Page:Outlines of Physical Chemistry - 1899.djvu/149

 of the solvent corresponding to one gram-molecule of dissolved sub- stance is expelled through the semipermeable wall, and this requires a work to be done equal to rt. The liquid forced out is then frozen,

and thus the system is deprived of I \ calories (x being the latent heat of fusion of one gram of solvent). The temperature of the solu- tion is then lowered to (t - A)° and the I grams of frozen solvent

n

dropped into it and made to fuse by furnishing the necessary heat. The heat of fusion formerly taken from the system, is thus given back to it, but at a lower temperature. The temperature of the system is then raised to t° and the initial state thus re-established. Here, again, we have a reversible cycle which leads to the equation :

�� �BT __ A

n

�From which

�A _ b t 2 n

�But

�A = c'.?.

�Therefore \

��or, substituting 2 for r as before, c' = ?2l

Remark.— In the chapter on Cryoscopy we made use of the con- stant c. As this refers to one molecule of substance dissolved in 100 grams of solvent, it is 100 times smaller than c' and, conse- quently, c = —.

D. Caution.— By combining the relationships found under a, b and c, we can find some very interesting connections between/ V m, n, g, A (for boiling or freezing point), t and s{also for boiling or freezing point). But these deductions are only exact for extremely dilute solutions whose densities can, without appreciable error be set equal to that of the solvent. Experiments can hardly be per formed on such dilute solutions, and so we must look for discrepancies between the theory and what is actually found.

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