METHOD FOR DETERMINING RESPIRATORY EXCHANGE IN MAN 597 



be subtracted from the observed barometric pressure to obtain the true pressure. The 

 vapor tension of water for various temperatures is given in Table II on page 598. 

 (2) The barometer tube lengthens or contracts with heat or cold, and therefore the 

 barometric readings must be corrected. The corrections for ordinary barometric read- 

 ings are found in Table III, page 598. The figure corresponding to the temperatures 

 is subtracted from the barometric reading in order to obtain correct baromjetie pres- 

 sure. 



In the above experiment, the correction for the barometer is 2.41 mm. (see Table III, 

 page 598), and that for vapor tension at 20 C. is 17.4 (see Table II, page 598). 

 Actual Barometric Pressure 747 (17.5 + 2.39) =727.21 mm. 



The coefficient of expansion of gases is taken as 0.003665) or 1/273; therefore the 

 volume of equals the volume at 1 divided by 1-0. 003665 t; and hence 



Vo = Vx273 = - , when Vo = Volume at and V = Volume at t. 



273 + t 1 + 0.003665 t 



VT* 

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



P pressure ; or working both corrections together, 



VPx273 VP 



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

 This formula applied to the present problem reads: 



Vo = 100x727.2 = 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.F., the factor is 

 0.89124; therefore: 



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



0.89124 x 4.57 = 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 Eeat value of 1 liter of O, at various respiratory quotients, it is found 

 that at a E.Q. of 0.87, 4.888 calories are expended; 16.28 liters of O 3 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 reference to the chart for 

 determining the surface area of man (page 576), this would be found to be 1.96 

 square meters. The caloric expenditure per square meter in the above case is therefore 



79 



= 40.3 calories. 



1.96 



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 : 



%N 2 78.7 78^8 78^9 79X) TJU 79^2 79J3 79A 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 80.3 80.4 80.5 80.6 



21.16 21.19 21.22 21.25 21.28 21.31 21.35 21.38 



