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Chapter 2
Energy
The concept of energy is key to thermodynamics. Most physical chemistry students already have a good intuitive understanding of energy. Instead of starting
with a macroscopic treatment of energy (i.e. heat and work) as most thermodynamics courses do, we will begin with a microscopic discussion of energy on
the molecular level.
2.1
Microscopic Energies
Lest the title of this section causes confusion, I should emphasize that whether
big or small, energy is energy and as such there is no distinction between microscopic and macroscopic energy. But by “microscopic” energy, I am talking
about the energy contents of a system on the molecular level, i.e. how energy
is contained in atoms and molecules.
The total energy content E of a substance is composed of two parts: (a)
the kinetic energy (KE) of the substance is due to the motions of all the
microscopic particles inside the system, and (b) the potential energy (P E) of
the substance is a function of the particles’ positions, such that
E = KE + P E.
(2.1)
If we know the instantaneous momenta (~
pi ) of each particles in the system and
its mass (mi ), we can easily write down the total kinetic energy as:
X
KE =
|~
pi |2 /2mi .
(2.2)
i
On the other hand, even if we know the position of each particle (~ri ), the
potential energy is a much more complex function of the particle positions and
it is often difficult to write down the precise functional form of the potential
energy. So we will be satisfied at the time being with the following:
P E = F(~r1 , ~r2 , . . . ~rN ),
1
(2.3)
2
2.2. THE INTERNAL ENERGY AS A STATE VARIABLE
without writing out the explicit functional form of F. The major contributions
to the total potential energy of a system come from the the bonded interactions
(due to the relative positions of bonded atoms) and the non-bonded interactions
(due to interactions between non-bonded molecules, including steric repulsions,
London dispersion forces, electrostatic interactions, hydrogen bonds, etc.) In
addition, if the particles in the system interacts with an external field (e.g. polar
molecules feeling an oscillating electric field coming from a laser light passing
through the sample, or nuclear spins aligning with a strong external magnetic
field), the potential energy will be augmented with contributions coming from
the particle-field interactions.
Since the microscopic energies, KE and P E, depends on the instantaneous
momenta and positions of all the particles, the total energy must be a function
of the microscopic state of the system. But since we know that the system
moves from one microscopic state to another almost continuously in time, you
may expect that the total energy of the system should be nearly impossible to
specify at any moment of time. However, there is at least one occasion where
we can know for sure the total energy of a system isn’t changing at all. This is
when we artifically close off the system to any transfer of energy to and from
the outside. In such a system, the total energy content of the system is fixed.
Even though the system is changing continuously from one microscopic state
to another, it is really only transiting among a subset of the microscopic states
which have the same total energy.
2.2
The Internal Energy As A State Variable
In Chapter 1, we said that even though the precise microscopic state of a system
with N particles requires on the order of 6N variables to specify, the number of
state variables needed to specify the macroscopic state of a system in equilibrium
is surprising small, usually just a handful. The most convenient state variables
are often chosen to be P , V , n and T .
Since we are studying energy in this chapter, we may ask if the total energy
of the system E may be used as one of the state variables instead. Moreoever,
since we have shown above that if we can fix the total energy of the system,
the microstates of the system are confined to moving among a subet of the
microscopic state having the same total energy E. We can try replacing one of
the state variables among P , V , n, and T for an extensive system by E. So
which one should E take the place of? To answer this question, we need to
figure out which state variables should be best selected to be the independent
variables.
We know that for an extensive system, three of the four state variables may
be chosen independently, while the fourth one is dependent on the first three.
To visualize independent and dependent variables, you can imagine that an
experiment comes with a control panel. The control panel has three knobs,
each one allowing you to dial in a certain value of each of the three independent
variables. For example, the three dials may be: n, V , and T . Then all other
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state variables, e.g. P , would have to be dependent variables, i.e. the value of
a dependent variable can not be chosen arbitrarily, but it must be a function of
n, V and T .
