General Chemistry

Question

A scuba diver that ascends to the surface too quickly can experience decompression sickness, which occurs when nitrogen that dissolves in the blood under high pressure, forms bubbles as the pressure decreases during the ascent. Therefore an understanding of the gas laws is an important part of a scuba diver's training. In fresh water the pressure increases by 1 atm every 34 ft below the water surface a diver descends. If a diver ascends quickly to the surface from a depth of 68 ft without exhaling, by what factor will the volume of the diver\'s lungs change upon arrival at the surface? Assume the atmospheric pressure at the surface of the water is 1 atm.

Answer 1

The pressure at 68 ft below the surface will be 1 atm original pressure + 1 atm/34 ft x 68 ft = 2 atm = 3 atm

Looking at it another way, when the diver reaches 34 ft. the pressure is 2 atm. When he/she reaches 68 ft, it is 3 atm, and so on.

Now the question becomes, how will the volume of the diver's lungs change in going from 3 atm pressure to 1 atm pressure (the pressure at the surface)? Assuming a constant temperature, we are now dealing with Boyle's Law where P1V1 = P2V2. This relationship tells us that if the pressure goes from 3 atm (P1) to 1 atm (P2), the volume (V2) has to become 3x the volume that it was at 3 atm. Or (3 atm)(V1) = (1atm)(V2)

V2/V1 = 3/1