I. Answers B and E are true statements, whereas A, C, and D are false.
II. Answers A, D and E are true statements, whereas B and C are false.
1. PAO2 = [760 × 3 - (47 + 40)]; PAO2 = 2193 mmHg.
The PO2 of the blood leaving the lung capillaries is assumed to be equal to PAO2. The physically dissolved oxygen: (0.022 × 2193 × 1000/760) = 63.5 ml STPD per l.
This is more than the arterio-venous oxygen difference of a resting person (50 ml STPD per l).
2. The chemically bound oxygen is (170 × 1.39) = 236.3 ml STPD per l.
3. The CO2 transport from tissues to lungs suffers from lack of oxygen free haemoglobin and thus of carbamino-binding sites.
1. The weight of a 30 m column of sea water equals (height * relative density * the acceleration of gravity). The calculation is as follows:
(30 m * 1033 kg*m-3 * 9.807 m*s-2) = 303918.9 (kg m s-2) m-2.
This is 303918.9 N m-2 or 303.9 kPa.
The pressure at the surface of the sea is 1 atm.abs or 101.3 kPa.
Accordingly, the total pressure at 30 m of depth is 405.2 kPa or 4 atm.abs.
2. The sum of the 3 fractions of gasses in the alveolar air is always 1, whether the diver is at the surface or at the bottom of the sea.
At the surface: FACO2 = 40/(760-47) = 0.056 and FAO2 = 100/(760-47) =0.1403.
Thus, FAN2 = 1 - (0.056 + 0.1403) = 0.8038. PAN2 = 0.8038*(760-47) = 573 mmHg.
At the bottom, the diver is assumed to maintain his metabolic rate and the same alveolar fractions as already calculated. Thus, PAN2 = 0.8038*(405.2/0.1333 - 47) = 2993 mmHg.
3. Without step decompression the diver develops decompression sickness, so it is not advisable to drag him up without the used of decompression tables and the relevant steps.