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 P_a1 in his lungs and an external pressure P_f=P_atmosphere+ρ_watergL

As he gets to the surface, the external pressure decreases on the lungs, so the lungs are allowed to expand further, and since he doesn't 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 P_a2. So, at the surface, P_atmosphere-P_a2=9.3kPa

and so I have three unknowns and two equations?
I was thinking maybe the pressure in the lungs stays the same, but that's only if you assume that his lungs can't 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 P_atmosphere-P_a2=9.3kPa

P_f=P_atmosphere+ρ_watergL

P_a2=P_a1=P_f

so comes out to L=0.95 m ...

Homework Statement

Homework Equations

The Attempt at a Solution

 

[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?