V - volume or rate of flow of a gas. VO2 - volume of oxygen consumed or volume o
ID: 3522002 • Letter: V
Question
V - volume or rate of flow of a gas.
VO2 - volume of oxygen consumed or volume of oxygen
consumed per minute.
VCO2 - volume of carbon dioxide produced or volume of
carbon dioxide produced per minute.
VI - volume of air inspired or volume of air inspired
per minute
VE - volume of air expired or volume of air expired
per minute.
FEO2 - fraction of oxygen in expired air.
FECO2 - fraction of carbon dioxide in expired air.
Please add formulas
1. While riding a stationary bicycle ergometer, the following steady-state values were obtained on a subject whose body weight was 74 kg:
VI = 40.16 L/min
FEO2 = 0.168
FECO2 = 0.04
Based on this information, calculate this subject's steady-state
(a) absolute VO2 in L/min
(b) relative VO2 in ml/kg/min
(c) VCO2 in L/min
2. While performing arm-crank ergometry, a patient's expired air was collected in a Douglas bag. The collection period was 40 seconds and the patient's weight was 85 kg. The following values were obtained on the air in the Douglas bag:
V = 12.0 L
FEO2 = 0.171
FECO2 = 0.036
Based on this data for this 40-second period, calculate the subject's
(a) VE in L/min
(b) absolute VO2 in L/min
(c) relative VO2 in ml/kg/min
(d) VCO2 in L/min
Explanation / Answer
1. While riding a stationary bicycle ergometer, the following steady-state values were obtained on a subject whose body weight was 74 kg:
VI = 40.16 L/min
FEO2 = 0.168
FECO2 = 0.04
Based on this information, calculate this subject's steady-state
(a) absolute VO2 in L/min
FEN2 = 1 - 0.168 – 0.04
FEN2 = 0.792
Absolute VO2 = VI x (0.2093 – [0.7904 x FEO2 / FEN2])
Absolute VO2 = 40.16 x (0.2093 – [0.7904 x 0.168 / 0.792])
Absolute VO2 = 1.67 L/min
b) Relative VO2 = (1.67 L/min / 74 kg) x 1000 ml/L
Relative VO2 = 22.57 ml/kg/min
(c) VCO2 in L/min
VCO2 = VI x ([0.7903 x FECO2 / FEN2] – 0.0003)
VCO2 = 40.16 x ([0.7903 x 0.04 / 0.792] – 0.0003)
VCO2 = 1.6 L/min
2.
V = 12.0 L
FEO2 = 0.171
FECO2 = 0.036
Based on this data for this 40-second period, calculate the subject's
(a) VE in L/min
VE in 40 sec = 12 L so in 1 min VE = (12/40)*60 = 18 L/min
b)
FEN2 = 1 - 0.171 – 0.036
FEN2 = 0.793
VI = (VE x FEN2)/FIN2
VI = (18x0.793)0.7904
VI= 18.06 L/min
Absolute VO2 = VI x (0.2093 – [0.7904 x FEO2 / FEN2])
Absolute VO2 = 18.06 x (0.2093 – [0.7904 x 0.171 / 0.793])
Absolute VO2 = 0.702 L/min
b) Relative VO2 = (0.702 L/min / 85 kg) x 1000 ml/L
Relative VO2 = 8.25 ml/kg/min
(c) VCO2 in L/min
VCO2 = VI x ([0.7903 x FECO2 / FEN2] – 0.0003)
VCO2 = 18.06 x ([0.7903 x 0.04 / 0.793] – 0.0003)
VCO2 = 0.7145 L/min
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