Now instead of n, V and T , could I have chosen n, V and P as the independent variables, and T would be the dependent variable as a result? Theoretically, this is perfectly acceptable, because the equation of state relates the four
variables to each other, and any one of the four can always be expressed as a
function of the other three. However, you will never see this combination used
as independent variables, because it is almost impossible in practice to control
P and V independently. To see why, imagine how you could control the pressure
of a gas. Typically, you would control the pressure by putting a certain mass on
the piston as shown in the drawing. When the system comes into equilibrium,
its internal pressure would match the applied pressure. But for the system to
adjust its internal pressure, its volume must be free to move; therefore, controlling both P and V independently at the same time is practically impossible.
The variables P and V are called a “conjugate pair”. Notice that a conjugate
pair has one variable being intensive and the other extensive. So we see that we
can use (n, P, T ) or (n, V, T ) as independent variables, but never (P, V, T ).
Figure 2.1: Controlling the internal pressure of a gas.
Now back to the question of which state variable should E replace. We
recognize that the most natural candidate would be T . This is consistent with
our experience that if we add energy to a system, its temperature generally goes
up. Moreover, E is extensive while T is intensive; therefore E and T meet the
requirement of a conjugate pair. That E and T are in fact the correct conjugate
pair (more precisely, it is E and 1/T ) is not so easy to prove, but it is certainly
reasonable. We will explore this in greater detail in Chapter 3.
So we see that (n, V, E) would be perfectly good independent variables to
use to describe the macroscopic state of a system at equilibrium. But what
is the relationship between a system with (n, V, E) as independent variables
and another one with (n, V, T )? Let’s say we have one system which has been
designed so I can independently control n, V and E. I dial in certain values
(n∗ , V ∗ , E ∗ ) and observe the the temperature comes out to be T 0 . Now, I
prepare a second system with the same chemical composition, but for this one
I have designed it so I can independently control n, V and T instead. I dial
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2.3. THE INTERNAL ENERGY OF IDEAL GASES
in (n∗ , V ∗ , T 0 ). If I observe the energy, then it should turn out to be E ∗ .
That means these two systems must be in the same thermodynamic state, even
though the system is controlled through different independent variables. For
the experimentalist, this fact is very useful, because designing an experiment
to control a certain state variable may not be so simple in practice (e.g. the
energy), but we can get around this problem by controlling its conjugate variable
instead if that turns out to be easier.
2.3
The Internal Energy of Ideal Gases
To get a good handle on the magnitude of the internal energy content of the
system, we will take a detour and consider two specific systems: a monatomic
ideal gas and an ideal gas of diatomic molecules.
2.3.1
Monatomic Ideal Gas
An example of a monatomic gas is Ne. Being a noble gas, Ne atoms have very
weak interactions with one another. Under normal conditions, Ne is close to
ideal. But a monatomic gas could also be a radical. For example, a sample of
Cl atoms is a monatomic gas. But Cl atoms obviously should not be very ideal,
because two Cl radicals can easily recombine to form a strong Cl-Cl bond.
We start with n mole of an ideal monatomic gas in a container with fixed
volume V and at temperature T . According to what we have learned in the
last section, this sample of gas should have a well-defined macroscopic energy
content E at equilibrium. We will try to calculate it.
In an ideal gas, the particles do not interact with each other. Therefore, each
particle travels at its own velocity. There is no potential energy, only kinetic.
The distribution of the velocities (or momenta) follow the Maxwell-Boltzmann
distribution, which could be easily determined experimentally:
|~
p|2
,
(2.4)
f (~
p) = C exp −
2mkT
where k is the Boltzmann constant 1.38 × 10−23 J/K, m is the mass of each
particle, T is the absolute temperature, f (~
p)δ~
p is the fraction of particles with
momentum between p~ and p~ + δ~
p, and C is a normalization constant so that the
integral over all p~ comes out to be 1. (The Maxwell-Boltzmann distributiion
can also be derived from fundamental principles of statistical mechanics, but in
this chapter we will take the Maxwell-Boltzmann distribution as experimental
fact.)
