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Collective response by Robert D. Freeman to the reply by Belandria to criticism of Belandria's paper on "Entropy Coupling" (J. Chem. Educ. 1995, 72, 116-118) The second law of thermodynamics is one of the most highly respected concepts in the pantheon of Science. Despite many attempts to overthrow
or subvert it, there has, as yet, been no documented, verifiable, repeatable
example of that having been accomplished. That fact does not prove
that the second law can never be broken, but it does send the clear
message that anyone who proposes a scheme for doing so should expect thorough
dissection of the scheme and should be prepared to supply full information
about the details of the scheme.
José Belandria (JB) has failed to provide those detailsboth in
his original paper (1) and in his response (2) to his critics
(3). Various thermodynamic deficiences in his paper (1) have
been described (3) but JB has offered essentially nothing in his
reply (2) that helps to resolve the issues.
In the interest of brevity, consensus, and closure, the Editor has asked
me to prepare, on behalf of all the critics (3) a reply to JB's
response (2). In doing so, I have attempted to find an approach
that focuses on the critical point and that elicits additional imput from
Belandria. Note that since JB has explicitly challenged the second law,
logically one can not appeal to the second law to show that JB is wrong.
All of the critics question how JB arrived at the indicated final pressure
in tank A (1). In response (2) he says: "To get
the final pressure I set up a thermodynamic model for the whole process
using eqs 33 to 42 and investigated". However, if one examines eqs
33 to 42, one finds that the final pressure in tank A is fixed at
405.32 kPa, and there is no indication of how that pressure was obtained.
Since the final pressure in A is a crucial point in the criticism,
it is useful to examine JB's "thermodynamic model" in some detail;
for reasons disclosed later, it is convenient to refer to this example
as the "connected tanks" case. In the reconstruction of his eqs
33-42 below, I have used parameters slightly different from those given
in JB's paper, so as to minimize confusion between the two calculations.
I assume exactly the same arrangement and process as that described by
JB (1), with these two changes: (i) the initial temperatures are
1400 and 400 K in A and B, respectively, and (ii) the initial
pressure is 1 bar (100 kPa) in both. Note that the final temperatures are
fixed at 1400 K by the specifications of the process, and that the final
pressure in B is readily obtained from the known increase in temperature
at constant volume. The only parameter not fixed by the initial
temperatures and pressures and by the process specifications is the final
pressure in A. With these new parameters, JB's eqs 3342 become:
Delta UA= 0
; [A is isothermal] (33b)
(34b)
(35b)
(36b)
wB = 0 [isochoric]
(37b)
wA = QA
[Delta UA = 0] (38b)
(39b)
(40b)
(41b)
(42b)
A comment is required with regard to eq 42b. Given JB's concept of
entropy creation/destruction, his eq 19 for Delta SgA
is logically consistent and eq 21 follows; however, his eq 27 for Delta
SgM is logically inconsistent, and is arthmetically incompatible
with eq 29. If in his eq 27 a minus sign is inserted before each integral
sign, these difficulties are eliminated. JB has acknowledged the problem
with his eq 27 (see below) and attributes it to a "transcription error"a
typo; eq 42b has the needed minus signs before the integral signs. Note
also that 42b is included in the list of equations above only to provide
1:1 comparison with JB's eqs 3342; Olivares and Colmenares (3)
provide a detailed discussion of the errors in JB's eq 42.
In eqs 33b42b, numerical values have been obtained for all thermodynamic
quantities except Delta SA and those which depend on
it. Delta SA, in turn, depends on the final pressure
in A. The analysis is straightforward to this point, but it can
not be completed without a value for PAf. How can one
get that value? From what is given above, one can not. To obtain
the correct value for the final pressure in A one must (i)
measure it, or (ii) specify the compression process in sufficient detail
that the final pressure may be determined. JB's specification (1)
of the compression process as "isothermally in a nonreversible way"
is not sufficient.
However, one can easily place some limits on the final pressure in A.
