Re: dielectric constant of water

[Pentcho Valev <pvalev@BAS.BG>]

On Mon, 25 Aug 2003 09:48:40 -0400, Bob LaMontagne
<rlamont@POSTOFFICE.PROVIDENCE.EDU> wrote:

> Pentcho Valev wrote:
>
> > In my view, the effect is due to the so-called clathrate water which is
> > formed in the vicinity of hydrophobic surfaces (in this case the surface
of
> > the wall of the bag) and is sometimes called "water of hydrophobic
> > hydration". When the wall is thin, there is polarization, i.e. "water of
> > polar hydration" is formed near the surface and you measure a high
> > dielectric constant. However if the wall is thick, water of hydrophobic
> > hydration predominates - it forms many layers with almost no
polarization.
> >
>
> Since the bag is on both sides of the water, wouldn't the polarization
> effects simply cancel (same sign - no net field)?

It seems you regard the polarization as somehow caused by the bag whereas I
think the field is entirely caused by the plates - the bag is presumably
neutral in this respect. However the role of the (hydrophobic) bag is
essential in creating many layers of water of hydrophobic hydration with
almost no polarization. This type of water arrangement is usually
underestimated - people are misled by the term "hydrophobic" - it seems
that nothing important could happen to water near a hydrophobic surface.
This is not the case - in fact, this type of arrangement is ice-like,
voluminous, strongly exothermic and competes very successfully with
polarization as long as the polarizing agent (the plate) gets more distant
(due to the thickness of the wall). Of course, these arguments of mine are
rather speculative for the moment - the effect could still have nothing to
do with water of hydrophobic hydration.

Pentcho