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CHAPTER TWO
INTERNAL ENERGY
AND
THE FIRST LAW OF THERMODYNAMICS
INTRODUCTION
In the last chapter we briefly discussed ways in which a system can interact with its environment, and how
this system can be characterized by an equation of state. This equation of state describes how the macroscopic
parameters of the system are related to one another. As a simple, specific example, we introduced the concept of
an ideal gas which can be completely characterized by only four macroscopic parameters: pressure, volume,
number of molecules (or moles), and temperature. By defining the specific volume, the number of parameters
necessary to specify a unique state of this ideal gas was reduced to three. Since these three parameters are related
to one another through the equation of state, any equilibrium state of an ideal gas can be uniquely specified once
the equilibrium pressure and specific volume of the gas is known (since the temperature is just a function of the
pressure and the specific volume).
By examining how the state variables are related to one another (through a set of partial differential
equations), we showed that even in the case where the equation of state of a system is not known exactly, an
approximate equation of state can be determined by experimentally measuring the change in one system
parameter with respect to another (holding all other parameters fixed). For these state variables, the measured
(or calculated) change in these variables is unique as the system changes from an equilibrium state E to an
equilibrium state F.
Thus, as a thermodynamic system changes from one state to another each of the system properties change in
a well defined manner. In addition, the principle of conservation of energy must also be satisfied. This means
that as the system changes from state E to state F, any change in the energy of the system must be equal to the
energy added to or removed from the system. A more precise statement of this principle is the first law of
thermodynamics. To properly express this law, we must first determine what we mean by the internal energy of a
thermodynamic system, and then we must look carefully at the ways in which energy can cross the boundary of
such a system. For a closed system energy can cross the system boundary as heat or as mechanical work. Unlike
the changes in the state variables, however, the change of these quantities as the system changes from a state E to
a state F depend upon the path taken from E to F, not just the end points.
The Internal Energy of an Ideal Gas and the Ideal Gas Law
A Molecular Model for the Pressure. One of the great successes of classical physics was the derivation of the
ideal gas law based upon the assumption that a gas is composed of molecules which move in relatively simple,
predictable ways. A simple form of this derivation follows.
Consider a gas of R monatomic molecules contained within a volume Z . We assume these molecules
interact only during actual collisions (something like hard spheres), ignoring any long-range interactions. This
assumption is reasonable at low pressures (and large specific volumes) since the average distance between
molecules is extremely large in comparison to the size of the molecules themselves. ÐRemember that the Van der
Waal's interaction potential was of the order "Î<' ÞÑ We assume, however, that these collisions are sufficient to
create a random distribution of the molecules, so that the number of molecules per unit volume 8 œ R ÎZ will be
the same at every point within the container. (For typical systems we can ignore the effect of gravity on the gas
which would tend to increase the density with depth. For large systems, however, this effect cannot be ignored.)
We now zoom in on a differential volume .Z , which contains .R œ 8 .Z molecules. The molecules
within this small volume all have different speeds and are moving in different directions, but we assume that all
directions are equally likely Ði.e., that there is spherical symmetry). Thus, the number of molecules within the
volume .Z with velocity vectors directed so that the molecules would strike the wall of the container within the
Chapter 2: Internal Energy and the First Law
2
area .E indicated in the diagram below is given by
.RH œ .R
.H
.E -9=)
.E -9=)
œ .R
œ 8 .Z
#
%1
%1 <
%1 < #
(2.1)
dV
θ
r
dΩ
dA
Fig. 2.1 The molecules within a small differential volume element .Z have velocity vectors pointing
randomly in all directions (spherically symmetric). We represent the number of molecules within .Z with
velocity vectors pointing toward a particular region of area .E on the container wall by .RH . This number
depends upon the size of the solid angle . H subtended by the area .E. The solid angle depends upon the
area .E, the orientation ) of the differential element .Z relative to the normal, and the distance < from the
differential volume to the wall.
The solid angle H is defined in a manner similar to the definition of the angle in radians. The angle ) in radians is
defined as the arc-length divided by the radius. Similarly, the solid angle H in steradians is defined as the area
perpendicular to the distance measured to the center of that area divided by the square of that distance (usually
denote as a radius). Since the area of a sphere is 41<# , the solid angle subtended by a sphere is
H œ %1<# Î<# œ %1.
Of all the molecules in volume .Z only a certain fraction of these have a velocity magnitude between @
and @ .@. We represent this fraction by the expression 1@ .@, so that the number of molecules within the
volume .Z with velocity between @ and @ .@ can be expressed by the equation
.R@ œ .R 1@ .@ œ 8 .Z 1@ .@
(2.2)
We know that the individual molecules are constantly making collisions which change both the magnitude and the
direction of their velocity vectors. However, we assume that the average number of molecules with a particular
magnitude and direction is always the same. This will be true, however, only if the size of the volume element
.Z is large enough to contain a very large number of molecules.
We express the number of molecules that have a particular speed ranging between @ and @ .@ and that
are also moving in a direction which would cause them to strike the area .E, by the equation
.RHß@ œ .RH 1@ .@ œ 8 .Z
.E -9=)
1@ .@
%1 <#
(2.3)
This relationship gives the number of molecules within a particular differential volume .Z with a speed @
(between @ and @ .@) moving in a direction which would allow them to strike the wall area .E. This particular
Chapter 2: Internal Energy and the First Law
3
differential volume is located a distance < away from the wall at an angle ). To determine all of the molecules in
the gas which have a particular speed @ and which are directed in such a way that they would strike the wall area
.E we must integrate over all the possible values of < and ). This means that we must express the differential
volume .Z in terms of spherical coordinates. Choosing the origin at the center of the area .E, and letting the D axis point into the volume containing the gas, the differential volume can be expressed as
.Z œ <# =38) .< . ) . 9
(2.4)
.E -9=)
1@ .@ <# =38) .< . ) . 9
%1 <#
(2.5)
giving
.RHß@ œ 8
This can be integrated from 9 œ ! to 9 œ #1, from ) œ ! to ) œ 1Î#, and from < œ ! to some maximum < value.
However, if we let .< œ @ .>, we can determine the number of molecules per second which actually strike the area
.E with a speed @. Making this substitution, we obtain
.E -9=)
1@ @ .> .@ <# =38) . ) . 9
%1 <#
.E .>
œ8
1@ @ .@ =38) -9=) . ) . 9
%1
.RHß@ œ 8
.RHß@
(2.6)
If we integrate over a particular area ?E of the container wall for a period of time ?>, we find that the number of
molecules striking the wall per unit area per unit time with a particular speed @ coming from a particular angular
direction designated by ) and 9 is given by
.RHß@
8
œ
1@ @ .@ =38) -9=) . ) . 9
%1
?E ?>
(2.7)
The boxed equation can be integrated over all angles and over all speeds to give the total number of
collisions with the wall per unit area per unit time
R
8 ∞
8
œ ( @ 1@ .@ œ Ø@Ù
?E ?>
% !
%
(2.8)
In the last equation, the integral of the speed over the speed distribution is just the average speed. If we knew this
average speed, we could use this last equation to determine the rate at which the molecules leak out of a hole of
area ?E (provided the hole is small enough that the random nature of motion inside the container is not
modified).
Likewise, the boxed equation can be used to determine the pressure exerted on the walls of the container.
By definition, the pressure is the perpendicular force per unit area acting on the wall surface. The force exerted
by the molecules on the wall is equal and opposite to the force exerted on the molecules by the wall. If we
consider a single molecule striking the wall, moving with a speed @ at an angle ) relative to the normal, the
perpendicular force exerted by the wall on the molecule is given by
JD œ
7@ -9=) " Ð " 7@ -9=)Ñ
# 7@ -9= )
?: D
œ
œ
?>
?>
?>
(2.9)
provided the collision is perfectly elastic. Thus, the differential pressure exerted on the wall of the container by
the molecules moving with a speed @ at an angle ) must be the force per unit area times the number of molecules
which strike the wall with a speed @ at an angle ), or
JD .RHß@
# 7@ -9=)
8
œ
‚
1@ @ .@ =38) -9=) . ) . 9 ?E ?>
?E
?E ?>
%1
JD .RHß@
8
.T œ
œ
1@ 7@# .@ =38) -9=# ) . ) . 9
?E
#1
.T œ
(2.10)
Chapter 2: Internal Energy and the First Law
4
To obtain the total pressure on the wall of the container, we integrate over all angles and speeds to obtain
T œ
1Î#
#1
8 ∞ #
8
#R "
=38) -9=# ) . ) ( . 9 œ Ø7@# Ù œ
Ø 7@# Ù
( 7@ 1@ .@ (
#1 !
