Thermodynamics and Controversy - Journal of Chemical Education

Chemical Education Today

Letters Thermodynamics and Controversy

An Exceptional Theoretical Process

I suspect that there may be a fourth law of thermodynamics: discourse on any thermodynamic topic increases spontaneously (and perhaps exponentially). This reflects the importance of thermodynamics, and its success. Albert Einstein called thermodynamics, “…the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown.” Even a hint of an exception to thermodynamic principles generates the closest possible scrutiny and considerable discussion. The letter that begins below and those that follow beginning on page 281 fill 13 pages in all. They consist of criticisms and a reply related to a three-page paper by José Belandria that appeared in February, 1995. The author’s reply to his critics is longer than the original paper, as is one of the criticisms. The critics have had their say, the author has responded, and, on behalf of all the critics, Robert Freeman has prepared a reply to that response. At this point we can consider this issue closed, and the Journal will publish no further comments on Belandria’s paper, nor on the criticisms of it that appear here. We thank all of the contributors to this discussion for the time and effort they have devoted to clarifying the situation.

The article by Belandria (1) has so many problems that we hardly know where to start. Neither the “exceptional” process nor the conditions are sufficiently rigorously defined to do a proper analysis. Thus, we are forced to start from scratch and define the changes of state which occur and the nature of the systems and their surroundings so that we can do meaningful calculations. Consider one mole of an ideal monatomic gas of constant Cv of 12.47 J mol{1 K{1 confined to an adiabatic container sealed by a frictionless weightless piston. This is container A and TAi = 1500 K, VAi = 123.086 L, and P Ai = 101.33 kPa. Container A is connected to container B by a rigid adiabatic wall. Container B is also adiabatic and rigid and initially contains one mole of an ideal monatomic gas. TBi = 373 K, PB i = 101.33 kPa, and VB i = 30.607 L. Replace the wall between A and B with a massless rigid diathermic partition. Now, reversibly and isothermally compress the gas in A until the temperature in B reaches 1500 K. The pressure in B is now 407.49 kPa. The heat absorbed by gas B (from A) is given by ∆U = nCv∆T = 14,054 J. Since all of the walls (except the separating partition) are adiabatic, then the 14,054 J must come from A. A simple calculation shows that VAf = 39.883 L and PAf = 312.68 kPa derived from the work done on A to produce 14,054 J. The entropy change in A is ∆S A = {14,054/1500 = {9.37 J mol{1 K{1 . The entropy change in Bi is obtained from ∆S B = nCv ln(TB f /TB i) = 17.35 J mol{1 K {1 . The entropy change of the universe is ∆Suniv = ∆SA + ∆S B = +7.98 J mol{1 K{1 for the work done on compressing the gas in A plus heating the gas in B. Please note that the only way the compression in A may occur isothermally is via a reversible process. This is the minimum work of compression. And, entropy changes must be calculated along reversible paths. If you assume the final state for A indicated by the author then 17,289 J are generated. Continuing, if you assume the author’s final state for B, then 14,054 J are absorbed. What happens to the excess heat? The author’s numbers are incompatible. There are no surprises here and no exceptions to the laws of thermodynamics.

John W. Moore Editor

Literature Cited 1. Belandria, J. I. J. Chem. Educ. 1995, 72, 116–118.

Rubin Battino (Emeritus) Wright State University Dayton, OH 45435 Scott E. Wood (Emeritus) Illinois Institute of Technology Chicago, IL 60616

letters continued on page 281

256

Journal of Chemical Education • Vol. 74 No. 3 March 1997


Letters An Exceptional Theoretical Process, continued from page 256 I am puzzled by several aspects of the paper by Belandria on “...Internal Entropy Coupling” (J. Chem. Educ. 1995, 72, 116–118). I’ll skip comment on obvious problems with significant digits; otherwise, I am questioning not the author’s calculations but his interpretation. Belandria postulates that the metal partition separating chambers A and B has “negligible mass” and negligible heat capacity; I have no quibble with that. Later, he interprets his results as indicating that “there is a creation of entropy in the metal partition M during the process.” While there certainly is an increase in entropy in the process, I would argue that the metal partition has nothing to do with “creating entropy”, except the obvious function of being a partition between two chambers. To elaborate: the entropy in chamber B increases because it receives energy (“heat”) and its temperature rises; nothing else happens in B, so calculation of energy and entropy change is straightforward given the initial and final temperatures. The situation in chamber A is more complicated; since compression of the gas in A is “nonreversible” but otherwise unspecified, the work done on the gas can not be calculated directly but must be inferred from the energy (“heat”) transferred to B. On the other hand, the entropy change of the gas in A can be calculated because entropy is a state function, and data (p,T,n) are given. Now, let us examine the entropy changes. The entropy of the gas in B increases (because the temperature rises and the volume is constant); the entropy of the gas in A decreases (because the temperature is constant and the volume decreases). The entropy change of the “universe” is just the sum of the changes in A and B: ∆SU = ∆S A + ∆S B = {11.53 + 17.39 = 5.83 J/K (Belandria’s numbers). But what about Belandria’s “creation/destruction of entropy”? Application of this term could be justifiable in the case of chamber A, wherein the decrease in entropy of the gas based on (p,T,n) data is greater than the decrease based on “heat flow” to chamber B (but see below). However, reference

to “creation of entropy” in metal partition M smacks of witchcraft. There is an increase in entropy as a result of energy flow through the partition, but the increase arises from the difference in temperature on the two sides of the partition and the partition is an innocent bystander in the process. Use of the term “creation” in this context not only raises troublesome questions (e.g., how can an object with no mass and no heat capacity “create entropy”?), it diverts attention from the real origins of the changes in entropy. Further, the term “creation” has a number of connotations that are undesirable in a scientific discipline. I believe that the terms “creation/destruction of entropy” are an unfortunate choice and cause more problems than they cure. Another puzzling aspect of this paper is Belandria’s claim that the process described is somehow “more efficient” than a corresponding reversible process. Just before his “Conclusion”, he states “the reversible work required for the same isothermal compression process is 17,288 J”—compared to 14,054 J for his irreversible process. This seems to me a totally invalid comparison: if such a reversible isothermal compression were done, the heat generated (17,288 J) would (i) raise the temperature of chamber B well above the posed equilibrium value of 1500 K, violating the second law (direction of heat flow), or (ii) raise the temperature of both chambers above 1500 K, violating the posed conditions of the process. The question to be asked, it seems to me, is: given the posed initial conditions, let the gas in A be compressed isothermally and reversibly to a final pressure such that the heat produced in A is exactly that required to produce the given results in B (i.e., 14,054 J); what is the final pressure? I obtain 312.7 kPa, compared to 405.32 kPa for Belandria’s irreversible process. Finally, the result quoted above— 14,054 J for the irreversible process and 17,288 J for the reversible—rings a warning bell. One of the standard results of thermodynamics is that getting from state X to state Y can be accomplished with minimal work if the reversible path is taken. Yet Belandria tells us that in his process the reversible path requires more work. How can that be? Well, he tells us nothing about the work done on the gas in A; he has simply postulated the initial and final conditions in chamber A without providing data about the change. It is

