"Statefunctionness" of entropy

[Pentcho Valev <pvalev@BAS.BG>]

"Carl E. Mungan" wrote:

> Please give an example of a complex system for which entropy is *not*
> a state function.




I am sorry for not being able to respond to all arguments  concerning
thermodynamics - I don't have a personal computer and in this public computer room
my time is limited. Still I hope my next analysis will be relevant to many of the
arguments raised.
     Consider a system able to produce two types of reversible work - e.g. the
system consists of two suspended and stretched springs. The maximum work the system
can do, ISOTHERMALLY, can be presented as

dWm = -F1 dL1  -  F2 dL2               /1/

where F > 0 is the force of contraction and L is the length of the spring. On the
other hand, L1 and L2 are independent variables which, under the conditions
specified, fully determine the state of the system. Hence in this case a state
function is a function of L1 and L2. I hope this is indisputable.
     Now we can ask: Is Wm a state function or not? If not, this would mean that
there exists a reversible (or close to reversible) cycle undergone by the system as
L1 and L2 change such that, at the end of the cycle, the net work done by the
system is positive, in violation of the second law. Also, since we do not doubt the
first law, this positive net work would have been done at the expense of heat
absorbed from the surroundings, i.e. at the end of the cycle the net heat absorbed
by the system is positive. In accordance with the definition dS=dQrev/T, this would
mean that the entropy is not a state function.
     All these sad conclusions would be true if Wm is not a state function, i.e. if
Wm is not a function of L1 and L2. There is a mathematical theorem stating that,
for any function f(x,y), the mixed partial derivatives d^2 f / dxdy  and d^2 f /
dydx  are EQUAL. If they are not, that means that f is not a function of x and y.
In our case, if Wm is a state function,

dF1 / dL2  =  dF2 / dL1                     /2/

The derivatives are partial - L1 is fixed (constant) on the left and L2 is fixed on
the right.
     For any system able to produce two types of reversible work, the partial
derivatives analogous to those in /2/ have physical meanings and are MEASURABLE.
This means that thermodynamics possesses an easy test for the validity of the
second law and the "statefunctionness" of entropy. If the experimentally measured
partial derivatives prove equal, so much the better for the second law and entropy.
If not (e.g. measurements show that one is positive and the other negative), the
second law is incorrect and the entropy is not a state function.
     In some cases we do not need a specially designed experiment - the information
is already available. In the present oversimplified case, since the springs do not
interact by definition, we have

dF1 / dL2  =  dF2 / dL1  =  0                        /3/

i.e. in this case we would not be right to doubt the second law and the
"statefunctionness" of the entropy. But how about cases in which spring-like
samples do interact? Let us now consider "chemical springs" that do interact. There
are two types of macroscopic contractile polymers which on acidification (decrease
of the pH in the system) contract and can lift a weight (see fig. 16A in D. Urry,
J. Phys. Chem. B, 101, 11007-11028, 1997 for visualization). They differ in that
those designed by Urry (U) absorb protons on stretching (as Lu increases), whereas
those designed by Katchalsky (K) release protons on stretching, as Lk increases
(see discussion on p. 11020 in the same paper).
     Let us assume that two macroscopic polymers, one of each type (U and K) are
suspended in the same system. At constant temperature, if the second law is correct
and entropy is a state function, we must have

dFu / dLk  =  dFk / dLu                       /4/

     The values of the partial derivatives in /4/ can be assessed from experimental
results reported on p. 11020 in the same paper. As K is being stretched (Lk
increases), it releases protons, the pH decreases and, accordingly, Fu must
increase. Therefore, the left partial derivative in /4/ is positive. In contrast,
as U is being stretched (Lu increases), it absorbs protons, the pH increases and Fk
must decrease. Therefore, the right partial derivative in /4/ is negative. If the
estimated difference between the two derivatives does not result from experimental
mistakes, the unavoidable conclusion is that the entropy is not a state function
for this particular system.
     The posting has become too long. If I could find time and energy, tomorrow I
will analyse an even more interesting (for physicists) system.

Pentcho