EXAMPLE. Ill 



calling p and p' the densities of the gas, we have, from (3) 

 of Art. 54, 



P-p = kp(l + at), 



P r -p' = Tcp' (1 + at') ; 

 , from (1) of Art. 48, we have vp = v'p'. 

 P' p' v l + at' 



P-p ' v'l + at' 

 which gives the value of P'. 



(1) 



COR. If a and o' be the densities of vapor under the two 

 conditions, we have 



p o(l + a 

 Dividing (1) by (2), we get 



p P' p' vo 



~~ 



v'o' Pp' pp' 



or, = ^ - , (3) 



vo Pp pp 



If Pp' > P'p, v'o' will exceed vo ; i. e., more vapor will 

 have been absorbed by the gas. But if Pp' < P'p, then 

 v'o' will be less than vo, and the gas must therefore, in 

 changing its volume and temperature, have lost a portion 

 of its vapor. (See Besant's Hydrostatics, p. 138.) 



EXAMPLE. 



Having given the pressures P and p of a volume v of 

 atmospheric air, and of the vapor it contains, to find the 

 volume of the air, without its vapor, at the same pressure 



P, the temperature remaining constant. 



P p 

 Ans. Volume of air = r> ^ 