clearly implied (second paragraph under “The Process”) that all of the “nonreversible” work done on the gas is converted to “heat” which flows to chamber B, so that this “heat” flow into B is a direct measure of the postulated “nonreversible” work. It is simply not possible for this amount of work to accomplish the stated compression of the gas in A. Belandria’s postulated data violate the second law; consequently, none of his results can be considered valid. I see no merit in this paper other than its being used as an “debugging” assignment for a thermodynamics class: Find what’s wrong. Robert D. Freeman Oklahoma State University Stillwater, OK 74078

t In the recent paper by José I. Belandria, J. Chem. Educ. 1995, 72, 116–118, halfway down the second column on p 118 one is arrested by the claim that “the compression process occurring in tank A is more efficient than a reversible compression process for the same change of state.” This cannot be true when, as here, the change of state is nothing but the isothermal compression of an ideal gas from 1 atm to 4 atm. On the author’s engineering convention, which makes w negative for a work input, the cited isothermal compression must satisfy the familiar relation w ≤ {∆A The inequality figures in the general case, while the equality applies to the special case of reversibility, where {∆A = wrev In the given isothermal change of state, the alteration (∆A) of the Helmholtz function is a constant eliminated by summing the last two relations to obtain the restrictive condition w ≤ wrev For the reversible isothermal compression from 1 to 4 atm, the author (correctly) gives as the requisite work wrev = {17.3 kJ. But the author’s mysterious irreversible “theoretical process” purports to achieve the same isothermal compression with work w = {14.05 kJ. These work terms stand, alas, in the forbidden relationship w > wrev

Vol. 74 No. 3 March 1997 • Journal of Chemical Education

281


Letters The author’s theoretical process is thus a thermodynamically impossible process. To see that it involves a transparent violation of the second law, imagine tank A as the cylinder of a Carnot engine in thermal contact with an immense heat reservoir at 1500 K. Beginning at 1 atm, let the author’s notional irreversible isothermal compression proceed to 4 atm with work input of {14.05 kJ. Let the gas then resume its original state by a reversible isothermal expansion yielding a work output of +17.3 kJ. Drawing on only a single heat reservoir, this indefinitely renewable closed cycle thus delivers a net work output of over 3 kJ per cycle. Thus, a reductio ad absurdum. One wonders where the author got his figure for the final pressure (P f) in tank A. That Pf = 4 atm is nowhere derived or justified in any way. It is simply announced— first in the diagram on p 116. The author calculates the heat required to warm tank B from 373 to 1500 K, and concludes (correctly) that this 14.05 kJ must also represent the magnitude of the work input to the isothermal system A. Unhappily, a {14.05 kJ work input just won’t suffice to compress the gas in A to P f = 4 atm. Using this work input in the most effective possible way (i.e., reversibly) we calculate P f for the isothermal compression from 8.314 (1500) ln(1/P f) = {14,054 ln(P f /1) = +1.127: Pf = 3.086 atm Hence this reversible isothermal compression proceeds with ∆SA = 8.314 ln (1 / 3.086) = {9.369 J/K Alternatively, we can calculate ∆SA from the heat expelled in that reversible isothermal compression, finding ∆SA = {14,054 / 1500 = {9.369 J/K Perfectly concordant with each other, these figures differ significantly from the author’s ∆SA = {11.53 J/K, which he derives from a supposed Pf = 4 atm. The difference is just the {2.16 J/K he alleges to be “destroyed” in tank A. Need we now credit any such entropy destruction? The 2.16 J/K discrepancy in ∆SA propagates itself in the author’s conclusion that ∆Suniv = +5.83 J/K, as against the present finding that ∆Suniv = ∆S A + ∆SB = +7.99 J/K. Whatever the numerical value, need we follow the author in attributing the rise in S univ to a creation of entropy in the diathermal membrane that separates A from B? A more prosaic alternative view finds the increase in S univ a simple consequence of the flow of heat from a hotter body A to a cooler B. Leonard K. Nash Harvard University Cambridge, MA 02138

tank A is more efficient than a reversible compression process for the same change of state” (1). In this note we would like to solve the “paradox” and to show that a careful analysis of Belandria’s proposal demonstrates that such a process is not feasible and that no violation of Prigogine’s formulation exists. Furthermore, we show the usefulness of Prigogine’s formulation and clarify some relevant aspects of such formulation. The Physical Unfeasibility of the Process The process in discussion is well described in ref 1. Briefly, it consists in the heating of a constant-volume system B, consisting of one mole of an ideal gas initially at 373 K and 101.33 kPa, which ends up in a final state of 1500 K and 407.49 kPa. The heat is provided by the irreversible isothermal compression of one mole of an ideal gas, system A, from an initial pressure of 101.33 kPa to a final pressure of 405.32 kPa. The two systems are coupled through a dividing metal partition M of negligible mass. In order to attain the equilibrium temperature of 1500 K, keeping its volume constant, system B should absorb a heat

Q B = ∆E = m B C vB ∆T = 14,053.69 J

The heat released from system A through the partition M is then QA = {QB = {14,053.69 J. Since the compression of the ideal gas in A is isothermal, the work that has to be done on it is wA = {14,053.69 J. This work can be provided in one, two, or infinite steps. When carried out in one irreversible step, an external pressure P ex has to be applied to a piston acting on system A. The work wA is then

w A = Pex(VA 2 – VA 1) = nRTA Pex 1 – 1 PA 2 PA 1

Pex = –

PA 2PA 1 wA ⋅ = 152.25 kPa nRTA PA 2 – PA 1

(3)

For simple mechanical reasons it is impossible to compress a gas up to 405.32 kPa by applying an opposing pressure as low as 152.25 kPa. In fact, the higher pressure to which system A can be compressed in one step is obtained by letting the external pressure be equal to the final pressure Pex = PA 2. Then

w A = nRTA 1 –

Pex PA 1

(4)

from which the maximum final pressure would be PA2 = Pex = 215.52 kPa instead of 405.32 kPa. The other extreme would be to carry out the compression in infinite steps, that is, reversibly. Then 2

wA =

PdV = – nRT ln 1

from which PA2 = 312.71 kPa.

