Solution of Atmospheric Nitrogen and Oxygen by Water. 335 



Hence the area of water air-surface was 56*34 sq. cm. 



When a gas is dissolving in a liquid, we may assume that 

 the rate of passage of the gas into the liquid is proportional 

 to the partial pressure of the gas and the area of liquid 

 exposed. Hence: — 



Rate of passage of gas into liquid = S . A . p, where 

 je> = partial pressure of gas, A = area of surface, and S = rate 

 of solution for unit area. 



As solution goes on, the gas in the upper layers of the 

 water escapes into the air, and the rate of its escape is pro- 

 portional to the amount of gas in solution ; hence if w = weight 

 of gas per c.c. in upper layer, then the rate of escape of gas from 

 liquid =f.w .A. This gives us as the net rate of solution — 

 S.A.p — f.w.A; and when equilibrium is reached, i. e. at 

 saturation, S . A . p=f. w . A or Sp=fu\ 



The value of u w" is generally unknown, since the gas 

 diffuses rapidly from the surface layer of the liquid and the 

 exact gas-content is uncertain. If we keep the liquid mixed, 

 we render A the area uncertain in general. But if a method 

 of mixing the liquid which would leave A still determinate is 

 possible, then w T e can calculate the rate of solution for a given 

 .area, assuming that the gas remains at constant density. 

 These conditions are complied with in the case of a cylindrical 

 bubble moving up a narrow tube. 



If V = volume of liquid and p = density of the gas (assumed 

 constant), the rate of solution is: — 



dVw . . 



-^ = SAp-f w A 



= SAkp-fwA, 

 dw _ SA 7 j. A 



••• Tt--i k P- fw r 



which z=a~-bw ... (1) when a = -rf-kp and £>=/.—, 



w— r '= Ce~ bt , iv = when i = 0, hence c= — - , 

 „\ w= j (1 —e~ lt ) } when £=oo , w = the saturation value. 



Equation (1) shows that plotting the values of , against 



2 A 2 



