Discrepant Balloons
ฝัง
- เผยแพร่เมื่อ 18 ธ.ค. 2012
- Connect two different-size balloons and predict which way the air will flow.
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Now i want an experiment with equally filled baloons
This is really good for explaining for how alveoli work for bronchopulmonary dysplasia patients too.
This experiment demonstrates Laplace's Law. Which explains that (without the presence of surfactant) smaller alveoli in the lung are more likely to collapse than larger alveoli.-- The relationship is explained by P = 2T/r.
what does the T express in this equation
@@marcopolo2028 superficial tension
@@Greekgod_17 than you
This video solved a question I’ve been struggling with for a while. Thank u!!
That was really interesting, thanks for sharing!
I have an undergraduate exam in a few hours in Laplace pressure and I literally didn’t get in until now. Thanks!
hi. writing this 10 years later (2022)....did you pass your exam?
Did You pass the exam?
Love this. Thanks, Bob.
beautifully demonstrated... convinced, i always thought from othe demonstrations that the small baloon was leaking outside
Law of Laplace beautifully explained thank you sir❤️
Very nice explanation
very educative .thank you regards
Thank you!
Mind-blowing
Great Video, where can i get a valve like the one used in the video?
Here's the write-up with all the materials you'll need: www.flinnsci.com/discrepant-balloons/dc91666/
what would happen if you tried to blow up both of the the balloons at the same time while they were connected?
amazing......
Thanks
Thanks sir
I'm not convinced, if we start with two equally filled balloon then the system should move to the same configuration (an empty plus a bigger balloon)
another inportant thing to look at is the air in the valve should move from the higher pressure to low pressure, since the pressure in the bigger balloon is higher because to fill it we need more and more pressure .
That's what pj sir explained today😁
Yeahh😎😎😎😂👍
thamk you
If you over-inflate the large balloon this won't work and they should reach equilibrium.
what dose surface area have to do with this ?!
it is about the tendency of pressure equalization.
Well, the answer is yes, the pressure is equalizing in both the balloons in the entire process, since the pressure is higher in the smaller balloon the air continues the flow from the smaller balloon to the larger balloon until the pressure is equalized, but the equalization of pressure implies that the energy is minimized since the energy of the system cannot increase as it would violate the second law of thermodynamics.
Why does the surface have anything to do with the energy of the system of balloons?
The answer is because when you blow up a balloon, its surface or material comes under stress (as it increases in volume) which arises due to intermolecular forces between the molecules that make up the material of the balloon. Naturally, when you blow up a balloon, The amount of work you do is spent in expanding the balloon's surface area which in turn increases the energy stored in the surface of the balloon as the surface area of the balloon increases.
So in the setup, the configuration which has the lowest energy will have a 100% probability of being the outcome (as is the case in all macroscopic processes), and since we have already established that the combined energy of the balloons is proportional to the combined surface area of the balloons, the resulting configuration would be that of the minimum energy configuration and hence should possess the minimum combined surface area.
Note: Here I am assuming that the process is isothermal, meaning there is no change in temperature and hence there would be no change in the average kinetic energy per molecule of the air inside the balloons and the tube.
For simplicity, you can imagine the energy stored in the material of the balloon (When it is expanded) as elastic potential energy.
Hope this helps. :)
@@rulersonicboom4737
Thanks a lot for the very useful replay.
Atmospheric pressure vs the pressure in each balloon has nothing to do with it?
No this should also work the same in a vacuum. The pressure inside the balloon is mostly due to the contracting force of the material of the balloon.
The equalization of pressure implies that the energy is minimized since the energy of the system cannot increase as it would violate the second law of thermodynamics.
Why does the surface have anything to do with the energy of the system of balloons?
The answer is because when you blow up a balloon, its surface or material comes under stress (as it increases in volume) which arises due to intermolecular forces between the molecules that make up the material of the balloon. Naturally, when you blow up a balloon, The amount of work you do is spent in expanding the balloon's surface area which in turn increases the energy stored in the surface of the balloon as the surface area of the balloon increases.
So in the setup, the configuration which has the lowest energy will have a 100% probability of being the outcome (as is the case in all macroscopic processes), and since we have already established that the combined energy of the balloons is proportional to the combined surface area of the balloons, the resulting configuration would be that of the minimum energy configuration and hence should possess the minimum combined surface area.
Note: Here I am assuming that the process is isothermal, meaning there is no change in temperature and hence there would be no change in the average kinetic energy per molecule of the air inside the balloons and the tube.
For simplicity, you can imagine the energy stored in the material of the balloon (When it is expanded) as elastic potential energy.
Hope this helps. :)
Now I understood the math behind it☺
Now add surfactant (i.e. dish soap) to the smaller balloon. Will the opposite happen? I would love to see a video demonstrating this on TH-cam. I predict the surfactant will decrease the surface tension on the small balloon, potentially enough to disrupt Laplace's law (note: this is why there is surfactant in the alveoli of human lungs. Keeps the small alveoli from collapsing due to their own surface tension. No lecithin surfactant in premature baby=collapsed lungs)
Your comment is really helpful. Thank you very much!!
Very Interesting! Thanks for sharing!