282

(2)

If we require this work to be {14,053.69 J and the final pressure of system A to be 405.32 kPa, as proposed by Belandria, a contradiction is reached, since then

t In a recent paper published in this Journal, Belandria (1) proposed a seemingly paradoxical process which, without violating the classical formulation of the second law, presented an internal entropy coupling. That is, internal destruction of entropy in one part of the system with coupled entropy creation elsewhere. This, in fact, violates Prigogine’s formulation of the second law (2). If true, the consequences of this exceptional process would be dramatic since, as the author concludes, “the compression process occurring in

(1)


PA 2 PA 1

(5)


No matter how the gas in A is compressed isothermally, to destroy a work of 14,053.69 J, the final pressure should be in the interval between 215.52 kPa and 312.71 kPa. Therefore the final pressure of 405.32 kPa used by Belandria (1) is unattainable on simple mechanical, nonthermodynamic arguments. Thus, the proposed process is not physically feasible. If one compresses the gas reversibly to the final pressure of 405.32 kPa, then the needed work would be, from eq 5, wA = {17,288.47 J. From a thermodynamical point of view, the system would have done (destroyed) more work than in a reversible process between the same states! Entropy Considerations In the frame of the classical formulation of the second law, the proposed process involves two coupled changes: the irreversible isothermal compression of system A and the heat transfer from the hot reservoir A to the constant-volume system B. The entropy changes corresponding to the initial and final states of ref 1 are simply

∆S A= nR ln

PA 1 = 1 mol × 8.314 JK –1mol–1ln 101.32 405.32 PA 2 = –11.53 JK –1

∆S B = C v ln

(6)

TB2 = 12.47 JK –1 ln 1500 = 17.36 JK –1 (7) 373 TB1

The massless metal partition M has a negligible heat capacity and therefore ∆SM = 0. The entropy change of the Universe is then

∆SU = (–11.53 + 17.36 + 0) JK –1 = +5.83 JK –1

∆ eS =

dQ T

(10)

and

dQrev (11) T where dQ is the actual heat transfer to the system, T its temperature, and dQrev is the heat transfer through a reversible process between the same initial and final states. The second law becomes ∆S =

∆ iS =

dQ rev – T

dQ ≥0 T

(12)

This is nothing else but the Clausius inequality. For an isothermal process, this is equivalent to Qrev ≥ Q and hence wrev ≥ w. This says that no internal entropy destruction is possible in any region of space. In fact, Prigogine clearly states that “We can therefore say that ‘absorption’ of entropy in one part, compensated by a sufficient ‘production’ in another part of the system is prohibited” (2). The evaluation of the entropy creation in systems A and B follows as in eqs 16 to 25 of ref 1; that is,

∆ e SA =

dQ A Q A = = –9.375 J K –1 TA TA

∆ i S A = ∆S A – ∆ e S A = ( –11.53 + 9.37) J K –1 = –2.16 J K –1

(13)

(14)

(8)

That is, the entropy increase due to the heat absorption in system B dominates the entire process giving ∆SU > 0, even though the final pressure in system A is unphysical. This is not a contradiction, since the second law says that a spontaneous process will always give ∆SU > 0, and conversely, that a physically or chemically possible process with ∆SU > 0 is spontaneous, but it does not imply or guarantee that if ∆SU > 0 the process will in fact occur. The second law does not give information on the dynamics of physicochemical feasibility of a given process. A classic example is the production of H 2O(l) from its elements H2(g) and O 2(g) at 1 atm and 25 °C. Thermodynamically that reaction is highly favored because ∆S U = ∆Gf /T > 0, but it is well known that the reaction does not occur spontaneously unless a catalyst or a spark initiates the reaction. Coupled processes, where a process not favored thermodynamically is coupled with one highly favored to make the first one to occur, are very important and well known: for example, in the metabolism of living beings. But, in those cases, the unfavored process must be physically or chemically possible. In Prigogine’s formulation the total entropy change of a given system is written in terms of the internal entropy production ∆iS and the entropy flow ∆ eS:

∆S = ∆ eS + ∆ iS

where ∆e S is defined as

∆ e SB =

dQ B = TB

C VB dTB = 17.36 J K –1 TB

∆ i SB = ∆ SB – ∆ e SB = 0

(15) (16)

However, the evaluation of ∆ i S for system M, the infinitesimal metal partition in ref 1, is erroneous. Its evaluation should be carried out carefully. By definition,

∆ e SM =

dQ M TM

(17)

Here we have stressed that the temperature inside the integral is that of the metal M. It is straightforward to realize that if the partition’s mass is negligible, heat capacity is also negligible and no heat would be absorbed, leading to ∆e SM = 0. Another way to look at this is to consider a system M with finite mass mM and specific heat capacity cM , and then let the mass tend to zero. We can separate the process occurring in M into two steps. The metal first absorbs heat {QA from system A, increasing its temperature from TM 1 to T9M . It then releases the heat completely to system B, decreasing its temperature to the final temparature TM 2. In that case –QA = mMcM (T M2 – TM1) + QB

(18)

(9)


283


Letters ∆ eS M =

T 9M TM 1

dQ M + TM

TM 2 T 9M

TM2 dQ M = m M c M ln TM TM1

the total entropy change of the metal heated from T M1 to T M2 is

TM2 ∆S M = m M c M ln TM1

Σn ∆ i Sn = ∆ i S A + ∆ i S B + ∆ i S M

(19)

(20)

= ( –2.16 + 0 + 0) J K –1 = –2.16 J K –1 < 0

If we apply the results of eq A7 (appendix) to evaluate ∆i S: 3

∆ i S = –2.16 J K –1 + Σ n

3

∆ eS M =

dQ A + TA

dQ B TB

Entropy Production of the Entire System We would like to stress that, while entropy is an extensive property and therefore the entropy increase of the Universe is the sum of the entropy changes for each system, namely

∆SU = Σ ∆Sl l

(23)

the so-called entropy production is not an additive property. The detailed demonstration of this is given in the appendix. In fact, for the process under discussion, since there are no heat flows into the entire system ∆e S U = 0

(24)

and the entropy production of the entire system is just equal to the entropy change ∆i SU = ∆SU = 5.83 J K–1 > 0

(25)

while the sum of the entropy production of each system in the process in discussion is negative; that is,

284

dQnm T

(27)

dQnm dQ B dQ M dQnm = + + T TA TB TM TB2 QA = + C v ln + 0 = 7.99 J K –1 TA TB1

(28)

Using this result in eq 28, we get ∆i S = (–2.16 + 7.99) J K–1 = 5.83 J K–1

(22)

an expression that has no thermodynamic foundation. Therefore, no entropy coupling is actually detected; the internal entropy destroyed in system A is not compensated by systems B and M. As we mentioned above, Prigogine’s formulation clearly states that the process in A, with ∆iSA = {2.16 J K{1, is not possible. This result is in agreement with our previous discussion where we showed that in fact the proposed final state for system A is unattainable on simple mechanical grounds. Belandria’s process is not exceptional; instead, it illustrates the beauty of Prigogine’s local formulation: while the global classic formulation gives ∆SU > 0 allowing the possibility of the coupled process, Prigogine’s local formulation give ∆i SB = ∆i S M = 0, since the heat transference is done reversibly; but ∆i SA < 0, which says that the proposed irreversible compression is not possible! To close this section, it is worth mentioning that if we used a correct mechanical value for PA2 of 215.52 kPa for compression in only one step, the ∆i SA = 3.09 J K {1 > 0. In fact, any value in the interval of 215.52 kPa to 312.71 kPa would give ∆i S A ≥ 0, as it should be.

