ImaHamster2 ... Re. "... this hamster finds it interesting that the answer to this thought experiment would be different if performed on the equator of a fast spinning asteroid." Why would it be different? Curious.
Q: Thanks for entering the discussion with such a helpful input. This conundrum has been put to hundreds of Forum members in Australia, and I thank you for increasing my Minority of One to a Minority of Two.
Chagur, early in this discussion part of the challenge was to reply with reasoning. Factors that might be relevant needed to considered to determine if they played a significant role. Centrifugal force opposing gravity was such a factor. If gravity were less or the cylinder significantly higher then centrifugal force would counteract gravity. The pressure gradient would be high at the bottom, low in the middle, and high at the upper end. On a spinning asteroid the effect would be exaggerated so that the very slight “high” pressure “gravity” bump at the asteroid surface would be insignificant, leaving a continuous gradient from low to high due to centrifugal force. (If Q wishes to view the gas through Boltzmann’s equations then “centrifugal force” should be interpreted in a statistical sense involving momentum transfer between the gas molecules and the moving cylinder walls. No individual gas molecule is constrained to rotate with the cylinder.) As the thread progressed Scaevola added simplifying assumptions such as initial pressure X = sea level pressure, no centrifugal force, no temperature variation, no variation of gravity with height, and vertical rather than slanted. With all these assumptions now in place this hamster is waiting for Scaevola’s explanation. Hoping to learn something interesting. (So far the most interesting discussion has been Orthogonal’s ambient pressure manometer. This hamster still doesn’t know what it would give as an upper pressure reading.)
Q adds an interesting twist with the kinetic molecular theory of gases. Yet, atoms and molecules are usually subject to Earth’s gravitational pull. When we take a cylinder of this length and turn it vertically (90 degrees) to stand on one end, its top end 28,000 ft above sea level, (a point we’ve been reminded of more than once) how is gravity not going to affect the pressure? The mass of the enclosed gas may be small compared to the container, yet... at what point has the force of gravity been nullified? A cylinder better than 5 miles long (high?), as Chagur points out.
Counterbalance: Your input is welcome. The wonderful thing about Gedankenexperimenten, Thought Experiments, is that you can arbitrarily remove all the complicating and misleading variables, and reduce the problem to its basic form. The wonderful thing about Forums, too, is that we can have FUN thinking about Science.. A hint to get you moving along the right path. Imagine the horizontal VLC filled with ball bearings, so small that they cease to be ball bearings completely, and are better thought of as just points of Mass. Fill it up with simply billions and billions of them, and seal the cylinder. These point sources of Mass have another quality; their collisions are completely elastic, so they bounce off of each other losing no energy at all. Since the VLC is filled at room temperature, there is a lot of these collisions going on;because that movement only stops when things get cooled down to Absolute Zero, and Room temperature is way above that. So what do you think will happen to these constantly colliding points of mass? Do you think that they would congregate in one part of the cylinder in preference to another while the cylinder was horizontal? Why then would they start to do so, when the cylinder is turned through 90 degrees? Even IF gravity had something to do with it, and it doesn't, wouldn't each miniscule ball-bearing point of Mass be affected equally? And if there was an equal effect on all the tiny point masses, why would the way things were before, change? And the status of the system previously was? Horizontal. And the pressures at horizontal were? Equal. So when the VLC is hoisted through 90 degrees, the pressures will be? ......... You say it.
Q Outstanding!!! Omni - Q - ous!!! Please Register or Log in to view the hidden image! So where the heck have you been in the other science discussions? Please Register or Log in to view the hidden image!
Quite true, Scaveola. One of the two main reasons I visit Sciforums. And given the content of both of my posts, I’m a bit surprised you might think I wasn’t having fun--and for the sake of having fun. Save the “dunce” cap, mate. Don’t think it’s going fit the head of anyone posting on this thread. Please Register or Log in to view the hidden image! Thx, Counterbalance
Scaevola, here’s a different thought experiment for ya. A whole horde of hamsters builds a tower 28,000 ft. high. Being great nest builders the tower is air tight and capped with a roof to keep out the snow. Noticing how difficult it is to breathe at 28,000 ft. This hamster closes the only window allowing access to the outside air. (A little bitty hamster sized window.) Now the tower is “sealed”. Does air (knowing the tower is now sealed) rush to equalize the pressure and save this hamster’s life? Or must this poor hamster expire?
