Re: O2

On Aug 6, 1:38 pm, Dallas <Cybnorm@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
On Mon, 06 Aug 2007 08:21:19 +0200, Mxsmanic wrote:
False, because the atmosphere is not confined.

I suspect what you are missing is that the atmosphere *is* confined... by
itself.

I'm not declaring this for a fact yet, it's just how I see the physics. It
will be interesting to see what RAP has to say about it.


While I hate to agree with MX, he is actually correct in this case
(mostly, except for the mass of the atmosphere being constant. The
mass is in a sufficient volume and changes are so slow that it can be
considered constant).

This is a bit contrived but may help to demonstrate the point, if we
remember that the atmosphere is a "fluid":

Imagine the atmosphere as water in a 5m length x 10m width x 2 meter
high pool. The pool does not have a cover on it, meaning that the
water is free to "expand" upwards. Now, imaging that the temperature
of the water in this pool is 0 degree C (ok, for sake of argument,
let's just pretend that water doesn't freeze at 0 C <g>), and that
it's density is ohhhhh, 1kg/m^3. Now, ignoring pressure of the
atmosphere (pretend our pool is in a vaccum), the pressure exerted by
the water alone perpendicular to the bottom of the pool is defined by:
P = rho*g*h

where rho = density of the water, g = acceleration due to gravity, and
height = height of the water column. Now, let's say that

So, if we measure the height of the water at 0C and we see that it is
1 meter, we get, in other words, the volume of the water at 0 C is 50
cubic meters:

P = 1kg/m^3 * 9.81 m/s^2 * 1 m, which is 9.81 Newtons per square meter
(Pascals). You'll notice the measurement units which is describes a
force (Newtons) directed upon an area (square meters). Now, heat that
water up to a temperature of 100 C and imagine that there is no loss
of mass due to evaporation. You still have 50 kg of water since no
mass was lost (we arrived at that because density = mass/volume).
Now, we know that density, temperature, and pressure are directly
related to each other. We also know that the mass of the water has
not changed, nor has the area of the bottom of the pool. Because of
this fact, we know that the final pressure MUST still remain at 9.81
Pascals (if we remember, that since F=ma, and that neither m nor a
have changed, force must remain constant, and since we haven't changed
the area over which the force is being applied, the pressure remained
constant.) Therefore, the only variable that has changed is the
density.

So, let's look at how we can compute the new density.

Now, we know using the combined gas law that PV/T = k. When comparing
the same thing at two different states, we can arrive to P1V1/T1 =
P2V2/T2. Now, plug in our numbers (we'll convert to Kelvin here to
prevent a divide by zero; it will cancel out just fine).

V2 = P1/P2 * V1 * T2/T1

So, plugging in and solving, we get V2 = 50m^3 * 373.15/273.15. That
means that the volume of the water should now be 68.305 m^3

Sooo, now we can figure out the density, which which is 50kg/
68.305m^3. Therefore, we have a rho of 0.732 kg/m^3

Now, we can plug that in to find the new height of the water, which
would be

9.81 Pa = 0.732kg/m^3 * 9.81 m/s^2 * h

Now, solving, we see that we get a height of 1.366 meters.

You'll notice that I did not constrain the volume of the system,
because that is how the atmosphere behaves (as a fluid, with no
constraint on it "above"). You'll also notice that I contrived the
example so that no mass was lost to evaporation (in the case of the
atmosphere: we are ignoring changes in mass due to external factors
such as composition changes and such). I am also not addressing the
localized differential of adding heat to say only one half of the pool
of water, I am assuming that the heat is added evenly. Different
energy transfers (e.g. sun on one side of the earth) would be the
cause of localized volume and pressure differences, current flows (and
the resultant kinetic energy), etc as the entire system tried to
stabilize itself. If you realize that temperature is a measure of
energy, and realize that the act of heating and cooling something
involves an energy transfer, AND that the laws of conservation of
energy exist, you can see why the stabilization occurs.

Constraining the volume at 50 cubic meters and finding the change in
pressure is left as an exercise for the reader. :)

Notice, there is NO mention in any of this about pressure exerted on
the water itself; the water is the exerter of pressure; it has a mass
that is exerting force on a surface. What you have to remember also
is that in both my contrived example and atmosphere itself is that it
is not a closed system; there is no "cap" at the top. One also has to
remember that the atmosphere is considered to be a fluid (albiet in
gaseous state) and is subject to external energy transfers (i.e. solar
energy) and that it is subject to composition changes.



.

Relevant Pages

  • Re: Modest Proposal - Common Interplanetary Booster
    ... the idea of using Mars water and a solar or nuclear power ... Pressurizing the mars atmosphere to grow crops would also ... atmosphere to Earth normal pressure, in two days, and keep it there. ...
    (sci.space.policy)
  • Re: world with less air pressure - survivable?
    ... provides a great pressure garment - you don't "explode in a vacuum", ... At 37 deg-C, the vapor pressure of water is 0.07 Atm, ... Why do you think a lower-density or lower-pressure atmosphere would ... but also the atmospheric temperature gradient. ...
    (rec.arts.sf.science)
  • Re: Reply to Eric Rowley -- Ozone
    ... Where did you get this idea that the water vapor caused ... other than you claiming that pressure causes high temperatures." ... the temperature would have been thousands of degrees. ... of the planet up there in the atmosphere. ...
    (talk.origins)
  • Re: Reply to Eric Rowley -- Ozone
    ... increased pressure would increase temperature. ... Bravo thought the water needed that degree of temperature to stay ... It isn't as simple as that, because the volume of the atmosphere ...
    (talk.origins)
  • Re: Question about an alternate abiotic origin of Martian "blueberries"
    ... > probably stems from the thin atmosphere - it has no mechanical erosive power ... The ripples are clearly formed by water since they are aligned ... all those fine laminations ... >> and with more water vapor there would be more ice clouds. ...
    (sci.geo.geology)
Loading