D O C U M E N T 1 9 2 M AY 1 9 2 8 1 8 7 small light source likewise do not appear so promising that I would venture to sug- gest to an experimental physicist that he invest his time in them.[3] The most expe- dient course would be to publish the idea as such in a scientific journal, taking note of the practical difficulties in implementing it.[4]— Respectfully yours, 192. From Walther Meißner [1] Berlin-Friedenau, Rotdornstr. 15 May 1928 Dear Professor, You will have been waiting for news about the results of the calculations concerning gas degeneracy and the law of corresponding states.[2] The computa- tion, which was initially very tedious, unfortunately has shown that making use of the van der Waals equation for determining the critical constants in the p,V,Tx diagram (Tx ) is not appropriate,[3] since the deviations of the resulting critical constants from the experimental values are too large to permit us to neglect the errors owing to the van der Waals equation. For example, according to your theory, for helium[4] [5] [6] and, strangely enough, accord- ing to Fermi,[7] . Thus, according to both theories, does not reach the value of , which it should have for helium[4] so that the law of corresponding states would remain valid at the critical point (i.e., so that takes on the normal R value). But this may just be a result of the inadequacy of the van der Waals equation.— If one determines the result for , using the given values of , , and , and furthermore for the point [8] , mak- ing use of the isotherms of Holborn and Otto,[9] then one finds, according to your theory, according to Fermi, without corrections for the quantum effects, that is, in the system, while at the same cor- responding-state point for nitrogen, where the quantum effects can be neglected, results. Strangely, thus has the same sign according to your T 1 V–1T–3 /2  = V k x V k 1 0,10 +  = T kx x T k 1 0,176 –  = px k p k 1 0,26 –  = Rx R----------- T kx x pxVx k k R 1 0,01 + = = V k x V k 1 0,095 –  = Tx kx T k 1 0,185 +  = px k p k 1 0,31 +  = Rx R 1,00  = Rx R 1,045  Rx px p p k x ---- - = px k Vx k T kx x tx T T kx x ------ - 2,85 = = vx V V k x ----- - 3,14 = = px 2,81 = px 2,89 = p  T px 2,72 = p px 3,22 = = px p –