Re: An Inflationary Account - #6 - Entropy as missing information

On Mon, 23 Jul 2007 03:10:05 GMT, "Perplexed in Peoria"
<jimmenegay@xxxxxxxxxxxxx> wrote:


"r norman" <r_s_norman@xxxxxxxxxxxx> wrote in message news:v938a3p3htr82sl45gnuucud3cv0so50jc@xxxxxxxxxx
On Sun, 22 Jul 2007 21:17:07 -0400, "Perplexed in Peoria"
<jimmenegay@xxxxxxxxxxxxx> wrote:


"Perplexed in Peoria" <jimmenegay@xxxxxxxxxxxxx> wrote in message news:TPSoi.12469$rL1.2887@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

"r norman" <r_s_norman@xxxxxxxxxxxx> wrote in message news:agm7a3pd9he11psm45f3q4bvo3c4m92hus@xxxxxxxxxx
[snip]
In a biological context (where delta G = delta H = T delta S), a
reversible reaction is one conducted at equilibrium which is defined
as a condition where delta G = 0. If delta H = 0, then delta S is
also 0.

Agree.

And I should have added that when an ion moves to a state of lower
potential (down a voltage gradient) delta H is definitely not zero.

The point is that the movement described involves two coupled
activities, one with negative delta G, the other with positive delta G
of equal magnitude. A charged particle can move 'downhill' from one
potential to another in two ways: In a dissipative, irreversible
process as occurs in a resistor, heat is produced. In a conservative
reversible process the energy is transformed to a different form, say
kinetic energy as occurs in an electric motor. In a conceptual motor
of 100% efficiency, no heat is produced. The energy released by the
electrical movement is coupled to do the work required to pump the ion
"uphill" in its concentration gradient. Or, alternative, if the ion
moves "downhill" from high to low concentration, the energy released
goes into doing electrical work, moving the charge up its potential
gradient. The equilibrium process is reversible and there is no
entropy change.

That is my understanding of it. I tried to make the parallel case of
a gas expanding in two ways, reversibly and irreversibly. In one case
there is an entropy change. In the other case there is none.

Your two cases are indeed parallel in my understanding. And in both
cases, in the reversible subcases, there is a flow of either heat or
work across the system boundaries. The entropy of the universe is
unchanged. But the entropy of the focal system may be changed.

In a system of ice and water at equilibrium, water molecules may
crystalize or melt at random. But equal numbers go each direction
at the same rate. Any one molecule, going in one particular direction
does not change G - i.e. delta G is zero. But it does change H and S.
And as it changes H, it generates or absorbs heat. And that heat
is assumed to cross the closed system boundary into or from the
surroundings. Balancing the heat flow is a change in the entropy
of the closed system as given by dS = dQ/T. But an equal and opposite
change in the entropy of the surroundings. And, because the entropy
of the isolated system does not change, the process is said to be
reversible. The melting or freezing of water is reversible if conducted
at 0 C. But the entropy change of the water system itself is dramatic.

Incidentally, I am no longer sure that the energy for the transfer of
charge across a gradient should be considered part of delta H. It may
be that for this case, the Gibbs formula should be written as
delta G = delta H + E delta Q - T delta S
I'm trying to check this, so far without luck. Any thermodynamics
geeks out there know which is right?

This last paragraph is very wrong. The E delta Q is part of delta G.
The free energy term consists of force time distance, surface tension
times area, pressure times volume, potential times charge, and
chemical potential times mole number. Converting one form of free
energy to another can be done with not change in heat content or
entropy.

I think your mistake is in improperly considering the case of the gas
expanding reversibly. Yes it does work but in my examples, the work
is performed INside the system, not on the outside world. Perhaps it
is better to consider a see-saw (teeter-totter?) or a pulley with two
weights. As one weight moves down in a gravitational field, potential
energy of gravity is released. If that weight simply crashes into the
ground, the energy is converted to heat and entropy goes up. However
if the energy is used to lift another equal weight an equal distance,
the energy is simply transferred from one location to another. Look
at the system consisting only of the first weight and it does work to
the 'outside'. But look at the system consisting of both weights and
no external work is done, no heat is produced, no entropy is changed.
Yet one weight was lowered and the other weight was raised. That is
the way to look at a coupled process. One side goes 'downhill', the
other goes 'uphill'. Energy is transferred from one form to another.
No heat. No entropy.

.

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