$
$
Z
#
!
!
(2.11)
The Maxwell-Boltzmann Velocity Distribution. To actually proceed further, we must know the
functional form for the distribution of molecular speeds within a monatomic ideal gas. The relative distribution
for the B-component of the velocity (with similar equations for the C- and D -components) is known as the
Maxwell-Boltzmann distribution, and is given by
0 @B .@B œ G/"7@B Î#5X .@B
#
(2.12)
where G is the normalization constant. You will notice that the exponential terms contains the B-part of the
kinetic energy divided by the factor 5X . This functional form for the distribution of molecular velocities can
actually be derived for a model gas of monatomic molecules using some of the statistical techniques which we
will develop later.
To evaluate the average value for the magnitude of the velocity and for the kinetic energy we need to
know how to evaluate Gaussian integrals of the form
O8 œ (
∞
M8 œ (
∞
#
B8 /"-B .B
(2.13)
"∞
and
#
B8 /"-B .B
(2.14)
!
where 8 œ !ß "ß #ß á . The techniques for evaluating these integrals is given in the appendix to this chapter. A
table of these integrals is given below.
Table 2.1: GAUSSIAN INTEGRALS
#
∞
Values of the integral of the form: M8 œ ' B8 /"-B .B
0
8
M8
"
#
!
1"Î# -""Î#
"
#
"
"
%
#
1"Î# -"$Î#
"
#
$
$
)
%
-""
-"#
1"Î# -"&Î#
-"$
&
If 8 is even:
O8 œ '"∞ B8 /"-B .B œ #M8
If 8 is odd:
O8 œ '"∞ B8 /"-B .B œ 0
∞
∞
#
#
We are now in a position to determine the average velocities and kinetic energies of a gas for which the
Maxwell-Boltzmann distribution is valid. First let's evaluate the normalization constant G . We do this by
Chapter 2: Internal Energy and the First Law
5
normalization
(
∞
"∞
where - œ
7
#5X .
0 @B .@B œ G (
∞
"∞
/"7@B Î#5X .@B œ #G (
#
∞
!
" " "
#
/"-@B .@B œ #G ” 1 # -" # • œ "
#
(2.15)
Thus, the normalization constant G is given by
G œ - # 1" # œ Ê
"
"
7
#15X
(2.16)
The distribution function for the B-component of the velocity, then, is given by
0 @B .@B œ Ê
7
#
/"7@B Î#5X .@B
#15X
(2.17)
f(vx )
and a plot of this function is shown in Fig. 2.2 below.
0
vx
Fig. 2.2 The Maxwell-Boltzmann distribution for the B-component of the velocity.
Notice that this function is symmetric about the point @B œ !. This means that the number of molecules
having a certain positive value of the velocity is equal to the number having that same negative value Ðjust as
many are moving to the right as are moving to the leftÑ. Also notice that this function has its maximum value for
@B œ !. This means that the most likely value of the B-component of velocity (and also the C- +8. D -components)
is zero! If we calculate the average value of the B-velocity using our defining relations,
Ø@B Ù œ (
∞
"∞
@B 0 @B .@B œ (
∞
"∞
@B Ê
7
#
/"7@B Î#5X .@B
#15X
(2.18)
we get zero, since the integrand is a product of an even and an odd function and we integrate over symmetric
limits.
The fact that the average value of the components of the velocity vector is zero does not mean that the
average of the velocity magnitude is zero. To understand this, we must look at the complete velocity distribution
function for all values of the components. This is given by the product of each of the individual velocity
distribution equations, or
0 Ð@B , @C , @D Ñ .@B .@C .@D œ ’
7 $Î# "7@# Î#5X
.@B .@C .@D
“ /
#15X
(2.19)
where @# œ @B# @C# @D# . This distribution function can be written solely in terms of the magnitude of the
velocity vector by making a transformation from three-dimensional Cartesian coordinates to three-dimensional
spherical coordinates. The differential volume .@B .@C .@D can be written in spherical coordinates in the form
.@B .@C .@D œ @# =38) . ) . 9 .@
(2.20)
Chapter 2: Internal Energy and the First Law
6
Now, since there is no dependence on the angles ) and 9 we can integrate over these variables and obtain a new
distribution function which depends only on the magnitude of the velocity vector:
1Ð@Ñ .@ œ ’
7 $Î# "7@# Î#5X
%1@# .@
“ /
#15X
(2.21)
g(v)
A plot of this function is shown in Fig. 2.3 below.
0 v
m
v
vrms
vavg
Fig. 2.3 A plot the the Maxwell-Boltzmann speed distribution function, showing the most
probable speed @7 , the average speed @+@1 or Ø@Ù, and the root-mean-square speed @<7= or ÈØ@# Ù.
The speed distribution function goes to zero as the velocity magnitude goes to zero, indicating that all
particles are moving with some velocity (even though the directions may actually cancel out). In fact, the most
probable speed of a molecule in the gas (for a given temperature) is given by
@7 œ Ê#
5X
7
(2.22)
Two other quantities which can be calculated from the speed distribution equation are the average speed
Ø@Ù, which is given by
Ø@Ù œ Ê
) 5X
1 7
(2.23)
and the root-mean-square speed, given by
@<7= œ ÈØ@# Ù œ Ê$
5X
7
Exercise 2.1. Begin with the speed distribution equation and derive the expression for the
most probable speed.
Exercise 2.2. Use the speed distribution equation to find an expression for the average speed
of the molecules.
Exercise 2.3. Beginning with the speed distribution equation derive the expression for the
root-mean-square speed.
(2.24)
Chapter 2: Internal Energy and the First Law
7
The Internal Energy and the Ideal Gas Law. We are finally in a position to determine the average kinetic
energy of the molecules in our model system. To do this we must evaluate the integral.
ØI538 Ù œ (
∞
!
"
7 $Î# "7@# Î#5X
#
%1@# .@
“ /
Œ 7@ ’
#
#15X
(2.25)
where you will notice that although the limits of integration for the velocity components go from minus infinity to
plus infinity, the limits of integration for the magnitude of the velocity go from zero to infinity. Evaluation of this
integral gives
ØI538 Ù œ
3
5X
2
(2.26)
The pressure, then, is given by
T œ
R 2
R# 3
R
ØI538 Ù œ
5X
Š 5X ‹ œ
Z 3
Z $ 2
Z
(2.27)
or
T Z œ R5 X
(2.28)
which is just the ideal gas equation!
We have derived the ideal gas equation using a simple model and making very few assumptions
(principally that of random motion and random distribution). But we have also gained a new piece of information
in the process. By introducing a specific model for the underlying atomic system, we have shown that the
pressure of the gas is a measure of the average kinetic energy of the molecules and this average kinetic energy is
simply a function of the temperature. Our simple model, however, has not included any intermolecular forces.
We would expect there to be some slight attractive force between molecules (as we mentioned in the discussion of
the Van der Waal equation) arising from an induced dipole interaction. And we would expect the average
potential energy to be a function of the average distance between the molecules. If the average distance between
molecules is large enough, then the effects of these intermolecular forces should be negligible. Thus, for large
specific volumes we would expect our model (the ideal gas model) to fairly accurately represent a real gas.
If we wish to include the fact that the molecules of a real gas do interact with each other, we can include
an average potential energy term, along with the kinetic energy of the molecules and express the total internal
energy of our gas by the equation
ØI> Ù œ ØOÞIÞÙ ØY Ù
where ØY Ù depends upon the average distance between the molecules, and, therefore, the average density of the
system. If the volume of the system is held fixed during some process, we would expect that the average potential
energy of the system would remain more or less unchanged. In this case, the total energy of the system would just
be equal to the kinetic energy of the system. For this reason, a gas undergoing processes at constant volume
should act much more like an ideal gas than one which occurs at constant pressure.
For simplicity, however, our model for an ideal, monatomic, gas will ignore intermolecular forces so that the
total internal energy of this gas is just the average translational kinetic energy of all the molecules making up the
system. Thus, if we add energy to this system in the form of heat or by doing work on the system, the
conservation of energy demands that the internal energy must increase. This means that the average kinetic
energy of the molecules increases, which shows up as an increase in temperature.
But what if we have a gas that is not monatomic? Since polyatomic molecules can vibrate and rotate as well
as translate, we must consider the energy associated with each of these types of motion in adition to the kinetic
energy of translation. Although we used the Maxwell-Boltzmann equation to find the average kinetic energy of
our system, we can make use of the generalized Maxwell-Boltzmann equation to determine the average energy of
any system. The generalize Maxwell-Boltzmann equation is given by
ØIÙ œ ( ( â(
∞
GI /"IÎ5X . 7
"∞
(2.29)
where I is the total energy of the system, and where d7 is a differential “volume” given by # .:3 .;3 , where the
Chapter 2: Internal Energy and the First Law
8
:3 and ;3 are generalized momenta and coordinate pairs (for example the angular momentum P) and the angle )).
