Re: Isobaric expansion

["Glenn A. Carlson" <gcarlson@MAIL.WIN.ORG>]

> Subject: Re: Isobaric expansion
> Date: Sat, 18 Mar 2000 16:26:46 -0800
> From: Leigh Palmer <palmer@SFU.CA>
>
> > Equilibrium thermodynamics does fail when applied to real
> > thermodynamic processes, but to different degrees.  Relatively slow
> > processes (slow compared to the relaxation time of the system) are
> > described very well by equilibrium thermodynamics.
>
> One might say they are described perfectly well. If the "failures"
> are not measurable in practice then they are not failures. Your
> conception of ideality is that of a philosopher and not that of a
> physicist.

"Perfectly well"?  Equilibrium thermodynamics cannot describe the
states of a gas *during* a free expansion.

"The actual free expansion is, of course, a complicated irreversible
process involving turbulence and gross nonuniformities of pressure and
temperature (to the extent that these quantities can be defined at all
for such a marked nonequilibrium situation).  Equilibrium conditions
prevail only in the initial and final situations.  Nevertheless, to
predict the outcome of the process, the only knowledge required is
that of the energy function E characteristic of equilibrium
macrostates of the system."  [Reif, "Fundamentals of Thermal and
Statistical Physics," p. 176.]

Equilibrium thermodynamics can describe the states of a gas during a
constrained expansion, but the measureable difference between the
actual states and the predicted states increases as the rate of change
of the process approaches the relaxation time.

"1. [Relaxation time] << [time of experimental interest]:  In this
case the system comes to equilibrium very quickly compared to times of
experimental interest.  Hence probability arguments concerning the
resulting equilibrium situation are certainly applicable.

2.  [Relaxation time] >> [time of experimental interest]: . . .

3.  [Relaxation time approximately equal to time of experimental
interest]: In this case the time required to reach equilibrium is
comparable to times of experimental significance.  The statistical
distribution of the system over its accessible states is then not
uniform and keeps on changing during the time under consideration.
One is then faced with a difficult problem which cannot be reduced to
a discussion of equilibrium."

[Reif, "Fundamentals of Thermal and Statistical Physics," pp. 93-94.]

> I believe that conventionally "equilibrium" in this context implies
> "thermodynamic equilibrium". I can't recall using the term "dynamic
> equilibrium". What does it mean?

In a reversible chemical reaction, "[a]t the point when the rate at
which product is being formed equals the rate at which product is
being decomposed, the reaction is said to have reached a state of
dynamic equilibrium.  The term "dynamic" is used because even though
no further net change occurs in the concentrations of rectants and
product once equilibrium has been established, molecules are still
being changed.  Reactant molecules continue to combine and product
molecules continue to decompose."  [Buell and Girard, "Chemistry: An
Environmental Perspective," p. 205.]  This is what I mean by dynamic
(as distinguished from static) equilibrium, i.e., there is no further
change in the thermodynamic properties of the gas, but molecules are
still moving about, colliding and transferring energy between them.

>                                     The term "static equilibrium", used
> in mechanics, often pertains to systems in which motion is present.
> It does not imply statsis, a point which must be made when teaching
> the topic.

"The term 'equilibrium' implies either that the object is at rest or
that its center of mass moves with constant velocity.  We deal here
only with the former, which are referred to as objects in "static
equilibrium'."  [Serway, "Physics for Scientists and Engineers," 4th
Edition, p. 337.]  Even accepting Leigh's definition of "static
equilibrium" to include motion or not, the use of the qualifier
"static" carries no meaning at all.  Otherwise, what, then, would be
an example of equilibrium which is not static?

The isobaric process discussed in freshman physics is case #1 above.
The free expansion is case #2.  A complete discussion of these cases
must include a discussion of how equilibrium thermodynamics succeeds
for case #1 and fails for case #2.  Otherwise, the student will leave
with the mistaken view that equilibrium thermodynamics applies
perfectly well to any real process.  This distinction between
equilibrium and nonequilibrium thermodynamics also explains the
difference between boiling (equilibrium or nearly so) and evaporation
(nonequilibrium).

Glenn A. Carlson, P.E.
St. Charles, MO