ILLUSTRATIONS OF THE DYNAMICAL THEORY OF GASES. 399 



and this must be zero, if the pressures are equal on each side of the plug. 



Now if Q lf Qt be the quantities transferred through the plug by the mean 

 motion of translation, Q l = p l V 1 = M,N 1 V 1 ; and since by Graham's law 



Q t ~ 



we shall have 



MjN^ F, = 3fjJV,v, F, = U suppose ; 



and since the pressures on the two sides are equal, -r- 2 = ~~J^> an( i the 



way in which the equation of equilibrium of the plug can generally subsist is 



when Z, = L t and /, = ?,. This implies that A = C and B?= D. Now we know 



^ 



that v,'B = v?C. Let K= 3 ; , then we shall have 



v, 



...................... (40), 



and - = = K (v 1 p l + v,p,) + - ........................... (41). 



The diffusion is due partly to the motion of translation, and partly to that of 

 agitation. Let us find the part due to the motion of translation. 



The equation of motion of one of the gases through the plug is found by 

 adding the forces due to pressures to those due to resistances, and equating 

 these to the moving force, which in the case of slow motions may be neglected 

 altogether. The result for the first is 



+N.BP.V, -fc (v,- r,)+-=o ...... (42). 



Making use of the simplifications we have just discovered, this becomes 



whence U= - d f _ .................. (44) . 



dx v '' 



