Lectures Notes on Chapter 2
In Chapter 2, we will concentrate on the concepts associated with the first law of
thermodynamics, that is, that energy must be conserved. While the law itself is fairly
intuitive for most people (you can't get something from nothing), how it applies to
chemistry and to the quantities of work and heat (which are the two forms of energy what
can be transferred in or out of a system) is not obvious.
Chap. 2.1
Let us start with a few definitions:

System: The system is the thing or region of interest.

Surroundings: The surroundings is everything else.

Energy: The capacity to do work. (Note that energy does not have to do work -- it
may dissipate as heat or it may just sit there as potential energy -- but it always
has the capacity to do work.)

Open: An open system is one in which matter can go in or out of the system.

Closed: A closed system is one in which matter cannot go in or out of the system.

Diathermic (or sometimes Diabatic): A diathermic system is one in which heat can
go in or out of the system

Adiabatic: An adiabatic system is one in which heat cannot go in or out of the
system.

Isolated: An isolated system is one in which neither matter nor heat can go in or out
of the system.

Heat: Heat is a form of energy which can transfer in or out of a system. It is
different from work in that heat is energy stored in the random motion of
molecules (see work below).

Work: Work is also a form of energy which can transfer in or out of a system. It is
different from heat in that work is energy stored in the organized motion of
molecules (such as moving a weight up or down).
The above definitions you need to remember and understand. I will use these words on
exams and expect you to know what they mean (I may even ask you to define them).
Chap. 2.2
This part of the chapter considers the first law of thermodynamics itself. For this, we will
speak of the internal energy of the system, U, which is just the total energy available in
the system at any given time. Since we have defined heat (q) and work (w) as the only
two possible forms of energy that can enter or leave the system (at least for a closed
system where the mass is constant), it follows that any change in the internal energy of
the system must be due to heat or work going in or out of the system.
Now we come to an important, but often confusing detail: work and heat are defined in
such a way that work and heat are positive if they result in a net increase in internal
energy in the system and are negative if they result in a net decrease in internal energy in
the system.
Consider an example. If I put my diathermic system in contact with surroundings that are
colder than it, heat will pass out of the system. Thus q is negative. If I put the diathermic
system in surroundings that are warmer than it, heat will pass into the system and is thus
positive. If I do work on the system, such as compression of a gas in the system, then the
work is positive because the internal energy of the system will increase. If I let the
system do work on the surroundings, such as expansion of a gas in the system, then the
work in negative.
Note that this is the sign convention of physical chemists. Engineers sometimes use a
different sign convention! Be careful. This brings us to the most important equation you
have seen thus far:
U  q  w
In other words, the change in internal energy in the system is just the sum of the work
done on (or by) the system and the heat transferred to (or from) the system. U can
either be positive, negative or zero. Remember that it is a change, not an absolute energy.
Chap. 2.3
Now let's consider the mathematical formalism we will use to describe work. In physics,
you probably learned that a work is performed if you push against a force, F, for some
distance, x (assuming the force does not change with distance):
w   Fx
You will notice that I have used a negative sign here. This is in keeping with our sign
conventions, as you will see below. Now, what if we have a system which consists of a
piston in a cylinder. How would we put work in terms of things we can easily measure
(pressure, volume…)?
Here we can see what is happening. As the piston
expands, it is pushing against a force. This force
comes from the atmospheric pressure (molecules
in the atmosphere hitting the back side of the
piston). Remember that pressure was a force per
area, so force is just a pressure times an area. To
convert to work, we just multiply the force by the
change in distance, x. This gives Pressure x Area
x Distance. But the Area times the distance
traveled is just the volume change, so we can say
that the work is just the negative of the Pressure
times the change in volume.
If the force (external pressure) is not constant, we can write a more general expression for
work:
VF
w    Pex dV
VI
Above, we considered what would happen is work was against a constant external
pressure, like the atmosphere, w = -PexV, but what if the work occurs reversibly? First,
what does reversibly mean? It means that the work is always occurring in a system very,
very close to the point where opposing forces are equal. In the case we are considering,
this is when the pressures are maintained the same on both sides. This happens, for
example, if there is some process going on inside the piston which is slowly increasing
the pressure and the piston is continually expanding, keeping the pressure equal on the
two sides. This could happen upon heating the gas inside. In this case, we can write Pin
= Pex and therefore for a reversible expansion:
VF
w    Pin dV
VI
The book just uses P and always considers P without a subscript to be the internal
pressure.
In the special case of a reversible expansion where the temperature is constant and the
gas is ideal (a reversible, isothermal expansion of an ideal gas), we can substitute the
ideal gas law for P, P=nRT/V:
VF
w
VI
V
nRT
dV  nRT ln F
V
VI
To consider a specific example of pressure volume work, see the tutorial.
Chap. 2.4 and 2.5
In the last section we related changes in two of the things we can measure, pressure and
volume, to one way energy can go in or out of a system, work. Here we relate the other
obvious observable, temperature, to the other way that energy goes in and out of the
system, heat. This turns out to be conceptually and mathematically much simpler. Since
temperature is already proportional to the kinetic energy of molecules and heat is just the
transfer of random kinetic energy of molecules from one place to another, we should
suspect that temperature and heat are directly related:
q  CT
This is true under conditions where changing the temperature does not cause some
chemical change in the system or some change in molecular interactions. It is not true if
something happens inside the system which increases or decreases the temperature. C is
called the heat capacity, since it is some measure of how much heat is taken up or given
off for a given change in temperature. Its value depends on the nature of the molecules
under consideration and on the conditions under which the heating and cooling were
performed. The simplest case is when we hold the volume of the system constant. In this
case there is no PV work to worry about. Only heating is occurring (heat is the only
source of energy change in the system). In this case, we use CV for our heat capacity and
call this the heat capacity at constant volume. You can look this up in tables for different
kinds of compounds (e.g. in the back of the book).
We know that U = w + q. At constant volume (assuming only PV type work), w = 0,
so for this case Uconst. vol. = q = CV T. Thus, at constant volume, the internal energy
change is just given by the product of the heat capacity and the change in temperature. A
more precise way of saying this (without assuming that CV is constant) is:
 U 

