From: henley@eng.auburn.edu (James Paul Henley)
Subject: Re: Thermodynamics FAQ
Organization: Auburn University Engineering
Date: Tue, 28 Jul 1992 12:16:53 GMT

o What is Thermodynamics?

Due to many misunderstandings about my Thermo FAQ, I'm going to take it
slow, one question at a time.  I realized that part of the problem is
in the definition of Thermodynamics.  Physicists and Chemists study 
molecular basis for Thermodynamics, and really they derived many of
the fundamental properties.  Engineers study large scale applications
of Thermodynamics, and in so doing need to make distinctions that don't
apply at the molecular level.  The difference is between microscopic and
macroscopic viewpoints.  Chemical Engineers, being Chemists and Engineers,
have to study both.  I intentionally avoided the microscopic considerations.
I have tried to give a layman's explanation of Engineering Thermodynamics
because very few members of t.o study Thermo from a macroscopic point
of view.  I'm sorry that some of the definitions that I use cause people
so much grief.  They were not intended to be controversial.  I didn't make
any of them up - they all came from the references that I cited.  In fact,
as much as possible, I tried to use direct quotes.

For those who don't understand the difference, let's take a simple
example.  Suppose you have a container.  The container can be divided into
two partitions of equal volume which we will call A and B.  If there is
no barrier between the two partitions, then air molecules can flow freely
from one side to the other.  What is the probability that all the air
molecules are in one partition?

Well, that depends.  

What if there is only one molecule?  Again, that depends.  If the container
is large enough that the probablility of the molecule being halfway is very
small, then the answer is 100%.

What is the probablity of the molecule being in partition A?  50%

OK, what about two molecules?  The probability of both molecules being in
partition A is 25%.  Doesn't this violate the second law of thermodynamics?
No.  The sample size is not large enough to smooth out random variations.
How do we increase the sample size?  One of two ways.

1)  Make a number of observations of the two molecules, keeping track of
    how many observations put Molecule 1 in partition A and how many
    observations put Molecule 2 in partition A.  At first, there will be
    a lot of scatter in the data, but over a long period of time, the
    scatter will smooth out, and we will find that each molecule will spend
    half of its time in partition A.  Now, given a long period of time,
    with no selective forces causing the molecules to go in a particular
    direction, if one of the molecules spent all of its time in partition A,
    that *would* violate the second law of thermodynamics.

This is statistical thermodynamics, and it is a *MICROscopic* study of the
Laws of Thermodynamics.  Since we are dealing with molecules, we measure
quantity in number of molecules - moles.


2)  Increase the number of molecules in the container.  If the concentration
    of molecules is high enough, we can use physical measurements like
    Temperature and Pressure to determine what the average molecule in our
    sample is doing.  If the sample size is large enough, and measurements
    tell us that all the molecules are in side A - in other words 0 pressure
    in side B and a measurable pressure in side A, then that would violate
    the second law of thermodynamics.

This is engineering thermodynamics, and it is a *MACROscopic* study of the
Laws of Thermodynamics.  Since we are dealing with physical measurements,
we measure quantity in mass:  kg, or lbm.

For any system that you want to apply a thermodynamic analysis to, you have
to use the appropriate means of studying the system - microscopic or
macroscopic.  When dealing with a very large system such as the earth, a
macroscopic analysis is the only one that makes sense.

Some distinctions on the macroscopic level break down at the microscopic
level.  The distinction between chemical energy and electrical energy for
example.

Also the border between mass and energy.  An electron in molecular orbit
is considered as part of the matter of the molecule, but let that electron
escape at a high velocity, and it acts more like energy.  So at the
molecular level there are only two types of systems - open and closed - 
because when you lose matter from an atom, you can't always distinguish
whether the escaping matter is mass or energy.  At the macroscopic level,
it is easy to distinguish.  Just define matter to be made of low velocity
atoms and molecules and pure energy to be made of high velocity sub-atomic 
particles.  We can draw a pretty clear line - a low velocity hydrogen ion
is mass.  Anything smaller, or a proton moving faster is energy.  At the 
macroscopic level it is very simple to make the disctinction.

