What is the partial pressure of O2 in the lungs if the atmospheric pressure (total pressure) is 755mmHG, the PN2 is 561mmHg, PCO2 is 0.29mmHg, and PH2O in the lungs is 41mmHg?
Ptotal = sum of partial pressures.
755 = pN2 + pH2O + pCO2 + pO2
Substitute the numbers and solve for pO2 in mm Hg.
Post your work if you get stuck.
So, am I adding all of them and then divided by 755?
No.
755 = 561 + 41 + 0.29 + pO2
Solve for pO2 of substitute X for pO2 and solve for X
To determine the partial pressure of O2 in the lungs, you need to subtract the partial pressures of other gases (N2 and CO2) and the pressure exerted by water vapor (H2O) from the total pressure.
The given information includes:
Total pressure (Patm) = 755 mmHg
PN2 (partial pressure of N2) = 561 mmHg
PCO2 (partial pressure of CO2) = 0.29 mmHg
PH2O (pressure exerted by water vapor) = 41 mmHg
To find the partial pressure of O2, you can use the formula:
PO2 = Patm - PN2 - PCO2 - PH2O
Substituting the given values into the formula:
PO2 = 755 mmHg - 561 mmHg - 0.29 mmHg - 41 mmHg
Calculating the above expression:
PO2 = 152.71 mmHg
Therefore, the partial pressure of O2 in the lungs is approximately 152.71 mmHg.