D O C U M E N T 1 0 7 N O V E M B E R 1 9 2 5 1 2 9 valid up to there from there onward, the entropy would remain constantly = 0. It is very peculiar that an analytic formula would yield this result. The course of the energy (E) and of the free energy is shown quali- tatively in the nearby drawing, for two different volumes . The dashed lines indicate the course it would classically continue along. In the pV-diagram, a curve is the limiting curve between the state of an ideal gas and one of almost complete degeneracy. Better said, the region lying to the left of it is not entered at all. As soon as an isotherm reaches the limiting curve from the right-hand side, it merges into the limiting curve.[4] The corner in the curves and is not really sharp but rather somewhat rounded the transitional re- gion is, as far as I can see, of the order of magnitude .— All of this is so strange that I am constantly suspecting that it’s wrong. But it seems to me—as a consequence of the initial posit—to be right after all. With best regards, yours respectfully and sincerely, E. Schrödinger Note:[5] The critical place always lies in the region of supersaturated vapor because, according to the Stern-Tetrode vapor pressure formula, the entropy for saturated vapor is still positive . –ΨT) ( V2 V1) ( p Const V3 5 -- - ------------- -= E T) ( p V) ( abscissa- 3N 2 ------ - ------------------- β 3N 2 ------ -= heat of vaporization T ----------------------------------------------- - = © ¹ § ·
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