ImaHamster: Yay! A counter-thought experiment! A perfectly valid option in a Scientific discussion. I say, no, there wont be a 'rush' of air to save little Hamster's life, but he will survive. If he can survive at 14000ft, that is. Why 14000 ft?, I hear you cry in unison. Consider this. In building and then sealing the Hampster Tower, (move over Don Trump), what they have effectively done is to take a 'core' of the atmosphere. Just like you take a 'core' of a tree to count tree rings. Then they have sealed the core container, and let the pressure equalise. (It WILL equalise,you know, just like you put in high pressure air at the valve side of your car tyre at the service station, but it doesn't stay congregated around the valve, does it?) Assuming the atmosphere to diminish inpressure at a constant rate with height, ( a very reasonable assumption), the average pressure will be.....? You say it. So the equilibrium pressure will be....? You say it. Another way to realise why Hamster won't die. Let's imagine, just before the Hamster Tower was sealed, Donald Trump, who was inspecting the tower at the ground level, and completely pissed off about being Trumped, expressed his opinion with a huge Fart, and left the building, miffed. The richest, smelliest Fart ever. Enough to blister paint on the wall. You could smell this one even at the tiniest concentration. Would Hamster, in the sealed container, with a very, very sensitive nose,EVER smell Don's fart? What do you think, and Why?
Hmmm…now this hamster can breathe but the air is too rich up here. Need a little less Trump. No problem. Scaevola says the pressure will equalize so the pressure at the top is now greater in Hampster tower than outside. Just open the window and let air out until the top pressure outside equals the top pressure inside and then close the window. With the window closed the pressure equalizes again. Hmmm…just by opening and closing this little hamster window the air is being “pumped” out of Hamster Tower. This hamster could continue opening and closing the window until the entire tower is pressurized at the “28,000 ft. level”. Interesting but too thin to breathe. Hmmm…let’s go to the bottom and open up a window. Air comes in until the bottom outside pressure equals the inside pressure. Close the window and the air redistributes, lowering the inside pressure. Open the window and let more air in. Close the window and let the air redistribute. Continue until the entire tower is pressurized to sea level. Who would have believed these tiny hamster muscles could change the entire tower pressure from sea level to the "28,000 ft." level and back merely by opening and closing these tiny windows? Put in a turbine and this hamster not only has fart free air but also an unending supply of electricity. PS This unreasonable hamster would not assume atmospheric pressure declines linearly with height.
Why is it that "simple" questions are the most difficult to answer? Thanks again scaevola for bringing this one to the table! Scaevola, I don't think you appreciate the gravity of this situation. Please Register or Log in to view the hidden image! I'll stand on the shoulders of Newton and repeat: (m1*m2)/r^2 . A molecule of air is overwhelmingly massive compared with a photon, yet even photons are deflected by gravity. Of course the product of m1*m2 for two molecules of air is negligible, but if m1 represents the mass of an air molecule while m2 represents the mass of the earth, the product is not negligible. The m1*m2 product of the mass of the Sun and that of the lightest atoms - hydrogen - produces an attractive force so great as to produce the nuclear fusion that enables all life on this planet! On the other hand, the Moon has no atmosphere partly because its mass is insufficient to retain light gases. I would remind you that the "weight" of a column of air on one square foot of the Earth's surface is 2117 pounds. Would you kindly explain to me why you believe that the gravitationally induced force of such magnitude could be considered negligible? Not at all; the gravitational force is a function of the distance-squared between a given molecule of air and the Earth's center. A molecule (miniscule ball-bearing) high in the sky is slightly less affected by gravity than a molecule at the surface of the Earth. Good catch Hamster. I wrote "average density" when I thinking "local density". Scaevola wrote that the cylinder was to be pressurized to "x psi". Of course "x" is a variable that can take on any real value, including zero. To simplify this problem, fill the horizontal cylinder to a pressure of 14.7 psia or 0 psig, which is the pressure of a standard atmosphere at the surface of the earth. You could "fill" the horizontal cylinder to this pressure by simply opening a valve at either end, and closing it when the pressures inside and outside the cylinder are equal. So, let's forget about the