Not complete. You forgot hysteresis. The balloon remembers whether you were blowing it up or letting it down. Try it. The one you let down has less pressure than one that is being blown up. Surface area doesn't matter.
The equalization of pressure implies that the energy is minimized since the energy of the system cannot increase as it would violate the second law of thermodynamics.
Why does the surface have anything to do with the energy of the system of balloons?
The answer is because when you blow up a balloon, its surface or material comes under stress (as it increases in volume) which arises due to intermolecular forces between the molecules that make up the material of the balloon. Naturally, when you blow up a balloon, The amount of work you do is spent in expanding the balloon's surface area which in turn increases the energy stored in the surface of the balloon as the surface area of the balloon increases.
So in the setup, the configuration which has the lowest energy will have a 100% probability of being the outcome (as is the case in all macroscopic processes), and since we have already established that the combined energy of the balloons is proportional to the combined surface area of the balloons, the resulting configuration would be that of the minimum energy configuration and hence should possess the minimum combined surface area.
Note: Here I am assuming that the process is isothermal, meaning there is no change in temperature and hence there would be no change in the average kinetic energy per molecule of the air inside the balloons and the tube.
For simplicity, you can imagine the energy stored in the material of the balloon (When it is expanded) as elastic potential energy.
Hope this helps. :)
@@rulersonicboom4737 You forgot hysteresis. Common mistake
@@sorova I am not familiar with the term hysteresis. Maybe you are correct about that part as I haven't learnt about it yet, but your statement that combined Surface Area of the two balloons, plays no role or cannot be the part of the explanation should be corrected according to my explanation. If my explanation was insufficient due to my lack of knowledge or not being familiar with hysteresis then I would appreciate if you would take the time to explain that to me in simple terms! As I would like to understand your point better. :)
It's real easy to come to the wrong conclusions. I have done it many times. :-(
Is there a law of conservation of surface area? No.
Air flows from high pressure to low pressure. The pressure in the big balloon is lower than the pressure in the small balloon because the rubber is stretched thin in the big balloon. Use your rig to actually measure the pressure in each balloon separately using a gauge that reads in the 1 or 2 psi range or you can rig a manometer from some clear plastic tubing and dark tea. We all feel the effort going down as we continue to blow up a balloon.
Paul Ohlstein Doing the math doesn't help if you're missing the physics.
+Paul Ohlstein Correct, it is related to pressure. Initially as the balloon is filled, the pressure increases. However, if the balloon is filled with a critical number of air molecules, the pressure actually begins to decrease as more air is added (due to the elastic properties of the rubber). It just so happens that in the video, the large balloon exceeds this number of molecules, and the small balloon does not. Thus, the smaller balloon is actually at a higher pressure. In order for the pressure to reduce in the small balloon so that it equals the pressure in the large balloon, the small balloon decreases in size since it is below the critical number of air molecules. Since those air molecules have to go somewhere, the only place to go is the large balloon in this closed system. This leads to the larger balloon expanding (and also decreasing slightly in pressure, but the smaller balloon decreases in pressure more so the pressures are equal).
If both balloons were blown up either below or beyond the critical number of air molecules (not one balloon on either side of the critical number), then they would in fact end up at the same size in this demonstration.
one would think that's a pretty important caveat to've included in the video, right?
btw, how is the critical number determined?
Hava basıncında karşımıza çıktı eğlenceli deney
I do this demonstration a lot. As a polymer physicist I can tell you his explanation is incorrect. There is an easy way to demonstrate why his explanation is incorrect here and to demonstrate the correct explanation.
What easy way?
@@samirshah848 The effect has due with the hysteresis of the rubber balloon. If you were to map the pressure inside the balloon as a function of the strain (size), You would see the stress (pressure) increase drastically in the balloon under initial inflation. When the balloon gets to a certain size there is a stress relaxation and the pressure drops as the balloon gets bigger. That initial elastic restoring force that makes the balloon difficult to inflate gets smaller after the 2nd or third time you blow it up. So, blow two balloons up to the same size (kind of big), then let some of the air out of one so that it is smaller. Then attach the balloons to the apparatus and open the valve. You will see that nothing happens. There are only two outcomes that can happen. Either the discrepant event occurs, or there is no observable change.
@@kirtpage3573 So the difference in stress of the elastic material is the driving factor and not the pressure difference of the air inside?
easy
this explains premature blue babies....
Air has no volume. Same amount of matter does not mean same volume. Bigger pressure makes air smaller. So not an argument.
this was explained wrong.
This was explained correctly. You can explain the process through equalization pressure in both the balloons or minimization of energy stored in the balloon due to surface tension. Assuming the process is isothermal, meaning that the temperature is constant which implies that the average kinetic energy of air molecules inside the balloons and the tube doesn't change throughout the process. Since energy stored in the balloon due to surface tension is directly proportional to Area, therefore the outcome of the experiment should be the configuration in which the energy of the system is minimum which would imply that the outcome should be of the configuration where the combined surface area of the balloons is minimum.
The formula for Energy Stored due to Surface Tension (for reference),
U = SA
U=Energy Stored due to Surface Tension,
S= Surface Tension (Property of the material),
A= Surface area of the membrane.
Hope I made my point. :)
This guys explanation lacks clarity and real understanding. Poorly done.
Mind-blowing