Since the kinetic energy of a particle is p2 /2m, we can easily calculate the
total energy of n mole of this gas
2
Z
|~
p|
|~
p|2
3
E = nNa C d3 p
exp −
= nRT,
(2.5)
2m
2mkT
2
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where Na is Avogadro’s number and R = Na k is the gas constant. We see that
the total internal energy of n mole of monatomic ideal gas at temperature T is
entirely due to the kinetic energy and is 3nRT /2, which is independent on V
as well as the mass of the atoms themselve. For examle, a mole of Ne and a
mole of Ar would have the same E at the same T , even though Ne is lighter
(which means at the same T , Ne atoms on the average travel faster than Ar).
Also, we notice that each monatomic gas particle has 3 degrees of freedom. The
inetgral above can be factored into a product of three integrals, over px , py and
pz . Each degree of freedom gives rise to a kinetic energy contribution of 21 nRT .
Notice also that the internal energy of an ideal gas has no direct dependence on
V . This is not generally true for real gases and fluids.
2.3.2
Diatomic Ideal Gas
Now we consider the internal energy of a diatomic gas like Cl2 . There is a
chemical bond between the two Cl atoms. We will assume that the bond is
rigid and the bond energy is D, i.e. the potential energy is −D when the two
Cl atoms are separated by the bond distance, and 0 when they are infinitely far
apart.
For each of the Cl atoms in a Cl2 molecule, it has 3 degrees of freedom. The
two Cl atoms together should have a total of 6. But since the bond is assumed
rigid, a Cl2 molecule only has 5 degrees of freedom. Computing the kinetic
energy as in the last section for n moles of diatomic molecules, we should get
KE = 52 nRT .
In addition to the kinetic energy, there is also potential energy due to the
chemical bonds. For n mole of molecules, we should have P E = −nD. The
total internal energy should therefore be:
E=
5
nRT − nD.
2
(2.6)
Comparing the size of KE and P E, you will see that the kinetic energy
content is typically much smaller than the potential energy content. The bond
energy of typical diatomic molecules is of the order of several hundred kJ per
mole, whereas the kinetic energy is only severl kJ per mole at room temperature.
Now if you have n moles of a diatomic molecule like Cl2 at some T , and
you want to know how much energy does it take to dissociate all the Cl2
molecules, you can simply compute the difference between the diatomic gas
and the monatomic gas. But since n moles of Cl2 dissociates to give 2n moles
of Cl, the difference is:
3
∆E = (2n)RT −
2
5
nRT − nD
2
=
1
nRT + nD.
2
You see that ∆E is predominantly due to the bond dissociation energy.
c 2009 by C.H. Mak
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6
2.4
2.4. MACROSCOPIC INTERNAL ENERGY
Macroscopic Internal Energy
Even though we have shown how to calculate the internal energy content of a
gas in the last section, we could rarely rely on this method to determine the
internal energy of a macroscopic substance. The reason is that the potential
energy is often very complicated, and it can not be determined but for the very
simplest systems. In addition, the assumption of ideality certainly does not
apply for most substances.
In practice, there is rarely any need to know the total energy content of a
macroscopic sample. Usually, we are more interested in the energy change,
or ∆E, that accompanies a chemical or physical transformation in the system.
∆E can be determined by monitoring the energy that goes into or comes out of
the system during this transformation. That this is true is stated by the First
Law of Thermodynamics.
1st Law of Thermodynamics: The change in the internal energy
of a system during a transformation is equal to the energy transferred
into the system.
The first law is simply a statment of the conservation of energy: If the energy
content of the system goes up, it must have come from the outside, i.e. the
system cannot create energy by itself. On the other hand, if the energy content
of the system goes down, it must have dumped energy to the outside.
2.5
Work and Heat Are The Two Ways to Transfer Energy
There are two ways to transfer energy to or from a system: either as heat (q)
or as work (w). In terms of these, the 1st law can be written as:
∆E = q + w.
(2.8)
Heat is the form of energy transferred due to a temperature difference (always
from the hot to the cold body), while work is the form of energy transferred due
to mechanical motions of the system.
