40* MOLECULAR FREE PATHS 435 



We then have for the momentum carried over in unit time in the 

 positive direction the expression 



Q,df = dfNm(lm/ir)*fdu ^e-^B Cdr e~^ds sins cos^f 2 ^ ; 



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from this the momentum carried over in the negative direction, 

 viz. 



Q 2 df= - dfNm(km/Tr)% f dw w 3 e- ftwla>2 B f dr e~^ r f ds sin s cos 2 s f ""cfy, 



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differs only in sign and in two limits of integration. Therefore the 

 pressure exerted (or the force per unit area) is given by the 

 formula 



p = 2Nra(&ra/7r)tr d<^ wV^'B ["dre-^'ds sins cos 2 r( 2 *eZ0, 

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which is obtained more simply in the case when df forms part of 

 the wall of the containing vessel. On carrying out the integrations 

 we obtain 



p = N/2k, 



for which may be written, in accordance with the formulae of 

 19*, 



p = 



A variation of this mode of carrying out the calculation, which 

 deserves mention, consists in our introducing the time in which a 

 path is traversed by a particle instead of the path itself. Among 

 the particles contained in the element of volume dx dy dz there 

 are 



47T- W(&ra)f e-* w "VdwB<r B ^ dx dy dz 



which continue for the interval t without collision their straight 

 path, which was begun with speed w, and then collide in the follow- 

 ing element of time dt. Since the state of the gas does not alter 

 with the time, there must come in just as many particles, which 

 after the lapse of the time t have acquired the velocity a;, in the 

 place of those which lose their former speed w. Therefore the 

 number of particles which in unit time proceed from the element 

 dx dy dz with speed w, and collide after the lapse of time t, is 

 ^-* Wtt 'Vdu> e-**Bt- l dtdxdy dz. 



Of these the number 



/rfe-^Mu e- zi Et~ l dt r~ 2 cos s dfdx dy dz 

 reach the element df if the time t is sufficient for the length r to 



F F 2 