If the compression is assumed to be reversible, PAf may be calculated
from:
wA = -RT ln(PAf/PAi)
(1)
The result is 2.92 bar, and Delta SA is -8.91 J/K,
whence Delta SgA in eq 40b is zero - as expected for
a reversible isothermal process. One might also ask what value of PAf
makes Delta Su equal to zero. From eq 39b, with Delta
Su set to 0, one obtains PAf = 6.55 bar
(4); for higher values, Delta Su is negative.
For any assumed value of PAf greater than 2.92
bar and less than 6.55 bar, Delta Su will be positive,
Delta SgA will be negative ("entropy is destroyed")
and Delta SgM will be positive ("entropy is created");
that is, JB's criteria are met. But which of these - if any (4)
- is the correct value of PAf?
In early December, I wrote to Belandria, presented the reconstructed
version of his example [i.e., the material in the paragraph surrounding
(33b - 42b) and half of the following paragraph], and solicited his answer
to these questions: (i) What is the value of PAf in eq
39b? and (ii) How did you arrive at your value? In his reply, Belandria
commented on the transcription error in his eq 27 (see above), then presented
some calculations that essentially duplicate those in the preceding paragraph
(which was not included in my letter to him), and then provided these comments:
"According to this [JB's] view, there is an infinite set of final
pressures between 2.92 and 6.55 bar where internal entropy coupling exists
and the compression process is more efficient than a reversible compression
for the same change of state. In this region total entropy change is greater
than zero and energy is conserved, and [the] process describes trajectories
not predicted by classical thermodynamics or Caratheodory's theorem. Finally,
to answer your question there is an infinite set of PAf values that
satisfy eqs 33b - 42b according to the above description.
In relation to the process depicted in [JB's original paper (1)]
I detected that internal entropy coupling occurs in a range of final pressures
between 312.75 and 811.09 kPa. In the article I selected one final-pressure-value
of this set, equal to 405.32 kPa, and designed the system published by
the Journal."
These comments provide the first indication from JB about how he obtained
the final pressure 405.32 kPa in his example (1). It seems clear
from these comments that JB believes that any value of PAf
is legitimate if his criteria are satisfied, that is, if Delta
Su is positive, Delta SgA is negative
("entropy is destroyed"), and Delta SgM is
positive ("entropy is created"). Parenthetically, if JB is correct,
one wonders how Nature "knows" which of the "infinite set
of PAf values" to chooseor is the value of PAf
a matter of chance?
In subsequent correspondence with JB, I posed another example, the "separated
tanks" case. The scenario is very similar to the "connected tanks"
case, above, and the initial conditions are the same. However, now tank
A and tank B are separated, and partition M is replaced
by MA in the end of tank A, and MB
in the end of tank B. Suppose that these two tanks are immersed
in an isothermal "thermal reservoir" or "heat bath"
at 1400 K, and are widely separated in that bath. To execute the process,
manipulate MB to permit heat to flow from the reservoir
into tank B until the temperature in B is 1400 K; "close"
MB. Then manipulate MA to permit heat
flow from A to the reservoir, and compress A until QA
= -QB; "close" MA.
For this "separated" example, the changes in tank B
are the same as in eqs 33b-41b, and the heat bath has undergone no change
(it lost 12.47 kJ to B, but the same amount was restored from A).
Further, QA and wA are the same. In
fact, all of eqs 33b-41b apply to this new example.
My question to JB was, in this "separated tanks" case, what
is the final pressure in tank A, and how did you obtain the answer?
JB agrees (letter dated March 11) that eqs 33b-41b apply to this case,
and states: "For the specified process there is not a temperature
gradient between tank A and the surrounding heat reservoir, and
to carry out the compression process under such a condition requires that
heat should be released from tank A to the heat reservoir in differential
amounts to avoid finite temperature gradients [i.e., the process must be
carried out in a way], similar to a conventional isothermal compression.
For this condition, Delta SgA is zero and the final pressure
for the separated tank A is 292 kPa [2.92 bar]." He also states
that "'isothermally' means that the temperature is constant or almost
constant throughout the process". He does not elaborate on the meaning
of "almost" in this context (e.g., ± 0.1 or ± 1.0
or ± 10.0 or ± ?? kelvin). Note that JB's value for the final
pressure in A is the same as that given above, following eq 1, for
a reversible, isothermal compression.
