Physics

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As a non-academic, I've been studying this stuff for 3 months. Appreciate any tips from pros as to where I'm off base:
There are 3 types of questions in the DK Workbook.
1.Basic knowledge stuff: "Reduce the temperature around a balloon and what happens?" A: It shrinks, because the lower temp. causes the molecules inside the balloon to collide less often, reducing the pressure.This won't happen with a tank because it is not a flexible container. Pretty simple.
2.Whole Increments stuff: "Something has a volume of 4 at 33 fsw (2 ata). What is it's volume at 99 fsw (4 ata). A: 2 (you can either divide 4 by 2 since at 4 ata it will have half the volume as at 2 ata, or you can take it to the surface to find it has a vol. of 8 and divide that by 4 to get that it would have a vol. of 2 at 4 ata). Again, pretty simple--basically OW Manual stuff.
3.Hard Ones: Not in whole increments...: Something has a vol. of whatever at 85 fsw. What is it's vol. at 192 fsw? Now the formulas--Boyle's-P1*V1=P2*V2. Charle's: P1*V1 over T1 = P2*V2 over T2. Plug in the givens and find the one part missing. Things to remember: Don't forget to include the one ata for the atmosphere. You need to take it to the surface for these problems. If dealing with temps., must be in absolutes (F + 460=R or C+273=K). Eliminate what you can-such as if you're only dealing with something that is affecting one tank the volume is obviously a constant, so just cross out "V".

Type 3 problems are the ones that seem to concern us. Not enough examples in the Workbook. Simple mistakes seem possible.
Re: PV=nRT....I've read about this. Not sure I really understand how it fits into these problems. Why would you and how would one use this instead of Boyle's & Charle's? Of course, I'm a musician....
 
3.Hard Ones: Not in whole increments...: Something has a vol. of whatever at 85 fsw. What is it's vol. at 192 fsw? Now the formulas--Boyle's-P1*V1=P2*V2. Charle's: P1*V1 over T1 = P2*V2 over T2. Plug in the givens and find the one part missing. Things to remember: Don't forget to include the one ata for the atmosphere. You need to take it to the surface for these problems. If dealing with temps., must be in absolutes (F + 460=R or C+273=K). Eliminate what you can-such as if you're only dealing with something that is affecting one tank the volume is obviously a constant, so just cross out "V".
You just have a slight terminology glitch there, although it may even be in your materials. (I don't have them to review, although if anyone ever catches me out diving and lets me take a peek, I would certainly do that.) Charles' Law is *not* P1 * V1 / T1 = P2 * V2 / T2. That's the "Combined Gas Law".

If you remember the Combined Gas Law, you can use a little mnemonic to remember who's who. For Boyle, you think "boil", so you grab the temperature out of the formula and hold it constant. For Charles, you think "Charles Schultz", who drew "Peanuts", so you grab the P out of the formula and hold it constant. (For Gay-Lussac, you think, "Well, I've already boiled peanuts, so all I've got left is the volume." Hey, no mnemonic's *perfect*... :biggrin: You don't need the name, but if you're wondering what your 3000 psi at 75F is going to be when your tank's locked in your 125F car, that's the one you'll use.)

Combined Gas Law:
P1V1/T1 = P2V2/T2

Boyle's Law:
Think "boil", so grab and hold the temperatures...
P1V1/[-]T1[/-] = P2V2/[-]T2[/-]

P1V1 = P2V2
at constant T

Charles' Law:
Think "Charles Schultz drew Peanuts", so grab and hold the Ps...
[-]P1[/-]V1/T1 = [-]P2[/-]V2/T2

V1/T1 = V2/T2
at constant P

Gay-Lussac's Law, a.k.a. the unnamed law about tanks in the sun:
Think "They didn't even bother to mention this", so grab and hold the leftover V...
P1[-]V1[/-]/T1 = P2[-]V2[/-]/T2

P1/T1 = P2/T2
at constant V

Re: PV=nRT....I've read about this. Not sure I really understand how it fits into these problems. Why would you and how would one use this instead of Boyle's & Charle's? Of course, I'm a musician....
In scuba classes, if you remember the Combined Gas Law, the Ideal Gas Law (PV=nRT) is not something you need to remember. Consider it a bit of trivia about how they came up with the gas laws you'll actually use. (It's like the information about decompression models. It's not necessary to be able to do partial differential equations, as long as you know how to apply the tables or use the computers that result from them.)


If I get a chance, I'll see if I can whip up a Gas Law Problem Generator page to randomly create example problems (and solutions to check your result against, of course). I already have a generator for problems based on the NAUI dive table data, so it would be a nice complement to that.

