424 Mr. W. Sutherland on the 



solution (see, for instance, Tait, Trans. Roy. Soc. Edinb. 

 xxxiii. and xxxv.). But the exact value of the constant 

 does not concern us in our present problem, which is to de- 

 termine the viscosity of a mixture of gases in terms of the 

 viscosities of the pure gases and not absolutely. Accordingly 

 we can avail ourselves of the above easy and convenient 

 method of evaluating the viscosity of a mixture. With v the 

 mean velocity of the molecules of the single gas, and v the 

 average number of collisions experienced by a molecule in a 

 second, then \=v/v, and our result for a single gas becomes 



rj = nmv 2 /6v (1) 



Now according to the kinetic theory for forceless, smooth, 

 perfectly restitutional, spherical molecules of radius a, and 

 mean square velocity 3/c 2 /2 (see, for instance, Tait, Trans. 

 Roy. Soc. Edinb. xxxiii.), 



v=2n(a-{-a) 2 7Ti(fc 2 + K 2 ^; .... (2) 



and by these two equations the viscosity is expressed solely 

 in terms of the given mechanical constants of the system of 

 molecules. 



In the case of a mixture of two gases, 1 and 2, let the pro- 

 portions be % molecules of 1 to n 2 molecules of 2 in unit 

 volume ; then the number of encounters that a molecule of 1 

 experiences in unit time with molecules 2 may be denoted by 

 iv 2 , and 



1 v 2 = 2w 2 (a 1 + a 2 ) 2 7r*(/e 1 2 + /e 2 2 )5; .... (3) 



(see Tait, Trans. Roy. Soc. Edinb. xxxiii.) ; and the number 

 of encounters per second of a molecule of 1 with other 

 molecules of 1 is 



1 v 1 = 2n 1 (a 1 + ajV(^ 2 + * 1 2 )* (4) 



Then v x being the average velocity of a molecule of 1, its 

 mean free path in the mixture is 



^i = V(i^i + i^) (5) 



Then if, as in the case of a single gas, we assume that a single 

 collision even between molecules of 1 and 2 suffices on the 

 average to bring them to the characteristic state of the layer 

 in which they collide, we can take the viscous action due to 

 the molecules of 1 as given by the above simple theory of a 

 single gas, namely n i m 1 v 1 2 w/6( l v l + l v 2 )D. But of course this 

 assumption is quite unjustifiable, except in the case where the 

 molecule of 1 has the same mass as that of 2, in which case 

 it becomes just as allowable as in that of a single gas, neither 

 more nor less. When m x is different from m 2 , then if the 

 two solid planes and the mixed gas between them were at 