higher average pressure in the cylinder! Let's fill the VLC to 0 psig while it's lying on its side. Now, seal the air inside by closing the valves. Raise the VLC to the vertical position. Q and scaevola maintain that the air pressure inside this long vertical cylinder remains equal from top to bottom. We all agree that the atmospheric pressure at altitude is less than at the surface of the Earth. So, if I were to crack open the top valve "B", the 0 psig internal cylinder air would escape from the valve into the rarefied upper atmosphere. Since air has escaped from the cylinder, by Q and scaevola's argument the pressure throughout the entire cylinder will be equally reduced to the same (below 0 psig) value. Lastly, crack open valve "A" at the bottom of the VLC to allow the now higher pressure (0 psig) surface atmospheric air to rush into the cylinder. Once again, by Q and scaevola's argument, the pressure at the bottom and the top will always remain equal; thus the VLC will continue to take in air at the bottom and exhaust it at the top. The VLC has become a perpetual motion engine! I believe that a sample of air captured in a cylinder, be it horizontal or vertical, responds the same as a column of free air of similar dimensions and position relative to the Earth. I believe that; air molecules, rocks, people, all particles having mass, in fact, are acted upon by Newton's law of gravitational attraction. The vector force of the Earth's gravity always points towards the center of the earth. If the concept of a gravitational force is bothersome, imagine the VLC in outer space, attached to a giant centrifuge. An equivalent force of angular acceleration produces the same effect as does gravity; that is, it "piles up" the air molecules into the far end of the cylinder. The local gas pressure is proportional to the local gas density. Michael
Orthogonal be careful. Opening those valves could end all life on earth. This hamster suspects the reason for these differing views is a book that made some statement like, “Gases expand to fill and equalize pressure in a closed container.” The book never bothered to explain the exceptions to that particular “rule”. This hamster wishes a little more time were spent teaching exceptions and a little less spent teaching rules.
Or ... That without a 'surface/atmosphere' interaction, there is no such thing as an apparent 'Coriolis force/effect'. Sorry about that Please Register or Log in to view the hidden image! Thanks scaevola, been waiting years to throw that little curve. Please Register or Log in to view the hidden image!
The lower in elevation a gas is whilst in a gravitational feild the more compressed it is. This scenario is impossible.
Continuous force Pressure can be defined and obtained from a kinetic viewpoint by deducing the mass, speed and directional change of individual molecules as they move about through the system (statistical thermodynamics). Pressure is, however, more clearly defined and measured, within the confines of the pressurized cylinder, from a continuum viewpoint (classical thermodynamics) through the mathematical representation: P=Fn/A. Fn is the force exerted perpendicular to the surface area, A, with A approaching infinitesimal size. From a mathematical view, the area is becoming a point, however the area is still regarded as a continuum from a molecular viewpoint therefore still possessing a sufficient population of molecules to define matter as continuous rather than discrete.
It is remarkable how the answers in this Canadian/ North American forum mirrors the answers given in the Australian forum. In each case, the vast majority of respondents gave the intuitive, but quite wrong answer that there would be greater pressure at the bottom of the Very Long Cylinder. The Second Law of Thermodynamics is the essential reading for the understanding of this problem., and I urge those other than Q, who clearly has grasped the concept, to read as much as necessary in this field until they too, KNOW that the pressures will be equal. This is a great Trans-Pacific exchange of ideas; why not visit the Australian site at www.abc.net.au/forums. Forums on Nature, Science and Computers there receive around a million posts per year, and the debate is interesting on a wide spectrum of subjects. All The Best from The Land Down Under......
scaevola, <i>In each case, the vast majority of respondents gave the intuitive, but quite wrong answer that there would be greater pressure at the bottom of the Very Long Cylinder.</i> Assertion does not prove your case. I'm afraid you'll need to <i>explain</i> why the answer given is "quite wrong". I suggest you go swimming in a pool with a deep end and a shallow end. Dive down to the bottom at both ends and compare the pressures on your eardrums. I think you will find that the pressure at the deep end is greater than at the shallow end. What's your explanation for that?
