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

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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!

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)

## 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.