If barometric pressure is 480mmhg, what is the partial pressure of 02 in inspired air? At this pressure, what % of 02 would you have to gave in the air you are breathing so that the inspired p02 would be normal? ( 160 mmhg)
To determine the partial pressure of O2 in inspired air when the barometric pressure is 480 mmHg, we need to know the composition of air and the partial pressure of other gases present.
Air is composed mainly of nitrogen (N2), oxygen (O2), carbon dioxide (CO2), and some other trace gases. The composition of dry air is approximately 78% nitrogen, 21% oxygen, and 0.04% carbon dioxide.
To calculate the partial pressure of O2 in inspired air, we multiply the barometric pressure by the fraction of O2 in the air:
Partial pressure of O2 = Barometric pressure * Fraction of O2
Given that the barometric pressure is 480 mmHg and the fraction of O2 is 0.21 (21%), we can calculate the partial pressure of O2:
Partial pressure of O2 = 480 mmHg * 0.21 = 100.8 mmHg
So, the partial pressure of O2 in inspired air is approximately 100.8 mmHg when the barometric pressure is 480 mmHg.
Now, let's calculate the percentage of O2 needed in the air you are breathing so that the inspired pO2 is normal (160 mmHg).
We can use the following equation:
% O2 = (Partial pressure of O2 / Inspired pO2) * 100
Given that the inspired pO2 is 160 mmHg, we can substitute the values to find the percentage of O2:
% O2 = (100.8 mmHg / 160 mmHg) * 100 = 63%
Therefore, in order to have a normal inspired pO2 of 160 mmHg, you would need approximately 63% of O2 in the air you are breathing.