Question

The total lung capacity of a typical adult is $5.0L$. Approximately $20\%$of the air is oxygen. At sea level and at a body temperature of $37°C$, how many oxygen molecules do the lungs contain at the end of a strong inhalation?

Short answer

The lungs contain $\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}$oxygen molecules at the end of a strong inhalation.

Step-by-step solution

: Given information and Formula used

Given : The total lung capacity of a typical adult : $5.0L$.

Percentage of Oxygen in air : $20\%$.

Body temperature : $37°C$

Theory used :

The ideal gas law is given by $pV=nRT$

Solve this for $V$we get : $V=\frac{nRT}{P}$

Another form of the ideal gas law, which refers to the number of molecules$N$of the gas rather than the number of moles, is :

$pV=\frac{N}{N_A}RT$
Where $N_A$is the Avogadro number. Solving for $N$:

$N=\frac{p{N}_{A}V}{RT}$

Finding out how many oxygen molecules the lungs contain at the end of a strong inhalation.

The pressure must be expressed in pascals, and at sea level the pressure is $1atm$. The volume of oxygen is $20\%$of the total lung capacity. So we can find the volume of oxygen gas by converting the unit of volume from role="math" localid="1648402333826" $$\begin{gathered} L\mathrm{to}m³: \\ p=(1atm)\frac{1.01325x{10}^{5}Pa}{1atm} \\ =1.01325x{10}^{5}Pa \\ \mathrm{and} \\ V=0.2(5L)\frac{1m³}{1000L} \\ =1x{10}^{-3}m³ \end{gathered}$$
Step 2
Temperature must be expressed in Kelvin, so we need to convert its unit from Celsius such :
role="math" localid="1648402474692" $$\begin{gathered} T_K=T_C+273 \\ =37°C+273 \\ =310K \end{gathered}$$
Now we substitute the values for $N_A,P,V,R\mathrm{and}T$into equation $N=\frac{p{N}_{A}V}{RT}$to get :

$$\begin{gathered} N=\frac{(1.013\times {10}^{5}Pa)(6.02x{10}^{23})(1x{10}^{-3}m³)}{(8.31J/molK)(310K)} \\ =\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathrm{molecules} \end{gathered}$$