THE MECHANISM OF RESPIRATION 283 



ten minutes, and the number of respirations be counted, the tidal air 

 can be calculated ; from the analysis the respiratory exchange can be 

 estimated as well as the composition of the expired air (see later, p. 317). 



One of the best forms of apparatus for the estimation of the gases 

 in inspired and expired air is shown in Fig. 151. 



The gas is measured in the graduated gas-burette A, provided 

 with a three-way tap. Surrounding the gas-burette is a water-jacket. 

 The whole is supported by a clamp and retort stand. The gas-burette 

 is connected by pressure tubing to the levelling tube B, which is held 

 by a spring clamp attached to the retort stand. A and B contain 

 mercury, and by raising or lowering B gas can be expelled from or 

 drawn into A. One of the connections of the three-way tap is used 

 for taking in the sample, the other connects the burette with an 

 absorption apparatus arranged as in the figure. 



The bulb E, filled with 20 per cent, caustic potash, absorbs C0 2 . 

 The bulb F, filled with alkaline pyrogallic acid solution, absorbs 

 O 2 . Alkali in G and H protects the pyro solution from the air. F is 

 emptied and refilled through K. The tap on tiie absorption pipette 

 places either E or F in connection with the gas-burette. There is 

 a control tube by which alterations in temperature or barometric 

 pressure during the analysis can be compensated for. 



The sample of gas is taken into the burette from the sampling 

 tube (Fig. 150), mercury being sucked into the tube to take the place 

 of the gas entering the burette. After measuring the amount of the 

 sample it is passed into E to absorb C0 2 . When a constant reading 

 is obtained, it is passed into F until all the oxygen is absorbed. 

 To take an example : Suppose the amount of gas taken in was 

 20-12 c.c. After absorption of carbon dioxide in E the burette 

 reading was 19-06; after absorption of oxygen in F it was 16-01. 



Then in 20-12 parts of the sample there is 20-12-19-06 = 1-06 of 

 C0 2 , in 100 parts therefore there are 



1-06x100 



^20- llT =5 ' 22 a PP r( > x - 



Also in 20-12 parts of the sample there are 19-06-16-01=3-05 

 parts of O 2 . 



In 100 parts therefore there are 



3-05 x 100 

 "26-12 =1 

 The percentage of sample, therefore, is 5-22 OO 2 ; 15-11 O 2 . 



26-12 ' U a PP rox - 



