Question

Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At sea level and at an average body temperature of 37 ℃, how many moles of oxygen do the lungs contain at the end of an inflation?

Answer 1

Concepts and reason

The concept required to solve this problem is ideal gas law.

First solve for the volume from the given percentage of the oxygen present in total volume. Finally, use the ideal gas law and rearrange it to calculate the number of moles of the oxygen.

Fundamentals

The ideal gas equation is,

$PV = nRT$

Here, P is the pressure, V the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

The gas molecule volume is 20% of the total volume that is,

$\begin{array}{c}\\V = \left( {20\% } \right)\left( {5.0{\rm{ L}}} \right)\\\\ = 1{\rm{ L}}\\\end{array}$

Rearrange the ideal gas equation to solve for n.

$\begin{array}{c}\\PV = nRT\\\\n = \frac{{PV}}{{RT}}\\\end{array}$

Use the rearrange ideal gas equation.

Substitute $1.01 \times {10^5}{\rm{ Pa}}$ for P, $1.0{\rm{ L}}$ for V, $37^\circ {\rm{C}}$ for $T$ and $8.314{\rm{ }}{{\rm{m}}^3} \cdot {\rm{Pa/K}} \cdot {\rm{mol}}$ for $R$ in the equation $n = \frac{{PV}}{{RT}}.$

$\begin{array}{c}\\n = \frac{{\left( {1.01 \times {{10}^5}{\rm{ Pa}}} \right)\left( {1.0{\rm{ L}}} \right)}}{{\left( {8.314{\rm{ Pa}} \cdot {\rm{L/K}} \cdot {\rm{moles}}} \right)\left( {37^\circ {\rm{ C}}} \right)}}\\\\ = \frac{{\left( {1.01 \times {{10}^5}{\rm{ Pa}}} \right)\left( {1.0{\rm{ L}}} \right)\left( {\frac{{1\,{{\rm{m}}^3}}}{{1000\,{\rm{L}}}}} \right)}}{{\left( {8.314{\rm{ }}{{\rm{m}}^3} \cdot {\rm{Pa/K}} \cdot {\rm{moles}}} \right)\left( {37 + 273{\rm{ K}}} \right)}}\\\\ = 0.039{\rm{ moles}}\\\end{array}$

Ans:

The number of moles of the oxygen in the lungs is 0.039 moles.

Answer 2

Never mind this question...i figured out my mistakes

Answer 3

Wait!! You might know how to do it, but I still don't. Help :(