J/mol·K x 25 K
= 1.45 kJ
The heat capacity at constant pressure, C, is greater than, or occasionally equal to, the heat
P
capacity at constant volume, C. The difference is small for solids and liquids (typically about 1.7
V
J/mol·K), but for an ideal gas the difference is appreciable and is easily calculated. The constant
pressure and constantvolume restrictions can be dropped from the derivatives of equations 6a
and 15 for the special case of an ideal gas, because both the energy and enthalpy of an ideal gas
are independent of pressure and volume. Taking 1 mol of gas,
or
(I.G.) C = C + R (16)
P V
IDEALGAS EXPANSIONS. The work performed on an ideal gas as it expands isothermally was
calculated previously for two important special cases (equations 9 and 10). Because the e nergy
of the ideal gas is independent of pressure and volume, the energy change must be zero in an
isothermal expansion. Thus we conclude that Q + W = 0; the work done on the gas is equal and
An exception is that heat of combustion is given as a positive number (although the
11
amount of thermal energy absorbed by the system is negative) and is therefore  H. If there is
any question concerning the convention followed, look for a reaction such as hydrogen plus
oxygen to give water, where we know energy is given off, so H is negative.
reaction
7/10/07 1 23
opposite to Q, the thermal energy absorbed by the gas in the expansion.
At constant temperature, the product PV is also constant for an ideal gas. Therefore, at
constant temperature, H = (E + PV ) = E = 0. Enthalpy change is also zero for any
isothermal expansion (or compression) of an ideal gas.
One often sees the generalization that a gas cools on expansion. That is generally not true
except when the gas does work on its surroundings (thus giving up energy to the surroundings)
with no compensating transfer of thermal energy to the gas (or for certain situations invol ving
nonideal gases). Obviously there is no cooling in an isothermal expansion.
PHASE CHANGES. The heat of fusion is the amount of thermal energy absorbed by a solid when it
melts (fuses) under constant pressure. From equation 14 this is identically H for the melting
process. Similarly, the heat of vaporization is H. Although E is nearly the same as
vap fusion
H, because of the small volume changes involved, the difference between E and H
fusion vap vap
can be appreciable, as was shown above. Tabulated values are invariably the enthalpy changes.
CHEMICAL REACTIONS. The heat of reaction, as tabulated, assumes constant pressure and is
therefore identical with the H of reaction. That is, it is assumed that the pressures associated
11
with each reactant and each product are the same before and after the reaction, so the total
pressure remains constant during the reaction.
The assumption of constant individual pressures is much less important when considering
heats of reaction than in some later applications. Keep in mind, however, in interpreting
thermodynamic processes, that we are not considering a process that starts with reactants, only,
and concludes with products, only. Rather, we assume the only change is the chemical reaction.
Thus we begin with a mixture of products and reactants and allow the reaction to proceed
without change in the physical conditions (temperature, pressure, or concentrations) of the
substances. One way of achieving this experimentally would be to have the total amount of
material large, and allow only a small part of the reactants to combine. (The measured values are
then calculated per mole of a reactant or a product.) We will find, in Chapter 4, an alternative
method is often to measure a reaction in an electrochemical cell, under equilibrium conditions.
We will return to this assumption of fixed conditions when we explicitly consider chemical
equilibria.
State Functions
We have seen that those properties such as energy, enthalpy, temperature, pressure, and
volume that depend on the state of a system are called state functions. If measurements
carried out in Boston, Brooklyn, Birmingham, and Berkeley are to be compared, we must be
A mole is a certain number of molecules or other particles (Avogadros number, equal
12
to 6.02 x 10 ). Thus it is necessary to indicate what kind of molecules are being counted ( e.g., H
23
is not the same as H ). In practice, the number of moles is determined from the mass and the
2
molar mass (or socalled molecular weight) with units of g/mol understood. At one tim e,
quantities such as gram mole and kilogram mole were defined, but under the modern SI
terminology there is only one mole (symbol mol). Nevertheless, a weight equal to the molar mass
number in pounds may appear under the title of pound mole.
Specific quantities are generally defined as dimensionless ratios of the (ext ensive)
13
properties of a substance to the (extensive) properties of another substance. Thus specific gravity
is the ratio of the density of a substance to the density of water (or, for gases, the ratio to the
density of air). Specific heat was originally the ratio of heat capacity of a substance to the heat
capacity of water. With Q measured in calories, the heat capacity of water is 1 cal/g·K, so the
specific heat was equal to numerical value of the heat capacity in cal/g·K. Over time, specific heat
has been taken to be equal to the heat capacity per unit mass (per degree), regardless of the units
of mass or energy. Therefore, units must now be specified.
