DISSOCIATION CURVE. 135 



there is a partial pressure of about 20 mm. Hg. oxygen in the 

 tonometer.* 



When the mercury has reached this level, or one near it, clip 3 

 is closed and the height at which the mercury stands very accurately 

 noted. Clip 2 is closed, after which the mercury is allowed to fall 

 to zero by opening 3. The tonometer is removed and rotated so 

 that the blood becomes spread out as a thin film on the walls, 

 after which it is placed in a water-bath kept about 40 C. in which 

 it is constantly rotated for about 15 minutes. 



On removal from the bath the pressure in the tonometer must 

 again be measured. For this purpose the tonometer is reattached 

 to A and the pump is turned on (with 3 closed) until the mercury 

 has risen to the level at which it previously stood. Clip 4 is closed 

 and 2 opened. If there has been no leak, and time has been allowed 

 for the tonometer to cool down, there will be practically no differ- 

 ence between the two readings. If a difference of more than 5 mm. 

 is observed it must be noted and the pressure prevailing in the 

 tonometer taken as the average between the two readings. 



Meanwhile 3 c.c. of freshly prepared weak ammonia water con- 

 taining a trace of saponin (0.5 c.c. aq. ammonia in 500 c.c. water) 

 has been placed in the blood gas bottle. A pointed glass tube 

 about 30 mm. long is now attached to the rubber tubing of the tono- 

 meter and this is removed from the barometer and held in a vertical 



*This is computed as follows: After suitable adjustment the standard baro- 

 meter in the room is read and from the reading is deducted the tension of aqueous 

 vapour at the temperature of the room (for Table, see page 277 of Appendix). 

 The difference gives the pressure in mm. Hg. of an atmosphere of dry air. Since 

 air contains 20.96 oxygen, the partial pressure of this gas in the tonometer must be 



20 96 

 equal to : ths. of the difference between the height to which the mercury is 



-LUU 



raised in B and the corrected barometer reading. Thus, suppose the room 

 barometer to be 753.4 mm. and the temp. 20 C., the corrected barometer reading 

 is 753.4-17.4 = 736 mm. 



Then--? 6 X736 = 154.2 mm. 2 



-LUU \ 



20 X 736 



Suppose a tension of 20 mm. 2 is desired, then = =95.45 mm. 



154 .2 



That is the mercury in the barometer must be raised to 736 95.45 or 640.55 mm. 

 above the level in the reservoir (R). 



