I: Answers B, D, and E are true statements, whereas A and C are false.
II: Answers A, C, and D are true statements, whereas B and E are false.
1. The blood alcohol-permille or concentration at 8 p.m. is (1.48 + 60 min * 0.0020) = 1.6 g kg-1 or 1.6 permille at the time of the traffic accident.
2. The alcohol intake is (150 ml * 0.40) = 60 g of alcohol (ethanol). The distribution pool for alcohol is (58 kg * 0.60) = 34.8 kg. Assuming instantaneous distribution, the theoretical alcohol- permille at time zero (7 p.m.) is: (60 g)/34.8 kg =1.72 g kg-1. The theoretical oxidation occurring in 2 hours is (120 min * 0.0020) = 0.24 g kg-1. At 9 p.m. the expected blood alcohol concentration is (1.72 -0.24) = 1.48 g kg-1. This is exactly what was found, so the statement seems to be correct.
1. According to the ideal gas equation VSTPD * 760/273 = 8 * (752 - 20)/(273+ 20). Isolation of VSTPD yields the value 7.1793 m3 or 7179.3 litres of oxygen. The survival threshold is FIO2 : 50/(752-20) = 0.0683. The total volume of oxygen available for survival is (7179.3 * (0.2093 - 0.0683) = 1012.28 l of oxygen. The survival period is:
1012.28/(3*0.190) = 1776 min or 29.6 hours.
2.The RQ is 0.83, so for each person the carbon dioxide output per min is (0.19* 0.83) = 0.1577 l STPD min-1.
3. The theoretical carbon dioxide accumulation amounts to (0.1577*3*1776) = 840 l STPD. The corresponding FICO2 is 840/7179.3 or 0.117 (11.7%) as an estimate.
4. The consequences of CO2 accumulation are CO2 intoxication and CO2 narcosis. Long before the theoretical accumulation should occur, the 3 victims have died in a terrifying state of anxiety by CO2 intoxication. Humans cannot survive in an atmosphere with 11.7% CO2.
1. The difference between pK and pH is (9.3 - 7.3) = 2 decades or 102. Thus, we have 100 times as much NH4+ as ammonia in the blood.
2. 12 600/0.46 = 27 391 mmol oxygen/day. RQ = V°CO2 / V°O2 ;
V°CO2 = 0.83×27.391 = 22.7 mol CO2/day.
3. The synthesis of urea in the liver is the major route of removal of NH4+:
2 NH3 + CO2 ---- CO(NH2)2 + water. Thus, NH4+ is eliminated by the use of half a mol of carbon dioxide daily for urea production.
1. FAO2 = 14.1 kPa/(101.3 - 6.2) kPa = 0.148.
FAN2 = 78 kPa/(101.3 - 6.2) kPa = 0.82.
FACO2 = 1 - (FAO2 + FAN2) = 0.032.
FACO2 = V°CO2/ V°A ; V°A = V°CO2 2/FACO2
V°A = 660/0.032 = 20 625 ml STPD/min.
The PACO2 is (0.032×95.1) = 3 kPa (22.5 mm Hg). The PaCO2 is assumed to be similar, when tension equilibrium is established; the normal venous mixing can only increase the tension to a small extent. Such a low PaCO2 is a clear indication of hyperventilation.
2. V°O2 = V°A [FAO2 (FAN2/FIN2) - FAO2]
V°O2 = 20.625×[0.2093×(0.82/0.79) - 0.148].
V°O2 = 1.427 l STPD per min.
3. R = 660/1427 = 0.46.
This R value is lower than the lowest possible RQ (0.7 by pure fat combustion).