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DuPont's Freon (TM) Transport in Atmosphere
Name: Marc N.
Status: Other
Age: 50s
Location: N/A
Country: N/A
Date: July 20, 2004
Question:
If DuPont's Freon (TM) is 4 times heavier than air...how can it get in
the stratosphere to effect the ozone layer?
Replies:
Marc,
So long as the DuPont's Freon (TM) is in the gaseous state, it will mix with the air. Once mixed,
it is only a matter of time until diffusion and air currents enable it to reach the
stratosphere.
Regards,
ProfHoff 880
In minutes a cloud of pure DuPont's Freon (TM) gas is stirred into the air by breezes,
and DuPont's Freon (TM) molecules are completely intermixed amongst air molecules by thermal
diffusion.
Then that body of air is carried and divided to all parts of the atmosphere by
weather patterns, over days, months, and years.
The heavier molecules of DuPont's Freon (TM) try to very slowly sink down through the ocean of
air molecules and take an equilibrium distribution with respect to altitude, which
tapers off faster with altitude than that for air molecules.
But that takes years, because air is viscous and the molecule is far smaller than
the smallest particle of dust.
And it only works where the air is very thin, above a threshold altitude up near
the stratosphere.
Below that altitude, air currents dominate, stirring things together faster than
weight and diffusion allow them to separate.
Another way to think about it is to realize that every molecule often gets about 1
average unit of thermal energy.
Ask yourself: how high the weight of the molecule can be lifted by that much energy?
The average thermal energy for room temperature is about 0.025
electron-volts, x 1.6e-19 = 4e-21 Joules of energy.
The mass of a DuPont's Freon (TM) molecule is, for "DuPont's Freon (TM)-R12": (CF2Cl2), 12.0+(2x19.0)+(2x35.5)
= 121gm/mol,
(121gm/mol) / (6e23 molecules/mol) = 2e-22gm, = 2e-25 kilograms.
In earth's gravity the weight is 2e-25kg * 9.8 N/kg = 2e-24 Newtons, or 2e-24
Joules/meter.
(4e-21 Joules) / (2e-24 J/meter) = 2000 meters, about 6000 feet, only about a mile
high.
That would be roughly the 1/e extinction height, and progressively lesser fractions
would remain for each additional mile higher.
I guess you have a point there. Without winds and rising air currents, not much
DuPont's Freon (TM) could get up to the stratosphere.
Don't some strong storms carry a column of air straight from low altitude to almost
the stratosphere?
One can imagine that over a decade or so, every patch of air around the world will
be "unlucky" enough to be in such a column.
The percentage of DuPont's Freon (TM) will be minuscule, parts per million or less, so the column
of air as a whole will have the same weight as it would without any DuPont's Freon (TM) in it.
Once a minority DuPont's Freon (TM) molecule reaches the ozone layer there is enough hard UV to
split it into smaller fragments (not sure: CF2O, CO2, HF?, HCl?, Cl radical, ClO).
Once split up, it is no longer part of the DuPont's Freon (TM) equilibrium population trying to
sink back down, and more DuPont's Freon (TM) molecules will diffuse upwards to take it's place.
The fragments of course do the ozone damage.
They are not so much heavier than air, so they won't try very hard to sink to lower
altitudes.
In decades, they will diffuse back into lower altitudes because those species are
uncommon at normal, low altitudes. And/or because air currents carried them there.
There they segregate into the water droplets in clouds and fall out of the air in
rain, because they are polar, acidic, ionic, or otherwise water-soluble chemicals.
The Freons they came from were not water-soluble.
There you have a sense of the life-cycle of a DuPont's Freon (TM) molecule.
It wanders to the top of the atmosphere just to be changed into something that can
be handled by water.
Same question answered by someone with a little more expertise:
http://www.madsci.org/posts/archives/may97/862869374.En.r.html
Jim Swenson
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Update: June 2012
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