RESPIRATORY EXCHANGE. 121 



(3) Actual Bar. Press. = 747 mm. Hg. 



(4) Corrected Bar. Press. ((3) -Factor in Table III) =747- (17.4+2.41) = 



727.19 mm. Hg. 



(5) Volume of Resp. Air at 760 mm. and C. ( (1) X Factor in Table IV) 100 X 

 0.89062=89.062 Litres. 



(6) Corrected Volume of Resp. Air 



(a) per mii1ute= 5. 937 Litres. 

 (6) per hour = 356.248 Litres. 



Calculations of 2 A bsorbed and COi Eliminated. 



(7) Volume per cent. CO 2 in spirometer air =4.0. 



(8) Volume per cent. O 2 in spirometer air = 16. 5. 



(9) Volume per cent. CO 2 and O 2 in spirometer air (7) (8) =20.5. 



(10) Volume per cent. N in spirometer air (100) - (9) =79.5. 



(11) Volume per cent. N in room air taken as 79 = 79.0. 



(12) Oxygen equivalent of (10) = (10X0.265) =79.5X0.265 =21.06. 



(This may also be obtained from Table I). 



(13) Volume per cent. O 2 absorbed (12) - (8) =21.06-16.5 = 4.56. 



(14) Volume percent. CO 2 eliminated (7) -(.03) =4.0-0.03 = 3.97. 



(15) Total O 2 volume (standard conditions) absorbed per hr. (13)X6(b) = 



16.24 Litres. 



(16) Total CO 2 volume (standard conditions) eliminated per hr. (14)X6(b) = 



14.14 Litres. 



The Caloric Value Calculated from the Gas Exchange. 



(Indirect Calorimetry). 



This can be done by using table V (Appendix) provided we know the non- 

 protein R.Q, which is given in the 1st 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 ab- 

 sorbed, 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, 9.35 X0.6 = 5.610 gm. CO 2 or (since 1 gm. CO 2 = 0.5087 1) 

 2.85 1 must be subtracted from 14.14 Litres giving 11.29 Litres, and similarly 

 8.45 X. 6 = 5.070 gm. O 2 or (since 1 gm. O 2 = 0.7 1) 3.5490 must be subtracted 



from 16.24 giving 12.69. The non-protein R.Q. is therefore - - =0.889. 



1 . * K ' 



Referring to table V we see that at R.Q 0.889 1 litre of O 2 equals 4.91 C. .'. in 

 Ihr. 4.91X16.24 = 79.73C were expended. As a matter of fact for many pur- 

 poses it is sufficiently accurate to use the uncorrected R.Q. for table V. 