I will wait until we come to a consensus on question 1 before furthur
discussion of the other questions.
------------------------------------------------------------------------------
o What is Thermodynamics?

  "Thermodynamics is defined as the study of energy, its forms and
   transformations, and the interactions of energy with matter." [1, p.5]

Energy can exist in a number of forms, electrical energy, chemical energy,
potential energy, kinetic energy, PV energy, mechanical energy, etc.  In
order to apply the laws of thermodynamics mathematically, which is the
only way to "prove" anything, you must have a definition of energy that
is consistent with the laws of Thermo. The Laws of Thermo describe
the Laws by which transformations in energy must abide.  They have never
been shown false, and they have been demonstrated so thoroughly, that
they are not considered theories, but laws.  The field of engineering
is based largely on these laws, and in most fields of engineering*,
proposed processes must first be shown to satisfy these laws to merit
furthur consideration.  In chemical engineering, they are necessary criteria
for chemical reaction equilibria.

* At Auburn University, all departments in the Engineering School except
  Computer Science require their students to take EGR 201 - Introductory
  Thermodynamics.  Applications of the Laws of Thermodynamics can be
  demonstrated in every field of engineering represented, including
  Agricultural Engineering.

W : Work - Energy transfer due to a force acting against an opposing force*.
           In the most general sense, work can be described as change.

    The opposing force may be such things as friction, electric potential,
    gravity, inertia, pressure, etc.

* I have amended my previous definition because "resistance" is too ambiguous.

Q : Heat Transfer - one of three different types of energy transfer:
   1) Conduction
   2) Convection
   3) Radiation

Enthalpy is a convenient property defined in terms of internal energy,
pressure, and volume:

H = U + PV

H : enthalpy
U : internal energy
P : pressure
V : volume

Gibbs energy is a measure of the amount of energy available to do work
(ie. to effect a change) in chemical processes.  To determine Gibbs energy
we take the enthalpy, and subtract out the disordered energy, ie. the energy
that is not available to do work:

G = H - TS

G : Gibbs free energy
S : entropy
T : temperature

So, what is entropy?

Entropy is a measure of the disorder of the energy of a system.  Ordered
energy is available to do work, disordered energy is not. So, mathematically
we see that entropy*temperature is the amount of disordered energy at that
temperature.  This is not *exactly* correct, but is a good first approximation
in visualizing entropy on a macroscopic level.  It is would be more precise to
say that the equilibrium entropy level times temperature gives the amount
of enery that cannot be used in achieving equilibrium.  Once equilibrium is
achieved, there is no more energy available to do work at the equilibrium
conditions.

One criterion for chemical equilibruim in a closed system is that the total
Gibbs free enery must be at a minimum, which means that entropy must be at a
maximum.

  In SI units:

     S (the extensive property) has units:    J/K
     s (the intensive property) has units:   J/(kg K)  or J/(kg mole  K)

The relationship is:  
     For mass units:  S = m*s (m is mass)
     For mole units:  S = n*s (n is mole number)

If we multiply S*T we get energy units - (J/K)*(K) -> J

Intensively:

For mole units: s*T*m - (J/(kg mole K)*(K)*(kg mole) -> J
Likewise for mass units: s*T*m - (J/(kg K))*(K)*(kg) -> J


[1] _Fundamentals_of_Engineering_Thermodynamics_, Howell and Buckius,
McGraw-Hill, 1987
[2] _Introduction_to_Chemical_Engineering_Thermodynamics_, _Fourth_Edition_,
Smith and van Ness, McGraw-Hill, 1987
[3] _Classical_Thermodynamics_of_Nonelectrolyte_Solutions_, van Ness and
Abbott,  McGraw-Hill, 1982


James P. Henley Jr.
Chemical Engineering Dept.
Auburn University