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October 6

Pressure of a gas

I can visualize pressure inside a liquid pretty easily: water molecules are touching each other and exerting forces on their neighbors. But I'm a little confused about how pressure in a gas works. The gas molecules pass usually through each other, so when people speak of a force by one part of the gas on another, are they speaking about the occasional collision between gas molecules or about some sort of momentum flux due to gas molecules moving through an imaginary surface? 74.15.136.9 (talk) 03:05, 6 October 2012 (UTC)[reply]

It doesn't matter much whether the gas molecules collide or not, though I think the mean free path in typical gases may be shorter than you suppose. Even if they never collided with each other, they'd still collide with the walls of the container (otherwise they'd keep on going and not stay in the container), and thereby exert force on it. Or, with no container, if you have a solid object immersed in the gas, the same arguments apply — the gas molecules collide with the solid object, bounce off it, and thereby exert force on it/transfer momentum to it. --Trovatore (talk) 03:08, 6 October 2012 (UTC)[reply]

Oh, actually I missed that you said "one part of the gas on another". In that case, I suppose it does matter (if the gas molecules really passed freely through the gas at large, they'd just keep moving through it until they hit a solid boundary, and then it wouldn't make much sense to talk about "one part of the gas"). But the article I linked says that under standard conditions (one atmosphere; not sure whether 0 or 25 degrees Celsius) the mean free path is only 68 nanometers. --Trovatore (talk) 03:13, 6 October 2012 (UTC)[reply]

In a cube of 68nm, there would be (I think) (68*10^-9)³*6.025*10²³*1000/22.4=8457 molecules. But that's if it was 0°C. If that gives any idea... Ssscienccce (talk) 04:23, 6 October 2012 (UTC)[reply]

Not sure just what your point is, but I guess that is a good intuitive guide to what's going on, assuming your figures are right. The cube root of that is about 20, so roughly speaking you're saying a molecule would pass 20 other molecules before bumping into one. --Trovatore (talk) 05:39, 6 October 2012 (UTC)[reply]

Just to get an idea, as you say; maybe I'm a bit obsessive-compulsive when it comes to calculations, judging by the amount of scratch paper riddled with numbers (if that's the right expression) going in the bin every day. Ssscienccce (talk) 07:53, 6 October 2012 (UTC)[reply]

The molecules in a gas do collide with each other, which allows the gas to undergo things like laminar flow and turbulent flow and allows pressure to equilibrate throughout the gas. In situations where the gas molecules do collide with the walls more frequently than with each other (such as small pores), Knudsen diffusion occurs, and the macroscopic notion of pressure breaks down somewhat.--Wikimedes (talk) 07:18, 6 October 2012 (UTC)[reply]

You can work out from the gas laws and kinetic equations and so forth (we did it in high school, I can't give the exact derivation from memory) that on average an individual oxygen or nitrogen molecule in the atmosphere at room temperature travels about 10cm before it hits another air molecule (mean free path mentioned above) and does so while travelling at 1,000 kmph. They may be small, but they are fast, and there's a lot of them. Grokking that should help your intuition. μηδείς (talk) 16:54, 6 October 2012 (UTC)[reply]

I think you're off by about seven orders of magnitude. Maybe you meant nm instead of cm? --Trovatore (talk) 17:48, 6 October 2012 (UTC)[reply]

I may certainly be wrong, I am going on memory from the 80's. But I did mean to say 10cm. Nanometers sounds way too small. I'll see if I can google a result. Hmm, yes, our own article says 68nm. I wonder if I am just remembering the number wrong, or if I have some other result in mind. Well, that should also put my 1,000kmph figure to doubt. μηδείς (talk) 17:52, 6 October 2012 (UTC)[reply]

Physical Chemistry by Atkins 6th edition p.30 gives a typical mean free path for N2 at 1 atm to be about 70nm. It also states that for N2 or O2 at 25C and 1atm the molecules travel at about 350m/s which is 1260km/h, so Medeis' speed is about right.--Wikimedes (talk) 18:30, 6 October 2012 (UTC)[reply]

Indeed, and it's not all that difficult to estimate this from first principles. The Bohr radius is 0.5*10^(-10) meters, so the elastic cross section for hydrogen molecules should be of the order of few times 10^(-20) m^2, let's take this to be 10^(-19) m^2. Then the mean free path L is the average distance a hydrogen molecule needs to travel before collding with another one. This means that within the volume swept out by the cross section of 10^(-20) m^2 over the distance L of 10^(-19) m^2 L , there should be one molecule on average.