2

Σn Σm

(21)

In the limit of mM → 0 all the entropy terms are negligible and ∆SM = ∆eSM = ∆iSM = 0, as expected. Belandria fails to get this result because he uses, instead of eq 19,

2

Σm

where

and the internal entropy creation is null ∆i SM = ∆SM – ∆e SM = 0

(26)

(29)

which is identical to ∆S, as expected, because the entire system is isolated and ∆e S = 0 with ∆SU = ∆S = ∆i S. Belandria (1) failed to recognize that the entropy production is not an additive property, as we show in the appendix. In order to obtain the expected result in eq 29, he was then forced to use eq 22 for ∆e SM, which corresponds numerically to the right hand side of eq 28. This led that author to make the massless partition responsible for most of the entropy production of the entire system, a result at first sight striking. Conclusions We have carefully analyzed the apparently exceptional process proposed by Belandra (1) and solved the “paradox” that would have led to an irreversible process more efficient than its corresponding reversible process. We pointed out that the process in tank A, destroying entropy (i.e., ∆i S A = { 2.16 J K {1), is not physically feasible, since the opposed external pressure needed to compress a gas must be higher than its internal pressure. Secondly, we have recalculated the internal entropy production in the metal partition to obtain ∆ i SM = 0. Therefore, actually, no internal entropy coupling occurs in Belandria’s process. Finally, we have shown that entropy production is not an additive quantity; that is, the entropy production of an entire system is not simply the sum of the entropy production of its parts. Instead, as given in the appendix, an additional term due to the heat exchange among the subsystems must be added. In conclusion, the theoretical process proposed by Belandria is very illustrative of the usefulness of Prigogine’s formulation of the second law. While the global classic formulation gives ∆S U > 0 for the coupled process in discussion, Prigogine’s local formulation says that processes occurring in systems M and B with ∆i S = 0 are reversible, and that the compression proposed for system A with ∆ i S A < 0 is not feasible. Appendix Here we will show how to calculate the entropy production of an entire complex system. Let us consider a complex system composed of N coupled subsystems. For convenience, let us split the heat transferred to system, dQn into



That is, the entropy production is not an additive property.

two parts N–1

dQn = d eQ n +

Σm

dQnm

(A1)

where de Qn is the heat transferred to subsystem n from the outside surrounding the entire system and dQ nm is the heat transferred to subsystem n from its surrounding neighbor subsystem m. The sum is over all the rest of subsystems. Clearly, by conservation of energy, dQnm = –dQmn N N –1

Σn Σm

dQnm = 0

By definition, the entropy production of a given subsystem n is

d i dSn = dSn –

dQn Tn

(A2)

where dS n is the infinitesimal entropy change of subsystem n and Tn its temperature. Similarly, the entropy production of the entire complex system is di S = dS – d e S

(A3)

where dS is the entropy change of the entire system, given by

dS = Σ dSn n

(A4)

since the entropy is an extensive property. Using the notation introduced in eq A1, the external entropy change d e S is, by definition, given as N

de S = Σ n

d e Qn Tn

(A5)

Summing over all subsystems in both sides of eq A2 and using eqs A1 and A4 N

N

N N –1

Σn d i Sn = dS – Σn d Te Qn n – Σn Σm

d Q nm Tn

(A6)

Then using eqs A3 and A5 and integrating over all processes occurring, we have that the net entropy production of the entire system, ∆i S, is N

N N –1

∆iS = Σ ∆ i Sn + Σ n n

Σm

dQnm Tn

(A7)

This result is physically meaningful. It says that the net entropy production should be obtained by adding not only the internal entropy productions of the individual susbystems, but also the entropy fluxes among them. Equation A7 is quite general and not well known. The second term on the right-hand side vanishes only when the temperature of the entire system is uniform and Tn = T for all subsystems. Therefore, in general N

∆iS ≠ Σ ∆ i Sn n

(A8)

Acknowledgment We would like to thank Professor Belandria for making a copy of his manuscript available to us and for kindly discussing his proposed process. Literature Cited 1. Belandria, J. I. J. Chem. Educ. 1995, 72, 116–118. 2. Prigogine, I. Thermodynamics of Irreversible Processes; Interscience: New York, 1997.

Wilmer Olivares and Pedro J. Colmenares Universidad de los Andes La Hechicera, Mérida, 5101, Venezuela In a recent article (J. Chem. Educ. 1995, 72, 116–118), J. I. Belandria maintains that the second law need not hold locally, that entropy can be destroyed in part of a macroscopic system as long as a larger amount is generated in some other part. But the author’s analysis is faulty—he assumes an impossible condition in the final state for his illustrative process. That impossible condition then gives rise to the “startling” behavior of the entropy. With respect to the details of Belandria’s process: the surroundings deliver 14.055 kJ of work to subsystem A; the energy equivalent of that work flows into subsystem B and increases the temperature of B from 373 K to 1500 K. The point the author failed to appreciate is that the path specified for the work interaction of 14.055 kJ uniquely determines the final pressure, P A(final), of the gas in subsystem A. Examples

Example 1 Let the wall of subsystem A be pierced by a thermally nonconducting rod that presses against a friction pad in subsystem A. Rotate the rod against the friction pad, doing 14.055 kJ of work in the process. The energy equivalent of the dissipated work finds its way into subsystem B and raises the temperature of B to 1500 K. The pressure of the gas in subsystem A remains at 1 atm. Example 2 Compress gas A by using a constant external pressure of 5 atm, and solve for ∆V A from the relationship {P(external)∆VA = 14.055 kJ. The final pressure, PA(final), turns out to be 1.29 atm. Example 3 Let a series of constant external pressures—say, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0 atm—each bring about the same change in volume ∆VA9. Solve for ∆VA9 from {(1.5 + 2.0 + 2.5 + 3.0 + 3.5 + 4.0)∆VA9 = 14.055 kJ. Then ∆VA = 6∆V A9 and PA(final) = 1.69 atm. Example 4 Compress gas A reversibly and isothermally: 14.055 kJ = R × 1500 ln (P A(final) / 1 atm) and PA(final) = 3.086 atm. This is the end of the line. For the given conditions, 3.086 atm is the maximum pressure that can be attained in subsystem A. The range of allowed values for P A(final) is


