Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 13 - Temperature and Kinetic Theory - General Problems - Page 388: 74

Answer

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Work Step by Step

Use the ideal gas law. The amount of gas, n, is constant, and so is the temperature. We now know from PV=nRT that the product PV is constant. The pressure at a depth of h meters is $P=P_o+\rho gh$, where $P_o$ is atmospheric pressure and $\rho$ is the density of water. $$(PV)_{under}=(PV)_{surface}$$ Solve for the volume at the surface. $$V_{surface}=V_{under}\frac{P_{under}}{P_{surface}}$$ $$ =V_{under}\frac{ P_o+\rho gh }{ P_o}$$ $$ =(5.5L)\frac{1.013\times10^5 Pa+(1000kg/m^3) (9.80m/s^2)(9m) }{1.013\times10^5 Pa }\approx 10L$$ The air takes up almost twice the volume that it did before, so quickly rising to the surface without exhaling would clearly not be advisable.
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