Question

Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snorkel tube and thus are at atmospheric pressure. In atmospheres, what is the difference between his internal air pressure and the water pressure against the body if the length of the snorkel tube is (a) 20cm (standard situation) and (b) 4.0m (probably lethal situation)? In the latter, the pressure difference causes blood vessels on the walls of the lungs to rupture, releasing blood into the lungs. An elephant can safely snorkel through its trunk while swimming with its lungs 4.0m below the water surface because the membrane around its lungs contains connective tissue that holds and protects the blood vessels, preventing rupturing.

Answer 1

Hint: Recall the expression for pressure at depth h. Remember that here we are asked to find the difference in pressure at depth h with the pressure at the surface of the liquid. But we know that this pressure at the water surface is the atmospheric pressure, so, accordingly solve the question.

Formula used:
Expression for pressure at depth h,
$P=P_a+\rho gh$

Complete answer:
In the question we are asked to find the pressure difference $\mathrm{\Delta }P$ between the internal air pressure and water pressure against the body of a man snorkeling with a tube of height h. We are given two different situations, firstly a standard situation where the length of the snorkel tube is 20cm and secondly a probable lethal situation where the height of the tube is 4.0m.

We know that the pressure difference directly depends on the depth at which the man is present which is equal to the length h of the snorkeling tube.
Let the internal air pressure of the man be given by$P$, also, we know that the pressure at the surface of the water is the atmospheric pressure which is given by,$P_a$
The pressure at depth h below the water surface is given by,
$P=P_a+\rho gh$
$\Rightarrow P-P_a=\rho gh$
From this expression, we know that the pressure P at depth h below the surface of liquid whose mass density is $\rho$ is greater than the atmospheric pressure by $\rho gh$. This excess pressure at a certain depth of liquid surface is called Gauge pressure.
So basically we are asked to find the gauge pressure for two different lengths of snorkeling tube.
Mass density of water,
$\rho =100kg{m}^{-3}$
Acceleration due to gravity,
$g=9.8m{s}^{-1^{}}$
(a) Standard situation:
For height $h=20cm=0.2m$
The pressure difference $\mathrm{\Delta }P$ is given by,
$\mathrm{\Delta }P=\rho gh$
$\Rightarrow \mathrm{\Delta }P=1000\times 9.8\times 0.2$
$\Rightarrow \mathrm{\Delta }P=1960Pa$
But we are asked to find the pressure difference in atmospheres. We know that,
$1atm=101325Pa$
Therefore,
$\Rightarrow \mathrm{\Delta }P={\displaystyle \frac{1960}{101325}}atm$
$\Rightarrow \mathrm{\Delta }P=0.019atm$
Therefore, the pressure difference is 0.019atm for the standard situation given.
(b) Probable lethal situation:
For height $h=4.0m$
The pressure difference can be given by,
$\Rightarrow \mathrm{\Delta }P=1000\times 9.8\times 4.0$
$\Rightarrow \mathrm{\Delta }P=39200Pa$
$\Rightarrow \mathrm{\Delta }P={\displaystyle \frac{39200}{101325}}atm$
$\Rightarrow \mathrm{\Delta }P=0.39atm$
Therefore, the pressure difference is 0.39atm in the probable lethal situation given in the question.

Note:
We should always find the final answer in units as given in the question. If not specifically mentioned you could avoid doing that. In this question when we substitute the given values in the expression, we are getting the pressure difference in Pascal, but we have to convert it into atmospheres as per the requirement of the question.