Question

A person takes 15 breaths 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. 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$= {\text{ }} - 6000{\text{ kj mo}}{{\text{l}}^{{\text{ - 1}}}}$ burnt in the body per day & how much heat is evolved. $\Delta {H_{com}}$ Of sucrose is. Take temperature to be throughout. [Assuming${{\text{V}}_{{\text{inhaler}}}}{\text{ = }}{{\text{V}}_{{\text{exhaler,}}}}{{\text{P}}_{{\text{atm}}}}{\text{ = 1atm}}$]
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

For the given question 1st calculate the amount of oxygen required to burn 1 mol of sucrose and after that determine the amount of energy which will be produced by the combustion of oxygen and sucrose.

Complete step by step answer:
From given data we can state the following,
${O_2}$​ involved in one breath:
\Rightarrow \dfrac{{{\text{volume of breathe inhaled in each breathe }} \times {\text{ oxygen%}}}}{{100}} \\\ \Rightarrow \dfrac{{448 \times 21}}{{100}} = 94.08{\text{ ml}} \\\
${O_2}$​ Exhaled from one breath:
\Rightarrow \dfrac{{{\text{volume of breathe exhaled in each breathe }} \times {\text{ oxygen%}}}}{{100}} \\\ \Rightarrow \dfrac{{448 \times 16}}{{100}} = 71.68{\text{ ml}} \\\
Therefore, ${O_2}$​ used in one breath = Volume inhaled – Volume exhaled = 94.08 −71.68 = 22.4 ml
As per the given question, 15 breath in one minute, So total number of breaths in 1 day
= 15 × 60 × 24 = 21600
Total amount of oxygen used in 1 day = Total number of breath × Volume of oxygen used in one breath = 21600 × 22.4ml = 483840ml
This ${O_2}$​ is used at $27^\circ C$ (300K)
So, Volume of ${O_2}$​ used at $0^\circ C$will be
$= \dfrac{{483840}}{{300}} \times 273 = 440294.4{\text{ ml}}$
The following reaction takes place while combustion of sucrose:
${C_{12}}{H_{22}}{O_{11}} + 12{O_2} \to 12C{O_2} + 11{H_2}O$
(Sucrose)
Molecular mass of sucrose is = 342 gm
At STP, 22.4 litre or 22400ml of ${O_2}$​ will be used in the above reaction.
So, in the above reaction 12 moles of oxygen is utilized
Therefore, 12×22400ml ${O_2}$​ is utilized to burn 1 mole or 342 gm sucrose.
Let 440294.4 ml ${O_2}$​ is used to burn “X” gm of sucrose
Therefore,
Amount of sucrose burnt:
$X = \dfrac{{{\text{Molecular mass of sucrose }} \times {\text{ volume of }}{{\text{O}}_2}}}{{{\text{Number of moles of }}{{\text{O}}_2}{\text{ }} \times {\text{ Volume at S}}{\text{.T}}{\text{.P}}}} \\\ \Rightarrow X = \dfrac{{342 \times 440294.4}}{{12 \times 22400}} = 560.19{\text{ gm}} \\\$
According to the given question:
$\Delta {H_{com}}$​ of sucrose = −6000 KJ/mol
Therefore, energy produced by 440294.4ml ${O_2}$​ with sucrose is:
$= \dfrac{{\Delta {{\text{H}}_{com}}{\text{ }} \times {\text{ Volume of oxygen}}}}{{{\text{Number of moles of }}{{\text{O}}_2}{\text{ }} \times {\text{ Volume at S}}{\text{.T}}{\text{.P}}}} \\\ = \dfrac{{6000 \times 440294.4}}{{12 \times 22400}} = 9828{\text{ kJ/day = 9}}{\text{.82 MJ/day}} \\\$
Therefore, the correct option is (D).

Note: Enthalpy of combustion: it is defined as the amount of heat energy produced on burning one mole of substance completely in the presence of oxygen. While solving the numerical the volume of oxygen should be calculated at 273K, the amount of oxygen should be taken in millilitres and the mass calculations should be done in grams, if not then conversions of units should be done.