160 ENERGY CHANGES INVOLVED IN SECRETION 



Method of Estimating the Work done against Osmotic Pressure in 

 separating each Constituent of a Secretion. The amount of work 

 done in separating each constituent of a secretion can easily be 

 deduced when the pressure of the substance in the lymph and 

 in the secretion are known, and the total volume of the secretion. 

 But such estimation must be made for each constituent of the 

 secretion separately, and the total work done is the sum of the 

 work 'done in the separation of each constituent. It leads to 

 quite a fallacious result to merely take the two depressions of 

 freezing point of the lymph and secretion respectively, calculate 

 the total osmotic pressure of lymph and secretion from these two 

 values, and then assume that the work done is the product of 

 the volume of the solution and the difference in pressure. For 

 the amount of volume energy change, as has been pointed out in 

 a previous chapter, depends upon the two pressures for each 

 constituent between which pressure has varied for that particular 

 constituent, and since in the formation of a secretion the same 

 ratio is not preserved between the pressures of the various con- 

 stituents as exists in the lymph, but one constituent is far more 

 compressed or concentrated than another, it cannot be taken 

 that the lymph is compressed or concentrated as a whole as it 

 were by a piston impermeable to all the dissolved constituents, 

 and the work done obtained from the total initial and final osmotic 

 pressures and the change in volume, but instead the work done 

 upon each pressure giving constituent must be taken separately, 

 and the total work calculated as the sum of all these fractions. 



As demonstrated in a previous chapter, the work done when 

 a grm. molecule of substance is compressed from pressure p l to 



pressure p^ is given by the expression RT log , and if Q be any 



Pi 

 other weight in grm. of the substance and M the molecular weight, 



Q 



then the number of grm. molecules will be ^r, and the expression 



for the amount of work done in changing the pressure of the 

 quantity Q grm. in solution at pressure p l to pressure p. 2 will 



If now there are any number of substances A, B, C ...... N in 



solution in the secretion in quantities Q a , Q^, Q c ...... Q n , and 



the molecular weights of the substances be M ft , M />? M c ...... M n , 



