If a scuba diver releases a 10-mL air bubble below the surface where the pressure is 3.5 atm, what is the volume (in mL) of the bubble when it rises to the surface and the pressure is 1.0 atm?

1 Answer
Aug 14, 2016

Answer:

Larger! This is a simple manifestation of Boyle's Law .

Explanation:

#P_1V_1=P_2V_2#

And thus #V_2=(P_1V_1)/P_2# #=# #(10*mLxx3.5*atm)/(1.0*atm)# #=# #35*mL#

This MARKED difference in volume illustrates the key rule in scuba diving, something the instructors hammer home to you from your very first lesson:

#"NEVER HOLD YOUR BREATH"#.

Many fatalities have occurred when scuba divers ascend without breathing out (even a 1-2 m ascent is dangerous); the pressure in their lungs can expand rapidly upon ascent. Of course at 30-40 m depths they are breathing air at 4-5 atmospheres. This is not so much a problem for those (few) free divers that can simply hold their breaths, as the gas in their lungs compresses upon descent and the volume of expansion is fixed.

Even when you are breathing properly at depth, divers can suffer disorientation and confusion due to the narcotic effects of nitrogen at pressure (i.e. "nitrogen narcosis").