and the Partition of Thermal Energy. 685 



3. Compressibility. 



To find the compressibility, we notice that the work done 

 against the molecular forces = ft per molecule for 1° rise of 

 temperature. Now the pressure at the surface = FN 2/3 . 

 Therefore the work done against it = So^EN 2 - 3 where a ± is 

 the coefficient of linear expansion. 



The work done against the inter-molecular forces per unit 

 volume = 3F67N = 3a 1 FN 2/3 , there being six molecules 

 arranged about each individual molecule, and 



ha = hi = c/. l l. 



Nj3=6ct 1 FN 2 < 3 . 



F= 7 A /J since NZ 8 = 1. 

 _ «0 _ 2*0 . ail 



«- 3F - —ft-~- 



/5N 

 Now the surface pressure = NF 2/3 = -x — . 



If the pressure be increased by one dyne without change 



/» 



of temperature, it increases by ^-^ th part of itself. 



/» 



.*. a diminishes by a^5^ P art °£ itself. 



or 



Sa 



a 

 SI 



I " 



0= 



6«i SI 



= ~~N/3~^* 

 a 6*i 



~'~T' W 



SI 18a t a 





cy- 2 OlO 



/3 2 = 



.36-! cR 



ft = 



=6 *V^- 



This is the equation which has to be substituted for (4) in 

 the former paper. It will be noticed that this value of ft 

 increases the value of F and diminishes the value of a in the 

 ratio of 24 : 19. 