In the absence of explicit details, one must analyze JB's claims from
the available evidence; even with two papers (1, 2) and two or three
letters available, it is still difficult to be sure about the basis for
JB's statements; this lack of precision and detail has been a major problem
from the beginning (1). The preceding paragraph seems to imply rather
strongly that, in JB's view, the final pressure in tank A is different
for the two casesseparated vs. connected tanksbecause of a difference in
the temperature gradient between tank A and its surroundings. Clearly,
JB is correct in saying that there is no (or an infinitesimal) gradient
in the separated-tanks case. However, the implication that there is
a gradient in the connected-tanks case contradicts the stated
specifications (1), namely, the compression of the gas in A
occurs isothermally. In the connected-tanks case, the gases in the two
tanks are at different temperatures; there must be a temperature gradient
somewhere. But if the gas in A is compressed isothermally,
as JB specifies, there can be no gradient within the gas, and the surface
of the partition M must be at the temperature of the gas. Hence,
the gradient must lie totally within the partition M and can have
no effect on the final pressure of the gas. Further, in an isothermal compression,
the path of the state of the gas, as represented on a P - V plot,
is - obviously - along an isotherm; as Battino and Wood have noted (3),
this is precisely the same path followed by the system (gas A) during
a reversible (and isothermal) compression; that is, isothermal compressions
and expansions are inherently reversible. Therefore, JB's specification
(1) that the compression of A occur "isothermally in
a nonreversible way" is self contradictory.
If the gas in A is compressed isothermally, as specified, then
the final pressure in A is the same for both cases, connected or
separated. JB's value for the final pressure in A for the connected
case (1, his Dec. 13 letter) is based on at least two contradictions:
(i) temperature gradients during an isothermal process; and (ii) an isothermal
compression occurring in a "nonreversible way". He also ignores
the fact that, independent of his "criteria" about "entropy
creation and destruction, there are constraints on the possible final pressure
in Acontraints that can not be hand-waved away with appeals to "entropy
coupling"; Tykodi (3) and Olivares and Colmenares (3)
have given simple, mechanical, nonthermodynamic arguments that show JB's
value for the final pressure in A (1) to be impossible. Finally,
Olivares and Colmenares (3) have argued persuasively that JB's application
(entropy "creation and destruction") of Prigogine's formulation
of "entropy production" is not valid.
Until Belandria provides a legitimate scheme for confirming his value
of PAf, the status of his "exceptional theoretical
process" must be placed somewhere between "simply wrong"
and "unproved and highly unlikely to be proved"; the critics
believe very strongly that the proper description is "simply wrong".
Finally, on behalf of all the critics (3), who are all experienced
thermodynamicists, I express our concern about the review process that
resulted in publication of JB's original paper (1). Our unanimous
opinion is that it should not be been publishedcertainly not in its present
form. An acceptable alternative form might have been foundfor example,
"Provacative Opinion"type, but even that is highly questionable.
Literature Cited
1. Belandria, J. I. J. Chem. Educ. 1995, 72, 116188.
2. Belandria, J. I., J. Chem. Educ. 1997, 74, 286.
3. Batino, R.; Wood, S. E.; Freeman, R. D.; Nash, L. K.; Olivares, W.;
Colemenares, P. J.; Tykodi, R. J. J. Chem. Educ. 1997, 74,
256, 281286.
4. Calculation of the value PAf = 6.55 bar when Delta
Su = 0 has been done solely to clarify what JB has done.
Note that in the calculation it has been implicitly assumed that all other
parameters remain the same as in the reversible isothermal compression
case (PAf = 2.92 bar; eqs 33b-41b and 1). Clearly, this
can not be, for it violates the first law. Compression of the gas in A
to a pressure higher than 2.92 bar requires additional work; therefore,
wA increases, which means the -QA must
increase (not stay the same), which means that QB
must increase, which means that the final temperature in B must
be higher than 1400 K (which also violates the specifications of the process).
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