7/10/07 1 24
certain that we are dealing with substances in substantially identical states. An important question
is how much information we must have about any system in order to determine its state, and
hence to fix completely its physical and chemical properties.
EXTENSIVE AND INTENSIVE PROPERTIES. If the amount of the system is important, it must be
specified, either by mass or by equivalent measure such as the number of moles. Some
12
properties, such as energy, heat capacity (C or C ), or volume, will be proportional to the
P V
amount of material. These are called extensive properties. Other quantities, s uch as pressure,
temperature, and density, are independent of the amount of material and are called intens ive
properties. Properties such as molar volume, density, and molar heat capacity are examples of
intensive properties derived from extensive properties.
13
Apart from the size of the system, the most obvious variable to control is the composition.
For a pure substance, the degree of purity and the nature of the residual impurities should be
known. For a solution, the concentration of each of the components is required (one less
concentration value than the number of components in the solution). Where there are possible
differences of phase (solid, liquid, or vapor, or more than one crystal structure) the phase studied
must be reported. In addition, each worker must be able to reproduce exactly such properties of
the system as temperature, pressure, volume, density, viscosity, and refractive index.
Fortunately, if each system can be assumed to be at equilibrium (in its most stable form, or
state), not all of these variables are required. If the composition and mass are known, it is in
general necessary to know only two additional variables (with occasional others as already
mentioned). For example, if the temperature and pressure are known the volume can be
calculated (assuming the necessary measurements have been previously made) whether the
substance is solid, liquid, or gaseous, and from this the density, viscosity, and refractive index
follow. To define the state of a gas it would be equally satisfactory to specify the pressure and
volume, or the volume and temperature. On the other hand, knowing that 1 g of water has a
∫
==
2
1
12
E
E
dEEEE
2
1
2
2
2
1
2
12
1
xxxdx
x
x
=
∫
( ) ( )
1122
,
,
,
,
22
11
22
11
yxyxxydydxxdy
yx
yx
yx
yx
==+
∫∫
∫
22
11
,
,
yx
yx
ydx
( )
++=+=
∫∫
1
3
12
3
2
2
,
,
3
3
1
3
3
1
3
2
1
22
11
xxxxdxxydx
x
x
yx
yx
7/10/07 1 25
volume of 1.000132 cm under 1 atm pressure does not determine whether the temperature is 0 C
3 o
or 8.1 C, because the water has a minimum volume, or maximum density, at 4 C. Nor could one
o o
determine the pressure accurately if given the volume and temperature of a liquid or soli d. This
demonstrates that some discretion is required when variables other than pressure and t emperature
are selected, even if extra variables such as electric or magnetic fields are not relevant.
EXACT DIFFERENTIALS AND LINE INTEGRALS. A property, or state function, can depend only on
the present state of a system. It cannot depend on history that is, on how the system arrived at
that state provided we limit the discussion to equilibrium states, so that there can be no strains
in solids, or other residual effects. It follows, therefore, that the change in such an equil ibrium
property, in going from one state to another, can depend only on the initial and final states.
Mathematically, this means that the change, described by an integral, depends only on the limit s of
the integral. For example,
Nothing need be known about the process except the energy of the initial state, E, and the
1
energy of the final state, E, to find E. Integrands of this type are called exact differentials.
2
Most integrals we will encounter are well defined in
themselves. For example,
Some, however, are not defined. For example,
cannot be evaluated without additional information. If y is
known as a function of x, then we can make a plot of y vs. x, as
in Figure 3. Then, for example, if y = x + 3,
2
( ) nRTnbV
V
an
P =
+
2
2
For a more thorough treatment of exact differentials and line integrals see advanced
14
calculus books. There is a direct mathematical test to determine whether or not a function of two
variables is an exact differential.
The constant b is a measure of the volume physically occupied by the molecules, and
15
thus not available to the gas. The constant a is best visualized as arising from the curved paths of
the molecules, because of mutual attractions. This decreases the frequency of wall collisions
7/10/07 1 26
We can say that Iy dx has been integrated along the line shown in Figure 3. This is called a line
integral.
14
Line integrals appear in thermodynamics when it is necessary to evaluate a quantity that
depends on the path. Letting y = P and x = V in the example above gives the familiar integral for
work,
W = IP dV
It is not sufficient to know initial and final states (the end points of the line); the exact path must
be known. This was demonstrated for the calculation of work in Figure 2. Thermal energy
absorbed by a system also depends on the path, so Q is given by a line integral.
