Diving Depth for 1/20 Atmospheric Pressure Differential

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A diver using a snorkel can only handle a pressure differential of 1/20 atmospheric pressure, which limits their diving depth to approximately 0.5 meters due to water density of 1025 kg/m^3. The calculations show that atmospheric pressure cancels out, indicating it doesn't affect the gauge pressure experienced by the diver. The key factor is the pressure difference between the outside and inside of the lungs, rather than the absolute pressure. The discussion emphasizes that a diver cannot effectively inhale through a snorkel beyond a shallow depth. Overall, the findings highlight the limitations of snorkel diving at greater depths.
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Homework Statement


A diver with a snorkle can handle a pressure differential of 1/20 atmospheric pressure between his the oustide of his lungs and the inside. How deep can he dive? The water's density is 1025 kg/m^3.


Homework Equations


P=P_0+pgh


The Attempt at a Solution


I assumed P_atm = 101325 PA and the density of air is 1.29 kg/m^3.

I came up with P_(outside lungs)-P_(inside lungs)=(1/20)P_atm

P(outside lungs) = P_atm + pgh = 101325 + 1025*9.8*h
P(inside lungs) = P_atm + pgh = 101325 + 1.29*9.8*h

Solving for h I got 0.5 meters... Just seems way to shallow.
 
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Your solution is correct; a diver would not be able to inhale through a snorkel at a depth of more than just a few feet.

By the way, you don't need to concern yourself with the pressure or density of the air. What matters here is the gauge pressure, which is the pressure experienced by the diver over and above that of the air he's trying to breath through the snorkel.

- Warren
 
chroot said:
Your solution is correct; a diver would not be able to inhale through a snorkel at a depth of more than just a few feet.

By the way, you don't need to concern yourself with the pressure or density of the air. What matters here is the gauge pressure, which is the pressure experienced by the diver over and above that of the air he's trying to breath through the snorkel.

- Warren


Just curious, how would you solve it differently then? In my solution the atmospheric pressures cancel out any way, so would your solution be the same as mine?
 
this may sound weird, but how or where did you get "101325" from?
 
louie3006 said:
this may sound weird, but how or where did you get "101325" from?

1 Atm = 101325 Pascals
 
This explains why when I tried to use one of those "noodles" (flotation devices for pools and lakes , mostly for fun not safety) which was about 5 - 6 feet long (Looks like a giant straw) I felt the air being sucked out of me no matter how hard I tried.
 
bpw91284 said:
Just curious, how would you solve it differently then? In my solution the atmospheric pressures cancel out any way, so would your solution be the same as mine?

Yep, the fact that the atmospheric pressure cancels out is evidence that it doesn't matter in the first place. :smile:

- Warren
 
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