Physics 1028 Chapter Notes - Chapter 12: Hot Air Balloon, Operating Temperature, Molar Mass

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Published on 24 Sep 2020
School
Western University
Department
Physics
Course
Physics 1028
Professor
Selection of Even–Numbered Problem Solutions
Chapter 12
P–12.2
First we consider the diver at a depth of 7.0 m below the water surface without the snorkel tube. The external
pressure on the diver’s body is given by Pascal’s law. We assume a fresh water lake with water density ρ =
1.0 g/cm³:
(1)
Note that we did not identify a value for the air pressure patm in this equation.
While diving to this depth, the diver’s body adjusts to the pressure p in this equation by being slightly
compressed until the internal pressure is in equilibrium with the external water pressure. This includes an
increase in the gas pressure in the lungs and an increase in the blood pressure.
If the diver now tries to use a 7 m long snorkel to connect to the air above the water surface in order
to breathe, the pressurized air in the lungs gets expelled upward through the tube to the lower pressure
airspace, which causes the pressure in the lungs to drop to air pressure patm. That leads to a pressure difference
between the inside of the lungs and the water outside the diver’s chest of:
(2)
This difference corresponds roughly to 70% of atmospheric pressure, which is sufficient to collapse the lungs
and force the still pressurized blood into the lungs. This is called a lung squeeze and is not a nice way to die!
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P–12.4
The pressure increase equals the applied force divided by the area. Note that the direction of the force is in
this case perpendicular to the piston surface, and thus, the pressure definition p = F/A is sufficient. The area
of a piston, sealing the opening of a cylinder, is a circular disc with A = π r² where r is the inner radius of the
cylinder. Thus:
(3)
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Document Summary

First we consider the diver at a depth of 7. 0 m below the water surface without the snorkel tube. The external pressure on the diver"s body is given by pascal"s law. We assume a fresh water lake with water density = Note that we did not identify a value for the air pressure patm in this equation. While diving to this depth, the diver"s body adjusts to the pressure p in this equation by being slightly compressed until the internal pressure is in equilibrium with the external water pressure. This includes an increase in the gas pressure in the lungs and an increase in the blood pressure. That leads to a pressure difference between the inside of the lungs and the water outside the diver"s chest of: This difference corresponds roughly to 70% of atmospheric pressure, which is sufficient to collapse the lungs and force the still pressurized blood into the lungs.

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