physics question

[tracydr]

P1xV1=P2xV2 is the easy formula.

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

Like Mike said. Thanks for such a nice, clear explanation!

 

[SC_Hoaty]

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

P1xV1=P2xV2 is the easy formula.

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

Didn't y'all just solve it for being 7 L at 25 feet? And assumed it had 1 ATM of gas in it at sea level?

My point is that the problem is not sufficiently stated. WHERE is it 7 Liters?

 

[eth727]

Thanks Peterbj7 that's all I need is to be criticized for asking a question.

 

[SC_Hoaty]

Didn't y'all just solve it for being 7 L at 25 feet? And assumed it had 1 ATM of gas in it at sea level?

My point is that the problem is not sufficiently stated. WHERE is it 7 Liters?

To explain further

Assume container is 7 L at surface, and filled with ambient air.

P1 = 1 ATM (absolute)
V1 = 7 Liters

at 25 ft

P2 = 1 + (25/33) = 1.76 ATM
V2 = V1 * (P1/P2) = 7 * (1/1.76) = 3.98 L

at 85 ft

P2 = 1 + (85/33) = 3.58
V2 = V1 * (P1/P2) = 7 * (1/3.58) = 1.96 L

In many physics "word problems" there can be extra information that is of no use to solving the problem. If the compressible container held ambient air at sea level, the stop at 25 feet has nothing to do with its volume at 85 feet.

I freely admit that this all assumes 7 liters of ambient air at sea level. A different problem statement would yield different results.

For example - if a compressible container is filled at 25 feet (with seawater) and then taken to 85 feet, it will still have the original volume, as the seawater can be considered incompressible at these depths.

It's all in the problem statement.

(Is the horse dead yet, or shall whippings continue?)

 

[ditch-diver]

~sigh~ there is always one in every class that wants to read too much into the question

The poor guy asked a question, based on similar questions asked in pretty much every OW diver course on how to do basic pressure/volume change equations WITHOUT using the P1V1=P2V2 formula. In 25+ years of teaching this stuff, I have found there are loads of people who do not "get" formulas. It is not the way their brain works. Sure you can hammer it into them and they can then fill in the blanks and get an answer but because of their thinking/learning methods they are not grasping what the formula is telling them. That is why I write it out they way I did, it helps those people understand and remember how to do it.

 

[kyjellyfish]

agreed - there are no such thing as stupid questions...

when i was studying engineering in college, i remember posting a question on an engineering forum. my motivation was to not only seek help but to engage in discussion with professionals i'd find myself working with once i graduated. i could've gone to my professor or ta but it's was so much more meaningful to get it from the folks out living and breathing the material

 

[peterbj7]

Thanks Peterbj7 that's all I need is to be criticized for asking a question.

I'm not criticizing you for asking a question or for anything else. I'm merely pointing out that from the nature of the question you asked I believe you have a deeper lack of understanding that you should first address. Giving you a "quick fix" answer to your question won't help you in the long run. I'm assuming you want to understand this subject and hence remember it long-term, not just learn an answer that you'll have totally forgotten in a month?

 

[Thalassamania]

The real point is that the container (if filled with a gas) has (1.76/3.58) times the volume that it actually had at 25 fsw.

 

[Mike Boswell]

Okay, now that we have covered Boyle's Law, we can move on.

If my flexible container holding 7L of gas at 25 fsw is cooled from 85 degrees F to 40 degrees F, what will the new volume be?

 

[Lead_carrier]

Okay, now that we have covered Boyle's Law, we can move on.

If my flexible container holding 7L of gas at 25 fsw is cooled from 85 degrees F to 40 degrees F, what will the new volume be?

It should still be big enough to hold a 6 pack. That's the important thing