of velocity of the particle, by w’, y’, 2’, u’, v’, w’ those of the 
molecules, we have: 
ed 2 {28s NAO ee) Nees 
Gre Wey uuw jt nn v—v | oe (»—«)| 
because a an Si 
1 yoke 
M ox 
terms together yielding the quantity g. If we exclusively think of 
The term u will represent — pu of equation (2), the other 
ea te Or 
forces of collision, it is easy to see that — — will always be negative. 
Ow - 
If there are also forces of attraction, this quantity can also assume 
a positive value, but it will yet be predominantly negative. ') To 
ORs . 
carry out the caleulation the better we shall assume that - 18 
Lb 
negative for all active molecules. 
OR, Ot, a ‘ 
. and PD being equally probably positive and negative, they 
oy dz ; 
will be zero on an average, and the terms with v and w will be 
small compared with those with w. Hence these terms are neglected. 
Also the terms with w’, v’, and 2a’ equally assume either sign, 
but w’, v’, and ww’ on the other hand being very large compared 
with w, the terms containing these quantities, may not be left out. 
As a simplification we equate the middle values of the terms with 
/ 
u’, v’, and w’ with each other. 
1) If e.g, we think a number of particles in a parallelepiped space, the density 
“at every point will be me ‘7, ¢.representing the potential energy of a molecule. 
OR = Ore 
We may further put Aree Dr Jf we now form the integral: 
Wi a 
okie eae 
zE | ne BT dx dy de 
a 
partial integration yields: 
DE 1 (Yde — 
_f Je ne PL dy dz — az (35) nye BT da dy dz. 
The integrated term represents the external pressure on the side-walls of the 
parallelepiped, which are 1 to the «-axis. If we disregard gravitation and such 
external forces, this term yields zero, hence we see that the integral presents an 
essentially negative value. 
