METHOD FOR DETERMINING RESPIRATORY EXCHANGE IN MAN 597 



air at and 760 mm. Hg. To do this two things must be taken into account. (1) 

 Since the expired air is saturated with water, the pressure due to water vapor must 

 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) Since the barometer tube lengthens or contracts with heat or cold, the baro- 

 metric 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 barometric 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 (17.4 + 2.4) = 727.2 mm. 



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

 volume at equals the volume at 1 divided by 1 + 0.003665 t; and hence 



"\r 



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



1 + 0.003665 t ' 



VT* 



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



760 



P pressure; therefore working both corrections together, we have 



760 (1 + 0.003665 t) 

 This formula applied to the present problem reads: 



Vo = 100X727.2 =89.14 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 the total volume of air breathed in 15 min. was: 



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

 and the absorbed O 2 , 



0.89124 x 4.57 = 40.7 liters of O 2 , or 16.28 L. per hour. 



The Caloric Value Calculated from the Gas Exchange (Indirect Calorimetry). 

 This can be done by using Table V but it is more accurate to know the nonprotein R.Q. 

 which is given in the first column of the table. This latter is obtained by deducting 

 from the total CO 2 eliminated, the CO 2 derived from protein (found by multiplying the 

 urinary N by 9.35) and by deducting from the total O 2 absorbed the O 2 required to 

 oxidise protein (found by multiplying urinary N" by 8.45). Suppose, for example, that 

 14.4 gm. N is excreted in the 24 hrs. urine; i.e., 0.6 gm. per hr. Then, 9.35x0.6 = 

 5.610 gm. CO 2 or, (since 1 gm. CO 2 = 0.5087 L) 2.85 L must be subtracted from 14.152 L 

 giving 11.302 L and similarly 8.45 x .6 = 5.070 gm. O 2 or (since 1 gm. O 2 = 0.7 L) 

 3.5490 must be subtracted from 16.28 giving 12.73. The nonprotein R.Q. is therefore 



11 302 



' =.88. Referring to Table V we see that at R.Q. 0.88 one liter of O, equals 



1^.730 



4.900 C. in 1 hr. 4.9 x 16.28 79.77 C were expended. As a matter of fact for most 

 purposes it is sufficiently accurate to use the uncorrected R.Q. for Table V. 



