564 METABOLISM 



Vx273 V 



V = 273+T = l + 0.003665 t ' When V = 



VP 



The volume of gas being inversely as the pressure, Vo _ _ , where V = volume at 



760 

 P pressure; or working both corrections together, 



VPx273 VP 



Vo _ 



760 x (273 + t)~760 (1 + 0.003665 t) 



This formula applied to the present problem reads: 



Vo = 100 x 727.2 _ _ = 89.2 liters. 



760 (1 + 0.003665x20) 



The latter calculation can be considerably simplified by using standard 

 tables which give constants for corrections of gas volumes. These are 

 easily obtainable and are given in part in Table IV. 



According to these tables for 20 C. and 727.21 mm. Hg B.P., the 

 factor is 0.89124; therefore: 



0.89124 x 100 89.124 liters, 0C. and 760 mm. Hg. 

 0.89124x4.57 = 40.7 liters ^of O 2 in 15 min., or 16.28 L. per hour. 



The Caloric Value Calculated from the Gas Exchange. By reference 

 to Table V giving the heat value of 1 liter of 2 at various respiratory 

 quotients, it is found that at a R.Q. of 0.87, 4.888 calories are expended ; 

 16.28 liters of 2 is therefore equivalent to 18.4 x 4.888 = 79 calories. 



The results must be calculated for surface area as well as body weight. 

 Suppose the subject weighed 85 kg. and was 170 cm. in height; by refer- 

 ence to the chart for determining the surface area of man (page 540), 

 this would be found to be 1.96 square meters. The caloric expenditure 



79 

 per square meter in the above case is therefore ~~ = ^ 



TABLE I 



THE PERCENTAGE OF OXYGEN WHICH is EQUIVALENT TO THE NITROGEN FOUND IN THE 



EXPIRED AIR 



To obtain the nitrogen in the expired air, add the percentage of CO 2 and O 2 found 

 and subtract the sum from 100. The table gives the percentage for O 2 corresponding to 

 this figure: 



78.7 78.8 78.9 79.0 79.1 79.2 79.3 79.4 79.5 79.6 79.7 79.8 



%O, 20.86 20.88 20.90 20.93 20.96 20.98 21.01 21.04 21.07 21.10 21.12 21.14 



79.9 80.0 80.1 80.2 S0.3 80.4 80.5 80.6 



21.16 21.19 21.22 21.25 21.28 21.31 21.35 21.38 _ 



TABLE II 



TENSION OF AQUEOUS VAPOR IN MILLIMETERS OF MERCURY 



To obtain the dry barometer pressure, subtract the mm. Hg. corresponding to the 

 temperature of the air from the barometer pressure at the time of the experiment: 



Temp. 15 16 17 18 19 20 21 22 23 24 25 

 Mm. 12.7 13.5 14.4 15.4 16.3 17.4 18.5 19.7 20.9 22.2 23.5 



