Question

A person is exercising, burning 850 Cal/hr. While doing this exercise, they breathe deeply, so the volume they take into and out of their lungs is 2 L with each breath. 20% of the air inhaled is oxygen, and 16% of the air exhaled is oxygen. How fast do they have to breath? Answer will be exact.

15.1 Breaths/min

11.7 Breaths/min

22.0 Breaths/min

36.7 Breaths/min

53.0 Breaths/min

Answer 1

1 mole of Glucose liberates 686.4525 calories when it reacts with 6 moles of oxygen at STP, according to the chemical equation of resipiration:

C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O

Now, It is given that 2L of air is inhaled, out of which 20% is Oxygen. At STP, 1mole of O₂ occupies 22.4 Litre of volume.

Our oxygen content inhaled is 20% of 2L= 0.4L

Oxygen Content exhaled is 16% of 2L = 0.32L

Hence volume of oxygen that reacts with glucose is = 0.4L - 0.32L = 0.08L

0.08 L of Oxygen will contain 0.08/22.4 =0.003571 Moles of oxygen.

As per the chemical reaction above, it takes 6 moles of oxygen to react with 1 mole of glucose to liberate 686.4525 calories. This is the whole process in 1 breath.

Now we have 0.003571 moles of oxygen per breath. That will take 0.003571/6 = 0.0005952 Moles of Glucose.

If 1 mole of glucose liberates 686.4525 Calories, then 0.0005952 Moles will liberate ( 686.4525 * 0.0005952) = 0.4082026 calories per breath.

The person burns 850 cal/hr = 850/60 = 14.1667 Cal/min.

Breath rate =( 14.1667 Cal/min )/( 0.4082026 calories per breath ).= 36.7 Breaths/min