For example, when wood is burned, the chemical bonds inside it are rearranged and the internel energy change is transferred out as heat. If I am near
the fire, I can feel the heat because the fire is warmer than me, and my internal
energy rises because heat is being transferred from the wood to me.
Work, on the other hand, involves a “force” and a “displacement”. For
example, I can trap a fix number of air molecules inside a bicycle tire pump by
covering the outlet with my finger and compress the air inside by pushing on
the plunger. When I compress, I feel a resistance; therefore, I certainly have to
do work on the system. When the gas has been compressed, its internal energy
content is higher. This kind of mechanical work is called pressure-volume (P -V )
work, because I apply pressure to the system, and its volume changes. P -V work
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can be done in the reverse direction. For example, when a liquid evaporates its
volume expands and can push on a piston to do work on the outside. This
is the basis of the steam engine. P is the force, and the change in V is the
displacement. As we have noted above, (P, V ) forms a conjugate pair.
While P -V work is rather familiar to us, other kinds of work are more subtle.
For example, if I have to add particles to a system, work is certainly involved.
Imagine for a big party, a room is packed with people. You come in through
the door and spot your friends on the side of the room. You want to blend in,
but to meet up with your friends, you must push through all the other people.
This is the same kind of work that is needed when a molecule is added to an
equilibrium state in order to arrive at a new equilibrium state. The force in this
case is called the “chemical potential” (µ), the displacement is the change in n.
Therefore, (µ, n) is another conjugate pair.
As another example, if you have a number of mobile ions in a solution and
you apply an electric field to the system. The positive ions will move towards
negative potential and the negative ions will move the opposite way. Work is
being done on the system. The force is the electric field and the displacements
are the changes in the concentrations of the ions.
Notice that heat and work are not state variables, i.e. whereas a system in a
well-defined macroscopic state has a certain internal energy content, it does not
possess a certain amount of heat or work. The common term “heat content” is
certainly a scientific misnomer. For example, your bank account has a certain
amount of money. You can transfer money out of your account by writing a
check. But your account does not have a certain amount of checks. In fact, your
account is unaware of the existence of a check until funds are drawn against it,
i.e. at the point of money being transferred out. As such, there is no change of
heat or change of work. Heat and work are already changes.
Moreoever, since the internal energy is a dependent function of e.g. (n, V, T ),
if I make a small change in each of the three independent variables, say dn, dV
and dT , I can determine the change in E as:
∂E
∂E
∂E
dn +
dV +
dT,
(2.9)
dE =
∂n V,T
∂V n,T
∂T n,V
but there is no such thing as dq or dw. To avoid confusion, small amounts
of heat and work are denoted by δq and δw instead. If you have a sequential
process where the system undergoes a sequence of transformations i = 1 to k,
each with small amount of heat δqi and work δwi , the total amount of heat and
work are
Z b
Z b
k
k
X
X
q=
δqi →
δq,
w=
δwi →
δw.
(2.10)
i=1
a
i=1
a
You need to specify each of the steps i in order to calculate q and w. If you
only know the initial macroscopic state a and the final macroscopic state b of
the system, you would not be able to determine q and w. In other words, q and
w are path-dependent. Contrast this with the internal energy: If you know Ea
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2.6. CONSTANT-VOLUME HEAT CAPACITY
of the initial state and Eb of the final state, ∆E is always given by Eb − Ea and
is therefore path-independent.
2.6
Constant-Volume Heat Capacity
When (n, V, T ) are used as independent variables, E = E(n, V, T ). As we have
seen above,
∂E
∂E
∂E
dn +
dV +
dT.
(2.11)
dE =
∂n V,T
∂V n,T
∂T n,V
The partial derivative (∂E/∂T )n,V = CV is called the constant-volume heat
capacity.
As a special case, using the microscopic energies derived above in Eqs.(2.5)
and (2.6) for an ideal gas, CV would be equal to 3nR/2 for a monatomic ideal
gas, and 5nR/2 for a diatomic ideal gas.
2.7
Calculation of P -V Work
The work done on a system is given by
δw = −Pext dV,
(2.12)
where Pext is the external (i.e. applied) pressure, but not the internal pressure.