The integral is taken over all the accessible values of these coordinate pairs. The normalization constant G is
determined as usual by requiring that
" œ ( ( â(
∞
G œ ”( ( â (
∞
G /"IÎ5X . 7
(2.30)
"∞
or
"∞
/"IÎ5X . 7 •
""
(2.31)
It turns out that for any quadratic term in the expression for the energy, the average energy associated with that
quadratic term is "# 5 X Þ This is a statement of the classical equipartition theorem. Thus, for the case where the
energy of the system is due only to translational motion in three dimensions, i.e., where
Iœ
:C#
:B#
:#
D
#7 #7 #7
(2.32)
we have three quadratic terms, or three degrees of freedom, so that the average energy is
ØIÙ œ
$
5X
#
(2.33)
In general, then, the average energy of a particle is given by
ØIÙ œ
Á
5X
#
(2.34)
where Á is the number of degrees of freedom (i.e., the number of quadratic terms in the expression for the total
energy) of the system. Thus, the temperature of a system is a measure of the average total energy of the
molecules of the system. This internal energy is a consequence of the fact that the system is composed of
microscopic molecules which can move about in many different ways. Each type of motion has associated with it
kinetic and/or potential energy. The simplest case is the one where we have a monatomic gas which has three
degrees of translational freedom. A diatomic molecule, however, can not only move about from place to place,
but it can rotate and vibrate as well. The average energy for each degree of vibrational and rotational freedom
(for each quadratic term in the total energy) is "# 5X , so that a diatomic molecule, which can translate in three
directions, rotate about three different axes, and vibrate along one internuclear axis with both kinetic and potential
energy terms, has a total of 8 quadratic terms resulting in an average total internal energy per molecule of 45X .
You should notice, however, that the derivation of the ideal gas law shows how the pressure of the gas
depends upon the average kinetic energy of the system, not the average total energy of the system. This means
that the ideal gas law equation is valid even for polyatomic molecules. The internal energy of a system of
polyatomic molecules, however, will vary depending upon the number of degrees of freedom in the system. And
remember, our model of an ideal gas assumes that intermolecular forces are negligible. For a real gas, there are
forces acting between the individual particles which will generally be a function of the average distance between
the molecules. Thus, for real gases, the pressure on the wall of the container will differ somewhat from that
derived for an ideal gas, and the total internal energy of the gas must take the interaction potential energy (which
depends upon the average distance between the molecules) into account.
Exercise 2.4. The generalized Maxwell-Boltzmann equation has the form
ØIÙ œ ( ( â(
∞
GI /"IÎ5X . 7
(2.35)
"∞
where . 7 is a generalized volume element in both position and momentum spaceß and where
G œ ”( ( â (
∞
"∞
/"IÎ5X . 7 •
""
(2.36)
Chapter 2: Internal Energy and the First Law
9
For example, assume that the total energy were a function of the position B and the momentum :B ß
i.e., I œ IÐBß :B Ñ œ :B# Î#7 "# 5B# . Then we would write
ØIÙ œ G ( (
∞
"
"
#
#
Ð:B# Î#7 5B# Ñ /"Ð:B Î#7 # 5B ÑÎ5X .B .:B
#
"∞
(2.37)
This integralß then, has the form
ØIÙ œ G ( (
∞
"∞
#
#
Ð+" ?#" +# ?## âÑ /"Ð+" ?" +# ?# âÑÎ5X .?" .?#
(2.38)
Carry out this integration process and show that for each quadratic term you get a factor of "# 5X .
This is the classical equipartition theorm.
THE CONCEPT OF HEAT
THE EQUIVALENCE OF MECHANICAL ENERGY AND HEAT ENERGY
Let us assume that we have two ideal gas systems, say E and F, each at a different temperature, with
XE 9 XF . This means, of course, that the internal energy of system E is greater than the internal energy of system
F, or %E 9 %F , where % œ IÎ7. If we bring the two systems together, we find that the hotter system will cool
off, while the cooler system will heat up, corresponding to a transfer of energy across the boundary of the systems
due solely to a temperature difference between the two systems. We use the term 2/+> for the amount of energy
transfered in this manner. The amount of heat U which is added to (or removed from) each system canß in many
cases, be expressed in terms of the temperature change ?X of either system by the equation
U œ 7- ?X
(2.39)
where 7 and - are the mass and the specific heat capacity, respectively, of the system of interest.
We find experimentally that the heat capacity of the gas depends upon whether the gas is allowed to
expand while heat is being added to the system. The heat capacity of a gas which is held at constant volume
(where the gas is not allowed to expand) is found to be larger than the heat capacity of the gas at constant
pressure. Thus the heat added to the system depends upon the particular process, not just on the change in
temperature. The fact that the heat capacity is different in these two cases, and the fact that an expanding gas
must do work on its surroundings would seem to indicate that the heat added to the system (the enclosed gas) and
the mechanical work done by the system are directly related to one another.
As mentioned in the last chapter, this equivalence of mechanical work and heat energy was first postulated
by Count Rumford, while supervising the boring of cannon in Bavaria. Prior to Rumford's time, it was widely
believed that the amount of heat which a system contained was somehow fixed (much like the mass or charge of a
substance), and that the amount of heat which a system contained was a state variable. Rumford observed,
however, that a seemingly infinite amount of heat could be removed from a cannon as long as mechanical energy
was being supplied to that cannon (i.e., as long as the boring process continued). To cool the cannon while it was
being bored, the barrel of the cannon was filled with water. The water boiled away and had to be constantly
replenished during the boring process, but the temperature of the water, and thus the cannon remained relatively
constant.
The observations of Count Rumford can be stated in the following way. For a system in a steady state (i.e.,
the cannon barrel being maintained at relatively constant temperature), the amount of heat U which was
transferred from the cannon (and which entered the water) was equal to the amount of mechanical work [ done
on the cannon (due to friction), or
Uœ[
(2.40)
Thus mechanical energy and heat energy are in some way equivalent. The relationship of one to the other is often
stated in the following way: the mechanical energy is associated with the organized motion of the particles
making up a system, while heat is associated with the random motion of these particles.
Chapter 2: Internal Energy and the First Law
10
A couple of examples where mechanical energy is changed into an equal amount of heat energy can be easily
demonstrated:
1.
2.
Consider an object which is sliding across the floor with a speed @9 . This object has
mechanical energy in the form of kinetic energy (organized motion of the entire object
moving in the same direction). As friction between the floor and the object slow the object
down, the object is heated up (a transformation of organized mechanical motion to
disorganized random motion of the molecules making up the object). The object eventually
comes to a complete stop, so that the mechanical energy in the form of kinetic energy has
now been transformed into heat energy which has been distributed to the object itself and to
the floor during the sliding interaction.
Likewise if a container filled with a gas or liquid is lifted a height h, and then released, the
potential energy in the system is then transformed into kinetic energy of motion as the
container falls. Both the potential energy and the kinetic energy of this system are due to
the organized motion of the system. However, once the container strikes the floor, the
contents of the system will be heated as turbulence is generated or pressure waves move
back and forth within the material, or as the material is compressed and deformed,
generating random motion of the particles which make up the system.
James Joule ( ¶ 1840) performed several different experiments to try and demonstrate the quantitative
relationship between mechanical energy and heat energy. One such experiment consisted of a stirring devise
which was placed in a water bath. The stirring device was set in motion as a mass 7 fell through a distance 2.
The gravitational potential energy in the system was transformed into heat energy as the water in the system was
stirred. Another, more accurate experiment, allows a current to pass through a wire which is submerged in water.
The rate at which energy is added to the water (in the form of heat) is equal to the energy loss in the resistor
(T œ M # V). Modern versions of Joule's experiments show that the relationship between the conventional units for
heat and mechanical energy is given by
4.186 Joules = 1 calorie
(2.41)
A Note on Notation. When a state variable, such as the volume, changes by some small amount, we will
express this by the exact differential .Z which is independent of the particular process. However, when a small
amount of heat is transferred across a system boundary, we will use the symbol $U since the actual amount of heat
transferred depends upon the actual process involved. Thus, the amount of heat $U transferred in some process is
not an exact differential. Likewise, we will express a small amount of work done on or by a gas in a given
process by the symbol $[ , since it is not an exact differential.
