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Ideal Gas Equation

The compositi...

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 $6200 {L day}^{- 1}$ , moles of $O_{2}$ used and $C O_{2}$ generated by the body at $37^{\circ} C$ and $1.00 b a r$ are:
(Assume ideal behaviour)

A

\begin{matrix} Moles of O_{2} used & Moles of C O_{2} generated 9.64 & 9.64 \end{matrix}

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B

\begin{matrix} Moles of O_{2} used & Moles of C O_{2} generated 12.04 & 12.04 \end{matrix}

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C

\begin{matrix} Moles of O_{2} used & Moles of C O_{2} generated 12.04 & 9.64 \end{matrix}

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D

\begin{matrix} Moles of O_{2} used & Moles of C O_{2} generated 9.64 & 12.04 \end{matrix}

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Solution

The correct option is C $\begin{matrix} Moles of O_{2} used & Moles of C O_{2} generated 12.04 & 9.64 \end{matrix}$
We know,
$P V = n R T$
$∴ n = \frac{P V}{R T}$
Given,
$P = 1 bar, T = 37 + 273 = 301 K, R = 0.083 {bar L mol}^{- 1} K^{- 1}$

$Moles of O_{2} used = Moles of [O_{2} (inhaled) - O_{2} (exhaled)]$
$∴ Moles of O_{2} used = \frac{1 \times 6200}{0.083 \times 310} [\frac{21}{100} - \frac{16}{100}] = 12.04 mol$

$Moles of C O_{2} generated = Moles of [C O_{2} (exhaled) - C O_{2} (inhaled)]$
$∴ Moles of C O_{2} generated = \frac{1 \times 6200}{0.083 \times 310} [\frac{4.03 - 0.03}{100}] = 9.64 mol$

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Similar questions

Q. The composition of air inhaled by a human is

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0.03 % C O_{2}

and that of exhaled air is

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4.03 % C O_{2}

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Ideal Gas Equation

Standard XIII Chemistry

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