Chapter 13, Problem 45P

To determine

To Calculate:The number of molecules in each breath.

Answer to Problem 45P

The number of molecules per breath is $19\text{molecules/breath}$ .

Explanation of Solution

Given info:

The volume, $V=2.0\text{L}$

Height, $h=10\text{km}=1.0\times {10}^4\text{m}$

Formula used:

According to ideal gas equation,

  $n=\frac{PV}{RT}$

Here, $P$ is the pressure,

  $V$ is the volume of breath,

  $n$ is the number of molecules,

  $R$ is the universal gas constant,

  $T$ is the Temperature.

Write the expression for atmospheric volume.

Atmospheric volume, $\text{V}\text{=}\text{4π}{R}_{\text{earth}}{}^{2}h$

Here, ${\text{R}}_{\text{earth}}$ is the radius of Earth,

  $h$ is the height.

Write the expression for Galileo molecules per meter cube.

  $\frac{\text{Galileo molecules}}{{\text{m}}^{\text{3}}}\text{=}\frac{\text{number}\text{of}\text{molecules}}{\text{volume}\text{of}\text{atmosphere}}$

Calculation:

The number of Galileo molecules in one of our breaths by following the ideal gas equation such that,

  $\begin{array}{c}n=\frac{\text{PV}}{\text{RT}}\\ =\frac{1.01\times {10}^5\text{Pa}\times 2\times {10}^{-3}{\text{m}}^3}{\left(1.38\times {10}^{-23}\text{J/K}\right)\times \left(300\text{K}\right)}\\ =4.9\times {10}^{\text{22}}\text{molecules}\end{array}$

The atmospheric volume is,

  $\begin{array}{l}{\text{V=4πR}}_{\text{earth}}{}^{\text{2}}\text{h=4×3}\text{.14×}{\left(\text{6}{\text{.38×10}}^{\text{6}}\text{m}\right)}^{\text{2}}\text{×}\left(\text{1}{\text{.0×10}}^{\text{4}}\text{m}\right)\\ \text{=}\text{5}{\text{.1×10}}^{\text{18}}{\text{m}}^{\text{3}}\end{array}$

Now, calculate the Galileo molecules per meter cube

  $\frac{\text{Galileo molecules}}{{\text{m}}^{\text{3}}}\text{=}\frac{\text{4}{\text{.9×10}}^{\text{22}}\text{molecules}}{\text{5}{\text{.1×10}}^{\text{18}}{\text{m}}^{\text{3}}}\text{=}\text{9}{\text{.6×10}}^{\text{3}}{\text{molecules/m}}^{\text{3}}$

The number of molecules per breath is,

  $\begin{array}{l}\frac{\text{Galileo molecules}}{\text{breath}}\text{=}\text{9}{\text{.6×10}}^{\text{3}}{\text{molecules/m}}^{\text{3}}\text{×}\frac{\text{2}{\text{.0×10}}^{\text{-3}}{\text{m}}^{\text{3}}}{\text{1}\text{breath}}\\ \text{=}\text{19}\text{molecules/breath}\end{array}$

Conclusion:

The number of molecules per breath is $\text{19}\text{molecules/breath}$ .