346 DIRECT PROPERTIES OF MOLECULES 123 



the pressure the distance apart p of two molecules is 

 diminished by 0'000016p. An atmosphere is the pressure 

 of 10334 kilograms per square metre, and it therefore exerts 

 on the area p 2 corresponding to a single molecule the force 

 10334 #p 2 absolute units, wherein the kilogram is taken as 

 unit of mass and the metre as unit of length, g being the 

 acceleration of gravity. Since this force diminishes p by 

 -00001 6p, the force required to diminish p by xp is 



10334</p 2 z/ 0-000016, 



if we may make the assumption (which is doubtless approxi- 

 mately correct) that the force is directly proportional to the 

 amount of approach. The work done by this force while 

 the molecules are approaching each other is found by multi- 

 plying the force by half the diminution of distance, and it is 

 therefore 



10334#pV/0-000032. 



Boltznzann compares this expenditure of work with 

 the energy of the molecular motion in water-vapour, in 

 order to determine the shortening of distance, measured by x, 

 which occurs at a collision of two molecules. Let m be the 

 mass of a molecule, and therefore, with a kilogram and a 

 metre as units, m = lOOOp 3 , since 1 cubic metre of water 

 weighs 1000 kilograms ; further, with our former notation, 

 wherein G represents the mean value of the molecular speed 

 as calculated from the energy, the sum of the kinetic energies 

 of the two colliding molecules is 



2.imG 2 = wG 2 , 



where by 28 the value of G for water-vapour at the 

 temperature is to be taken as 614 metres per second. 



If we equate the value of the energy so determined to 

 the above value found for the expended work, and also put 

 g = 9-81, we have 



1000 x 614 2 p 3 = 10334 x 9'81p 3 z 2 / 0-000032, 



whence we obtain 



x = 1/2-9 = 