Thus, if we calculate the change in pressure of a gas as the gas is carried from state E to state F we can
write
F
?TEF œ TF " TE œ (
.T
(2.42)
E
where the symbol ?T represents the difference in pressure for two different states of the system. However, if we
allow heat to flow into our system as the system changes from state E to state F, we cannot write
***WRONG***
?UEF œ UF " UE œ (
F
$U
***WRONG***
E
because $U is not an exact differential, and the amount of heat added to the system depends on the process, not
just the endpoints. For this reason, we do not use ?UEF to represent the amount of heat added to the system
during a process (this notation implies the difference between two numbers associated with states E and F), but
rather UEF . Similarly, we do not write ?[EF to denote the work done as a system goes from state E to state F,
but [EF .
The Distinction Between Heat and Work
Chapter 2: Internal Energy and the First Law
11
In the last section we considered two systems at different temperatures that were brought in contact. The
internal energy of each system changed as energy flowed across the boundary between these systems. When
energy is transferred across the boundary of a system solely as a result of a temperature difference, we say that
heat energy (or just heat) flows from one system to the other. One might think of this as a transfer of microscopic
or random motion energy, due to a difference in the vibrational or translational energy on either side of a
boundary between two systems. This kind of energy transfer is known as conduction.
Energy transfer across a system boundary which is associated with organized motion is called work.
This organized motion, or work can take on many different forms. For a closed system, the most obvious example
of this work is the motion of a movable boundary between a system and its surroundings. As this boundary
moves, the system volume increases or decreases, and energy is either added to or removed from the system. For
an open system, one where we allow mass to enter and leave the system, we have work which is associated with
the motion of the fluid throught the system, even if the boundaries of the system are fixed.
Another example of organized motion across a system boundary (work) is the organized motion of
electrons moving across the boundary of a system through a wire. We call this electrical work. The energy
transfered across the boundary is given by
#
["# œ ( Z M .>
"
since $[ /.> œ Z M where V is the potential difference and I is the current. There are a number of other processes
in which energy is added to or removed from a system via interactions of the components of the system with
external magnetic and electric forces. These interactions are also designated as work, and must also be taken into
account for a general system. To keep things relatively simple, however, we will be working with simple systems
in which these effects are often neglegible.
When working problems, it is often extremely important to precisely specify the system of interest and
the system boundaries. To illustrate how the choice of the system and system boundaries can change the
appearance of a problem, consider the example below.
Example 2.1
This example illustrates the importance of clearly designating the system of
interest. First consider a well insulated oven heated by an electric coil. We want to
consider the air in the oven as our system. The boundary of our system is the inside walls
of the oven (which we consider adiabatic - allowing no heat to be added to or removed
from the air) and the outer surface of the heating coil. Since the heating coil is outside the
system, it is part of the surroundings, so when the heating coil heats up due to electrical
heating, the temperature difference between the coil and the air causes an energy transfer in
the form of heat from the coil to the air.
Now we want to consider the oven and coil as our system. The boundary of the
system is now the inside wall of the oven and constitutes an adiabatic boundary which
prevents heat from entering or leaving the system. When the oven is on, energy is being
added to the air in the oven since the temperature of this air rises, but this cannot be due to
heat transfer from the surroundings to the system. In this case, the energy transfer is
caused by electrical work being done as electrical charges are moved through the heater
element due to a potential difference. The energy added to the system is given by
#
["# œ ( Z M .>
(2.43)
"
since $[ Î.> œ Z M where V is the potential difference and I is the current. In this latter
case, then, the air temperature of the air in the oven increases, not due to a temperature
difference between the system and its surroundings, but due to mechanical work being
done on the system due to external forces.
T -.Z Work For Simple Gases In Closed Systems
Chapter 2: Internal Energy and the First Law
12
In order to more clearly understand the implications of the equivalence of heat and work, let's examine
how work can be done on or by an ideal gas. Consider our idealized cylinder containing an ideal gas (Fig. 2.4).
Remember that the lid is tight enough to prevent any of the gas from escaping, but loose enough to provide a
frictionless fit.
If the gas in the container is allowed to expand quasi-statically by a small amount .B then the amount of
mechanical work which is done by the system on the surroundings is expressed by the equation
$[ = [T38> E] .B ,
(2.44)
where .B increases as the volume increases, and where T38> is the internal pressure (which, because the process is
quasi-static, is the same at all points in the gas). Now in this quasi-static process, the external pressure at the
boundary T,9?8.+<C is opposite in sign but equal in magnitude to the gas pressure, T38> , inside the container (i.e.,
T,9?8.+<C = " T38> ). The boundary pressure will be equal to the atmospheric pressure To plus the pressure due to
the weight of the lid (763. 1ÎEÑÞ We typically drop the subscript on T38> and simply write T for the internal
pressure.
P
B
dL
A
L
P
VB
VA
V
dV
(a)
(b)
Fig 2.4 As the gas in the cylinder expands, the movable piston rises an amount .P as shown in (a). The
work done by the gas during this process is . w [ œ J .P œ T E .P œ T .Z as shown in (b). The total
work done as the gas is carried from state E to state F along the path indicated is the area under the
curve.
Thus, Equation 2.44 can be written in terms of the volume change, as follows:
$[ œ T E .B œ T .Z
(2.45)
where .Z œ E .B is the differential volume change in the gas. Thus, the work done as the system expands
reversibly (or quasi-statically) from state E to state F is given by
B
[AB œ ( T .Z œ ØT Ù (ZB " ZA )
(2.46)
A
which is just the area under a T -@ curve describing the appropriate process (see Fig. 2.4b). In this last equation,
ØT Ù is the average pressure during the process. We note that if the volume increases (i.e., if .B is positive, ZB 9
Chapter 2: Internal Energy and the First Law
13
ZA ) there is work done by the system, which is positive, whereas if the volume decreases (i.e., if .B is negative,
ZB B ZA ) there is work done on the system, and this is negative. The sign convention associated with work
arises from the attitude that energy being derived from a mechanical system is desirable - a positive thing, while
energy being consumed by a mechanical system is undesirable - a negative thing.
ÒR 9>/ À $ [ /B>/<8+6 œ " $ [ =C=>/7 Ó
The Work Done by a System is Path Dependent As mentioned earlier, the work done by a system as it
moves from state E to state F depends upon the particular process involved (or the path on the T ß @ß X surface
which is traced out). This means that the differential amount of work $[ done in a given process depends upon
the path and is not unique. Changes in the pressure, temperature, and volume of the system, however, do not
depend upon the process, since these are state variables.
To fully grasp the implications of the path-dependent nature of the work done by a system, we will consider
the work done by an ideal gas as it is carried through the different processes shown on the T -Z diagram in Fig.
2.5. First, let's consider the process whereby the gas in a cylinder undergoes an isobaric (constant pressure)
process as it expands from state E to state F. The work done by the gas in the process is given by
F
F
[EF œ ( T dZ œ T ( dZ œ T ÐZF " ZE Ñ œ T ?ZEF
E
(2.47)
E
or
[EF œ 8VXF " 8VXE œ 8VÐXF " XE Ñ œ R 5ÐXF " XE Ñ œ R 5 ?XEF
(2.48)
which is just the area under the T -Z curve. Since the final volume is greater than the initial volume, the work
done in this case is positive (it is work done on the surroundings by the system). Since the system is doing work
on the surroundings (i.e. energy is being transferred across the boundary of the system) the system must be
loosing energy during this process. However, the temperature of the system is increasing (notice the isothermal
lines) during this process. This means that although the system is loosing energy by doing work on its
surroundings, the internal energy of the system (which can be measured using the temperature) must be
increasing. The change in internal energy must be given by
?I œ R ”
Á
Á
5 ?XEF • œ R 5 ?XEF
#
#
(2.49)
where Á $ for a real gas is three dimensions. By comparing Equ. 2.48 with Equ. 2.49 we see that the increase
in internal energy must be greater than the work done (i.e., the mechanical energy lost) by the system.
P
TB > TA
A
PA
B
TB
C
PB
TA
VA
VB
V
Fig. 2.5. A T " Z diagram for an ideal gas as it is carried through a cyclic process: EpFpG pE.
Process EpF is isobaric (a constant pressure process), while process FpG is isochoric (or isometric,
Chapter 2: Internal Energy and the First Law
14
i.e., a constant volume process), and process G pE is isothermal (i.e., a constant temperature process).
The work done in each process if the area under the curve. The work is positive if the volume increases;
negative if the volume decreases during a process (and it is zero if the volume does not change).
Next consider the isochoric process from state F to state G . In this process, the pressure drops as the
temperature of the system drops, but the volume remains constant. Thus, no work is done by (or on) the system
during this process! However, the internal energy of the system must change since the temperature of the system
changes! This means that energy, in the form of heat, is being transferred out of the system. We express this
mathematically by writing
.I œ $ U
(2.50)
where .I is the change in internal energy and $ U is the amount of heat added to (or removed from) the system.
