When a person snorkels, the lungs are connected directly to the atmosphere through the snorkel tube and thus are at atmospheric pressure 1) In atmospheres, what is the difference$\mathbf{∆}\mathit{p}$between this internal air pressure and the water pressure against the body if the length of the snorkel tube is 20 cm (standard situation) &2) In atmospheres, what is the difference$\mathit{֏}\mathit{p}$between this internal air pressure and the water pressure against the body if the length of the snorkel tube is 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. As depicted in Figure, an elephant can safely snorkel through its trunk while swimming with its lungsbelow the water surface because the membrane around its lungs contains connective tissue that holds and protects the blood vessels, preventing rupturing.

Step by step solution

01

The given data

Density of water is$p=998\mathrm{kg}/{\mathrm{m}}^3$

02

Understanding the concept of hydrostatic pressure

We can use the formula for hydrostatic pressure in terms of density of water, gravitational acceleration, and length of the snorkel tube to find the pressure difference. Hydrostatic pressure is that, which is exerted by a fluid at any point of time at equilibrium due to the force applied by gravity.

Formula:

Net pressure applied on a fluid surface,$∆p=pgh$ (i)

03

a) Calculation of pressure difference between the internal air and water pressure against the body if snorkel tube is 20 cm

Using equation (i) and the given values, the pressure difference can be given as:

$$\begin{gathered} ∆p=998\mathrm{kg}/{\mathrm{m}}^3\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.2\mathrm{m} \\ =1956.08\mathrm{Pa} \\ =1956.08\mathrm{Pa}\times \frac{1\mathrm{atm}}{1.01\times {10}^2\mathrm{Pa}}(\because 1\mathrm{atm}=1.01\times {10}^5\mathrm{Pa}) \\ =0.019\mathrm{atm} \end{gathered}$$

Hence, the atmospheric pressure is 0.019 atm

04

b) Calculation at pressure difference between the internal air and water pressure against the body if snorkel tube is 4.0 m

Using equation (i) and the given values, the pressure difference can be given as:

$$\begin{gathered} ∆p=998\mathrm{kg}/{\mathrm{m}}^3\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 4\mathrm{m} \\ =39121.6\mathrm{Pa} \\ =39121.6\mathrm{Pa}\times \frac{1\mathrm{atm}}{1.01\times {10}^2\mathrm{Pa}}(\because 1\mathrm{atm}=1.01\times {10}^5\mathrm{Pa}) \\ =0.39\mathrm{atm} \end{gathered}$$

Therefore, the atmospheric pressure is 0.39 atm

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!