What I want to see is the amount of time it takes to re-carbonate the beer at 3ATA. THAT would be useful math.
That's a kinetics problem, which is a completely different area of physics than the simple thermodynamic state equations we're using, but the basic answer is that if you have the beer chilled and you really agitate the mixture, it takes very little time, indeed. I don't brew beer, but I do carbonate my own beverages. I can turn ice cold water into *super*carbonated soda in about 10-15 seconds of vigorous shaking. For more information, see Carbonating at Home with Improvised Equipment and Soda Fountains, PM me, or start a new thread. (We'd welcome one in the Dork Divers forum, I'm sure. :biggrin:)
 
If you don't mind I would like to print this out and use it as study material. I now have a warm and fuzzy. The last problem really help a great deal.
 
Joe that is absolutely a great idea. I'm sure that there are more "C" students like more out there that need just a little more......how would you say "special coaching":D

Clay you are awesome:wink:
 
Thanks for all the help, just passed the physics test and missed 2, two stupid ones that is!!!, Everyone's input really made the difference!!!
 
Thanks for all the help, just passed the physics test and missed 2, two stupid ones that is!!!, Everyone's input really made the difference!!!

:cheers: Nice to get that one out of the way....it's all gravy from now on....
 
Missed only two on the Physiology as well, but then missed 3 on the scuba skills and environment, might have something to do with me never doing PADI O/W, as I'm FAUI? oh well I passed at least!
 
Amascuba: I have 2 questions re your post on physics (updated Jan.27/08). In your example of raising a 180 pound object that displaces 3 cubic fsw--as stated, 3 times 64= 192 pounds of up force. So the object is already positively buoyant, no? You wouldn't take 180 from 192 to find you need add 12 pounds because you're already 12 pounds over, no? Question 2 is about Boyle's Law:Your example compares volumes (P1*V1=P2*V2). But I don't think you can have differing volumes because a tank is an inflexible container and only pressure can change, No?
 
In your example of raising a 180 pound object that displaces 3 cubic fsw--as stated, 3 times 64= 192 pounds of up force. So the object is already positively buoyant, no? You wouldn't take 180 from 192 to find you need add 12 pounds because you're already 12 pounds over, no?
That question wasn't about raising the object, but rather "How would you determine how much weight you would need to add to the object to make it neutrally buoyant in the water?"

You are correct that it displaces 192 pounds of seawater, giving you 192 pounds of buoyancy. As you're trying to find out how to make it neutral (not how to float it to the surface), you subtract the 180 pounds of weight it already has to find that you need to tie on 12 more pounds to balance out the buoyancy.

Question 2 is about Boyle's Law:Your example compares volumes (P1*V1=P2*V2). But I don't think you can have differing volumes because a tank is an inflexible container and only pressure can change, No?
The volumes (80 cubic feet and 40 cubic feet) are actually *not* the internal volume of the cylinder. When referring to scuba tanks in the "American style", the listed volume is the "expanded volume" of the full-service-pressure contents of the cylinder. (A tank with 80 cubic feet of actual internal volume would require somewhere near 2.5 tons of lead (minus the weight of the tank) just to sink it, and if you made it a sphere (for the smallest possible size), it would be over five feet in diameter! No telling how long it'd take to fill. :biggrin:)

If you remove the valve and fill it with water, you'll find the actual internal volume of an 80 cubic foot scuba tank with a service pressure of 3000 psi is right at 0.4 cubic feet. As you note, this actual volume of the cylinder does not change (except in exceedingly small amounts due to the minuscule amount of flexibility present in even a "rigid" metal cylinder, but that's another thread).

What the post was then referring to actually has nothing to do with the cylinder and everything to do with the *contents* of the cylinder once released. It would be more readily apparent what was actually being said if it were phrased otherwise. Dumping all the air from an 80-cubic-foot cylinder into a bag will give you 80 cubic feet of air at the surface but only 40 cubic feet of air at 33 feet

(As an aside: Elsewhere in the world where they refer to, for example, an 11-liter cylinder, they are in fact referring to the true internal volume of the cylinder. If you took the valve off an 11-liter cylinder and filled it with water, that's how much would fit. As 1 bar is quite close to 1 atmosphere, 1.01325 bar = 1 atm, referring to cylinders by internal volume simplifies some of the math considerably. On the other hand, referring to cylinders by "expanded volume", American-style, makes a different part of the math easier. It's 10 of one, half a score of the other.)
 
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