🚀 go-pugleaf

> In 1824 Sadi Carnot advanced a false premise and deduced a conclusion
> later called "the second law of thermodynamics":
>
> Kelvin's version: It is impossible to perform a cyclic process with no
> other result than that heat is absorbed from a reservoir, and work is
> performed.
>
> This implies that, when two INTERACTING heat engines absorb heat from
> a reservoir (the surroundings) and perform reversible work, the
> following equality of partial derivatives holds:
>
> dF1 / dX2 = dF2 / dX1
>
> where F1 and F2 are work-producing forces and X1 and X2 are the
> respective displacements.
>
> The above implication is more or less an oxymoron since the equality
> clearly characterizes NON-INTERACTING heat engines. At least in two
> cases (see  http://www.wbabin.net/valev/valev2.htm ) the partial
> derivatives are not equal and therefore the second law of
> thermodynamics is false.

Some day physical chemistry professors may realize that "students
learning by rote" is good (makes professors' life serene) but
"students thinking critically" is better. They may even discover that
Katchalsky's and Urry's Polymers (Pk and Pu) as described in

http://pubs.acs.org/doi/abs/10.1021/jp972167t
J. Phys. Chem. B, 1997, 101 (51), pp 11007 - 11028
Dan W. Urry, "Physical Chemistry of Biological Free Energy
Transduction As Demonstrated by Elastic Protein-Based Polymers"

are the perfect tool for an easy verification of the second law of
thermodynamics. On acidification ([H+] increases) both Pk and Pu
contract and can lift a weight. This means that:

dFk / d[H+]  >  0   (Lk constant)

dFu / d[H+]  >  0   (Lu constant)

where F is the work-producing force of contraction and L is the length
of the polymer.

Pk and Pu differ in that Pk RELEASE protons on stretching whereas Uk
ABSORB protons on stretching:

d[H+] / dLk  >  0

d[H+] / dLu  <  0

If a Katchalsky's polymer and a Urry's polymer are suspended in the
same solution, it follows from the above four inequalities that:

dFk / dLu  <  0    (Lk constant)

dFu / dLk  >  0    (Lu constant)

Clearly the second law of thermodynamics is false. If it were true, we
would have:

dFk / dLu  =  dFu / dLk

Pentcho Valev
pvalev@yahoo.com