If .I is positive, then $ U must also be positive; if .I is negative, then $ U must be negative. Thus, we define the
heat which passes through the system boundary as the amount of energy which is lost (or gained) by a system as
the temperature of the system drops (or rises) when there is no mechanical interaction between the system and its
surroundings: heat is the amount of energy transfered solely as a result of a temperature difference.
Finally, the system is returned to the original state E by means of an isothermal compression (G Ä E).
During this process, mechanical work is done on the system to compress it. This means that energy is being
added to the system by the mechanical interaction of the system with its surroundings. The work done by the
system in this process is given by
E
[GE œ ( T dZ œ 8VX (
G
E
dZ
œ 8VX lnÐZE /ZG Ñ
Z
(2.51)
G
where T œ 8VX ÎZ for an ideal gas. Again, this is just the area under the curve. But this area is negative, since
ZE B ZG . Since the temperature remains constant for this process, we know that the internal energy of the system
(an ideal gas) must remain the same. Since work is being done on the system (i.e., energy is being added to the
system by the surroundings), and since the internal energy is not changing, the system must be loosing energy in
the form of heat.
The First Law of Thermodynamics for Closed Systems
We can summarize our discussion to this point by a simple mathematical expression, called the first law of
thermodynamics:
.I œ $ U " $ [
(2.52)
which is basically a statement of the conservation of the total energy for a system. In this equation, .I is the
differential change in the internal energy of the system. The equation expresses the fact that the internal energy
of a system is a state variable, and that this state variable can be increased (or decreased) by an addition (or
removal) of heat from the system and by allowing the system to do work on the surroundings (or to be worked on
by the surrounding). By convention, the inexact differential $U is the differential amount of heat added to the
system: it is positive for heat added to the system; negative for heat removed from the system. Likewise, the
inexact differential $[ is the work done by the system: it is positive if work is done by the system on the
surroundings [corresponding to an expansion of the system], and negative if work is done on the system
[corresponding to a compression of the system]. This form of the first law assumes: 1) that the system is not
moving as a whole, so that any changes in the kinetic or the gravitational potential energy of the system as a
whole are ignored; and 2) that the system is closed, so that no mass can enter or leave the system.
In the case of a closed system, where the gravitational potential and kinetic energy of the sytem as a
whole must be considered, the first law can be written in the form
$ U " $ [ œ .I>9> œ .I .IOMR .IK"T SX
where .I is the internal microscopic energy, .IOMR is the kinetic energy of the system as a whole, and
.IK"T SX is the gravitational potential energy of the system as a whole. As an example, you might consider a
block of wood. This block of wood could be perched upon the top of a table where it is subsequently knocked off
Chapter 2: Internal Energy and the First Law
15
and falls to the floor. We might want to look at the energy changes in that system which might cause an increase
in the temperature of the block as a result of the fall. As a general rule, however, we will be considering only
stationary systems.
For the case of a stationary system which is open (such as a water heater), mass is allowed to pass
through the system. There will usually be some net energy transport across the system boundary as this mass
enters and leaves the system. In the case of a hot water heater, the cold water entering the system obviously has
less internal energy per unit mass than the hot water which leaves the system. Thus, there is a net energy transport
out of the system due to mass transport. For a brief treatment of open systems, refer to the appendix at the end of
this chapter.
Now, look back at Fig. 2.2. As the system moves from state E to state F to state G and then back to
state E, the net work done in the process is positive, since the positive work of expansion is greater than the
negative work of compression. If, however, we had retraced our steps from state G back to F and then back to E,
the work done by the system would have been zero! Again we see that the amount of work which one can extract
from a system is dependent upon the process and not just the endpoints. This means that the work which is done
on or by a system is not due to a change in a state variable: there is not a certain amount of work which is stored
in a system.
This same observation can be made about the amount of heat which can be removed from a system. As
Count Rumford observed while boring the cannon, the mechanical energy supplied to the cannon (i.e., work done
on that system) must be equal to the heat energy removed from the cannon, since the water boiled at constant
temperature (and so the internal energy of the system could no change). This means that the amount of heat
added to or taken from a system must also depend upon the process and not just the endpoints. Heat is not a state
variable. Mathematically, this means that the heat added to (or taken from) a system as well as the change in
mechanical energy of the system cannot be expressed in terms of exact differentials. One way of expressing this
is
* $[ Á !
* $U Á !
and
(2.53)
And so we write $ U and $ [ , respectively, for the differential amount of energy which crosses the boundary in
the form of heat and mechanical energy to remind ourselves that this symbol stands for a small change, not an
exact differential.
At the beginning of this chapter, we argued that the internal energy of an ideal gas containing R particles is
given by
I œ R”
Á
5 X•
#
(2.54)
which depends only on the temperature of the gas! Thus, the internal energy of an ideal gas is a state variable,
since the temperature is a state variable. If we take an ideal gas from state E to state F the internal energy of the
gas will change from IE Ä IF . If we then return to state E, thus producing a cyclic process, the internal
energy of the system will again be IE ! Thus, the internal energy of an ideal gas is a function of the state of the
system and can be expressed as an exact differential! This is expressed mathematically by writing
* .I œ 0
(2.55)
The integral around any closed loop of an exact differential is always zero! This means that for cyclic changes in
any system, the net change in the internal energy of that system is zero!
We have demonstrated that the internal energy of an ideal gas is a state variable, since it is only a
function of the temperature of the system. However, it should be obvious that the internal energy of any real gas
is a function only of the state parameters, or properties of the system. Thus, for real gases the internal energy may
be a function of the pressure and/or volume as well as the temperature, but the change in internal energy of a
system in a cyclic process will always be zero. The dependence of the internal energy of the system on the
various system parameters is an extremely important issue and will be examined carefully throughout our study.
It often seems strange that the exact differential .I can be equal to the sum (or difference) of two
inexact differentials, $ U and $ [ . But this fact leads to a very interesting and important consequence. Since the
Chapter 2: Internal Energy and the First Law
16
energy of a system is an exact differential, Equ. 55 gives
* .I œ * $ U " * $ [ œ !
(2.56)
* $U œ * $[
(2.57)
or
for a cyclic process. That is, in any cyclic process, the net amount of work done in the process must equal to the
net amount of heat added to the system.
Heat Transfer for Reversible Processes. Although the work done depends upon the process, the differential
amount of work done as a gas expands (or contracts) can be written, for quasi-static (or reversible) processes as
$[ œ T .Z
(2.58)
where both T and Z are state variables! The first law can, therefore, be written
$U œ .I T .Z
(2.59)
as long as the changes are reversible! In this case the heat transferred across the system boundary in a process
EpF is given by
UEF œ (
F
E
.I (
F
T .Z
(2.60)
E
Notice that although the right hand side of this equation is expressed in terms only of state variables, the amount
of heat added to the system still depends directly upon the particular process the system undergoes. For example,
consider an ideal gas, where .I œ $# R 5 .X œ $# 8V .X ß and where T Z œ 8VX Þ For a constant pressure
process, where TE œ TF œ To, we have
UEF œ
$
8VXF " XE T o ZF " ZE #
(2.61)
or, since To .Z œ 8V .X for constant pressure this can also be written
UEF œ
$
&
8VXF " XE 8VXF " XE œ 8VXF " XE #
#
(2.62)
However, for a process where the temperature is constant, we have .I œ !, giving
UEF œ (
F
E
8VX
ZF
.Z œ 8VX 68Œ Z
ZE
(2.63)
Classical thermodynamics is primarily about the determination of the heat transferred across system boundaries
and the work done during different processes based upon the changes in the state variables of a system.
Constant Pressure Processes and the Change in Enthalpy. When a gas is heated at constant pressure, the
internal energy of the gas increases (as the temperature changes) while the gas also expands. The total heat added
to the system in this process is found by integrating the first law equation
$U œ .I T .Z
(2.64)
to obtain
UEF œ (
F
E
.I (
F
T .Z
(2.65)
E
Now, if the pressure is constant ÐTE œ TF Ñ this last equation becomes
UEF œ IF " IE T ÒZF " ZE Ó œ IF " IE TF ZF " TE ZE
or, if we define a new state variable called the /8>2+6:C by the equation
(2.66)
Chapter 2: Internal Energy and the First Law
17
L œ I TZ
(2.67)
UEF œ LF " LE œ GT .X
(2.68)
we have
Thus, the change in enthalpy of a system is equivalent to the heat added to the system at constant pressure.
