Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 18 - Exercises and Problems - Page 343: 58

Answer

$.026 \ J$

Work Step by Step

We will call the process of inflating the lung isothermal, for the temperature is constant. Thus, the work done is equal to the area under the curve, which we approximate to be: $W = (.5)(1.5\ L)(.035 \ Pa)=.026 \ J$
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