# Fluid problem, pressure in lungs

#### kahwawashay1

"A novice scuba diver practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth L and swimming to the surface. He ignores instructions and fails to exhale during his ascent. When he reaches the surface, the difference between the external pressure in his lungs is 9.3kPa. From what depth does he start?"

Ok so at depth L (starting point), the diver has pressure Pa1 in his lungs and an external pressure Pf=PatmospherewatergL

As he gets to the surface, the external pressure decreases on the lungs, so the lungs are allowed to expand further, and since he doesnt exhale, the number of air molecules in the lungs remain the same, and since the volume becomes bigger, the pressure in the lungs must drop to Pa2. So, at the surface, Patmosphere-Pa2=9.3kPa

and so I have three unknowns and two equations?
I was thinking maybe the pressure in the lungs stays the same, but thats only if you assume that his lungs cant stretch? But in this case his lungs could rupture, which i think is the moral of the problem, so if lungs can rupture then they can stretch and so cannot have equal pressures at depth L and at surface?

So if the pressure inside the lungs is the same at the beginning and end, then when the diver is at depth L, since he expands his lungs to the max, then the air in his lungs must exert the same pressure against the water as the water exerts against the lungs, right?

So then Patmosphere-Pa2=9.3kPa

Pf=PatmospherewatergL

Pa2=Pa1=Pf

so comes out to L=0.95 m ...

1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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#### kahwawashay1

update:

I found solutions manual and 0.95 is the right answer. So can someone just please explain how the pressure in the lungs is the same at both points? why the lung pressure does not decrease as the diver goes up?

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