Off-gassing of Water in a Vacuum

off-gassing effects of water in a vacuum

Yokota@aquadevice.com

Reboiling Water in Vacuum

name         James
status       other
grade        other
location     MD

Question -   We put tap water in a glass jar, sealed the jar and
applied vacuum to approximately 28".  The water boiled rapidly for
about 50 to 60 seconds, then it slowed to a stop. That same jar of
water will  no longer boil at 28" of vacuum.  If we replace the
water with fresh water the same thing occurs, it boils the first
time but then will not boil after that. What change occurs to the
properties of the water (which remains in liquid form) that
prevents it from boiling the second time?
---------------------------------------
Two things: the dissolved gases that were in the water escape, and
the water gets colder, so that its temperature is below the boiling
point at 28" (I presume you mean inches of mercury?) vacuum.  That
temperature is about 37 C, which is probably warmer than your tap
water to begin with.  Most likely, then, the boiling you observe is
actually outgassing.  Either way, you should notice that the "spent"
water you remove from the vacuum chamber is cooler than when it went in.

Richard Barrans
Department of Physics and Astronomy
University of Wyoming
===================================================================
James,

What you are seeing is not really water boiling. Rather, gases
trapped within the water sample is outgassing. The bubbles you see
is not water turning into water vapor, but air that is escaping from
the liquid water. The boiling point of water at 29.5" (atmospheric
pressure at sea-level) is 100 degC. Reducing the pressure down to
28" will minimally change the boiling point of the water to
approximately 99 degC. Thus, unless you happen to be heating the
water to this temperature, the water is not really boiling.

Greg (Roberto Gregorius)
===================================================================
The information you have provided is incomplete.  Did it "boil" just
by pulling a vacuum or did you have to apply heat?  Did you continue
to pump on it to maintain the 28" of mercury vacuum?

If you were not applying heat and you were maintaining vacuum I
would suggest that it never boiled but that you were pulling
dissolved air out of the water.  Once all of it had come out, the
"boiling" stopped and could not be repeated.

To find out if this is the case:

A
  1    bring it back to atmospheric pressure in air
  2    stir or shake it vigorously in air for a few minutes
  3    pump to 28" of mercury vacuum & observe the results

B
  1   seal it off at 28" of mercury vacuum
  2   stir or shake it vigorously under vacuum for a few minutes
  3   pump to 28" of mercury vacuum & observe the results

If it "boils" again in experiment A but not in experiment B then it
is likely you are not boiling (vaporizing the water) but pulling
dissolved air out of the water.

Greg Bradburn
===================================================================
James-
    Only two things can be happening.
     1) dissolved air can be out-gassing. This is irreversible, only happens
once. Just like what you observed.
     2) the water, by boiling, will be cooling down. This is reversible.
           In an hour or so, if you close off the vacuum port but
 let in no air,
          the water will have warmed back up to room temperature,
          and you could show boiling again by opening the vacuum again.

When water is heated on the stove, the air bubbles leaving are usually much
smaller than the water vapor-bubbles of boiling.
Think about it: by reducing the heating rate, the boiling bubbles would get
a little smaller.
    And by reducing atmospheric pressure, the dissolved-air out-gassing
bubbles would get bigger.
The perceptible difference between out-gassing and boiling would be reduced,
     conceivably even obliterated or reversed.
When water is cooled by vacuum-evaporation (boiling at room temperature),
   it gets cooler than room temperature, but probably not as cold as
freezing.
   Room temp. is 20-25C, freezing is 0C, so the ambient is only 20C warmer
than the boiled water.
   Probably less.  And the jar is glass not metal, so it has poor thermal
conductivity.
   So the heat-flow from room air into the beaker will be a lot slower
       than it is for a pot over high heat on the stove.

To cause repetition of the dissolved-air out-gassing,
one would re-saturate the water with air.
Close off the vacuum line, insert a tube to the bottom
(maybe with many small holes like the tip of a fish-tank bubbler),
and pump air through it for some time.
Not sure, I would start with an hour, but less might be enough.
Opening the lid and waiting overnight might be enough too.
Stirring would have some rate in between.
The water will re-saturate with air,
and appear to boil when you again evacuate it.
If you have not dissolved as much air as it had before, the bubbling will
seem less.

I'd look up the vapor pressure of water, if I were you. Get a table.
There is a little graph at Wikipedia: Vapor_Pressure:
http://en.wikipedia.org/wiki/Vapor_pressure,  halfway down the page.
You could use a more precise pressure-gauge too.
29.9"Hg -28"Hg =1.9"Hg absolute pressure.
2"Hg absolute pressure is more than the vapor pressure of water at room
temperature.
I think your vacuum pressure might not be low enough to force actual
boiling at room temperature.
Maybe all you have seen is the out-gassing stage that always happens before
boiling,
and then the action stopped because you did not have enough vacuum to do the
next stage.
Tell you what, put a hot-plate under that beaker (set too low to boil a
water drop),
   or a couple of big light-bulbs beside it, to warm it gently.
If you warm your water up, at some temp (~50C?) it will boil continuously,
(and your vacuum pump will have to cope with a lot of re-liquefied water in
its exhaust.)
You can learn your real vacuum pressure from that temperature the
vapor-pressure table.

Jim Swenson
===================================================================
   It is difficult to analyze exactly what is going on without
knowing more of the detailed experimental conditions. For example
the "capacity" of your pump -- not just the final pressure, but
what volume per unit of time the pump can remove volatile
components. Without that information, I would suspect that the
initial "boiling" you see is not actually boiling, in the sense of
removing converting liquid water to water vapor. Rather, in the
absence of more details, I would suspect that what you are
observing is the removal of the components of air dissolved in the
water. These would include mainly carbon dioxide and oxygen. This
would be consistent your observation that there is no "boiling" the
second time you carried out the evaporation process.
   You could test this hypothesis by taking the water that has had
the air removed and shaking it vigorously to re-dissolve the
components of air. If just shaking the water, from which the
components of air have been removed, and then allowing those
components to redissolve -- and then you see the "boiling" reoccur
-- you could be pretty much on the right track that the "boiling"
is not the evaporation of water itself, but more the de-gassing of the water.
   You could then confirm this observation by boiling the water at
its normal boiling point -- which would remove the atmospheric components --
then re-do the boiling experiment. If you do not let this water stay
round long, you would not expect to see the "boiling" of dissolved gases.
   You could further confirm this hypothesis by freezing the water
a couple of times before carrying out the "boiling" experiment.
Gases are not very soluble in ice, so you would expect that a
couple of freeze/thaw cycles before the "boiling" would remove the
dissolved gases and you would not expect to see any bubbles that
you are calling "boiling".
    These kinds of experiments are how science works. Observation:
 Test(1) -- Hypothesis(1) -- Test(2) -- Hypothesis(2) -- and so on.

Vince Calder
Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: