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I have a question about breathing gas underwater. Here goes: As we go down, the quantity of gas breathed increases (because the pressure goes up) So for a diver that's going down at a speed of 18m/min until he reaches 60 meters underwater (It would take him 3 minutes and 20 seconds to do so). How would I calculate the amount of gas he's breathing during this 3 minutes and 20 seconds, knowing that the atmospheric pressure is 100kPa and the water's temperature is about 20C and that he's breathing rhythm is about 18L of gas per minute?

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I tried to calculate the average amount of gas breathed by using the quantity of air breathed every minute before the dive and the quantity of air breathed every minute 60 meters underwater. At 0m, I could use the perfect gas formula, PV=nRT, where P = 100kPa; V = 18L; n = ?; T = 293K\[n = (\frac{100*18}{8.314*293})=0.73mol\]At 60m, P = 700kPa; V=18L; n=?; T=293K\[n=\frac{700*18}{8.314*293}=5.17mol\]and calculate the average, is that a good way to do it? (\(Average =(5.17+0.73) /2=2.95mol/minute\\2.95mol/minute*3.33 minutes =3.94mol\))
Do you know if my work is correct for this kind of situation?

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if air intake is a function of depth you would need an integral to do it right
Well, intake is a function of pressure which is a function of depth
I don't think your assumptions are correct. I would expect both respiration volume and frequency to decrease as the pressure of O2 increased. The urge to breath is driven by the accmulation of CO2 in the blood. Therefore I would suspect you do not breathe a fixed volume, but rather a fixed number of moles of O2 per minute. I would expect THAT to stay constant as the diver descended, so that the actual volume of O2 consumed at the higher pressure would steadily decrease.
Volume is related to the quantity of air, greater pressure doesn't mean taking little breathing. that's what I was told :/
because the first way I did was to calculate with a fixed quantity of gas
I honestly don't know about that, but IF air intake is a function of pressure, which is a function of depth, then you need to integrate air intake as a function of pressure for the trip.
No, that's silly. It's well known that prebreathing pure O2 tremendously reduces the urge to breathe for a while afterward. But this has nothing to do with your method of solving the problem -- it was just a throught I had.
Pressure is calculate with the formula P = pgh, where h = height, g = gravitational acceleration and p is the density of the medium we are going into. So \[\large \int_{0}^{60}(pgh)dh\]?
you need a term for air intake A (per unit time) as a function of pressure A(P(t))dt... I forget hydrostatics :P
@Carl_Pham, the diver is breathing normal air during his trip, but as he goes down, the partial pressure of O2 will go up, so he would breath more oxygen than he would at the surface?
Ah, we integrate wrt to time?
Oh, ok ok I see :P
or you could make pressure a function of depth A(P(h))dt
I would integrate wrt if your function is air intake per unit time I suppose you could do air intake per meter dived, but that seems like a tricky function
If air intake is a function of depth then how fast you are moving will affect where you are at a given moment, and how much air you have taken in up to that point
He's breathing at a rate of 18m/min, so integrating wrt time would be more appropriate I think, so I would have to compute a formula for pressure wrt time, is that correct?
you could do that if you take the divers path as fixed I guess, but pressure could be wrt depth without too much trouble
something like\[dP=\rho gdh\]\[dA(P)=f(\rho gh)\rho gdh\]A is some function of pressure f(P)
again, not really sure... haven't done fluid mechanics in forever it would be nice to assume air intake is directly proportional to pressure
Air intake.. that would be his breathing speed?
then you could have\[dA\propto dP=k\rho gdh\]
air intake as a rate I was thinking, but then you have to find the total air intake which would require integration wrt time, so... you would have to make h a function of t and integrate wrt time in the end anyway I suppose
Alright, I'll try it out, thanks!
Good luck. Thinking about it makes me remember why I didn't like fluid mechanics all that much.
In case of failure, I'll call you :P
\[A\propto P(t)=k\rho gh(t)\]\[dA=k\rho gh'(t)dt=k\rho gvdt\]something like that for starters I think. See ya!
the above if pressure is directly proportional to depth
I mean air intake propto depth

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