Solubility of Radium Emanation. 845 



Method 1. In the mixing bulb V is the volume of the 

 liquid, and A — V the volume of the gas. After equilibrium 

 there will be (A — V)c units of emanation in the gas and 

 VcS in the liquid. The gas is expanded from a volume of 

 A — Y to a volume of T — V, so that the density is reduced 



m the ratio m - w ="■&.• 



Therefore, on closing the stopcock there will be in the 

 mixing bulb (A — v x )kc units of emanation in the gas, and 

 zy.'S in the liquid. Hence the total emanation in this bulb 

 equals (A— r^/'c + rjcS. 



Similarly, the emanation carried by the gas into the 

 sampling bulb will be (B — r. 2 )kc. and by possible small drops 

 of liquid which have been blown over into this bulb v 2 cS. 

 The total equals 



(B-t^c + tyS. 



Since the ratio of the emanation in the two bulbs equals r 

 we have 



(A-r^k + rS _ n v 



(B-r 2 )/c + r 2 S~ ? > ' ' ' * W 



the c cancelling out. From this formula S can be calculated. 

 As an example a determination at 14° 0. is given ; the 

 following are the data : — 



A = 33-18 c.o. 

 B = 18*87 „ 

 V = 18-72 „ 

 i\ = 18*66 „ 

 »j*=0-03 „ 

 h= 0-430 

 r= 1*456 

 Therefore S = 0'299. 



ca 



Method 2. In this method, although the expulsion of liquid 

 uses the gas in the mixing bulb to expand, all of it remains 

 in the bulb. Therefore, on closing the stopcock there will 

 be in this bulb (A— V)c units of emanation in the gas, and 

 t?icS in the liquid. The total will be (A— V)c+iyS. 



The emanation going to the sampling bulb will simply be 

 r 2 cS. Hence in this case 



( A-V) + ri S =r 



v„s • ■ ■ y-j 



