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Postby biostudent84 » Thu Apr 07, 2005 7:05 pm

Poison wrote:take a glass of water and drop some ink in it. Dont shake the glass. wait and see how the ink diffuses. Now tell me. do you need energy for this?


I saw a really neat video in my physics class about this last night. There is a hypothesis that says that if there is no motion in a fluid, then diffusion will not happen.

What they did was take two hollow cylinders, and placed one inside the other. Then, in the space between the two, they filled with glycerin. A line of dye was placed longitudinally from the bottom to the top of an area of glycerin...kind of like a candy cane's stripe, only straight instead of spiral.

The entire apparatus was turned X times clockwise. The line of dye mixed into the glycerin similar to the ink in a glass of water example. However, when the apparatus was again turned X times, only counter-clockwise, then the ink actually "unmixed" out of the glycerin, and made a relatively straight line of ink again.

I'm not saying this is what's happening in water...but it is food for thought. (Just don't eat the glycerin...you might digest it a little too well ;) ;) )

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Postby Poison » Thu Apr 07, 2005 7:24 pm

biostudent84 wrote:I saw a really neat video in my physics class about this last night. There is a hypothesis that says that if there is no motion in a fluid, then diffusion will not happen.
Kyle


I mean you do not need to shake it or something like that. :)
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Postby canalon » Thu Apr 07, 2005 8:16 pm

biostudent84 wrote:I saw a really neat video in my physics class about this last night. There is a hypothesis that says that if there is no motion in a fluid, then diffusion will not happen.


Be carefull! Because even if nothing is moving, you still have motion in glass of water: the brownian movements of water molecules. It's only at 0 kelvin that you do not have any movements at all.

Besides in this experiment the energy given to water molecule is coming from everywhere, (light, air...) This energy is enough to allow diffusion in water.

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Postby biostudent84 » Thu Apr 07, 2005 8:44 pm

Yep! That's why they used glycerin instead of water in this expirement. But it is interesting though. Theoretically, if you could get all motion of the water to stop and have it still remain a liquid, then diffusion would not happen. Right?
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Postby canalon » Thu Apr 07, 2005 8:50 pm

biostudent84 wrote:Yep! That's why they used glycerin instead of water in this expirement. But it is interesting though. Theoretically, if you could get all motion of the water to stop and have it still remain a liquid, then diffusion would not happen. Right?


Probably, but I wonder if you can have a liquid without any motion... I guess that without motions you could only end up with some cristal.

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Postby mith » Thu Apr 07, 2005 10:46 pm

@canalon
A crystal....like ice? :D
At 0 degree kelvin it's a bose-einstein condensate btw.
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Postby canalon » Fri Apr 08, 2005 2:00 am

mithrilhack wrote:@canalon
A crystal....like ice? :D
At 0 degree kelvin it's a bose-einstein condensate btw.


Like ice, yes :D
Hey, if the molecules do not move relatively to one another, and thus allowing diffusion, I do not think that you can have a liquid....
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Postby mith » Fri Apr 08, 2005 2:28 am

well everyone knows there's not diffusion if you put two solids together
*Has a knack for stating the obvious :D 8)
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Postby biostudent84 » Fri Apr 08, 2005 5:37 am

Lol! Yeppers! That's why I inserted that little "Theoretically" word in there ;). It's possible for it to happen, if only we were able to get everything else to exactly what we wanted. Which we can't :)
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Postby 2810712 » Fri Apr 08, 2005 11:12 am

Hey , that example is interesting!
Glycerin also has brownian motion, [ all liqids ] that turning expt, just tells that we can reverse the energitically favoured process can be reversed by providing energy of opposite sign.
BUT HOW CAN THAT LINE REMAIN UNDIFFUSED AND STILL STABLE [ not energitically stable,]???
One more q. about ur eg. if we provide energy in any other form then ? ? ? eg. if we heat the gycerin to give same amount os energy, [ here we cannot determine the sign of energy , so it is distributed over all molecules, so the line will probably not reform.

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