Question

The composition of air inhaled by a human is $21\%O_2$ by volume and $0.03\%C{O}_2$ and that of exhaled air is $16\%O_2$ and $4.03\%C{O}_2$. Assuming a typical volume of $6200{\text{L day}}^{-1}$, moles of $O_2$ used and $C{O}_2$ generated by the body at $37{}^{\circ }\text{C}$ and $1.00bar$ are:
(Assume ideal behaviour)

• A. $\begin{array}{cc}\text{Moles of}{O}_2\text{used}& \text{Moles of}C{O}_2\text{generated}\\ 9.64& 9.64\end{array}$
• B. $\begin{array}{cc}\text{Moles of}{O}_2\text{used}& \text{Moles of}C{O}_2\text{generated}\\ 12.04& 12.04\end{array}$
• C. $\begin{array}{cc}\text{Moles of}{O}_2\text{used}& \text{Moles of}C{O}_2\text{generated}\\ 12.04& 9.64\end{array}$
• D. $\begin{array}{cc}\text{Moles of}{O}_2\text{used}& \text{Moles of}C{O}_2\text{generated}\\ 9.64& 12.04\end{array}$

Answer 1

The correct option is C $\begin{array}{cc}\text{Moles of}{O}_2\text{used}& \text{Moles of}C{O}_2\text{generated}\\ 12.04& 9.64\end{array}$

We know,
$PV=nRT$
$\therefore n=\frac{PV}{RT}$
Given,
$P=1\text{bar},T=37+273=301K,R=0.083{\text{bar L mol}}^{-1}{K}^{-1}$

$\text{Moles of}{O}_2\text{used}=\text{Moles of}\left[O_2\right(\text{inhaled})-O_2(\text{exhaled}\left)\right]$
$\therefore \text{Moles of}{O}_2\text{used}=\frac{1\times 6200}{0.083\times 310}[\frac{21}{100}-\frac{16}{100}]=12.04\text{mol}$

$\text{Moles of}C{O}_2\text{generated}=\text{Moles of}\left[C{O}_2\right(\text{exhaled})-C{O}_2(\text{inhaled}\left)\right]$
$\therefore \text{Moles of}C{O}_2\text{generated}=\frac{1\times 6200}{0.083\times 310}\left[\frac{4.03-0.03}{100}\right]=9.64\text{mol}$