For example, when compressing a gas, the applied pressure from the outside
must be at least slightly bigger than the internal pressure for the compression
to occur. In the drawing below, we show an example where a fixed number
of moles of a gas is trapped inside a piston. The initial state values of the
pressure, volume and temperature are P1 , V1 , and T1 . Imagine a constant
external pressure Pa > P1 is applied to the piston from the outside. In order
to maintain the volume at its initial value V1 with Pa > P1 , we must insert
stops to prevent the piston from falling. When the stops are released, the gas
is then compressed at the external pressure Pa until the pistol hits the lower
stops. At this point, we wait for the system to settle in a new macroscopic state
with final pressure, volume and temperature P2 , V2 and T2 . For this example,
the external pressure Pa is still bigger than P2 , so if we remove the lower stops,
the piston will continue to fall until the internal pressure is finally equal to Pa .
But for the first part of this process, the work done on the system is syimply
w = −Pa (V2 − V1 ) = −Pa ∆V . Notice that we need a negative sign to make the
work done on the system positive for a compression. Notice also that the work
done does not necessarily have anything to do with the internal pressure.
In the second example, we will carry out the compression from initial V1 to
the same final volume V2 , but we will do this so that the gas stays as close to
equilibrium as possible during the transformation. We start with the system at
initial state (P1 , n1 , T1 ). With an external pressure just slightly larger than P1 ,
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Figure 2.2: PV work in a irreversible compression.
we compress the gas to Va . Then we reduce the external pressure to something
just slightly larger than Pa . With this applied pressure, we again compress the
gas slightly to Vb . Then we reduce the pressure again to something just slightly
larger than Pb . We repeat the process in small steps, until we reach the final
volume V2 . At this point, the external pressure is just slightly larger than the
internal pressure P2 .
Figure 2.3: PV work in a reversible compression.
If we carry out this process in more steps, then the external pressure in each
step can be kept closer to the internal pressure, but it still has to be slightly
larger than the internal pressure. In the limit of an infinite number of steps,
the external pressure can be kept essentially equal to the internal pressure at
all points during the process. In this case, the process is said to be reversible,
because but taking the same steps in the reverse direction, the gas can be
expanded from V2 back to V1 and it would arrive at exactly the same state as
it was initially. In contrast, the system in the first example can not exactly
retrace its previous steps in the reverse direction because the system goes out
equilibrium momentarily during the process and is thus irreversible. When a
process is carried out reversibly, the system remains in equilibrium throughout
the process. In other words, the process runs along a path on the equilibrium
manifold (i.e. the equation-of-state surface).
If the compression is carried out reversibly, then the external pressure must
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2.8. HEAT AND WORK IN A FEW SPECIAL EXPERIMENTS
Figure 2.4: A reversible process runs along a path on the equation of state
surface.
be equal to the internal pressure throughout the process, and the work becomes:
Z
w=−
2
P dV,
(2.13)
1
where P is now the internal pressure, and the limits of the integration refer to
the initial and final states. If you know the equation of state of the substance,
you can express P as a function of (n, V, T ) and perform the integral over V .
Notice that this is a special case, because work always refers to the applied
pressure.
2.8
Heat and Work In A Few Special Experiments
In this section, we will show how to calculate q and w for processes with certain
special conditions.
1. Constant-V Processes
If a process occurs at constant volume, w must be 0. Then ∆E = q alone.
Furthermore, if the process occurs in small steps so that the system stays
at equilibrium throughout (and n is fixed),
δqV = dE = CV dT,
(2.14)
where qV is the constant-V heat transfer. The total heat transferred is
RT
therefore qV = T12 CV dT . Notice that in general CV does depend on T .
Only in special cases where CV is temperature independent (e.g. in an
ideal gas) would we be able to simplify it further to qV = CV (T2 − T1 ).
Eqn.(2.14) is the reason why CV is called the “heat capacity” since it is
usually measured by monitoring the heat transferred during a constant-V
reversible process.
