Question

In a basal metabolism measurement timed at 6 minutes, a patient exhaled $52.5L$ of air measured over water at $20$. The vapour pressure of water at ${20}^{\circ }C$ is $17.5torr$. the barometric pressure was $750torr$. The exhaled air analyzed $16.75$ volume % oxyegen and the inhald air $20.32$ volume % oxygen. Both on a dry basis neglecting any solubility of the gases in water and any difference in the total volumes of inhaled and exhaled air, calculate the rate of oxygen consumption by the patient in$mL$ (STP) per minute.

Answer 1

$P=750-17.5=732.5torr$

$T={20}^{\circ }C=293K$

Inhaled (35) $O_2=52.5\times 0.2032$

Exhaled (: 5) $O_2=52.5\times 0.1675$

so at STP $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\frac{732.5\times 52.5(0.2032-0.1675)}{293}=\frac{760\times V}{273}$

$\Rightarrow V=1.683L$ for 6 min.

oxygen consumption per min (4 2: 48 $5^{+}$ :)$=\frac{1.683}{6}=\frac{0.28L}{min}=280mL/min$