eth727
Guest
a 7 liter collapsible container is taken from 25ft seawater to 85ft seawater. What is the new volume? What is a easy formula to do this problem without all the p1 p2+v1 v2?
physics question
- Thread starter eth727
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What's the container filled with?
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You gotta know what is in it. I assume air.
You gotta know if its 7 liters at the surface, or if it contains 7 liters at 25ft
Then you gotta figure out what the pressure change between 25 and 85ft seawater.
also the formula (P1)(V1) = (P2)(V2) is possibly the simplest way to go about it for accurate calculations. All you have to do is organize the information, determine the question you are trying to answer, and then answer it appropriately.
You gotta know if its 7 liters at the surface, or if it contains 7 liters at 25ft
Then you gotta figure out what the pressure change between 25 and 85ft seawater.
also the formula (P1)(V1) = (P2)(V2) is possibly the simplest way to go about it for accurate calculations. All you have to do is organize the information, determine the question you are trying to answer, and then answer it appropriately.
Or fill the container with an incompressible fluid. Then it still holds the same 7 liters.
Remember (P1)(V1)=(P2)(V2) assumes that you have a gas, and that the temperature is constant. It is derived from the ideal gas law (PV=nRT). For the depth range for your example, most common gases are still exhibiting "ideal" behavior.
Remember (P1)(V1)=(P2)(V2) assumes that you have a gas, and that the temperature is constant. It is derived from the ideal gas law (PV=nRT). For the depth range for your example, most common gases are still exhibiting "ideal" behavior.
As it stands the problem is unanswerable. If the OP doesn't realise that then (s)he should go back over even more basic work.
Is the air under pressure (like a 7 liter basketball)? You need to know both the volume (7 liters) and pressure at some known conditions. If you have an "absolute" pressure measurement, that compensates for ambient pressure (whether on the surface or at any particular depth of seawater) then you can jump straight into P1V1 = P2V2.
Example - Open 7 liter container at the surface and fill it with air. Now you have 1 ATM of pressure in a 7 Liter container. Take another identical container, and fill it from a compressor so that it now has 2 ATM of pressure inside (most gages will indicate 1 ATM, plus the one that exists at sea level = 2 ATM). Take them both down with you and observe they do not compress equally, but they do compress proportionally.
It is generally easiest to understand what is going on at the surface (where we know there is 1 ATM of ambient pressure) and work the rest from there.
Seriously, by asking this question you are showing that you really don't understand the whole question of gas expansion. Getting the answer to the question you intended to ask is not the point - you must study this whole subject much more. It's nit difficult, but either you missed something vital or (more likely) the person teaching you didn't understand it either.
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I am guessing here, but it sounds like a typical basic diving question so without going through all the stuff about where does it have seven Litres, what shape is it, what is it filled with.., to answer what i think you are asking....
take the ATA of your current depth, divide by the ATA of your new depth, then multiply by the original volume.
(1.76/3.56) X 7
0.49 X 7
3.43L
take the ATA of your current depth, divide by the ATA of your new depth, then multiply by the original volume.
(1.76/3.56) X 7
0.49 X 7
3.43L
Mike Boswell
Contributor
P1xV1=P2xV2 is the easy formula. It is Boyle's Law, i.e. PV=K http://en.wikipedia.org/wiki/Boyle's_law.
People get a little confused by pressure calculations. Just remember that one atmosphere is 33 feet of seawater and that there already is one atmosphere at sea level. So if you take any water depth and divide it by 33, and add 1, you will know how many atmospheres of pressure you have. For example, at 99 feet you have 1+ 99/33 equals 4 atmospheres.
Assuming the container holds gas, you solve this type of problem as follows: You know P1 which is 1+25/33, and P2 which is 1 + 85/33, and you know V1 which is 7 L. You can now solve for the unknown variable V2.
So, V2 = V1 x (P1/P2) = 7 x (1+25/33)/(1+85/33) = 3.44 L
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