Question

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, failing to exhale during his ascent. At the surface, the difference $\Delta $ p between the external pressure on him and the air pressure in his lungs is $9.3kPa$ . From what depth does he start? What potentially lethal danger does he face?

Answer 1

Hint: Here in this question, it is said that The diver is diving at a distance I in the swimming pool. As he intake oxygen from the tank and left his tank beyond the pool. After which he comes upwards in the pool so a pressure difference takes place. Here our external pressure decreased because we were moving in the upwards direction in the pool but the diver would not be able to exhale at that time because the pressure on his lungs got constant. Now, here we have to find the depth by which the diver is diving upwards.



Formula used:
$p = {p_ \circ } + \rho gL$

Complete step by step solution:
As we know that from the above conclusion that,
Here, when the diver fills his lungs at depth L, the air pressure inside his lungs is higher than usual because of the external pressure on him.
As we know the equation as,
$p = {p_ \circ } + \rho gL$
Where $\rho $ represents the density of water. The external pressure on him drops as he ascends, reaching atmospheric pressure ${p_ \circ }$ at the surface. His blood pressure also falls till it returns to the surface. The air pressure in his lungs, however, stays at the level it was at depth L since he does not exhale. The pressure differential $\Delta p$ at the surface is
$\Delta p = p - {p_ \circ } = \rho gL$
Here in this question, we need the value of $L$ so rearranging the formula as we need,
$L = \dfrac{{\Delta p}}{{\rho g}}$
As by putting all values from the question we get,
$L = \dfrac{{9.3 \times 1000}}{{1000 \times 9.81}}$
By doing further solution we get,
$L = 0.95m$
Hence, we get the depth where he starts from is $L = 0.95m$ .
This is not profound! However, the diver's lungs are enough to burst from the $9.3kPa$ pressure difference (or around $9\% $ of atmospheric pressure) and release air into the depressurized blood, which subsequently transports the air to the heart, killing the diver. There is no risk if the diver follows directions and exhales gradually as he ascends, allowing the pressure in his lungs to equalize with the external pressure.

Note: While solving such questions, always convert all the needed physical quantities into the same standard units in order to avoid any calculation mistakes and kilopascal is the bigger unit of pressure which is thousand times larger than a pascal.