The number density of hydrogen molecules at room temperature and a pressure of p = 1 atmosphere is n = p/(k T) = 2.5*10^(25) m^(-3). So, within the volume of 10^(-20) m^2 L there are 2.5*10^(25) m^(-3)* 10^(-19) m^2 L = 2.5*10^(6) L/m molecules, but L being the mean free path this should equal 1. This means that L = 4*10^(-7) meters. Count Iblis (talk) 19:58, 6 October 2012 (UTC)[reply]

Maybe Medeis was thinking of Rutherford scattering? Alpha particles will travel several centimeters in air before being deflected by a nucleus. Ssscienccce (talk) 08:31, 7 October 2012 (UTC)[reply]

I am afraid we may just have gotten the wrong answer somehow when we did the derivation as a class, since I remember being struck by the results being on a visibly imaginable scale, which 70nm is definitely not. Our chem teacher was on the ball, quite a good mathematician, so I can't see her making such a large error, or letting us remain in one. (I do remember getting a speed of around 1,250 kmph, and I just rounded it down for the example. I think the Rutherford result is a good guess as to what may be confusing me, or the fact that there's on the order of one hydrogen atom per cubic decimeter in deep space.) I'll have to ask some other students of hers if they remember doing the derivation.

I assume that the speed of sound is a result of the speed of the particles derived here? μηδείς (talk) 21:34, 7 October 2012 (UTC)[reply]

What kind of tech advance will we need to have GPS-locatable wedding rings?

That's the one thing you dread the most - losing a wedding ring. Sometimes you forget it in a public restroom, or during a picnic in an open field out there, you forget you took it off and now the field is too big for you to spot it.

That's why I would hope for my and her wedding ring to have a GPS chip, so that once the ring is lost, you go to "ringlo.st" and enter your information, then the GPS satellite tracks the ring down for you.

If anyone's concerned about "powering" the chip, I would hope that piezomechanics allow for kinetic movements to keep it powered up.

A. So how much farther do the necessary components need to miniaturize in order to fit comfortably in a wedding ring?

B. How much might it cost?

C. Would it be waterproof?

D. Are there other issues you'd like to bring up with this hypothetical GPS-locatable wedding ring?

Thanks. --70.179.167.78 (talk) 15:03, 6 October 2012 (UTC)[reply]

A) I think the electronics are mostly almost small enough already, with a couple exceptions: the battery (which it would need to remain powered when taken off) and the antenna (I would also include the display and keyboard, but presumably this would lack those).

C) I don't see waterproofing it being a major problem.

D) First, it wouldn't be "solid gold" anymore, so would be lighter and not feel like a quality ring. This would also make it more fragile. Finally, putting any type of technology on it means it could become obsolete, let's say if GPS units all switch to different frequencies.

A more reasonable approach which could be taken right now are to put passive RFID tags in them. These don't require a battery, but only work at a short distance, with a scanner you'd carry around. So, it would work great if you lost your ring around the house, but not if you lost it someplace random outside the house. However, the only reasons I can see for taking your wedding ring off outside the house are at a jewelry store when having it cleaned/adjusted, at a clinic for a CAT scan and such, and when engaged in adultery. Hopefully that's few enough places where you could bring the scanner. I don't quite understand why you'd take it off in a public restroom (unless this is where one engages in adultery). :-) StuRat (talk) 15:30, 6 October 2012 (UTC)[reply]

Another reason to take your ring off is when working with some dangerous machinery. I wirk with people in a factory, and some of them remove rings when working because of the danger of having the ring trap in active machinery. Some of them have lost fingers because of rings. 217.158.236.14 (talk) 08:19, 8 October 2012 (UTC)[reply]

As a workaround, could you just buy a fine chain, worn on your neck or attached to your jeans, and put the ring on it whenever you take it off? – b_jonas 19:27, 6 October 2012 (UTC)[reply]

Or just never take it off. My ring has not left my finger in over forty years, and now I cannot physically remove it.--Shantavira|^{feed me} 19:54, 6 October 2012 (UTC)[reply]

I agree that, currently, GPS tracking is infeasible, and probably will be for your lifetime due to power source constraints (unless you are content to replace/recharge the power source rather often). Another unsolicited alternative: get a dozen of these [1], declare them identical in the sense of "wedding ring," and always have a spare if you lose one ;) They also have safety benefits; I know someone who lost his left ring finger because he was wearing his wedding ring... SemanticMantis (talk) 21:16, 6 October 2012 (UTC)[reply]

Good idea. If you must have a silver/gold/platinum/bejeweled wedding ring, keep it in a vault, and wear something disposable from day to day. StuRat (talk) 21:33, 6 October 2012 (UTC)[reply]

Hi, OP here. No one answered how well piezo-recharging would work on rings.

E. Wristwatches are recharged by the kinetic movements of the wearer all the time, so why couldn't rings?

F. Also, if GPS changes to different frequencies, wouldn't the wearers get a mail/email notification letting them know of this, and to get it changed at the local jeweler / electronic store as soon as possible? --70.179.167.78 (talk) 23:41, 6 October 2012 (UTC)[reply]

E) It's not enough to broadcast a signal to a cell phone tower, which is what your system would require (or, even worse, broadcasting to a satellite).

