92 Mr. W. Sutherland on the Law of Attraction 



mass operated on, which is 



^a^ + 'bPbVjv 

 Therefore for the cooling effect of the mixture we have 



V 2 V 2 / V 2 V 2 



*A?A^A + «B^B + 2C V * A P A ^ A ^bPb^^B. 



To show with this formula how the cooling for a mixture of 

 and 1ST may be less than for either of the gases alone under the 

 same circumstances, let us suppose that 6 is less both than 6 A 

 and 6 B ; we w 7 ill see whether the supposition leads to a possible 

 or impossible conclusion. 



Let us denote 



V V 



— h by D, and -J?. 1 by D', remembering that D + D' = l ? 



Vj Vi & 



also s aPa by a, s bPb by /3. 



Then the supposed inequalities become 



«D 2 A + /3D' 2 B + 2CDD' \/a^6 A 9 B < (aD + /3D')0 A , 



0D' 2 B + kDW a + 2CDD' */«/36» A 6> B < (*D + 0D')0 B ; 



. • . «D0 A (D-1) + /3D' 2 B -/3D'6 A + 2CDD' V*£d A 6 B < 0, 

 or 



-aDD'0 A + /3D' 2 B -/3D / A + 2CDD / Va^6 A e B <0' t 



.-. -aD0 A + /3D'0 B -j80 A + 2CD*/«/30 A A <O. 

 Similarly 



-/3D / 6' B + «D^ A -a6' B + 2CD / v/^5 B <0; 



.-.adding -j80 A -«0 B + 2C(D + D') v / a/3(9 A 6> B <0. 

 If C = l, this becomes, since D + D'=l, 



-( V|8T A - V<) 2 <0, 



which is possible. If C is less than 1, the inequalities can also 

 still exist together. 



Thus that the cooling effect for a mixture of two gases 

 should prove less than that for either of the constituents has 

 been shown to be a possible consequence of the theory of 

 molecular attractions. 



By means of Thomson and Joule's experimental numbers 

 for mixtures of air and C0 2 in different proportions, we propose 

 finally to calculate the values of C obtainable from the cooling 