285


Letters thus 1 ≤ PA(final) ≤ 3.086 atm. By assuming a pressure of 4 atm for P A(final), the author has imposed an impossible condition on the process— thereby causing the entropy to behave in the “weird” way he describes. R. J. Tykodi University of Massachusetts Dartmouth North Dartmouth, MA 02747-2300

t Belandria replies: I appreciate this opportunity to demonstrate that there is no fundamental problem with my paper as some readers believe. Instead it shows a transformation that fits general thermodynamic restrictions and exhibits an internal entropy coupling that generate an unexpected behavior not seen in conventional thermodynamic systems. To analyze letters on my article I am going to answer first comments that are common across all letters and then comments that are specific for each author. A general comment related to feasibility considers that the process described in Figure 1 of the article is not permitted by thermodynamic laws. In this respect thermodynamics suggests that for a process to be feasible temperature must be greater than 0 K, energy must be conserved, and the total entropy change of the universe should be equal to or greater than zero (1–5). I have demonstrated in the article that the process sketched in Figure 1 fits the above requirements. Therefore, its operation meets general thermodynamic requisites and it should be feasible. In any case, an intuitive view of the dynamics of the process suggests that once the adiabatic film covering the metal partition is removed, transition starts with a spontaneous heat transfer between tank A and B, caused by a temperature gradient across the metal separation. Simultaneously, compression begins at a controlled rate to keep isothermal conditions in A. Then transformation may continue to reach the final state according to prediction. It is convenient to say that the process shows a set of conditions where internal entropy is simultaneously created and destroyed in different parts of the universe. This is an interesting behavior suggesting the existence of an internal entropy coupling not seen before in common systems. Under these conditions the process is more efficient than a conventional reversible operation. Now, the important fact here is that this unexpected universe may exist because it meets general thermodynamic requirements. Otherwise, articulation of thermodynamic laws should be reviewed to consider this case. Another common comment considers that the process of Figure 1 is not allowed by thermodynamics because it is impossible that the isothermal compression process with internal entropy coupling requires less work than a conventional reversible isothermal compression for the same initial and final state. To this objection, it is interesting to detect that by linking together a nonreversible isothermal compression with a heat transfer between two tanks it is possible to find a fea-

286

sible set of conditions where the nonisothermal compression work input is less than the work required by a common reversible isothermal compression for the same initial and final states. This result is unexpected from the point of view of classical thermodynamics, but irreversible thermodynamics suggests that such a behavior may occur as a consequence of the simultaneous production and destruction of internal entropy in different parts of the universe. Here an oriented heat transfer between tank A and tank B produces or creates enough internal entropy to drive a simultaneous nonreversible isothermal compression in tank A with destruction of internal entropy. Some authors believe that production of internal entropy causes a loss in capacity to do work (1, 3). Then, by analogy, destruction of internal entropy may increase the ability of the system to produce work. In this context the net result of simultaneous production and destruction of internal entropy is a gain in capacity of the system to do work relative to the corresponding reversible isothermal compression. During the operation energy is conserved and the total entropy change of the universe is greater than zero, fitting general thermodynamic requirements. Behavior exhibited by this process implies that irreversibility under internal entropy coupling conditions may enhance the ability of a system to do work relative to an equivalent reversible operation for the same change of state. I have further confirmed this by designing a feasible thermodynamic cycle with internal entropy coupling resulting in a cycle of greater efficiency than an equivalent Carnot cycle operating between the same temperature levels (6, 7, J. I. Belandria, unpublished). This finding is unusual and reveals an extraordinary feature of internal entropy coupling systems that suggests the possibility of designing feasible thermodynamic cycles more efficient than conventional classical ones by introducing steps involving simultaneous production and destruction of internal entropy. All the letters estimate the final pressure reached by a conventional reversible isothermal compression at 1500 K using as work input the value required in Figure 1 and find 3.086 atm. They argue correctly that a reversible compression cannot reach the final pressure of 4 atm obtained by the system sketched in Figure 1. This is true because the process described in the article is more efficient than a conventional reversible isothermal compression as a consequence of the simultaneous production and destruction of internal entropy, as I explained earlier. Some letters express opinions in relation to specification of the final state. For example, Nash wonders “where the author gets his figure for the final pressure in tank A”. Tykodi says “he assumes an impossible condition in the final state for his illustrative process”. Olivares and Colmenares state that “the final pressure of 405.32 kPa used by Belandria is unattainable”. And Freeman considers that “he has simply postulated initial and final conditions without providing data about the change”. To get the final pressure I set up a thermodynamic model for the whole process using eqs 33 to 42 and investigated the changes of state permitted by thermodynamic laws, keeping energy constant and the total entropy change of the universe equal to or greater than zero. Surprisingly, I detected a set of conditions allowed by general thermodynamic restrictions where internal entropy is simultaneously created and destroyed in different regions of the universe, and the



work required for the nonreversible isothermal compression is less than the value expected from classical thermodynamics. Then I selected at random one of these exceptional changes of state and designed the process shown in Figure 1 of the article. I have found that there is an infinite set of such states and several transformations that permit an internal entropy coupling process. The model indicates that the total entropy change of the universe is less than zero for temperatures in tank A below 940.14 K, keeping constant other variables. Therefore, the process is not allowed by thermodynamics in this range of temperature. Here, internal entropy destruction in tank A is greater than production of internal entropy by heat transfer. For temperatures in tank A above 940.14 K the total entropy change of the universe is greater than zero and the process should be allowed by the second law of thermodynamics. It is possible to see that for temperatures between 940.14 K and 4948.20 K the process shows an interesting and unexpected behavior suggesting an internal entropy coupling. In this range, internal entropy is simultaneously produced and destroyed in different regions of the universe, and the nonreversible compression in tank A is more efficient than a reversible isothermal compression. At 940.14 K the production of internal entropy is equal to destruction of internal entropy and efficiency reaches its maximum value. A thermodynamic cycle operating in the region of almost cancellation of internal entropy shows a greater efficiency than an equivalent Carnot cycle working between the same temperature levels (6, 7, J. I. Belandria, unpublished). For temperatures in tank A greater than 4948.20 K, internal entropy is created in all regions of universe and the system operates according to classical thermodynamics expectations. From this analysis it seems that in some relatively simple interconnected systems, general thermodynamic restrictions may allow the theoretical existence of a region with simultaneous creation and destruction of internal entropy. Under certain conditions close to equal production and destruction of internal entropy, the process exhibits a superefficiency, as Tykodi expresses. In this transformation, thermal death is retarded or avoided by the internal entropy coupling process, keeping variation of the total entropy of the universe as low as possible. In some letters Nash, Tykodi, and Olivares and Colmenares comment about entropy destruction. Tykodi opines that “there is never entropy destruction”. Nash asks “if we should give credit to entropy destruction” and Olivares and Colmenares follow Prigogine’s statement that “we can therefore say that absorption of entropy in one part, compensated by a sufficient production in another part of the system is prohibited”. To this question, irreversible thermodynamics suggests the possibility of simultaneous creation and destruction of internal entropy in relatively complex systems. These transformations have been detected in multireaction and biological systems, thermodiffusion, and active transport of ions, and in thermomechanical closed systems (5–8). In these systems there is at least a process that creates internal entropy coupled to a simultaneous transformation that destroys internal entropy. The destruction of internal entropy is not expected to take place by itself in a single process but can be made to occur by coupling it with another simultaneous process that creates enough internal entropy to compensate in-