Consequences of the First Law Applied to Ideal Gases
In this section we will use the first law to determine some important relationships that arise from our model
of an ideal gas. One must remember that it is our model which allows us to determine the equation of state of the
ideal gas and the equation for the internal energy. We present these examples here as an illustration of the nature
of thermodynamics and how one can determine important relationships among the various state variables and
other experimentally measurable quantities of a system.
In the case of an ideal gas composed of monatomic molecules, each molecule has three degrees of
translational freedom
I>9> œ
:C#
:B#
:#
D
#7 #7 #7
(2.69)
Applying the classical equipartition theorem to this system gives a total of three degrees of freedom, so that the
average total energy per molecule should be given by I œ $# 5X .
In the case of an ideal gas composed of diatomic molecules, each molecule has three degrees of
translational freedom and three degrees of rotational freedom, as well as two degrees of vibrational freedom (one
kinetic and one potential),
I>9> œ
:C#
P#C
:B#
:#
P#
P#
" # "
D B D .†< 5 < " <o #
#7 #7 #7 #MB
#MC
#MD
#
#
(2.70)
Applying the classical equipartition theorem to this system gives a total of eight degrees of freedom, so that the
average total energy per molecule should be given by I œ %5X .
The ideal gas law relation, however, is derived from the average translational kinetic energy of the
molecules, and does not depend upon the type of molecules making up the system. This means that we can use
the relation
T Z œ 8VX œ R 5X
(2.71)
for ideal gases composed of monatomic as well as polyatomic molecules. This relationship is limited primarily by
the assumption that there are no intermolecular forces present within the system.
A. Heat Capacity of an Ideal Gas for Constant Volume (Isochoric) Processes
Consider a gas confined inside a cylinder with a movable piston. If the piston is somehow locked in place,
so that it cannot move, the volume of the gas cannot change, and the system can do no mechanical work on its
surroundings. In this case, the first law simplifies to give
.IZ œ $ UZ
(2.72)
where we have used the subscript notation to indicate that the volume of the system is fixed. Using the definition
of the heat capacity, we can express the heat added to the system by the equation
$UZ œ GZ .X œ .I
(2.73)
Now this equation is valid for any system at constant volume. But the internal energy is a state variable, so that a
change in the system from state E p state F is given by
?IEF œ (
F
GZ .X
(2.74)
E
Here we have another interesting relationship between a system parameter and what appears to be a process
variable. Although the heat capacity at constant volume can be measured only for heat transfers occuring at
Chapter 2: Internal Energy and the First Law
18
constant volume, this measured quantity GZ can be used to determine the change in the internal energy of the
system between any two states E and F no matter what the process! This is because the internal energy of the
system is a state variable, dependent only on the endpoints!
For an ideal gas the internal energy is given by
Iœ
Á
R 5X
#
(2.75)
where Á is the number of degrees of freedom for our system ($ for monatomic molecules, ) for diatomic
molecules). Using this relationship in the equation above, we obtain
$ U œ GZ .X œ
Á
R 5 .X
#
(2.76)
or
GZ œ
Á
Á
R 5 œ 8V
#
#
(2.77)
which is the heat capacity at constant volume for an ideal gas. The heat capacity is clearly an extensive variable,
depending upon the size of the system (or number of particles in the system). We can make the heat capacity an
intensive variable by dividing by the number of moles of gas in the system. This gives for the specific heat
capacity at constant volume
-@ œ Á# V
-@ œ $# V
-@ œ %V
for a general ideal gas
for an ideal, monatomic gas
for an ideal, diatomic gas
(2.78)
B. Heat Capacity of an Ideal Gas for Constant Pressure (Isobaric) Processes
In this case, we allow the piston to move freely so that the pressure inside the cylinder must be equal to the
pressure outside (we neglect the fact that the piston would normally weigh something). This is, therefore, a
constant pressure process. For such a process the first law gives
.I: œ $ U: " $ [: œ $ U: " T .Z:
(2.79)
For an ideal gas, we find that the work done in a given process
$[: œ T .Z: œ 8V .X:
(2.80)
which means that the work done in an isobaric process is given by
[EpF œ T ÐZF " ZE Ñ œ 8V ÐXF " XE Ñ
(2.81)
The change in the internal energy of an ideal gas is given by
.I: œ
Á
8V .X:
#
(2.82)
so that the heat added to an ideal gas during an isobaric process if given by
$ U: œ .I: T .Z: œ
Á
Á
8V .X: 8V .X: œ 8” "•V .X:
#
#
(2.83)
But, by definition,
$ U: œ G: .X: œ 8”
Á
"•V .X:
#
(2.84)
Chapter 2: Internal Energy and the First Law
19
giving
G: œ 8 Á# "‘V
or
-: œ Á# "‘V
(2.85)
Thus, we have
-: œ Á# "‘V
-: œ &# V
-: œ &V
for a general ideal gas
for an ideal, monatomic gas
for an ideal, diatomic gas
(2.86)
Thus, for an ideal gas, we see that -: 9 -@ , and in fact
-: " -@ œ ”
Á
Á
" •V " V œ V
#
#
(2.87)
This means that more heat is required to raise the temperature of the gas at constant pressure than is required to
raise the temperature the same amount at constant volume. The reason for this should be obvious. During the
constant pressure process, the volume is allowed to expand, so that the system does work on the surroundings.
Thus some of the energy added to the system goes into doing work rather than increasing the internal energy (i.e.,
the temperature).
[Note: Later, when we look more closely at the relationships which can be determined from the
combined first and second law and from a knowledge of the equation of state and of the internal
energy, we will show that the heat capacity at constant pressure is always larger than the heat
capacity at constant volume for any substance, and in fact, is given by
-: " -@ œ T " @
(2.88)
in general. Now for an ideal gas, this equation becomes
-: " -@ œ V
(2.89)
in agreement with what we have shown above.]
C. Adiabatic Processes for an Ideal Gas
An adiabatic process is one in which there is no heat added to or removed from the system (i.e., where
$U œ !). This means that from the first law we have
$ U œ .I $ [ œ .I T .Z œ !
(2.90)
Now since the internal energy of an ideal gas depends only upon the temperature, we see immediately that any
change in volume of the system must result in a change in the temperature of the system when $U œ !. This
means, contrary to what you might have thought, that .X Á ! even though $ U œ !! [This adiabatic cooling
(where the temperature decreases as a gas expands even though no heat is gained or lost by the system) is what
makes an air conditioner work!]
Now for any reversible process, the work done during the process is a result of a volume change, which may
occur due to a change in pressure or due to a change in temperature, or both, giving
$[ œ T .Z œ T ”Œ
`Z
`Z
.T Œ
.X •
`T X
`X T
(2.91)
We can use our definitions of the isothermal compressibility and the isobaric coefficient of thermal expansion, ,
and " , respectively, to write this last equation as
$ [ œ T c " ,Z .T " Z .X d
(2.92)
Chapter 2: Internal Energy and the First Law
20
Now for an ideal gas we have " œ 1/X and , œ 1/T , giving
$[ œ T ” " Z
.T
.X
.T
.X
Z
• œ TZ ” "
•
T
X
T
X
(2.93)
For an adiabatic process, then, we have
.I œ " $ [ œ T Z ” .T
.X
"
•
T
X
(2.94)
but
.I œ
Á
8V .X
#
(2.95)
which can be written as
.I œ 8-@ .X
(2.96)
Notice that if we write the change in internal energy in this form, it would be valid for polyatomic molecules as
well as monatomic molecules.
The first law, then, for an adiabatic process can be expressed by the relationship
8-@ .X œ T Z ” .T
.X
"
•
T
X
(2.97)
TZ
• .X
X
(2.98)
or, moving the .X terms to the same side of the equation,
Z .T œ ”8- @ From the ideal gas equation we know that the second term in the brackets is just 8V, giving
Z .T œ c8 -@ Vd .X œ 8-: .X
(2.99)
since -: " -@ œ V for an ideal gas. Now, Z œ 8VX ÎT , which gives us
.T
.X
œ -:
T
X
(2.100)
.T
-: .X
œŠ ‹
T
V X
(2.101)
V
or
Since -: " -@ œ V, we can write this last equation solely in terms of -: and -@ to give
.T
-:
.X
œŒ
T
-: " -@
X
(2.102)
We find that for most real gases (although not for the ideal gas system), both -: and -@ vary with temperature, but
the ratio of specific heats,
-:
œ#
(2.103)
-@
is nearly a constant (varying much more slowly with termperature). For an ideal gas, this ratio is given by
Á " ‘V
-:
#
œ #Á
œ"
-@
Á
V
#
(2.104)
Chapter 2: Internal Energy and the First Law
21
or
# œ " Á#
# œ &Î$ œ "Þ'''
# œ &Î% œ "Þ#&
for a general ideal gas
for an ideal, monatomic gas
for an ideal, diatomic gas
(2.105)
If you examine the appendix you will see that the tabulated ratio of specific heats for real, monatomic gases is
very nearly 1.666 at room temperature, whereas for diatomic molecules (CO, N# ß O# ß and H# ) this ratio is 1.40 at
room temperature, somewhat higher than the value of 1.25 predicted by our simple model. The agreement
improves somewhat as the temperature of the diatomic gases increases. This would seem to indicate that some of
the degrees of freedom in the system are not “turned on” at the lower temperatures! We will come back to that
idea later on in our studies. [Note: In many engineering texts, the ratio -: Î-@ is given the symbol 5 , as is the case
in our appendix.]
