21. In the laboratory (Energy Expenditure &Exercise), we measured heart rate, ve

Question

21. In the laboratory (Energy Expenditure &Exercise), we measured heart rate, ventilation rate and volume, oxygen consumption, and carbon dioxide production during exercise. In principle, the rate of oxygen consumption (VO2) reflects the difference in oxygen between inspired air and expired air, recall that for simplicity, V here refers to rate (and should have a dot over it) and not volume. The volume of oxygen consumed equals the volume of air times the percent of air comprised of oxygen. The rate of oxygen consumption takes time into account. However, because temperature, pressure, and water vapor affect air volumes, measured air volumes are not all comparable to one another; to be comparable, they need to be corrected to a common set of conditions, such as STPD (standard temperature and pressure, dry). In simplified form, the equation for oxygen consumption is: where VE -the rate of expired air (volume/time), FIO2 -fractional concentration of inspired oxygen, and FEO2 fractional concentration of expired oxygen. FIO2 is constant in outside air (our lab software uses the value 0.2093). So: VO2-VESTPD (0.2093- FEO2). Outdoors, the fractional concentration of oxygen in the air is 0.2095. However, in a building in which people oxygen in the air. are consuming oxygen, we will use 0.2093 as the fractional concentration of The formula for correcting VE to STPD is: VESTD (VEATPS/time) x (Pb- PH2o)/760) x (273 /(273+ T)) where VEATes is the volume of expired air saturated and at ambient temperature and pressure, P is the barometric pressure, PH2o is the saturated water vapor pressure, 760 is the designated standard pressure (in mm Hg), 273 is degrees K, and T. is ambient temperature in degrees C. Units are not shown here for simplicity From these equations for VO2, it should be apparent that oxygen consumption is dependent on ventilation rate (VE) and oxygen extraction by the lungs (FIO2 -FEO2). Now use the equations above and the empirical values below (from our aerobic exercise lab experiment) to calculate VO2 (the rate of oxygen consumption). Be sure to show your calculations, including the units. Check that the units are right (it's easier than it might seem). VEATPS 53.7L Time interval-1 min Po = 767 mm Hg PH20 = 19.8 mm Hg T.- 22°C FEO2 0.163

Explanation / Answer

From the above given information & equations,

VE(stpd) = VE(atps) / time ×((P(b) - P(h2o)/760× (273/(273+T(a))).

Given that VE(atps) is 53.7 L.

Time is 1 min

P(b) is 767 mm Hg

P(h2o) is 19.8 mm Hg

T(a) is 22'c

FEO2 is 0.163.

So now VE(stpd) = (53.7/1) × (( 767- 19.8)/760) ×( 273/(273+22)).

VE(stpd)= 53.7 × ((747.2/760) ×( 273/ 295)).

VE(stpd) = 53.7× (0.9831) × (0.9254).

VE(stpd) = 48.85 litre/min & 0.814 in litre/second ( conversion of minute to seconds).

Now V(O2) = VE(stpd)( FIO2 - FEO2)

V(O2) = 48.85× (0.2093-0.163)

V(O2) = 48.85 × 0.0463 = 2.2617 Litre / min which is 226.17 millilitre/ min ( conversion of litre to millilitre).

Hence V(O2) is 226.17millilitre/min.

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