# 4.6: Examples of Particular Models of Thermodynamics


## Engines and Refrigerators

An extremely relevant application of simple thermodynamics is in analyzing engines and refrigerators. These devices, which either convert work into heat or heat into work, surround us in our everyday lives. The basic principle on which they operate is the cyclic thermodynamic process.

In a cyclic thermodynamic process a system (the refrigerator or engine) moves from an initial state, through a set of other states back to the initial state. The net change of the internal energy of the system over this cycle is zero. When completed, the system has returned to the exact state from which it started. Since the change in the internal energy is zero, the first law of thermodynamics tells us that the net work into or out of the system during a cycle must be equal to the net heat out of or into the system. In equations:

$\Delta U = Q + W = 0$

$Q = - W$

How do you determine the work done on the system in a cyclic process? This is very simple if you have a PV diagram of the process that you are interested in. It turns out that the absolute value of the work done in a complete cyclic process is equal to the area enclosed by the process on a PV diagram. Using the equation above you can then determine the heat into or out of the system.

It is also possible to directly determine the heat into or out of a system during a cyclic process. Just like work is the area under the curve of a Pressure vs. Volume diagram, the heat can be determined as the area under a different curve–the Temperature vs. Entropy curve. It follows that since the net work is the area enclosed in a PV diagram, the net heat into or out of a system is the area enclosed by the TS diagram describing its process. You will get a chance to try using this to calculate the heat that is given off in a cyclic process.

A system going through a cyclic process ends up in precisely the state that it started in, so the change in the entropy of the system is equal to zero. Zero you say? Isn’t this a violation of the 2nd Law? Well, no. While the change in the entropy of the system over one cycle is equal to zero, the entropy of the universe has unquestionably increased. As it turns out, the 2nd law is intimately connected with the maximum efficiency that can be achieved in cyclic processes such as engines and refrigerators.\

## Relating Empirically determined ∆H’s to Particle Models of Thermal and Bond Energies

We are now in a position to make sense of the exact relationship between the measured heats of formation of chemical substances, the heats of vaporization and heats of melting in physical phase changes and the bond energies and thermal energies defined from a particle model. We begin by noting that the act of measurement of the various heats, the ∆H’s, as well as measurements of heat capacities, CV and CP, are all thermodynamic processes. We need to bring our thermodynamic tools to bear on these measurement processes.

We start by writing down the 1st Law of thermodynamics and then applying it to a particular measurement process. We will use the determination of the heat of vaporization of some substance as an example, and from this analysis, find the exact relationship between the bond energy and the measured heat of vaporization. The change in bond energy could be calculated from parameters of the pair-wise potential energy, the particle packing geometry and spacing of the particles in the liquid and compared to the predicted value as obtained through an empirical determination of the heat of vaporization.

The heat of vaporization is normally determined and reported as a constant pressure measurement. Fortunately, the measurement is easiest to make under these conditions. Also, at constant pressure, the boiling point temperature is constant, which is also what we typically want.

Starting with the 1st Law and then putting in what we know, we get:

$\Delta U = Q + W$

$\Delta E_{thermal} + \Delta E_{Bond} = \Delta H_{vap} + -P \Delta V$

where we used the result that Q is equal to ∆H in a constant pressure measurement and used the expression for work that is applicable to fluids. We now need to be more specific about the substance whose heat of vaporization we are measuring. Suppose it is one of the noble elements (He, Ar, Ne, etc.). This information allows to determine both the P∆V term as well as the thermal energy term. (Yes, there are changes in the thermal energy as the substance vaporizes, as we will shortly see.)

First the P∆V term. We assume the ∆Hvap measurement was made at one atmosphere, which is standard. We would expect the substance to be well represented as an ideal gas when it is in the gas phase. We also know that the volume of simple liquids increases by about a thousand (particle spacing increases by about 10) when becoming a vapor. Thus, ∆V is simply V for the vapor phase to about 1/10 of a percent. Then, using the ideal gas relationship, the work term becomes -P∆V ≈ -PV = -RT (assuming one mole of substance). The thermal energy is directly proportional to the number of active modes in both phases. In the liquid phase there are six active modes per atom, while in the gas phase there are only three. Therefore, ∆Ethermal = -3/2 RT. Substituting these two expressions into the 1st law and rearranging, we get:

$\Delta E_{Bond} = \Delta H_{vap} -P \Delta V - \Delta E_{thermal}$

$\Delta E_{Bond} = \Delta H_{vap} - RT + \dfrac{3}{2} RT$

$\Delta E_{Bond} = \Delta H_{vap} + \dfrac{1}{2} RT$

Note that the two factors tend to cancel each other out. To get a sense of the magnitude of the error that is made if the value for ∆EBond is simply taken to be the heat of vaporization, we can compare the size of ∆Hvap with ½RT at 21.7 K, the boiling point of Neon. ∆Hvap = 1.73 kJ/mol for Neon while ½RT = 0.088 kJ/mol. Using just the value for the heat of vaporization would underestimate the size of the change in bond energy by only about 5%.

This example nicely shows how it is possible to make sense of thermal phenomenon–measurement of a heat of vaporization and the relation to bond energy changes, by simultaneously bringing to bear several of the models in this chapter. This is often necessary to make sense of many thermal phenomena.

# Contributors

This page titled 4.6: Examples of Particular Models of Thermodynamics is shared under a not declared license and was authored, remixed, and/or curated by Dina Zhabinskaya.