   Chapter 10, Problem 40P

Chapter
Section
Textbook Problem

During inhalation, a person’s diaphragm and intercostal muscles contract, expanding the chest cavity and lowering the internal air pressure below ambient so that air flows in through the mouth and nose to the lungs. Suppose a person’s lungs hold 1 250 mL of air at a pressure of 1.00 atm. If the person expands the chest cavity by 525 mL while keeping the nose and mouth closed so that no air is inhaled, what will be the air pressure in the lungs in atm? Assume the air temperature remains constant.

To determine
The final pressure in the lungs.

Explanation

Given info: The initial pressure is Pi=1.0atm . The initial volume is Vi=1250mL . 525 mL is added to the volume.

The final volume ( Vf ) is,

Vf=(Vi+525)mL

At constant temperature,

PiVi=PfVf

On Re-arranging,

Pf=Pi(ViVf)

Substitute (Vi+525)mL for Vf

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