ternal entropy destruction. Freeman, Nash, and Olivares and Colmenares discuss the creation of internal entropy across a metallic partition. Freeman says that reference to creation of entropy in the metal partition “smacks of witchcraft” because an object with no mass and no heat capacity cannot create entropy. He explains that there is an increase in entropy as a result of energy flow through the partition, but the increase arises from the difference in temperature on the two sides of the partition. Nash indicates that “the increase in Suniv is a simple consequence of the flow of heat from a hotter body A to a cooler B”. And Olivares and Colmenares calculate the creation of internal entropy due to heat flow through the metallic partition, concluding incorrectly that there is no creation of internal entropy in this transition. Now, in my article, system M is the metal partition of negligible mass surrounded by an imaginary surface representing the boundaries of system M. Surrounding system M are tanks A and B at different initial temperatures. Therefore, there is a heat flow from tank A to B through limits of system M. System M is receiving heat Qa at a constant temperature TA and expelling it to a source at a variable temperature TB without accumulation of energy or entropy. It is evident, as Nash and Freeman say, that there is a creation of entropy that may be attributed to heat flow from A to B arising as a consequence of the difference in temperature on the two sides of the partitions. I would like to say that the entropy balance expressed by eq 27 of the article is based on this consideration and I have assumed for convenience that internal entropy creation is located within the boundaries of system M. On the other hand, Olivares and Colmenares try to make rigorous demonstrations to estimate the creation or production of internal entropy in the process of Figure 1 and find erroneously that there is no production of internal entropy by heat flow through the partition. They attempt to evaluate entropy flow through the metal partition using eq 17 of their comments. They did not realize that their procedure evaluated the variation of entropy inside the metal partition rather than the entropy flow generated by heat transfer from tank A to tank B. Next, they concluded that eq 19 represents entropy flow across the metal partition. Indeed, eq 19 takes into account the variation of internal entropy inside the partition and does not represent the entropy flow caused by heat transfer from tank A to tank B through barrier boundaries. This is demonstrated when they calculate the entropy change of the metal partition and obtain eq 20, which is identical to eq 19, confirming my hypothesis. When eqs 19 and 20 are introduced in the entropy balance expression given by eq 21, they obtain zero production of internal entropy. Obviously, Olivares and Colmenares are not correct because everybody knows that internal entropy is produced when heat transfer takes place across a finite temperature difference. Freeman, Nash, and conventional engineering thermodynamic textbooks confirm opinion suggesting that entropy should be produced by the flow of heat from A to B through a metal partition (4, 5). From this wrong conclusion, Olivares and Colmenares consider that no entropy coupling exists and the process is not allowed by thermodynamic laws because the internal entropy destroyed in tank A is not compensated by production of internal entropy. This is not true, and whatever conception we select to analyze the process of Figure 1, we find that


287


Letters creation of internal entropy should occur as a consequence of heat transfer across a finite temperature difference. Also, when they calculate the total entropy production of the universe with eq 25 they find 5.83 J K{1. By a simple balance if {2.16 J K{1 is destroyed in tank A, then 7.99 J K{1 should be created in some other place. According to the discussion, such creation must be attributed to heat flow across the metal partition because nothing else produces entropy in my process. Therefore they do not solve the “paradox” affecting general validity of Prigogine’s formulation as they say. Next, they comment that my eq 27 used to estimate creation of internal entropy is not correct and does not have thermodynamic foundations. To this I would say that such an equation comes from a simple entropy balance in a metal partition and its surroundings. Some thermodynamics textbooks present similar entropy balances in related cases, demonstrating the validity of my calculations (4, 5). Tykodi comments that for system A to stay isothermal during compression means that the work interaction with the surroundings must be mechanically reversible. Similarly Battino and Wood say that “the only way the compression in A may occur isothermally is via a reversible process”. In relation to this, most laboratory and industrial isothermal operations are irreversible. Some conventional textbooks show examples of irreversible isothermal processes invalidating the above argument (9, 10). By controlling compression force and heat transfer we may reach irreversible isothermal conditions without serious difficulties. Now, I am going to discuss specific comments of each letter. Tykodi points out that my notation could be improved by using IUPAC rules. This may be true, but when I wrote this paper I had in mind engineering convention and I used it for simplicity and customary reasons. In any case, results and consequences of my work are independent of any arbitrary or conventional notation system. In a third point Tykodi assumes that I treat the uptake of heat by system B as a reversible process. Indeed, when I analyze tank B I do not make any previous assumption about reversibility, but entropy balance suggests that transition in B occurs without production or destruction of internal entropy. Since production of internal entropy is zero in B, then it appears that an event in tank B occurs as if it were reversible as Tykodi thinks. Tykodi indicates that the only irreversible part of my process is the heat transfer across the finite temperature between systems A and B. This picture is not correct because the process in tank A is also an attainable nonreversible isothermal compression as explained previously. Then he lists results for an imagined process consisting of a reversible compression in tank A, a reversible heating in tank B, and an irreversible heat transfer between A and B. Such values are correct for his assumption but not for the process described in my article, composed of a nonreversible isothermal compression in tank A, a constant volume heating in tank B, and an irreversible heat transfer between A and B across a metal barrier. His calculation does not consider the internal entropy coupling occurring in the cited process. Tykodi believes that entropy can only be produced and never can be destroyed. This statement is true in all systems described by classical thermodynamics, but my process is an interesting exception of this behavior. He argues that final state cannot be reached adiabatically from initial state for my selected path. To this consider-