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2. Reversible Isothermal Expansion of An Ideal Gas
R2
According to Eq.(2.13), the reversible work is w = − 1 P dV . When this
formula is applied to the isothermal (constant T ) compression of an ideal
gas,
Z V2
Z V2
nRT
V2
P dV = −
w=−
dV = −nRT ln
.
(2.15)
V
V1
V1
V1
The same formula of course works for an expansion too.
3. Adiabatic Processes
An adiabatic process is one where there is no heat transfer. This can be
achieved experimentally by thermally insulating the system. In this case,
q = 0, and ∆E = w alone.
4. Reversible Adiabatic Expansion of An Ideal Gas
When an ideal gas is expanded adiabatically from volume V1 to V2 , δq = 0
so dE = δw = −Pext dV . But in contrast with an isothermal expansion,
the temperature is not constant here, since work is done by the system but
there is no heat transfer, so the internal energy of the system decreases,
which should give rise to a temperature drop. This temperature drop can
be calculated for an ideal gas, since the internal energy of an ideal gas is
just CV T , which depends only on T . If we can measure the work done by
the system, then
∆E = CV (T2 − T1 ) = w,
(2.16)
and this allows us to calculate the final temperature of the system. This
formula works for both irreversible and reversible adiabatic ideal gas processes.
Furthermore, if the expansion is carried out reversibly, we know that δw =
−P dV which is also equal to dE. Using Eq.(2.9) and with dn = 0, we see
that
∂E
−P dV =
dV + CV dT,
(2.17)
∂V n,T
For an ideal gas, the energy is a function of T only, so (∂E/∂V )n,T = 0,
and the heat capacity is independent of T , therefore
CV dT = −
nRT
dV
V
or
CV
nR
dT = −
dV.
T
V
(2.18)
Integrating the left side from T1 to T2 and the right side from V1 to V2 ,
we get
nR/CV
T2
V1
=
.
(2.19)
T1
V2
Finally to obtain the work, we use the fact that the process is adiabatic
to get w = ∆E = CV (T2 − T1 ).
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2.9. ENTHALPY
Again, this result is a special case applicable to only ideal gases. It is not
generally valid for real gases. But if the temperature and volume changes
are small, CV could be approximated by a constant even for a real gas.
5. Constant-P Processes
If a process occurs at constant external pressure Pa and the internal pressure P is equal to Pa throughout the process, then w = −(P V2 − P V1 ) =
−P ∆V . But since P is the same at the beginning and at the end of the
process, we can also write this as w = −∆(P V ). Since P and V are
state functions, the work involved in such constant-P processes can be
computed easily by knowing just the initial and final states of the system. Moreover, since q = ∆E − w, the heat involved in such constant-P
processes is simply given by q = ∆E + ∆(P V ) = ∆(E + P V ).
2.9
Enthalpy
As we have shown in the last section, the heat evolved during a constant-P
process is given by ∆(E + P V ). Constant-P processes are quite common in
chemistry; therefore, the function E + P V is especially important. It is given
the symbol H and is called the enthalpy. Since E, P and V are state functions,
H is also a state function. H is extensive, because E and V are extensive.
Many chemical reactions are carried out in an open container (e.g. a beaker)
exposed to atmospheric pressure from abovee. At first sight, this may not
appear to be a self-contained system, because molecules can escape from the
container. Certainly, if our system is a gas, this experimental setup wouldn’t
make much sense because the gas will all escape into the atmosphere above
before it has a chance to settle into an equilibrium macroscopic state. The
constant-P assumption would therefore be more appropriate for a liquid or a
solution in an open container. But even in that case, it is often not immediately
clear to students why this would constitute a constant-P system, because it
is still an open system. The simplest way to resolve this inconsistency is to
think of the interface between the liquid with the atmosphere above it as an
inpenetrable membrane, like the latex balloons are made out of. In order for
the membrane to not burst, it must maintain equal pressure on both sides. If
the internal pressure is larger than the external pressure, the interface will move
to allow the system to expand to reach a new equilibrium state to equalize the
two pressures.