F) Perhaps, but plan on doing this every 5-10 years, as cell phone technology constantly changes. You'd also probably have to replace the guts, as a new frequency might require new electronics and a new antenna. Think about it this way, what 50 year old portable electronic technology is still used today ? Very little, for good reason. See many people with mono AM transistor radios ? StuRat (talk) 00:08, 7 October 2012 (UTC)[reply]

For future navigation I should note that the question on kinetic reclamation (charging by hand movements) is now being archived at Wikipedia:Reference_desk/Archives/Science/2012_October_2 though it is still visible above for now. Wnt (talk) 21:40, 7 October 2012 (UTC)[reply]

It's amazing that nobody has yet pointed out that GPS satellites do not track things. They broadcast a signal that GPS devices use to calculate their location. To make this work, the ring would have to include a GPS tracker and something comparable to a cell phone, so that it could call the owner to give its location. Looie496 (talk) 05:11, 7 October 2012 (UTC)[reply]

I did, in my response to E, above. StuRat (talk) 06:27, 7 October 2012 (UTC)[reply]

What's being described is essentially a cell phone, in that it can be tracked and communicated with (or from) at any time. People of most religions might do without the keypad and the address book, but there would be little savings in omitting the ability to communicate voice since microphones and speakers have become very small. So the question collapses to one of when a cell phone would be possible as a ring. This might depend on some genuinely unknown factors, such as the hazards of terahertz radiation and the feasibility of using it as a routine communications medium (since it could theoretically have a shorter antenna, etc.). Wnt (talk) 21:36, 7 October 2012 (UTC)[reply]

Religions ? StuRat (talk) 21:39, 7 October 2012 (UTC)[reply]

Antenna efficiency drops off very quickly below 1/4 wavelength - a wedding ring is most probably too small to house a GPS receiver antenna large enough to work. Even GPS receivers with full size antennas don't work indoors. It's cellular transmission antenna would be even less efficient though might still be able to work close to a cellular base station, but GPS satellite signals are very weak even outdoors with wide open skies. Roger (talk) 22:00, 7 October 2012 (UTC)[reply]

That's why I mentioned terahertz... though I have a wild suspicion that there's some metamaterial based solution to that problem that will turn up sooner or later if it hasn't already. (by analogy to microscopes beating the diffraction limit) Wnt (talk) 20:47, 8 October 2012 (UTC)[reply]

It would be more feasible to equip your environment with recording video cameras to track your ring's and your own every movement. They could be attached to your clothing or body in case you visit rest rooms. Or you could have movement sensors attached to your hands. Presumably the pattern of movement between your hands is characteristic of pulling the ring off. Then software could alarm you if you step more than a meter away from the ring. All this would be possible with today's accelerometers and cameras. Graeme Bartlett (talk) 23:23, 7 October 2012 (UTC)[reply]

Those don't sound practical, to me. Perhaps the RFID tag and scanner combo could be set up so that you always carry the scanner with you, and an alarm goes off whenever the ring moves out of range (a few yards/meters). StuRat (talk) 21:06, 8 October 2012 (UTC)[reply]

Frog/toad suggestions

Something I half-remember from a wildlife documentary from years and years ago. It's a frog or toad that lays her eggs and then places them in indentations on her back - and her skin actually grows over and covers the eggs completely. Then the tadpoles develop inside the eggs without hatching, and emerge from her back as fully-formed froglets...

Does this creature sound familiar to anyone? As I say, it was a long time ago that I saw this on TV, so my memory might be playing tricks on me - but I do seem to remember watching this as a kid and being freaked out by the skin part. --Kurt Shaped Box (talk) 23:28, 6 October 2012 (UTC)[reply]

Surinam toad. Deor (talk) 23:59, 6 October 2012 (UTC)[reply]

Shudder. That is still as disturbing to me now. --Kurt Shaped Box (talk) 00:05, 7 October 2012 (UTC)[reply]

Well done. I was going to suggest the Midwife toad, but it doesn't really fit the bill. Perhaps someone could add a brief mention of the Surinam Toad to our Frog article? Alansplodge (talk) 00:05, 7 October 2012 (UTC)[reply]

The mere thought of that makes me reach for my backscratcher. StuRat (talk) 04:55, 7 October 2012 (UTC) [reply]

Beautiful! evolution at its best. Richard Avery (talk) 07:19, 7 October 2012 (UTC)[reply]

It reminds me of those pictures that were doing the email rounds a few years ago that purportedly showed some horrible skin disease - but in actuality were elements of lotus seed pods 'shopped onto various body parts. --Kurt Shaped Box (talk) 10:46, 7 October 2012 (UTC)[reply]

There's a general term "marsupial frog" referring to this method of incubation. Wnt (talk) 21:44, 7 October 2012 (UTC)[reply]

Actually the Neobatrachian marsupial frogs have a pouch into which the eggs are deposited, whereas in the unrelated Mesobatrachian Suriname Toad the eggs are laid on the back and the skin of the back swells around them, enclosing each separately. μηδείς (talk) 01:38, 12 October 2012 (UTC)[reply]