Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 18 - A Macroscopic Description of Matter - Exercises and Problems: 45

Answer

(a) The pressure in the vacuum chamber is a fraction of $1.3\times 10^{-13}$ of the atmospheric pressure. (b) The number of molecules is $1.23\times 10^{11}$

Work Step by Step

(a) We can convert the pressure in the vacuum chamber to units of atm. $P = (1.0\times 10^{-10}~mm~Hg)(\frac{1~atm}{760~mm~Hg})$ $P = 1.3\times 10^{-13}~atm$ The pressure in the vacuum chamber is a fraction of $1.3\times 10^{-13}$ of the atmospheric pressure. (b) We can find the volume of the cylinder. $V = \pi~R^2~h$ $V = (\pi)(0.20~m)^2(0.30~m)$ $V = 0.0377~m^3$ We can find the number of molecules. $PV = NkT$ $N = \frac{PV}{kT}$ $N = \frac{(1.3\times 10^{-13}~atm)(\frac{1.013\times 10^5~Pa)}{1~atm})(0.0377~m^3)}{(1.38\times 10^{-23}~J/K)(293~K)}$ $N = 1.23\times 10^{11}~molecules$ The number of molecules is $1.23\times 10^{11}$.
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