From: fsmith@morgan.ucs.mun.ca (Frank Smith)
Subject: Re: Possible Gas Absorption Experiments
Summary: Error limits and other comments  
Keywords: palladium deuterium calorimetry pressure  
Organization: Memorial University of Newfoundland
Date: Tue, 15 Sep 1992 18:47:47 GMT

In article <920914152522.24a011a8@FNALD.FNAL.GOV> ames!FNALD.FNAL.GOV!DROEGE writes:
>One thing missing from Frank's summary are error limits on the measurement.  
>Since by my estimate they generate about 5000 joules in 10 minutes, this is a 
>power level of order 10 watts and they should be able to measure this to 0.1% 
>with their type of calorimeter.  What I do not understand, is how they know 
>the D/Pd ratio very accurately since it seems to be done with a pressure 
>measurement.  Possibly pressure gages have kept up with electronics since I 
>left school, but I doubt it.  Perhaps they heat the container after the 
>experiment and measure the evolved D?  I do notice that they give the energy 
>measurements to one more decimal place than the D/Pd ratios.  
>
The quoted error is a standard deviation of 0.25 kJ/mol D for 298 K. They do
some expts at 249 K and find the enthalpy of absorption is -19.1 kJ/mol D with
sigma of 0.29, while at 195 K it is -20.2 kJ/mol D with sigma of 0.49. They
comment that the rates of absorption of hydrogen were fast at 195 K as they
were at 298 K. As to pressure gauges, they are based on some electrical
principle that for the present escapes me, but I shall consult a colleague
about this. I do not believe they are of low accuracy.
>One thing that I do not understand is the evolution half of the experiment.  
>How is the H/D removed from the Pd so quickly?  I assume they put a vacuum in 
>the dosing volume.  My experience at atmospheric pressure is that it takes 
>many days for a charged sample to outgas.  I guess the thin foil is the 
>answer.
>
Yes, Tom you are right. Using Pd as a thin foil, cut into small pieces and
mixed in with essentially inert copper, they get low diffusion paths and hence
fairly rapid equilibration. As I understand the experiment, the pressure of D2
is gradually raised in the deuterium absorption direction and gradually lowered
in the desorption (evolution, they call it) direction. No vacuum would be
applied, but from a knowledge of the pressure and the container volume they
could work out the number of moles of D2 taken up, double it to get moles of D
and then compare to the known moles of Pd. I suspect that you use a smaller
mass of Pd than they do? Their main interest, by the way, is the plateau
region. But they do mention agreement of their calorimetric data with the
van't Hoff plots of Wicke and Nernst, Ber. Bunsenges.Phys. Chem. 68(1964) 224.
I must look at this work again as there is some allusion to there being no
excess heat production! The book by R. Lasser,"Tritium and Helium-3 in Metals",
Springer, Berlin,1989 sounds interesting as well.
>While they could try to get absorption heat as a function of D/Pd ratio by 
>comparing evolved heat and pressure, time lags in such a system would make it 
>very difficult to unscramble the data.  I assume that they have done many 
>experiments each with a different starting pressure in the dosing volume in 
>order to get enthalpy of absorption as a function of D/Pd ratio.  Frank, 
>please, what is the highest D/Pd ratio measured? 
>
Again you are right to an extent: they quote 29 experiments with D2 at 298 K
over the two-phase region. This would seem to mean 29 calorimetric measurements
with the D2 dose increased slightly in each, but remaining in the alpha-beta
composition range. The highest H/Pd ratio in the paper is 0.81 at 269 K, at
which temperature they did no D2 experiments. Otherwise, it is as I wrote last
time: D/Pd = 0.755, although higher ratios were measured for H/Pd.
>One thing that I could do would be to make a continuous measurement of 
>enthalpy of absorption as a single sample is charged.  The Flanagan experiment 
>looks like it gives the Pd a sudden shock.  If this is damaging to the 
>structure then a slow charge might give a different answer.  It seems that 
>slow charging results in higher final D/Pd ratios in electrolytic cells.  
>
I don't think a shock is involved in their experiments as they are essentially
reversible (apart from the hysteresis). But you are introducing D in an atomic
form whereas their experiments involve it dissociating first, so they are
likely to be slower than you, contrary to my question last time.

Frank R. Smith 

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