Question

We know that standard pressure is one atmosphere, or 760 millimeters of mercury. This pressure results from the weight of gas molecules in the atmosphere. As a diver enters the water, he is subject to both water pressure and air pressure. Because water is much denser than air, the pressure increases rapidly as the diver descends. At the depth of 34 feet in fresh water, the diver is experiencing 2 atmospheres of pressure (one from air pressure and one from the 34 feet of water). For every additional 34 feet the diver descends he will be under an additional atmosphere of pressure.
Since water pressure is proportional to depth, how many atmospheres of pressure would a diver experience at 102 feet? Why wouldn't this pressure squash the diver? Answering this second question may be easier if you think of the reason a person on land is not squashed by one atmosphere of pressure. Explain your answer in detail.

Answer 1

old question but will respond so this can be closed

like the question states, at sea level the diver experiences 1 atm of air pressure

for every 34 feet, he gains an additional 1 atm

so to find out how many atm he gains by diving, we can divide 102 feet/34 feet to get total water pressure in atm. add 1 to this, to account for air pressure.

now, as to why the diver doesn't get "squashed" by all this pressure? well, the internal pressure inside our bodies "cancels out" the external air pressure so the net pressure is essentially neutral.

reference: http://scienceline.ucsb.edu/getkey.php?key=1711