M. R. Bunsen on the Law of Absorption of Gases. 193 



The following is the method of solving this problem : — A 

 sufficient quantity of the gas to be examined is collected in the 

 absorptiometer, and its volume, pressure and temperature ob- 

 served. 



If the originally observed volume reduced to 0° be called V, 

 and its pressure P, we obtain the equation 



l = Vf + YV < 6 > 



Three absorptions of the gas are made with the volumes of 

 water //,, /i 2 and h 3 , and the corresponding volumes and pressures 

 reduced to 0°, V-J^v ^ T ^o> V 3 P 3 , observed for a constant tempe- 

 rature t. From these determinations, according to equation (6), 

 we obtain the following, in which « signifies the absorption- 

 coefficient of the first gas, and B that of the second at the tem- 

 perature t : — 



1 = 



y 



(V. + aW^ + zSW 



1=7^W + 



V 



(V 2 + «/* 2 )P 2 (V 2 +/3.4 2 )P g J 



1= ,„ ■* + 



.'/ 



(Vg+^3)P 3 (v 3+/ s/yiy 



From these four equations the unknown quantities x, y, a, and 

 B are easily found. The two last are the ordinatcs of absorption 

 of two gases for the temperature-abscissa t. If the numerical 

 values of these are calculated, it is easy to find in the table the 

 gas which has the same absorption-coefficient for corresponding 

 temperatures, and thus the nature of the mixture is determined. 

 The values of x and y give also the relative proportion between 

 the constituents. The determination of a and B is, in the case 

 of two gases, not difficult. If we place PV = «, P,V, = «„ 

 P 2 V 2 =fl 2 , P 3 V 3 =tf 3 audP,//j = 6 l , P 2 /i 2 = 6 2 , ~P 3 h 3 =b 3 , we obtain 

 first, — 



a + 8 = g i Z, A( a — a \ ) ( &a ~ h) - ggfl A( g ~ ff g) (h — h) + a 3 b A( a ~ %) ( b \ — h z) 

 , _ a A (a — g a) - a A ( a ~ g J -h b a,( a z— g 3 ) A 



and when the expressions on the right of tlicsc equations be 

 represented by A and 13, — 



« + /3=A (20) 



«-/3=± v / A s -4B. (21) 



Phil. May. S. 4. Vol. ( J. No. B8. March 1 855. O 