It is worth noting that P dV is a change (in P and V, because P changes with V), so P dV is an
inexact differential. W is an amount, not a change in work or any other property, so it is at best
misleading to describe W as a differential (exact or inexact). Careful terminology can be an aid to
understanding. Careless terminology is a hindrance to understanding.
EQUATION OF STATE. Knowledge of two properties, or state functions, is generally sufficient to
know, in principle, all other state functions (for that equilibrium state, if there are no extraneous
variables such as electric or magnetic fields). For example, if the pressure and t emperature of n
moles of an ideal gas are known, the volume can be found from the equation of state, V =
nRT/P, or PV = nRT. Any substance, whether an ideal gas or not, follows an equation of state
relating the pressure, volume, and temperature, but the true equation of state may be a very
complicated function and, indeed, may not be known accurately.
Several functions have been proposed as good approximations to the equations of state for
real gases. The bestknown is that suggested by van der Waals:
In this equation, a and b are constants that depend on the particular gas to be described.
When the volume is large and the temperature is high, the terms involving a and b are negligible
and the gas behaves as an ideal gas; but as the gas is cooled and/or compressed, the behavior
deviates more and more from the idealgas law, until eventually the gas condenses to a liqui d or
solid.
15
without changing their effect, and therefore decreases the pressure on the walls, requiring the
supplement to P to fit the ideal gas equation form; see Phys. Teach. 34, 248249 (April, 1996).
The reader with some knowledge of special relativity may recognize that the total
16
energy of any system is measured by its mass, multiplied by the square of the speed of light.
However, it would be necessary to measure masses about a million times more accurately than is
now possible to be able to determine energies to the accuracy required in thermochemistry.
7/10/07 1 27
It should also be possible to relate the volume of any liquid or solid to its temperature and
pressure, or to express such other properties as refractive index, heat capacity at constant volume
or pressure, thermal conductivity, heats of vaporization or fusion, or vapor pressures of solids or
liquids, in terms of the temperature and pressure. Some of these equations will be encountered in
later chapters.
Thermochemistry
The application to chemical reactions of the principles developed thus far is called
thermochemistry. In particular, the heats of reaction are measured and tabulated and from these
and from measured heat capacities the enthalpy changes are calculated for other rea ctions or for
other experimental conditions.
HESSS LAW. The enthalpy change for a chemical reaction, such as the oxidation of sulfur dioxide
to sulfur trioxide
2 SO (g) + O &6 2 SO (liq)
2 2 3
can be expressed as the difference between the enthalpies of the initial and final st ates.
H = H  H
reaction final initial
= H(2 SO )  H(2 SO )  H(O )
3 2 2
There is no way within thermodynamics of measuring an absolute energy, or an absolute
16
enthalpy. Only energy, and enthalpy, changes can be determined. However, knowing that these
energy and enthalpy changes depend only on the initial and final states, it is possible to add and
subtract chemical reactions and add and subtract the corresponding enthalpy changes. That is, we
may quite arbitrarily select a reference energy and/or enthalpy level and measure all values from
that arbitrary level. In particular, it is possible to tabulate heats of formati on, the enthalpy
changes in the reaction of the elements to form each compound, and from these to calculate
enthalpies of other reactions. This principle is known as Hesss law.
The reactions for the formation of the gases SO and SO from the elements are
2 3
S + O  SO
2 2
S + 3/2 O  SO (liq)
2 3
7/10/07 1 28
The measured enthalpy changes for these reactions at 25 C and 1 atm pressure are 296.90 kJ/mol
o
and 437.94 kJ/mol. Subtraction of the first reaction from the second gives
SO + ½ O &6 SO (liq)
2 2 3
and subtraction of the enthalpy changes gives 141.04 kJ/mol, which is the heat of reaction for the
oxidation of SO to SO (liq).
2 3
Exactly the same elements, in the same quantities, always appear on both sides of a chem ical
equation (which is why reactions as written are called equations). Subtraction of t he elements
from both sides of an equation will yield, on each side, product minus reactants for the reacti ons
of formation of each of the substances appearing in the original equation. In the example above,
the original equation was SO + ½ O &6 SO. Subtract 1 mol of S and 3/2 mol of O from each
2 2 3 2
side. The equation can then be written as the formation of each compound (i.e., of SO, O, and
2 2
SO ) from the elements.
3
(SO  S  O ) + (½ O  ½ O ) &6 (SO  S  3/2 O )
2 2 2 2 3 2
and therefore
H = H (SO )  H (SO )  H (½ O )
reaction form form form3 2 2
= 437.94 kJ/mol 296.90 kJ/mol  0
= 141.04 kJ/mol(SO liq)
3
(Notice that the heat of formation of any element, in its standard state, is necessarily zero.)
