Question

An athlete takes 20 breathes per minute at room temperature. The air inhaled in each breath is 200 mL which contains 20% oxygen by volume, while exhaled air contains 10% oxygen by volume. Assuming that all the oxygen consumed is used for converting glucose into $C{O}_2$ and $H_{2}O\left(l\right)$, how much glucose will be burnt in the body in 1 hour and what is the heat produced?

(Room temperature $={27}^{\circ }C$ and enthalpy of combustion of glucose is $-2822.5kJmo{l}^{-1}at{0}^{\circ }C$)

• A. a) $180g$ b) $=458.66kJheat$
• B. a) $360g$ b) $=358.66kJheat$
• C. a) $280g$ b) $=478.66kJheat$
• D. None of the above

Answer 1

The correct option is A a) $180g$ b) $=458.66kJheat$

$C_6{H}_{12}{O}_6+6O_2\to 6O_2+6H_{2}O$
In 1 min, the volume of oxygen used for the combustion of glucose is $20\times 200\times \frac{10}{100}=400mL$
In 1 min, the volume of oxygen used for the combustion of glucose is $60\times 400=24000mL=24L$
The number of moles of oxygen consumed are $n=\frac{PV}{RT}=\frac{1\times 24}{0.0821\times 300}=0.97\simeq 1mol$
1 mole of glucose requires 6 moles of oxygen.
Hence, the number of moles of glucose that undergo combustion are $\frac{1}{6}$
The mass of glucose that undergoes combustion in 1 hour is $\frac{180}{6}=30g$
The heat produced is $\frac{-2822.5}{6}=470.4kJ$