## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 1 - Exercises - Page 41: 135

#### Answer

$\underline{18.2\text{ atm}}$

#### 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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