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```----- Original Message -----

> > Obviously, dive planning would be simpler if you could ignore
> > the Helium at
> > this point.  The question is if you safely can.
> >
> >
>
>
>
> I like to talk with FACTS, so here it is:

The same facts that decompression _theory_ is based on correct?

>
> First of all there is the "DALTON'S LAW" which defines that "THE TOTAL
> PRESSURE OF A MIXTURE OF GASES IS THE SUM OF THE PARTIAL PRESSURE OF EACH
> GAS"
> The following equation clearly states the above law.
>
> P total = P1 + P2 + P3 + .......Pn
> (More info on Dalton's law in
> http://mot.cprost.sfu.ca/~rhlogan/dalton.html )
>
> Second, the "HENRY'S LAW" defines the solubility of gases in a given
liquid.
> According to this law, "THE SOLUBILITY OF A GAS INTO A LIQUID IS EQUAL TO
> IT'S PARTIAL PRESSURE" (More info on Henry's law in
> http://www.chem.ualberta.ca/courses/plambeck/p101/p01182.htm )

Ok. I'm still with you here....

>
> Combining these two laws we have the following results:
>
> 1) For the first dive our 'experimental diver' uses a mixture of 10%
Oxygen,
> 40% Nitrogen and 50% Helium. (In other words our diver breaths a mixture
of
> 50% Air and 50% Helium)
>  In the depth of 50m the TOTAL PRESSURE is 6 ATM, which is the sum of 3
ATM
> of Helium, 2,4 ATM of Nitrogen and 0,6 ATM of Oxygen.
> We know that the Oxygen is metabolized from the divers body, so the rest
of
> the gasses are dissolved into the bloodstream (and transferred into the
body
> tissues).

Uh.... you can be saturated with O2 also and the excess that is not
metabolized is then an inert gas under the increased partial pressures....

> As our 'experimental diver' stops it's dive and starts the ascent, the
> partial pressure of these gases is decreasing. As a result, the dissolved
> gasses start to decreased (escaping from the lungs during breathing
cycles).

Yep.....

>
> 2) In the second dive our diver uses a mixture of 64% Nitrogen and 36%
> Oxygen.
>  In the depth of 20m the TOTAL PRESSURE is 3 ATM which is the sum of 1,92
> ATM of Nitrogen, 1,08 ATM of Oxygen AND 0 ATM OF HELIUM (Because there is
NO
> HELIUM in the mixture).

>  Now according to Henry's law, since the partial pressure of Helium is
ZERO
> IN ANY DEPTH, the dissolved helium is still escape from the lungs, until
it
> reaches the Helium's partial pressure witch is ZERO. In other words our
> diver will continue to release Helium from his body, nomater what the
> pressure of the Nitrogen is.

But what about the He that has not yet escaped the tissues and gone back
into solution from the first dive assuming that some of the tissues became
supersaturated during the first dive? If the diver is not completely clear
of this remaing He then what remains is at an increased partial pressure
during the second dive.

I understand your examples with the gas laws sited but I still don't get the
He that remains after the SI from the first dive.

Also, can you explain, in layman's terms M values and how they play a part
in these examples?

Mike

>
>
> As we can understand from the above example, the answer to the
Huntzinger's
> question is that "YES, WE CAN SAFELY IGNORE THE REMAINING HELIUM IN THE
> SECOND DIVE".
>
>
> Further more:
> A) From the Dalton's law, we can see that two different gasses CAN coexist
> in a solution in their maximum dissolved quantity.
>
> B) From the Henry's law we can see that the dissolving of a gas into a
> liquid depends from the type of the gas also. (It's easier to dissolve
> Helium in a liquid rather than Nitrogen, because Helium molecules are
> smaller than Nitrogen's)
>
> C) Because of the above laws (and their example) it is possible and the
> common rule to  the Trimix divers, to use different mixture of gasses,
> including Nitrox and some times pure Oxygen during their Decompression
> stages in order to free their bodies from the remaining Helium before
> ascending to the surface.

So you are saying that after dive one above, when the diver exits the water
he is He free from that dive?

Mike
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