An entirely equivalent way of obtaining the same numbers is to consider the enthalpy of each
compound on a scale taken with reference to the elements. Such enthalpy values are called
standard enthalpies of the compounds; they are identical with the standard enthalpies of
formation.
Hesss law can often be applied to find heats of reaction that could not be directly measured
experimentally. For example, the reaction of two molecules of ethylene, C H, to form
2 4
cyclobutane, C H, would not readily occur quantitatively under conditions conducive to
4 8
measurement of the heat of reaction. But both ethylene and cyclobutane can be burned in
oxygen, and subtraction of these reactions gives the reaction equation desired.
2 C H + 8 O &6 4 CO + 4 H O
2 4 2 2 2
C H + 8 O &6 4 CO + 4 H O
4 8 2 2 2
Subtraction of the second from the first gives
2 C H &6 C H
2 4 4 8
and, therefore, subtraction of the H for the second combustion from the H for the first
combustion gives H for the condensation reaction. Heats of combustion (equal to  H )
reaction
are comparatively easy to measure and are often tabulated.
KIRCHHOFFS LAW. The heat of reaction at a temperature other than that given in a table can be
( ) ( )
∫ ∫
++=
1
2
2
1
products1reactants2
(17)
T
T
T
T
PP
dTCHdTCH
( )
[
]
∫
+=
2
1
) reactants(products12
T
T
PP
dTCCHH
7/10/07 1 29
found by calculating enthalpy changes along an arbitrary path. The total enthalpy change is
independent of this choice of path. The method is known as Kirchhoffs law.
Assume that H is known for a reaction at a temperature T and the H at another temperature,
1
T, is to be found. Starting with the hot reactants at T (Figure 4), the reaction could be carried
2 2
out isothermally to obtain products at the same temperature. An alternative path would be to
cool the reactants to the temperature T, carry out the reaction isothermally at T, and warm the
1 1
products to T. The heat of reaction at T is already known and if the heat capacities at constant
2 1
pressure are known, the enthalpy changes can be calculated for the processes of cooling reactants
and warming products. This path must give the same H as the isothermal reaction at T.
2
or,
because interchanging limits of an integral will change the sign,
If the difference in heat capacities is independent of temperature, this may be rew ritten in the form
H = H + [ C  C ](T  T ) (18)
2 1 P 2 1p (products) (reactants)
For example, given that the heat of reaction for rhombic sulfur burning in oxygen to yield
sulfur dioxide gas is  296.9 kJ/mol at 25 C (298 K), find H at 95 C (368 K). The heat
o o
capacities are given in Table 2. Insertion of the numerical values into equation 18 gives
Table 2 HEAT CAPACITIES'
Average values (in J/molK) for temperature ranges indicated
Compound C Temperature, C
P
o
He 20.8 200 up
H 28.8 25 to 200
2
O 29.4 25 to 200
2
H O(g)* 36.4 25 to 200
2
SO (g) 41.9 25 to 200
2
S (r) 23.7 25 to 200
S (m) 25.9 95 to 120
*For rough calculations it is sufficient to set
C (steam) = C (ice) = ½ C (liq H O).
P P P 2
This property associated with state functions has sometimes been confused with a
17
conservation principle. Enthalpy is not conserved.
7/10/07 1 30
H =  296,900 J/mol + (41.9 29.4  23.7) x 70 J/mol
368
=  296.1 J/mol
Sometimes there will be a phase transition during the warming or cooling process. Sulfur has
a phase change at 95 C, at which point rhombic sulfur goes to monoclinic sulfur; the monoclinic
o
sulfur melts at 119 C. The enthalpy changes are 11.78 and 39.24 kJ/mol. The heat of reaction
o
for liquid sulfur burning in oxygen to form SO at 119 C (392 K) can be calculated as follows (see
2
o
Figure 5).
H =  H  C (m) (119  95)  H  C (r) (95  25)  C (O ) (119  25)
392 fusion P tr P P 2
+ H + C (SO ) (119  25)
298 P 2
H =  39,240  25.9 x 24  11,780  23.7 x 70  29.4 x 94  296,900 + 41.9 x 94 J/mol
392
=  349.0 kJ/mol
Note that temperature differences can be found without conversion to the Kelvin scale.
Both Hesss law and Kirchhoffs law are simply applications of the principle that c hanges in
a state function, such as the enthalpy, are completely determined by the initial and final s tates.
