Question

A person takes 15 breathes per minute.The volume of air inhaled in each breath is 448ml and contains 21% oxygen by volume.The exhaled air contains 16%of oxygen by volume.If all the oxygen is used in the combustion of sucrose,how much of the Sucrose is burnt in the body per day& how much heat is evolved.$△H_{com}$ of sucrose is =$-6000kjmo{l}^{-1}$.take temperature to be 300 K throughout.[Assuming $V_{inhaledair}=V_{exhaledair},P_{atm}=1atm]$

• A. 7.822 Mj/day evolved,460 gm
• B. 7.822 Mj/day evolved,560 gm
• C. 9.822 Mj/day evolved,460 gm
• D. 9.822 Mj/day evolved,560 gm

Answer 1

The correct option is D 9.822 Mj/day evolved,560 gm

From given data,
$O_2$ involved in one breath$=\frac{448ml\times 21}{100}=94.08ml$
$O_2$ exhaled from one breath$=\frac{448ml\times 16}{100}=71.68ml$
$O_2$ used in one breath$=94.08-71.68=22.4ml$
$15$ breath in one minute, so total no. of breaths in $1$ day $=15\times 60\times 24=21600$
$O_2$ used in $1$ day $=21600\times 22.4ml=483840ml$
This $O_2$ used at ${27}^{o}C\left(300K\right)$
So, Volume of $O_2$ used at $0^{o}C=\frac{483840}{300}\times 273=440294.4ml$
$C_{12}{H}_{22}{O}_{11}+12O_2\to 12{CO}_2+11H_{2}O$
$\left(sucrose\right)$
Sucrose molecular mass $=342$
At $STP,22.4litre$ or $22400ml$ of $O_2$ is used in above reaction.
So $12\times 22400ml$ $O_2$ is used to burn $1mole$ or $342gm$ sucrose.
so, $440294.4ml$ $O_2$ is used to burn $xgm$ of sucrose
$x=\frac{342\times 440294.4}{12\times 22400}=560.19gm$
Given,
$\Delta H_{com}$ of sucrose $=-6000kJ/mol$
now, $440294.4ml$ $O_2$ with sucrose gives energy
$=\frac{6000\times 440294.4}{12\times 22400}=9828kJ/day=9.82MJ/day$