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:

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):

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:

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:

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:

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:

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:

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:

but since the pressure is constant and equal inside and outside (such as in an open flask), we can see that

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.

We can see that

for small changes. At constant pressure this becomes:

Thus, comparing this expression to the expression above for U at constant pressure, one can see that

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

Incorporating the idea of enthalpy above (again at constant pressure):

which can be written more exactly as (without assuming that the heat capacity at constant pressure is independent of temperature):

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:

Where the bar indicates a molar quantity. Alternatively one could write the whole equation in terms of molar quantities, for example:

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.

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:

Adiabatic ideal gases:

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:

Chap. 2.9

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 theheat 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:

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