17
This principle is combined with the equation arising from the first law that shows that i f the
pressure is constant, the enthalpy change will be equal to the heat absorbed by the system. Thus
the heat of reaction, by which we mean H (at a particular temperature, pressure, and
reaction
concentrations of reactants and products), is only equal to the heat absorbed i f the reaction
proceeds at constant pressure (and at the specified temperature and concentrations). It is
sometimes more convenient to carry out a reaction at constant volume. Then the heat absorbed i s
not equal to the heat of reaction (that is, to H ), but it is still determinate because heat
reaction
absorbed equals the change in energy when the system follows a constantvolume path and
because E is fixed by the initial and final states.
reaction
7/10/07 1 31
The experimental determination of a heat of reaction is called calorimetry. A typical
calorimeter (Figure 6) consists of a reaction chamber, surrounded by a layer of water, e nclosed
by sufficient insulation to prevent heat loss to the surrounding laboratory. The reactants, at r oom
temperature, are placed in the reaction chamber, the calorimeter is closed, and the reaction is
initiated by an electrically heated wire or other controlled energy source. The react ion will
normally be exothermic and the reaction chamber will therefore become quite hot, but the hea t is
conducted into the surrounding water layer so that the products, and the water, reach a final
temperature only slightly above the initial temperature. The E is the same as if the entire
reaction
process had occurred at the initial temperature even though the materials may have becom e quite
hot during the course of the reaction. The heat given off by the reaction is calculated by
observing the temperature rise of the water, using the condition that all heat given off by the
reaction must have been absorbed by the water. Small corrections are required for the cha nge of
temperature, from the initial room temperature, of the products, and for the small amount of
energy added by the hot wire or other initiation method. Some systems may also require a
correction for changes of concentrations during the reaction.
7/10/07 1 32
Problems
1. At room temperature the heat capacity ( C ) of most solid elements (except the very light ones)
V
is about 3 R. Specific heat is the heat capacity per unit mass, or the value in J/g·K. (For solids the
difference between constant volume and constant pressure conditions is small and may be
neglected here.)
a. Find the specific heat of Pb (molar mass = 207.19).
b. Find the specific heat of Cu (molar mass = 63.546).
c. Find E for 25 g of Cu when it is warmed from 20 C to 35 C.
o o
2. Estimate the final temperature when 10 cm of iron (density 7.86 g/cm ) at 80 C is added to 30
3 3 o
ml of water at 20 C.
o
3. Find the work done when 3 mol of N gas at 4 atm pressure expands slowly at 25 C against a
2
o
constant pressure of 1 atm.
4. Calculate the work done when 0.25 mol of SO at 27 C and 1 atm expands reversibly and
2
o
isothermally to a final pressure of 0.20 atm. Find Q for the gas during this process.
5. Calculate the final temperature if an ice cube (25 g) at  5 C is added to a cup of coffee (150
o
ml, or 2/3 cup) at 90 C, neglecting heat loss to the surroundings.
o
6. The heat of vaporization of benzene, C H, is 30.72 kJ/mol and the normal boiling point is
6 6
80.1 C.
o
a. What is E if the vapor is an ideal gas?
vap
b. Part of the thermal energy absorbed by a liquid as it vaporizes to a gas is required to
perform work on the atmosphere as the substance expands to a vapor. What fraction of the totalQ goes into work against the atmosphere when benzene (density about 0.88 g/cm ) vaporizes at
3
its normal boiling point?
7. Find E at 25 C for the oxidation of S (rhombic) to give SO (gas) and for the oxidation of S
o
2
to give SO (liq). The heats of reaction are  296.81 and  441.0 kJ/mol. Assume gases are ideal.
3
8. Calculate H and E for the reaction, at 25 C,
o
SO (g) + ½ O 6 SO (g)
2 2 3
The heat of vaporization of SO at 25 C is 43.14 kJ/mol.
3
o
9. Manganese can be prepared by a thermite process,
3 Mn O + 8 Al  6 9 Mn + 4 Al O
3 4 2 3
The standard enthalpy of formation of Mn O is  1387.8 kJ/mol and for Al O,  1657.7 kJ/mol.
3 4 2 3
Find the amount of thermal energy given off by the reaction as written above, starting with the
reactants at room temperature and ending with the products at room temperature.
10. The interconversion of graphite and diamond does not occur at room temperature. Explain
how you could determine H for this transition by measurements in the laboratory.
11. Calculate the heat of vaporization of water at 50 C.
o
12. Find H for the hydrogenation of ethylene to produce ethane,
C H + H 6 C H
2 4 2 2 6
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