Part A) Describe the trend in mean oxygen concentration seen in the data entered
ID: 70197 • Letter: P
Question
Part A) Describe the trend in mean oxygen concentration seen in the data entered in Table 1.
Part B)
Air is humidified as it passes through the anatomical dead space, becoming fully saturated with water. If 6.3 kPa (47 mm Hg) of H2O is added, it displaces the equivalent pressure of inhaled air. This must be considered when calculating the partial pressure of oxygen. Calculate (to the nearest whole number) the partial pressure of oxygen for each of the runs in Part I using the following formula [use 101.3 kPa (760 mm Hg)]:
Partial Pressure [O 2 ] = (Barometric Pressure – 6.3 kPa)(Mean Oxygen Concentration)
(a) Run 1
(b) Run 2
(c) Run 3
Part C)
In normal breathing, the addition of CO2 and H2O to air in the alveoli reduces the partial pressure of oxygen to approximately 13.3 kPa (100 mm Hg).
Subtract the partial pressure of O2 that you calculated for Run 3 from the value you calculated for Run 1 in question 4.
Part D)
The average amount of air breathed in and out during normal respirations is approximately 500 mL (this is called tidal volume). The volume of air inhaled and exhaled during deep breaths is approximately 4000 mL (this is called vital capacity).
Use these values to explain the difference in trends seen in Table 1 and Table 2.
Explanation / Answer
Part A) Describe the trend in mean oxygen concentration seen in the data entered in Table 1.
Answer) Here in this table we can see the mean O2 concentration (%) is high in case deep breathing when compared to normal breathing, the use of PVC pipe has resulted in less mean O2 concentration (%) in normal breathing. But the use of 15 cm pipe has resulted in increased mean O2 concentration in deep breathing, but the use of 30cm PVC pipe has resulted slightly less mean O2 concentration. So with 15cm long PVC pipe is giving maximum mean O2 concentration in deep breathing.
Part B)
Normal Breathing:
Run 1: For CO2-O2 Tee alone, Partial pressure [O2]=(101.3-6.3kPa)(16.39)=1557 kPa
For CO2-O2 Tee+15 cm PVC pipe, Partial pressure [O2]=(101.3-6.3kPa)(15.10)=1434 kPa
For CO2-O2 Tee+30 cm PVC pipe, Partial pressure [O2]=(101.3-6.3kPa)(13.76)=1307 kPa
Deep Breathing:
Run 1: For CO2-O2 Tee alone, Partial pressure [O2]=(101.3-6.3kPa)(17.31)=1644 kPa
For CO2-O2 Tee+15 cm PVC pipe, Partial pressure [O2]=(101.3-6.3kPa)(16.95)=1610 kPa
For CO2-O2 Tee+30 cm PVC pipe, Partial pressure [O2]=(101.3-6.3kPa)(16.32)=1550 kPa
Part C)
The addition of CO2 and H2O to air will reduce the partial pressure of O2 by 13.3 kPa
Run 1: For CO2-O2 Tee alone, Partial pressure [O2]=(101.3-13.3kPa)(16.39)=1442 kPa
For CO2-O2 Tee+30 cm PVC pipe, Partial pressure [O2]=(101.3-13.3kPa)(13.76)=1211 kPa
=1442-1211=231
Normal Breathing:
For CO2-O2Tee alone, Partial pressure [O2]= 1557 kPa
For CO2-O2 Tee+30 cm PVC pipe, Partial [O2]=1307 kPa
=1557-1307=250 kPa
Yes, it will affect the amount of oxygen bound to hemoglobin. With increasing pressure, the oxygen saturation is also increasing.
Part D)
In Table 1, in normal breathing the mean O2 concentration (%) obtained was less when compared to mean O2 concentration obtained in deep breathing. This shows the in deep breathing, the maximum inhalation of air has increased the intake of O2 level and with the attachment of PVC pipe it has further increased in deep breath, but in normal breath it has reduced.