Question

question 11
air density is 1.29 kg/m³ at sea - level and comprises 23% o₂ by mass. suppose an adult human being breathes an average of 15 times per minute, and each breath takes 400 ml of air. (a) what mass of o₂ is inhaled each day? (b) how many o₂ molecules is this? (note: the mass of one o₂ molecule is 32 u. one (1) u = 1.66053892173×10⁻²⁷ kg). show your work for credit.

Explanation:

Step1: Calculate volume of air inhaled per day

First, find the volume of air inhaled per minute. Each breath is 400 mL = 0.4 L, and 15 breaths per minute, so volume per minute $V_{min}=15\times0.4 = 6$ L. In a day (24 hours with 60 minutes per hour), the volume of air inhaled per day $V = 6\times24\times60=8640$ L. Since 1 L = 10^{- 3}m^{3}, $V = 8.64$ m^{3}.

Step2: Calculate mass of air inhaled per day

Given air - density $
ho = 1.29$ kg/m^{3}, using the formula $m=
ho V$, the mass of air inhaled per day $m_{air}=
ho V=1.29\times8.64 = 11.1456$ kg.

Step3: Calculate mass of O₂ inhaled per day

Since air comprises 23% O₂ by mass, the mass of O₂ inhaled per day $m_{O_2}=0.23\times m_{air}=0.23\times11.1456 = 2.5635$ kg.

Step4: Calculate number of moles of O₂

The molar mass of O₂ is $M = 32$ g/mol = 0.032 kg/mol. Using the formula $n=\frac{m}{M}$, the number of moles of O₂, $n=\frac{2.5635}{0.032}=80.1094$ mol.

Step5: Calculate number of O₂ molecules

Using Avogadro's number $N_A = 6.022\times10^{23}$ molecules/mol, the number of O₂ molecules $N=n\times N_A=80.1094\times6.022\times10^{23}=4.8242\times10^{25}$ molecules.

Answer:

(a) The mass of O₂ inhaled each day is 2.56 kg (rounded to two - decimal places).
(b) The number of O₂ molecules is $4.82\times10^{25}$ molecules (rounded to two - decimal places).