312-315] Conduction of Heat 261 



element dv is exactly equal to that of the collisions in which one of the 

 molecules before collision has velocities within these limits. 



The number of molecules which at any instant are moving with a velocity 

 between c and c + dc in the element dv is 



V 



h s m 3 



4>7rc 2 e~ hme * dc ............... (590), 



7T 



so that the number of collisions experienced by these molecules in time dt is, 

 by 251, equal to 



dt x (expression (590)) .............. . ...... (591), 



where 



This, then, is the number of molecules which, in time dt, experience a 

 collision in the element dv, and leave it with a velocity between c and c + dc. 

 All directions are equally likely. Now the element dxdy subtends a solid 

 angle 



cos 6 



at the element dv, so that the chance that, if a molecule escapes collision, it 

 will pass through the element dxdy, is 



(593). 



Multiplying expressions (591) and (593) together, we obtain 



yn^nyiS 

 -^- 6 c 2 e~ hmc * dc ...... (594). 



This is the number of molecules which leave the element dv in time dt 

 with a velocity between limits c and c + dc, in a direction suitable for passing 

 through the element of area dxdy. 



315. We have next to calculate the probability of a molecule describing 

 the free path r from dv to dxdy without collision. 



We cannot use the analysis of 286 to determine this probability, for in 

 the present problem the state of the gas varies from point to point as we 

 proceed along the path of the molecule. If, however, we denote by f(l} the 

 fraction of the whole which travel a distance at least equal to I without 

 collision, we obtain, just as in 286, the differential equation 



(595), 



in which B e is now evaluated at a point at distance Z along the path. The 

 solution of this equation is 



e~? Scdl ..... ...................... (596). 