  CV
 T V
This expression is a partial derivative. If you have not dealt with such things before, a
partial derivative is just the derivative of some function (in this case U) with respect to
some variable (in this case T) holding some other parameter is constant (in this case V).
The other important case for heating is heating at constant pressure. In this case, heat put
into the system not only results in a temperature increase. It also results in work being
done do to expansion of the system. As the system heats up, the volume will increase in
order to keep the pressure constant. This results in PV work. The change in internal
energy is just:
U const .P  q  w  q  Pex V
but since the pressure is constant and equal inside and outside (such as in an open flask),
we can see that
U const .P  q  PV
We do a great deal of chemistry in open flasks. Thus, we are interested in the kinds of
energy changes that can occur under such conditions. Notice that PV is itself a state
function as long as P is constant. In fact, the product PV is a state function whether P is
constant or not. This suggests the definition of a new state function called enthalpy.
H  U  PV
We can see that
H  U  PV  VP
for small changes. At constant pressure this becomes:
H  U const .P  PV
Thus, comparing this expression to the expression above for U at constant pressure, one
can see that
q  H
Remember that this is only true at constant pressure.
So, at constant volume (and no electrical work), the internal energy change is equal to the
heat. At constant pressure, the enthalpy change is equal to the heat.
Let us return now to the issue of the change in temperature with heat at constant pressure.
We now realize that when heat is put into a system at constant temperature, some of the
energy goes into raising the temperature and some of the energy goes into the expansion
of the gas. Therefore, in general, it takes more heat to raise the temperature by one
degree at constant pressure than it does at constant volume. The heat capacity at constant
volume is called CP and we can write that
q  C P T
Incorporating the idea of enthalpy above (again at constant pressure):
H  C P T
which can be written more exactly as (without assuming that the heat capacity at constant
pressure is independent of temperature):
 H 

  CP
 T  P
Before we continue, let's say a word about heat capacities in general. Heat capacities (at
either constant volume or constant pressure) are normally reported as molar heat
capacities. That is, the heat capacity of one mole of material. The expressions we have
used thus far are for total heat capacities. That is, the heat capacity of the whole system.
We could rewrite these expressions in terms of molar heat capacities by substituting:
CP  nCP
Where the bar indicates a molar quantity. Alternatively one could write the whole
equation in terms of molar quantities, for example:
 H 

  CP
 T  P
This states that the change in the molar enthalpy (the enthalpy per mole of material) with
temperature is equal to the molar heat capacity at constant pressure. Watch out for this or
it will get you!
An important aside:
Before we leave heat capacities, I would like to consider the internal energy dependence
on temperature in an ideal gas. There are some important lessons to be learned which I
do not think the book explains in enough detail.
Const.V
U  q  CV T
Const.P
U  q  PV
 nRT 
 C P T  P