For constant-P processes, we have shown that q = ∆H. For constant-P
processes carried out in small steps, dH = δq. Heat transfer is easy to measure
experimentally. If you want, you can also measure the volume change to get δw.
But as we will show in later chapters, this is unnecessary if you only want to
figure out whether the process is naturally spontaneous or not. For this purpose,
you actaully do not have to know w explicitly.
Since H is the constant-P heat transfer, it is most natural to think of H as
a dependent function on (n, P, T ), instead of (n, V, T ) for the internal energy
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E. The term constant-P should be understood in the sense that the internal
pressure P of the system is constant throughout the process, and it is maintained
by keeping it equal to the external pressure. Therefore, the pressure P in a
constant-P process is an adjustable independent variable, which I can control
by varying the external pressure.
2.10
Constant-Pressure Heat Capacity
Since H is a state function, we can write down its differential:
∂H
∂H
∂H
dn +
dP +
dT,
dH =
∂n P,T
∂P n,T
∂T n,P
(2.20)
Just like (∂E/∂T )n,V is called the constant-volume heat capacity CV , (∂H/∂T )n,P
is called the constant-pressure heat capacity CP . As we have shown above,
dH = δq at constant P and fixed n, so
δqP = CP dT,
(2.21)
where qP is the constant-P heat transfer.
As a special case, using the microscopic energies derived in Eq.(2.5) for a
monatomic ideal gas, H = E + P V = 3nRT /2 + P V = 5nRT /2, where we
have made use of the ideal gas law P V = nRT . Therefore, CP = 5nR/2 for a
monatomic ideal gas. For a diatomic ideal gase, CP = 7nR/2.
2.11
Enthalpy Change Associated With A Chemical Reaction
2.12
Hess’s Law and Standard Enthalpy of Formation
2.13
Determining Enthalpies at Non-standard
Temperatures
So far, we have been discussing ∆H for a reaction at constant P and T . What
if a reaction happens at constant P but T changes during the reaction? If the
temperature change occurs in small steps, we can take Eq.(2.21) and integrate
it over the temperature range, getting
Z T2
∆H =
CP (T )dT.
(2.22)
T1
The temperature-dependence of CP has been measured for a large number of
substances and are available either tabulated or in the form of an approximate
equation.
c 2009 by C.H. Mak
Copyright 2.14.
14
IS THERE A MICROSCOPIC INTERPRETATION OF HEAT AND WORK?
In addition to the heat transfer due to the temperature change, if there
is a phase transition between T1 and T2 , the enthalpy of fusion (for a solidliquid transition) or the enthalpy of vaporization (for a liquid-vapor transition)
must also be included in ∆H. During a phase transition, T of course remains
constant, so we would not use the heat capacity at the transition temperature.
2.14
Is There A Microscopic Interpretation of
Heat and Work?
To wrap up this chapter, we return to the microscopic description of energy
and ask a common question: On the level of atoms and molecules, what is
the difference between heat and work? Many students have the misconception
that heat has something to do the motions of particles. While it is true that
a system that has mobile particles which possess kinetic energy, the amount of
kinetic energy is related to the temperature of the sample, heat has nothing to
do with motions specifically. For example, as we have seen for an ideal gas, E is
directly related to T , and since transferring heat into the gas will generally raise
its temperature, heat leads to an increase in kinetic energy. But as a matter
of fact, putting work on the system can also increase the internal energy and
thus the kinetic energy of an ideal gas. Therefore, the correlation between heat
transfer and a rise in the kinetic energy is not at all unique. In fact, we will show
in the next chapter that a system that has absolutely no motions and therefore
no kinetic energy can also absorb heat.
The first law actually makes no distinction between heat and work. Together, they contribute to the change in the total internal energy of the system.
But microscopically, the internal energy of the system cannot be conveniently
partitioned into heat and work, and that is the reason why q and w are not
state functions. But as we learn about the second law in the next chapter, we
will see that heat and work do have a distinction in how they affect the entropy
of the system.
c 2009 by C.H. Mak
Copyright