Production of Vacua, and their Spectroscopic Examination. 235 



the liquid hydrogen experiments, might be joined together and filled 

 with oxygen or nitrogen at atmospheric pressure, and simultaneously 

 exhausted with the mercurial pump to a small fraction of an atmo- 

 sphere, and then sealed off from the pump and each other. One of 

 these two identical tubes could then be subjected to the hydrogen 

 cooling, following the directions already given, and the two vacuum 

 tubes now compared. If there was a marked difference in resistance 

 to the passage of the discharge in the frozen tube, then something 

 must have condensed, and by a few tentative trials a limit might 

 be reached when the initial exhaustion was unaffected by the hydrogen 

 cooling. Such experiments have not yet been made. The presence of 

 any vapour of mercury would require to be carefully eliminated, other- 

 wise the method would not be satisfactory. Tubes that are prepared 

 without taking special precautions to exclude organic matter and water 

 from the glass, deteriorate, especially with electrodeless tubes after the 

 discharge has taken place for some time. 



The rapidity with which the vacua are attained is such as theory 

 would suggest; assuming a hole of a square millimetre in section 

 through which the air rushes into the condenser, and that a velocity of 

 current between 600 and 700 feet a second is attained, then a vessel of 

 20 c.c. capacity could be reduced in pressure in 1 second to 1/10 of the 

 initial pressure, and if the same rate is continued at the end of 

 60 seconds to ( T V) 60 - Sir George Stokes has been good enough to 

 consider the problem and writes as follows : 



" Let V be the volume of the vessel, A the area of an aperture by 

 which the air is conceived as rushing out with the velocity v t p the 

 density of the air in the vessel at the time t, D the initial density, that 

 is, the atmospheric density. 



" Then, according to our hypothesis, Av.dt is the volume of air, and, 

 therefore, Avp,dt, the mass of air, which rushes out in the time dt. 

 But this equals the loss of mass of air in the vessel during the time dt, 

 and, therefore, 



Avp .dt=- Vdp, 



-a differential equation of which the integral is 



p = De- At? */ v . 



" Suppose now V to be 20 c.c., or 20,000 c.mm., A to be the area of 

 -a circle of 1 or 2 mm. diameter, say 2 sq. mm., v to be 333 m., or 

 333,000 mm., t (in seconds) to be 60 ; then 



- = 5254 xlO 434 . 

 P 



VOL. LXIV. 