288

ation I have explained that the process fits general thermodynamic requirements. Therefore, the overall adiabatic path selected is allowed by thermodynamics and may represent an exception of Caratheodory’s theorem for adiabatic processes (11). At the end of his letter Tykodi opines that referees did not review the article well. For me it is difficult to think that referees of a universally known journal did not read my work carefully and critically. I guess that reviewers felt that the article was interesting enough to be published independent of notation and nonconventional ideas that may generate an stimulating discussion about thermodynamic topics. Now, I will consider Nash’s comments. He states in his first paragraph that “the change of state taking place in tank A is nothing but the isothermal compression of an ideal gas from 1 to 4 atm”. To this I would like to say that the change of state taking place in tank A is something more than a conventional solitary isothermal compression. Indeed, the process illustrated in Figure 1 of my article represents an internal entropy coupling system in which an oriented heat transfer between tank A and tank B produces enough internal entropy to drive a simultaneous reversible isothermal compression in tank A with destruction of internal entropy. During the process heat is released to tank B, where the temperature varies from 373 to 1500 K and both tanks are covered externally by an adiabatic wall. From this outline it is possible to visualize that the process described in Figure 1 is not equivalent to a conventional isothermal compression as Nash considers. It is obvious, as he explains, that the minimum work required “by nothing but an isothermal compression at 1500 K from 1 to 4 atm” is {17.3 kJ. This performance corresponds to a conventional reversible isothermal compression releasing heat to a constant-temperature heat reservoir at the same temperature of the system equal to 1500 K. Classical thermodynamics postulates that the above value is the minimum work required for the best isothermal compression system designed by man for the given change of state. However, if we link an irreversible isothermal compression with a heat transfer between two tanks as described in Figure 1, it is possible to find a feasible set of conditions where the work input is less than the work required by a conventional reversible isothermal compression. In this sense, I have demonstrated that under internal entropy coupling it is possible to design a feasible irreversible isothermal compression with a work input of {14,054 J, which is less than the {17,289 J required by a reversible isothermal compression for the same initial and final states. This result is unexpected from the point of view of classical thermodynamics as Nash claims, but irreversible thermodynamics suggests that this behavior may occur as a consequence of the internal entropy coupling process as explained earlier. Nash tries to use the process in a closed cycle to conclude that the process described in my article violates the second law of thermodynamics, but he makes a wrong assumption that invalidates his reasoning. He writes “imagine tank A as the cylinder of a Carnot engine in thermal contact with an immense heat reservoir at 1500 K. Beginning at 1 atm, let the author’s notional irreversible isothermal compression proceed to 4 atm with work input of {14.05 kJ. Let the gas then resume its original state by a reversible isothermal expansion yielding a work output of 17.3 kJ”. Here, Nash is imagining a process different from the one described in



Figure 1 of my article. It is possible to see that tank A cannot be used as the cylinder of a Carnot engine in thermal contact with a heat reservoir at 1500 K as Nash thinks. It can be seen that during the process heat is transferred from tank A to tank B, which is a nonisothermal heat sink and its temperature varies from 373 to 1500 K. Also, both tanks A and B are covered externally by an adiabatic wall. Therefore, the cycle imagined by Nash does not fit the requirements of the geometry of the process described in the paper and his cycle does not work. It is evident that the process does not violate the second law of thermodynamics because the total entropy change of the whole universe is greater than zero. He continues and calculates the entropy change for his conventional reversible compression and finds a value of {9.369 J K{1 and concludes that the difference between the value given in my article and his value is the entropy destroyed, equivalent to {2.15 J K{1. He calculates the total entropy change for the conventional isothermal compression using the work input of Figure 1 and a final pressure of 3.086 atm and obtains a value of 7.99 J K{1 for his compression model, which is different from the system represented in Figure 1, where the total entropy change of the universe is equal to 5.83 J K{1. Therefore, there are not discrepancies in my conclusion, as he asserts, because all my calculations are correct and consistent with the system expressed in Figure 1; and his results are valid only for his system, which represents a different situation. Other specific comments appear in Battino and Wood’s letter. They assume a reversible compression in tank A and a reversible heating process in tank B and calculate correctly the required pressure and total entropy change of such a system. Then, they estimate the heat transferred from tank A considering a reversible isothermal compression and find 17,289 J. Next they compare this value with 14,054 J taken by tank B and ask “what happens to the excess heat?” Well, the numbers are incompatible because they compare heat intake of tank B with heat release from a reversible isothermal compression in tank A. They should make comparison using the actual heat released by the nonreversible isothermal compression described in the article, which is 14,054 J. In this case, the numbers are compatible. They said finally that there are no surprises here and no exceptions to the laws of thermodynamics. In relation to this, I have explained my ideas in the beginning of this rebuttal letter. Olivares and Colmenares start their letter considering that the process is physically unfeasible because according to classical thermodynamics for a reversible isothermal compression in tank A, pressure should be equal to or less than 312.71 kPa. As I explained earlier, the process meets general thermodynamic requirements. Then it should be feasible, being more efficient than a conventional reversible compression. The internal entropy coupling process allows the existence of a path with a final pressure greater than the value expected from simple mechanical arguments, and the system does more work than a reversible process between the same states. They continue to explain that a total entropy change greater than zero does not imply that a process will in fact occur because it depends on the dynamics of transformation. They suggest as an example that the formation of water from its elements at 25 °C and 1 atm is highly favored thermodynamically; however, it does not take place spontaneously un-

