You do not need to remove the gas.
I know you have heard of Bremsstrahlung, even if the word is almost unspellable to Anglos. Thank heavens for spell checkers and Wiki vids. Here is a little video that tells you why Bianchini would see tritium, if it was there. https://www.youtube.com/watch?v=yYLzarnlcUE It is not the beta decay which is seen - but instead it is the secondary gammas aka Bremsstrahlung . and yes - they would be on the edge of detectability, but a signal should show up above background on his meter - especially when the Rossi device is disassembled, as it is in the Penon report. From: Jed Rothwell Jones Beene wrote: No. That is not correct. Tritium would have already have been detected by Bianchini if it was there . . . I do not think so. Tritium would be trapped inside the cell. The decay product is a low energy beta. If a little tritium leaks out of the cell it is not likely to reach the detector, which only covers a small amount of the surface surrounding the cell. The only way Bianchini could detect this would be if Rossi makes a cell with a high quality tube and connectors to the cell contents and allows Bianchini to sample the gas. That is also the only way anyone could detect an increase in deuterium or any other gaseous nuclear product. This is a very difficult and involved thing to do. You have to purge the tube and other hardware. You have to use Swaglok connectors and you have to pay fanatical attention to cleanliness. If you touch any part of metal where the gas will flow, your fingerprint will contain more hydrogen than all of the reaction products from several days of high temperature heat production. Consider this: assuming the ratio of heat to helium is the same as plasma fusion, a Pd-D automobile that runs for a year, producing as much heat as the average gasoline burning automobile, will consume roughly 1 g of D2O. That's 48 million miles per gallon of D2O. - Jed
