Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 18 - Exercises and Problems - Page 331: 35


a) 1.49 mm b) $1.07\times10^{-7}\ J$

Work Step by Step

a) Atmospheric pressure is 760 mmHg. Thus, we find that these pressures, which are measured in terms of how much more than atmospheric pressure they are, can be given by: $P_0=840 \ mmHg$ $P_f=885 \ mmHg$ Since volume and pressure are inversely related, it follows: $ d =2 \sqrt[3] {\frac{840(.76^3)}{885}}=1.49 \ mm $ b) We know the following equation for the work done: $W = P_1(\frac{4}{3}\pi r^3)ln(\frac{P_1}{P_2})$ Thus, converting 840 mmHg to 111,963 Pa, it follows: $W = (111,963)(\frac{4}{3}\pi (.00076)^3)ln(\frac{885}{840})=1.07\times10^{-7}\ J$
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