less a catalyst or a spark initiates reaction. Now, if we analyze intuitively the dynamics of the process described in Figure 1 of my article, it can be deduced that the process starts spontaneously when the adiabatic film on the metal partition is removed. The removal of the adiabatic film is the act or impulse that initiates a spontaneous heat transfer from tank A to tank B because of the temperature difference across the metal barrier. Simultaneously, compression starts at a controlled rate to keep isothermal conditions in A. Then, the system may continue to reach a final state according to model prediction. In any case, experimental evidence would be necessary to verify my hypothesis; but theoretically, the process starts. Next, they said that I failed to recognize that entropy production is not an additive property as they show in eq A7. Although this may be generally true, for the process it is additive as I will now show. They find in eq 29 a value of 5.83 J K{1 for total entropy production according to their reasoning. Now, if I assume additivity, the total entropy production for the whole universe will be obtained summing up my eqs 21, 25, and 27 and I get 5.83 J K{1 , which is the same value found by them. Since the values coincide my hypothesis is correct. Other specific comments appear in Freeman’s letter. He considers that the term creation/destruction has a number of connotations and should not be used in a scientific discipline. In this respect I think that any term used to describe a process must have a physical or intuitive feeling to understand better its behavior. I have selected for the article a name that reflects the nature of a process involving simultaneous creation and destruction of internal entropy. The term creation/destruction gives us a stimulating view of some special transformations of nature and describes at different levels the events taking place in the article, where many interpretations are possible. Freeman states that another puzzling aspect is my claim that the nonreversible compression process is more efficient than a reversible isothermal compression for the same initial and final states. He says this is an invalid comparison because if such a reversible compression were done the heat generated would raise the temperature in tank B above 1500 K or raise the temperature of both chambers, invalidating the posed conditions of the process. To this argument I would say that my intention is to compare the process taking place in tank A in my model with a conventional reversible isothermal compression occurring in a tank surrounded by an isothermal heat reservoir. This tank is not connected to tank B as Freeman thinks. I found that the nonreversible isothermal compression in Figure 1 of my article requires a work input of {14,054 J. Now, if gas is compressed between the same states under isothermal reversible conditions in a tank immersed in an isothermal heat reservoir it would require a work input of {17,289 J. Evidently, the process taking place in tank A is more efficient than the corresponding isothermal reversible compression occurring in a tank releasing heat reversibly to isothermal surroundings. This seems to me a reasonable comparison to measure the efficiency of the process. He comments that the question to be asked is “given the posed initial conditions, let the gas in A be compressed isothermally and reversibly to a final pressure such that the heat produced in A is exactly that required to produce the given results in B. What is the final pressure?” He finds 312.7


289


Letters kPa, compared to 405.32 kPa found in the process. Obviously, these results mean that the compression process represented in Figure 1 is more efficient than a reversible isothermal compression. This behavior is a consequence of the internal entropy coupling process previously explained. Freeman also says that it is not possible to accomplish the stated compression of the gas in A. However, I have demonstrated that the process fits general thermodynamic requirements and should be feasible through the path described. Finally, Freeman states that there is no merit in this paper other than its being used as a debugging assignment. I would be happy if this paper were used as a debugging assignment, because I feel people will find a new vision of the universe and thermodynamic fundamentals. I consider that this paper has drawn attention from readers around the world generating an interesting discussion about the existence of internal entropy coupling processes and the implications of their extraordinary behavior. The article has interesting aspects and presents a stimulating situation that deserves to be discussed at different levels. I think this is relevant whatever interpretation we assign to it. Literature Cited 1. Smith, J. M.; Van Ness, H. C. Introduction to Chemical Engineering Thermodynamics; McGraw Hill: New York, 1975. 2. Modell, M.; Reid, R. C. Thermodynamics and its Applications; Prentice Hall: Englewood Cliffs, NJ, 1974. 3. Balzhiser, R. E.; Samuels, M. R.; Eliassen, J. D. Chemical Engineering Thermodynamics; Prentice Hall: Englewood Cliffs, NJ, 1972. 4. Sandler, S. I. Chemical and Engineering Thermodynamics; Wiley: New York, 1977. 5. Prigogine, I. Introduction to Thermodynamics of Irreversible Processes, 3rd ed.; Interscience: New York, 1967. 6. Belandria, J. I. XLIV Meeting ASOVAC; Coro, Venezuela, 1994. 7. Belandria, J. I. I Congress of Mechanical Engineering; Mérida, Venezuela, 1994. 8. Dickerson, R. E. Molecular Thermodynamics; Benjamin: New York, 1969. 9. Abbott, M. M.; Van Ness, H. C. Thermodynamics: Theory and Problems; McGraw Hill: New York, 1969. 10. Daniels, F.; Alberty, R. A. Physical Chemistry; Wiley: New York, 1966. 11. Spalding, D. B.; Cole, E. H. Engineering Thermodynamics; Edward Arnold: London, 1967.

both 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 33–42 become:

José Iraides Belandria Universidad de los Andes Mérida, Venezuela

t 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 details—

290


∆UA = 0 [A is isothermal] (33b) ∆U B = C V ∆TB = 12.47 J/ (mol K) (1400 K – 400 K) = 12.47 kJ

(34b)

Q B = ∆U B = 12.47 kJ (35b) [since WB = 0; see below] Q A = –Q B = –12.47 kJ (36b) [all heat into B comes from A] wB = 0

[isochoric] (37b)

wA = QA

[∆UA = 0] (38b)

∆Su = ∆S A + ∆S B = – R ln (Pf / Pi) A + C V ln (T f / Ti) B = – R ln (Pf ) A + 15.62 J/K ∆S gA = ∆S A –

∂Q A /TA = ∆S A + 8.91 J/K

(39b)

(40b)


∆S gB = ∆S B –

∂Q B/TB = 0 [see JB's eq 25]

∆S gM = ∆S M – – =0+

∂Q A /TA–

∂Q A /TA +

(41b)

∂Q B/TB

∂Q B/TB

(42b)

= –8.91 + 15.62 = 6.71 J/K A comment is required with regard to eq 42b. Given JB’s concept of entropy creation/destruction, his eq 19 for ∆SgA is logically consistent and eq 21 follows; however, his eq 27 for ∆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 33–42; Olivares and Colmenares (3) provide a detailed discussion of the errors in JB’s eq 42. In eqs 33b–42b, numerical values have been obtained for all thermodynamic quantities except ∆S A and those which depend on it. ∆S A, 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 P Af. 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, P Af may be calculated from: w A = –RT ln (P Af/PAi)

(1)

The result is 2.92 bar, and ∆SA is {8.91 J/K, whence ∆SgA in eq 40b is zero—as expected for a reversible isothermal process. One might also ask what value of PA f makes ∆Su equal to zero. From eq 39b, with ∆Su set to 0, one obtains PA f = 6.55 bar (4); for higher values, ∆Su is negative. For any assumed value of PAf greater than 2.92 bar and less than 6.55 bar, ∆Su will be positive, ∆SgA will be negative (“entropy is destroyed”) and ∆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 PA f? 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 ∆Su is positive, ∆SgA is negative (“entropy is destroyed”), and ∆S gM is positive (“entropy is created”). Parenthetically, if JB is correct, one wonders how Nature “knows” which of the “infinite set of PAf values” to choose— or is the value of PA f 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, ∆S gA 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


291


Letters 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 cases—separated vs. connected tanks— because 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

292

and destruction, there are constraints on the possible final pressure in A—contraints 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 P Af, 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 published—certainly not in its present form. An acceptable alternative form might have been found—for example, “Provacative Opinion”–type, but even that is highly questionable. Literature Cited 1. Belandria, J. I. J. Chem. Educ. 1995, 72, 116–188. 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, 281– 286. 4. Calculation of the value P Af = 6.55 bar when ∆S u = 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 (PA f = 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).


Thermodynamics and Controversy - Journal of Chemical Education

Recommend Documents