558 SMITHSONIAN MISCELLANEOUS COLLECTIONS VOL. 51 



For this initial condition we have given the data 



Po> *■ 02> ^2> *■ iV ""\. 



whence we know 



-r f g ^2 T _ t _ & ^l 



1 i2~ 2 02 ~j^~ l h\~ l i\ ~^r~ 



^p ^p 



With these values by the help of equations (i), (2) and (5a) we 



compute the following equation except the arbitrary constant 



'_ _ C v 2 



(P + I ).- - • j^r K (A> T o 2 - Pi T i2 + Pi T it - p h T hl ) + Constant. 



For the final condition p' i = p h + (p — p x ) whence the tempera- 

 tures of the two masses at the boundaries become 



( - r T/= T, l 





H 02 \ >> / b 



PJ " 2 m \Po 



(P + I ).- Cv g ■ y^ (p e r - p\ 1\ + A- r ia - Ph r h2 ) + Constant. 

 We find the heights or depths of the masses i' and 2' from 



h' = - p (7' - r. ); ti = - (T' { - T' h ) 



1 p v 01 *1 ' ' 2 £ 2 2 



The sum of the two masses is — — wherefore the available ki- 



g 

 netic energy of a unit mass is 



(P +T) - (P +T) e 



\ V*=g 



Po - Ph 



Numerical example. Let h x = h 2 = 2000 meters and let the other 

 data for this problem be the numbers indicated by a star (*) as fol- 

 lows (the pressures being in millimeters of mercury and the tem- 

 peratures absolute Centigrade) : 