In terms of # , the relationship between system parameters for an adiabatic process can be written
.T
- : /- @
.X
# .X
œŒ
œ
T
- : /- @ " 1
X
#"" X
(2.106)
This equation can now be integrated to obtain
ln Œ
T#
X#
#
ln Œ œ
T"
X"
#""
(2.107)
or
(# " "Ñ ln Œ
T#
X#
œ # ln Œ T"
X"
(2.108)
which gives
Œ
# ""
T#
T"
œ
Œ
œ
Œ
#
X#
X"
(2.109)
or
Œ
""#
T"
T#
#
X#
X"
(2.110)
from which we can obtain
T ""# X # œ const.
(2.111)
By using the ideal gas law, and solving for either T or X and substituting in this last equation, we can show that
T Z # œ const.
(2.112)
X Z #"" œ const.
(2.113)
and
The framed equations are the relationships between T ß Z ß and X which are valid for an ideal gas carried through
an adiabatic process. The equations are also approximately valid for most gases over a reasonable range of
temperatures were # is approximately constant. These three equations will play an important role in our
discussions of the efficiency of heat engines and the second law of thermodynamics.
D. Heat Capacity of an Ideal Gas for Constant Temperature (Isothermal) Processes
Chapter 2: Internal Energy and the First Law
22
If we use the traditional equation for the heat added to a system in terms of the heat capacity we might write
$U œ GX .X
(2.114)
but since .X must be zero since we are assuming a constant temperature process, we would be tempted to say that
no heat is added to the system during this process. However, this would be wrong! It is true that for an ideal gas,
the internal energy (given by I œ $# R 5X ) cannot change since the temperature remains fixed. However, if we
look at the first law for a reversible process (where $[ œ T .Z ) we have
$UX œ .IX T .ZX
(2.115)
and we see that even if .I œ ! we are still left with the equation
$UX œ T .ZX
(2.116)
From the ideal gas equation, we have, for an isothermal process,
T Z œ 8VX Ê Z œ
8VX
8VX
Ê .ZX œ "
.TX
T
T#
(2.117)
giving
$U œ "
8VX
.TX
T
(2.118)
which allows us to determine the heat added to the system during an isothermal process provided we know the
change in the pressure of the system. Thus, the heat added to the system as the system changes from state E to
state F is given by
UEF œ [EF œ " 8VX ln Œ
TF
TE
œ 8VX ln Œ TE
TF
(2.119)
[Note: Since T Z œ 8VX Ê T œ 8VX
Z , we see that for an isothermal process (where X doesn't
change) the ratio of pressures is the inverse ratio of volumes! Thus, this last expression can also be
written as
UEF œ [EF œ 8VX ln Œ
TE
ZF
œ 8VX ln Œ TF
ZE
(2.120)
We have just illustrated that heat may be added to or removed from a system at constant temperature. An
important class of processes which occur at constant temperature are phase transitions. However, the concept of
a phase transition is foreign to the concept of an ideal gas, since phase transitions occur because of intermolecular
forces!
NOTE: The equation
$UX œ GX .X
(2.121)
would seem to imply that GX œ ∞ for an isothermal process since .X œ ! and $ U Á !!
But what this really means is that the heat capacity for an isothermal process is undefined.
Appendix 2.1
The Evaluation of Gaussian Integrals
To evaluate the average value for the magnitude of the velocity and for the kinetic energy we need to
know how to evaluate Gaussian integrals of the form
O8 œ (
∞
M8 œ (
∞
#
B8 /"-B .B
(A2.1.1)
"∞
and
#
B8 /"-B .B
(A2.1.2)
!
where 8 œ !ß "ß #ß á . The techniques for evaluating these integrals will be developed in this appendix.
To evaluate the particular case were 8 œ !, we can square O! to obtain
∞
O!# œ (
/"-B .B(
#
"∞
∞
#
/"-C .C
(A2.1.3)
"∞
where we have used a different dummy variable in the second integral. This can be written in the form
O!# œ (
∞
∞
(
#
#
.B .C /"-ÐB C Ñ
(A2.1.4)
"∞ "∞
which looks like an integral over the area .B .C in the B-C plane. If we make a change in variables to plane, polar
coordinates we can express B# C# œ <# and the differential area .B .C can be written < . 9 .<, giving
#1
O!# œ (
!
∞
(
#
. 9 < .< /"-<
(A2.1.5)
!
where we can immediately integrate over the angle 9 to obtain
O!# œ #1(
∞
#
/"-< < .<
(A2.1.6)
!
This last integral can be readily evaluated by making a change in variables of ? œ <# , giving
O!# œ #1(
∞
!
∞
∞
1
" "-<#
/"-?
/
#< .< œ 1( /"-? .? œ 1”
• œ
#
"
!
!
(A2.1.7)
from which we obtain
O! œ (
∞
"∞
/"-B .B œ Ê
#
1
-
(A2.1.8)
Since the integrand in O! is even, the integral from ! p ∞, which we denoted by M! , is just half the value of
O! , or
M! œ (
!
∞
#
/"-B .B œ
" ∞ "-B#
"
" 1
.B œ O! œ Ê
( /
# "∞
#
# -
(A2.1.9)
Now, let's examine the other moments of this integral. The integral O" must be zero, since the integrand is odd
and the integral is over symmetric limits. The integral M" , however, is not zero, since the limits of integration are
not symmetric. However, M" is easy to integrate directly, using a simple change in variables of B# œ ?:
M" œ (
!
∞
∞
B/
"-B#
" ∞
" ∞
" /"-?
"
#
.B œ ( /"-B #B .B œ ( /"-? .? œ ”
• œ
# !
# !
# "- !
#-
(A2.1.10)
Chapter 2: Internal Energy and the First Law
24
The next moment integral, O# , can be written
O# œ (
∞
"∞
B# /"-B .B œ (
#
∞
"∞
"
`
` ∞
`
#
#
Š/"-B ‹ .B œ " ( /"-B .B œ " O!
`` - "∞
`-
(A2.1.11)
or
O# œ "
1
`
" " "$
Ê œ 1#- #
`- #
(A2.1.12)
As before the integral from zero to plus infinity is just half this integral, so we have
M# œ
" " "$
1#- #
%
Using these same techniques again and again, we can evaluate the higher moment integrals.
(A2.1.13)
Chapter 2: Internal Energy and the First Law
25
TABLE OF VARIOUS GAUSSIAN INTEGRALS
Values of the integral of the form: M8 œ '0 B8 /"-B .B
_____________________________________________
∞
#
8
M8
______________________________________________
" "Î# ""Î#
0
# 1
______________________________________________
"
""
1
# ______________________________________________
" "Î# "$Î#
2
% 1
______________________________________________
"
"#
3
# ______________________________________________
$ "Î# "&Î#
4
) 1
______________________________________________
5
-"$
______________________________________________
If 8 is even:
' ∞ B8 /"-B# .B œ #M8
"∞
If 8 is odd:
' ∞ B8 /"-B# .B œ 0
"∞
Appendix 2.2
The First Law for Open Systems
For a stationary system which is open (such as a water heater), mass is allowed to pass through the
system. This will normally result in a net energy transport across the system boundary. For such a system, we no
longer have a controlled mass, but a controlled volume which is defined by the system boundary. The first law
for such a system can be expressed by the equation
?U " ?[ I7 38 " I7 9?> œ ?IGZ
(A2.2.1)
where ?U is the heat added to the system across the system boundaries (other than the inlet and outlet); ?[ is
the work done by the system on its surroundings (perhaps by the movement of a movable boundary); I7 38 and
I7 9?> is the energy transported across the system boundary as a result of mass flow at the inlet and the outlet,
respectively; and ?IGZ is the total change in the internal energy of the control volume.