James R: Your swimming pool analogy demonstrates that liquids in unsealed containers, like water in a swimming pool, have certainly got a pressure differential. That is also true of the atmosphere; pressure reduces with height. But gases in sealed cylinders act in a counter-intuitive way. The very properties of the gaseous state make the very fast equilibration of pressure in a sealed cylinder inevitable and Thermodynamically essential. The input from Q is well worth re-reading. Let's head back to the swimming pool, and submerge an evacuated cylinder vertically in it. It has the usual magically accurate pressure gauges at each end. It is filled with pool water by opening both ends, and resealing them. The cylinder is at 90 degrees to the pool floor, one end at the pool floor, the other just under the water surface.. Surely we agree that you have effectively taken a 'core' of the swimming pool. like coring for tree rings. The values of the gauges will be determined by the depth of the pool and the density of the water, and the value of G at that point on the Earth. The values are not important, whether they are equal or not is. When you pull the cylinder out of the swimming pool,and lie the cylider down beside the pool, do the end-gauges read equal or different? If you now stand the cylinder upright, 90 degrees to the floor, do the gauge readings change from the horizontal values?
scaevola, Take your cylinder full of water and align it vertically. Let's say the pressure gauge at the top end reads p0 whilst immersed in the pool. The gauge at the bottom reads p0 + rgL, where r is the water density, g is the gravitational constant and L is the length of the cylinder. Now, seal the ends of the cylinder and remove it from the pool. Aligned vertically outside the pool, the gauges read exactly the same as they did before. The cylinder contains some water under pressure, and that pressure varies along the height of the cylinder. Turn the cylinder so it is horizontal. Now, the pressure on the top side will be p0 and the pressure on the new bottom side will be p0 + rgW, where W is the cylinder width. -------------------- <b>Note for other contributors to this discussion</b> scaevola originally posted his pressure ideas on the Australian Self-Service Science Forum. I am already engaged in discussion there, and would like to keep it all in one place, so I will not participate further in the current thread. I recommend anybody who is replying here to first look at the link below. The specific link to the Australian discussion is: <a href="http://www2b.abc.net.au/science/k2/stn/posts/topic604049.shtm" target="_blank">scaevola's pressure thread</a> This will get you only the relevant thread, with no navigation links to the rest of the forum. A link to the top level of the Australian forum is: <a href="http://www2b.abc.net.au/science/k2/stn" target="_blank">Self-Service Science Forum</a> All visitors are very welcome!
One more thought experiment: Find an extra-terrestrial source of air (just humor me). Build a giant air compressor here on earth with a hose long enough to reach the extra-terrestrial air source. Begin pumping air into our atmosphere until the pressure at sea level has been doubled. Does everyone in this thread agree that by merely doubling the earth's atmospheric pressure at sea level, the normal negative pressure-with-altitude gradient due to gravity will remain unchanged? Now, construct scaevola's very tall cylinder. Let the air inside the cylinder come to equilibrium with the atmosphere (including the usual gradient-with-altitude) before you place the top on the cylinder. Do we all agree that the act of placing a lid on a cylinder containing air in a condition of static equibrium does not alter the conditions of equilibrium inside the cylinder? Now that the cylinder is sealed, reverse the giant air compressor and return our atmosphere to exactly the same conditions as before. We are left with a tall sealed vertical cylinder containing twice the air pressure than exists in the surrounding atmosphere. This is what scaevola asked us to create in his original thought experiment. However, contrary to scaevola's assertion, this cylinder exhibits a decrease of pressure with altitude. If you disagree with my conclusion, would you be so kind as to point out to me exactly where I went wrong in this thought experiment? BTW, I'm starting to run low on these thought experiments Please Register or Log in to view the hidden image! All the best from the land up over, Michael
