Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 41: 135


\[\underline{18.2\text{ atm}}\]

Work Step by Step

The area of the face mask is \[125\text{ c}{{\text{m}}^{2}}\]. \[1\text{ c}{{\text{m}}^{2}}={{10}^{-4}}\text{ }{{\text{m}}^{2}}\] Thus, the area of the face mask in meters is \[125\times {{10}^{-4}}\text{ }{{\text{m}}^{2}}\]. The relation between pressure and force is as follows: \[P=\frac{F}{A}\] Here, P is pressure, F is force applied, and A is area. Thus, pressure is calculated as follows: \[\begin{align} & P=\frac{2.31\times {{10}^{4}}\text{ N}}{125\times {{10}^{-4}}\text{ }{{\text{m}}^{2}}} \\ & =1.85\times {{10}^{6}}\text{ N/}{{\text{m}}^{2}} \end{align}\] Now, \[1\text{ Pa}=1\text{ N/}{{\text{m}}^{2}}\] and \[1\text{ atm}=101,325\text{ Pa}\] So, \[\begin{align} & P=\left( 1.85\times {{10}^{6}}\text{ N/}{{\text{m}}^{2}} \right)\left( \frac{1\text{ atm}}{101325\text{ N/}{{\text{m}}^{2}}} \right) \\ & =18.2\text{ atm} \end{align}\] The pressure on the face mask is \[\underline{18.2\text{ atm}}\].
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