The physical fitness of athletes is measured by "VO2max

Question

," which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a VO2max
of 45 mL
O2/kg
body mass/min
, but a world-class male athlete can have a VO2max
reading of 88.0 mL
O2/kg
body mass/min.
Calculate the volume of oxygen, in mL
, consumed in 1 hr
by an average man who weighs 184 lbs
and has a VO2max
reading of 48.0 mL
O2/kg
body mass/min.

Answer 1

To calculate the volume of oxygen consumed in 1 hour by the average man, we first need to convert his body weight from pounds to kilograms:

184 lbs = 184 lbs x 1 kg / 2.2046 lbs ≈ 83.46 kg

Next, we calculate the volume of oxygen consumed in 1 minute:

VO2 = 48.0 mL O2/kg body mass/min x 83.46 kg = 3997.28 mL O2/min

Finally, we convert this value to mL consumed in 1 hour:

3997.28 mL O2/min x 60 min/hr = 239,836.8 mL O2/hr

Therefore, the average man would consume approximately 239,836.8 mL of oxygen in 1 hour.