Question

The normal respiratory rate for a human being is 15.0 breaths per minute. The average volume of air for each breath is 505 cm3 at 20◦C and 9.95 × 104 Pa. What is the volume of air at STP that an individual breathes in one day? Give your answers in cubic meters.

Answer 1

Answer : The volume of air at STP that an individual breathes in one day is $9.98m^3/day$

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of air = $9.95\times 10^4Pa$

$P_2$ = final pressure of air at STP = 1 atm  = $101.325\times 10^3Pa$

$V_1$ = initial volume of air = $505cm^3=505mL=0.505L$

$V_2$ = final volume of air at STP = ?

$T_1$ = initial temperature of air = $20^oC=273+20=293K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{(9.95\times 10^4Pa)\times (0.505L)}{293K}=\frac{(101.325\times 10^3Pa)\times V_2}{273K}$

$V_2=462L=462cm^3$

The volume for a single breath at STP is, $462cm^3$

Now we have to determine the volume of air at STP that an individual breathes in one day.

To determine the volume of air inhaled in one day, use the number of breaths per minute. Now convert minutes into hours. Then, convert hours into days. Finally, convert cubic centimeters into cubic meters, we get:

conversion used :

$\frac{462cm^3}{1\text{ breathe}}\times \frac{15.0\text{ breathe}}{1min}\times \frac{60min}{1hr}\times \frac{24hr}{1day}\times \frac{1m^3}{100cm^3}$

= $9.98m^3/day$

Therefore, the volume of air at STP that an individual breathes in one day is $9.98m^3/day$