A text-book of physical chemistry / by R.A. Lehfeldt.

  • Robert Alfred Lehfeldt
Date:
[pref. 1899]
Catalogue details

Licence: Public Domain Mark

Credit: A text-book of physical chemistry / by R.A. Lehfeldt. Source: Wellcome Collection.

Provider: This material has been provided by University of Bristol Library. The original may be consulted at University of Bristol Library.

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    but the density of the vapour = where v as usual is tlic molecular volume, and this by the characteristic equation is = Hence p ^jZR or = p m p RTD' (n) — p \ For the same reason as before, the fraction ——— repre- P senting the relative lowering of vapour pressure should, exactly, be log - log ]\ = • Combining this vnth the result just found for the change in boiling point, we have mP mlt „ ItD or P = BTD RT,T As examples of this result take for aqueous solutions : I = 536-4 calories = 536-4 x 42,000,000 ergs ; Z) = 0-959 (at the boiling point) ; 273+ 100. „ 586-4x42,000,000x0-959 Hence P = t = 57,900,000^ dynes per sq. cm. = 57-9t atmospheres; i.e. a rise of one degree in the boiling point implies an osmotic pressure of 57-9 atmos. For benzene, the latent heat being much smaller, the change of boiling point is much greater. Z = 93-4 calories ; D = 0-815 ; T = 273 + 80 ; 93-4x42,000,000x0-815 353 = 9,060,000« in C. G. S. units = 9-06< atmospheres. Keturning to the practical problem of determining molecular weights, we may use the elevation of the boiling
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