As a mass enters or leaves a control volume, the energy transported across the system boundary at the
inlet or outlet is given by
73 /7 3 œ 73 Ð/ /538 /:9> A0 69A Ñ3
(A2.2.2)
where 73 is the mass crossing the boundary at the inlet or outlet, /7 is the total specific energy transfered across
the boundary at the inlet or outlet, where / in the specific internal energy of this mass due to random molecular
motion (this would include translational, vibrational, and rotation molecular motion, as well as interaction
poential energy between particles), /538 is the specific kinetic energy of mass 73 as a whole, /:9> is the potential
energy of mass 73 as a whole, and A0 69A is the flow work done on the mass as it enters or leaves the control
volume.
To understand the concept of flow work, consider a volume Z which is just entering the control volume
through a pipe of cross secional area E. The volume is given by Z œ EP, where P is the distance from the
control volume to an imaginary piston. The work done by this imaginary piston as the volume Z is pushed into
the control volume is given by
[0 69A œ J P œ T EP œ T Z
(A2.2.3)
Thus, the flow work per unit mass of a mass entering the control volume A0 69A is just the product of the pressure
in the inlet and the specific volume (or inverse density) at the inlet:
A0 69A œ T @ œ T Î3
(A2.2.4)
The total energy transported across the system boundary, then, is given by
73 /7 3 œ 73 Ð/ /538 /:9> T @Ñ3
(A2.2.5)
Since flow work must always be present in an open system, we find it convenient to combine the specific internal
energy of the gas / (due to the random motion of the molecules) with the product T @ and introduce a new
quantity called the enthalpy defined by the equation
L œ I TZ
(A2.2.6)
Using this definition, the energy transported across the boundary of an open system is given by
I7 3 œ 73 c/ T @ /538 /:9> d3 œ 73 c2 /538 /:9> d3
(A2.2.7)
The expressions for the kinetic and potential energies of the mass moving across the system boundary are given
by
I538 œ
"
7i #
#
and
I:9> œ 71D
(A2.2.8)
Chapter 2: Internal Energy and the First Law
27
The first law, therefore, for an open system with a controlled volume can be expressed as
?U " ?[ 73 Œ23 i3#
i#
1D3 " 7/ Œ2/ / 1D/ œ ?IGZ
#
#
(A2.2.9)
where 3 stands for the inlet and / for the exit. This equation allows for multiple inlets and outlets. In the case of a
steady state, the energy of the control volume is not changing, and this equation can be written
?U " ?[ œ 7/ Œ2/ i/#
i#
1D/ " 73 Œ23 3 1D3 #
#
(A2.2.10)
Appendix 2.3
An Example of Non-T -.Z Work:
Calculating of the Work Done by an External Magnetic Field
and the Cooling by Adiabatic Demagnetization
for a Paramagnetic Salt
The Work Done by an External Magnetic Field
As an illustration of a system which is not an ideal gas system, we consider the work done by an external
magnetic field acting on a paramagnetic salt. When a paramagnetic salt is in the region of a magnetic field, the
first law change in energy of the system as the sample is further magnetized (not including the change in energy of
the applied field itself) is given by
.I œ $ U .o [. `
(A2.3.1)
where [ œ FÎ.o . The differential work done by the system in the magnetization process is therefore given by
.[ œ " .o [. `
You should notice here that one must be somewhat careful in assigning the proper sign (positive or negative) to
the “work” done by the system. A better approach might be to simply write out the change in internal energy of a
system as the extensive parameter (the magnetic moment in this case) is changed. In the case of the expansion of
a gas, the extensive variable Z increases when the internal energy of the system decreases, whereas in the case of
the paramagnetic material, the internal energy of the system increases as the magnetization of the material
increases!
The total change in internal energy due to the magnetization of the system is given by
Q0
?E ` œ " ? [ œ ( . o [ . `
(A2.3.2)
Q3
This “work” can be plotted on an “indicator diagram” in which [ is plotted against `, just as we did for the
expansion of an ideal gas. The work done in the magnetization process is just the area under the curve of [ vs.
`. The equation of state for this material can be approximated, in the case of low magnetic fields, by the Curie
law
`œW
[
X
(A2.3.3)
where W is the Curie constant which must be determined for each paramagnetic material. We can solve for the
field [ in terms of the magnetization, and obtain
[œ
`X
W
(A2.3.4)
giving, for the total change in energy of the system (or the total work done by the system) due to magnetization,
`0
?E` œ " [30 œ ( .o
`X
.`
W
(A2.3.5)
`3
For an isothermal process, this gives
`0
?I`
.o X
œ " [30 œ
(
W
`3
`
` .` œ
.o X ` # 0
º
W
# `3
(A2.3.6)
Chapter 2: Internal Energy and the First Law
29
or
?I` œ " [30 œ
.o X
Ð`#0 " `#3 Ñ
#W
(A2.3.7)
We can use the Curie equation to eliminate X in this last equation to give
?I` œ " [30 œ
.o ` 0 [0
. o ` 3 [3
"
#
#
(A2.3.8)
which is valid only for isothermal processes.
Typically, there may be several ways in which a system can interact mechanically with its surroundings.
We express this in the general form:
.IQ /-2 œ " .[ œ „ ]3 .\3
(A2.3.9)
3
where ]3 is a generalized force and .\3 is a generalized displacement, and where the sign is chosen to give an
increase in the internal energy for an increase in the generalized displacement. For example, if a paramagnetic
material is a gas, there could be a change in mechanical energy due to any change in magnetization, and also due
to any change in pressure.
Cooling by the Adiabatic Demagnetization of a Paramagnetic Salt
We stated earlier that the change in internal energy of a paramagnetic salt is given by
.I œ $ U .o [ . `
(A2.3.10)
if we neglect the mechanical energy change (“work") due to changes in pressure and volume. We can solve for
$U in this equation and obtain
$ U œ .I " .o [ . `
(A2.3.11)
We will assume that the internal energy is a function only of the temperature and of the magnetization, so that
`I
`I
Ñ` .X Ð
ÑX . `
`X
``
(A2.3.12)
`I
`I
.X ”Œ
" .o [ • . `
`X `
`` X
(A2.3.13)
.I œ Ð
The heat added to the system, then, is given by
$U œ Œ
For a so-called “perfect paramagnetic substance", the internal energy is a function only of the temperature, and
this last equation simplifies to give
$U œ Œ
`I
.X " .o [ . `
`X `
(A2.3.14)
Now, for constant magnetization, we have
$U` œ Œ
`I
.X`
`X `
(A2.3.15)
from which we can define the heat capacity at constant magnetization
G` œ
.U`
`I
œŒ
.X`
`X `
(A2.3.16)
The heat added to the system, then, for a perfect paramagnetic substance can be written as
$ U œ G` .X " .o [ . `
(A2.3.17)
Chapter 2: Internal Energy and the First Law
30
For an adiabatic process, where $U œ 0, this equation gives
G` .X œ .o [ . ` œ .o Œ
`X
.`
W
(A2.3.18)
which can be integrated to give,
(
X
G`
.o `# " `#o .X œ
X
#W
(A2.3.19)
Xo
This equation further simplifies if G` is a constant, giving
G` lnŒ
X
.o `# " `o# œ
Xo
#W
(A2.3.20)
or
lnŒ
X
.o `# " `#o œ 0 ˆ`# " `#o ‰
œ
Xo
#WG`
(A2.3.21)
where 0 œ .o Î#WG` . From this equation we get
#
#
X0 œ Xo /0Ð`0 "`o Ñ
(A2.3.22)
This last equation demonstrates how a paramagnetic material can be used to cool a sample by adiabatic
demagnetization. A large magnetic field is initially impressed upon a sample which is then cooled using
conventional refrigeration techniques to as low a temperature as possible. The magnetic field is then turned off,
so that `0 œ !, and the temperature drops exponentially.
This last equation can also be written in the form
#
#
X" /"0`" œ X# /"0`# œ const.
(A2.3.23)
The temperature can be eliminated from this last equation by using the Curie law
`œW
[
[
Ê X œW
X
`
(A2.3.24)
to give
W
[ "0`#
/
œ const.
`
(A2.3.25)
or
[ "0`#
/
œ const.
`
(A2.3.26)
which gives
[ œ E` /0`
#
(A2.3.27)
as the equation for an adiabatic magnetization of a perfect paramagnetic substance such that G` is constant. The
work done by the paramagnetic salt in an adiabatic process can now be calculated using the equations above.
Problem for later: Sketch on an indicator diagram a Carnot cycle for a perfect paramagnetic
substance where G` is a constant. Then calculate the work done and the heat added to the system
in each part of the cycle. Finally, calculate the efficiency of this Carnot cycle.