 P 
U  C P T  nRT
 C P  nRT
The important point here is that for an ideal gas, the change in internal energy is always,
always, always zero if the temperature does not change -- but only for an ideal gas!
Since U = w + q, this means that for an ideal gas, if T does not change, w = -q.
In an isothermal ideal gas w = -q.
You can also get an idea why CP = CV + nR for an ideal gas from the above expression,
though this is not a proper proof (we will see this in the next chapter).
Chap. 2.6
Now let's consider what happens in a ideal gas system that allows no heat to transfer
across the walls (adiabatic). In an adiabatic system, the only change in internal energy of
the system that can happen is through work, so U = w. Thus, the expressions we came
up with for the change in internal energy are going to give us the work.
For a constant pressure adiabatic system, w = U = (CP - nR)T = CV

But this is confusing. We are now using the heat capacity at constant volume to calculate
the internal energy change and the work at constant pressure. Let's go through this
another way. The only way to change the temperature of an ideal gas in an adiabatic
system is through work. This requires a volume change. Since the internal energy
change does not depend on path, lets do this in two steps. First we change the volume
holding the temperature constant (this requires some heat transfer, but don't worry, it will
cancel in a second). The change in the internal energy for this is zero since there is no
energy change. Next, we change the temperature holding the volume constant (which
will have just the opposite heat exchange with the environment, zeroing the total amount
of heat transferred). The internal energy change here is CVT. Thus the change in
internal energy and the work done in our adiabatic system is CVT, as above. This, by
the way, is also one way to prove that CP = CV + nR. The above argument is general.
For all ideal gases the change in internal energy is always the heat capacity at constant
volume times the change in temperature. For adiabatic ideal gases, the work is the heat
capacity at constant volume times the change in temperature.
All ideal gases:
U  CV T
Adiabatic ideal gases:
w  CV T
Chap 2.7
There are many times in chemistry when we are interested in reactions which occur at
constant pressure, and particularly under these conditions, it is useful to consider the
enthalpy change (the heat change at constant pressure) that occurs during the reaction.
Reactions that produce heat are releasing energy into the environment and are called
exothermic. Reactions that remove heat from the surroundings are called endothermic.
We can write down enthalpy changes associated with any process by subtracting the
enthalpy in the initial state from the enthalpy in the final state, but before we do this, we
need to have some convention about how to report them. This convention is called a
standard state and we will use this over and over again from here on out. We normally
report enthalpies (and later free energies and entropies) at the standard state. For most of
chemistry, the standard state is one mole of the pure form of the material at 1 bar of
pressure and the specified temperature.
This said, we can now write down the enthalpy change for any process in this state, and
then calculate enthalpy changes at other states as needed (more on this later). For
example, I can say that the enthalpy change associated with vaporizing one mole of water
at 100 C is 40.6 kJ. Or I could state the enthalpy change for converting 1 mole of A into
2 moles of B in the chemical reaction A 2B. Enthalpies also add. If I take two
processes, A  B and B  C and add them up chemically, I get A  C. I can also add
their enthalpies. The enthalpy of B  A is just the negative of the enthalpy for A  B
(this makes sense -- in one direction heat goes in and in the other direction heat comes
out).
Chap. 2.8
Chemists have also devised a bookkeeping scheme for calculating the enthalpies of new
reactions. You can simply use the enthalpies of formation for each of the products minus
the enthalpies of formation for each of the reactants (products are made and reactants are
used up) starting from standard materials. This is very well explained in the book and
follows directly from the arguments about adding reactions together above. Try this with
a few simple reactions just for practice. The illustration in the book is:
2 HN3 l   2 NO g   H 2O2 l   4 N 2  g 
 R H    f H  H 2O2 , l   4 f H   N 2 , g   2 f H  HN3 , l   2 f H   NO, g 
 892.3
Chap. 2.9
kJ
mole
If we know the standard enthalpy at one temperature, how do we get it at another? Recall
that the change in enthalpy with temperature just depends on the heat capacity at constant
pressure. So, we really just need to look at how the heat capacity at constant pressure
changes between the initial and final states of the process and then multiply this by the
change in temperature:
 R H  T2    R H  T1   T2  T1  R CP
The heat capacity difference is the sum of the product heat capacities minus the sum of
the reactant heat capacities weighted by the stochiometric coefficients in the chemical
reaction. So for A  2B it would be
 R C P  2C P,B  C P, A