446 SMITHSONIAN MISCELLANEOUS COLLECTIONS VOL. 51 



also form the adiabatic equation for the diminution of the quan- 

 tity of vapor with temperature. We thus obtain 



e x T M r M r 



l0 V x = W « log r + AR * T X ~ AR T X > 



If we substitute 



M r 

 A~R ' T = a 



and collect together the constant terms there results 

 log x + a x + m log T — log e m — constant 



The coefficient a which is contained as a factor in the coefficient 

 a can be taken from the same table 5, column 5, as a function of /. 



The quantity of moisture £ which enters the value of m n in 

 equation (7) is the sum of the vapor x and the water y, therefore 

 the difference £ — x = y gives the quantity of the condensed water. 



The rain stage attains its upper boundary when the temperature 

 has fallen to o° C. If now the liquid water has remained in the air 

 then it enters the isothermal hail stage in which the" water freezes 

 to ice. 



As an example we will now follow the mass of air, hitherto 

 considered during its further expansion within the rain stage. 



For the point of saturation, which is the initial point of the rain 

 stage, we had found t = 17.0 ; p = 733. 2 mm ; e = i.44 mm ; 

 p' = 7i8.8 mm ; £ — x x « 12.5 grams. 



During the rain stage the connection between pressure and tem- 

 perature is given by the equation of the adiabat (9) : from the table 

 4, column 5, we have in this case w n = 3.62. 



If now the expansion goes on further until the temperature cools 

 to io° C. then a = 76.0 and equation (9) in this case becomes 



76.0 

 log p' - - , = 2.6573 



From this we obtain the partial atmospheric pressure p' = 606. 3 mm 

 and p = 615.5, since e = 9.2 mm when t = io°. 



The amount of the vapor still present is 9.4 grams, hence the 

 vapor that has been condensed to water is y = £ — x = $.1 grams. 



The volume now amounts to V = 1.005 cbm. and the specific 

 volume v = 0.993 cbm. For further expansion to o° C. or to the 



