OK, we are agreed that P1V1/T1 =P2V2/T2 even applies to state changes of gas atoms. Normally Anything that effects the volume of the overall population but here is where COE meets Casimir effect
Which allows monatomic gas to translate "Freely" with very little opposition to fractional states, We can ignore the argument about this being relativistic or not and just treat this as a property of a Pd membrane that allows monatomic gas to pass but is a barrier to diatomic gas. My posit is that when f/h1 becomes f/h2 it finds itself surrounded by barriers where it must remain locked until it is again disassociated by random thermal energy. Unlike normal h2 this f/h2 does have one additional energy gradient that it can tap to help it translate back to normal h2. The sum of vacuum fluctuation energy that determined it's fractional value is decreasing as it leaves the cavity boundaries and enters the lattice, that differential wants to reshape the orbitals of the f/h2 but is prevented from doing so by the diatomic bond. I am humbly suggesting these f/h2 or f/d2 can take up positions normally occupied by a single d1 and that defects and cavities can be considered the pump houses where fractional diatomic bonds are utilized as containment vessels. Regards Fran Jones Beene Wed, 12 May 2010 20:20:58 -0700 -----Original Message----- From: Abd ul-Rahman Lomax > Yes, I understood that. The heat, though, doesn't come from expansion of hydrogen. Wrong. Some of the heat does come from expansion of hydrogen. Of course much more comes from combustion. >When air initially at a moderate temperature is expanded through a valve, >its temperature decrease because it starts out below the inversion >temperature of the constituent gases, and the expansion will cause a >temperature reduction as the result of the Joule-Thomson effect. Any gas >expanded at constant enthalpy will experience a temperature decrease ONLY if >is below the inversion temperature, however, and if above it will usually >experience a temperature increase. > I was not assuming endothermy from expansion, but from evaporation (more like sublimation in this case). There is no difference. > It's simply the reversal of the heat released from absorption. If hydride/deuteride formation is exothermic, and it is, then de-formation is endothermic. Wrong. You are missing the balance point. The balance is between the energy used to pressurize the gas before loading and the net energy returned by both hydride formation and hydride release. If the energy needed to compress hydrogen for loading a tank is say 10 W-hrs, then that can be balanced exactly against 5 W-hrs of exotherm for deuteride formation and another 5 W-hrs of exotherm for expansion. Get it? > Note that failure to account for the heat of formation of palladium deuteride could be a possible source of error in the calorimetric analysis of CF experiments. You finally got something right! > But it would not explain heat after death. Partly wrong. It can explain some of it, but usually there is much more heat after death than can be explained by expansion above the inversion temperature. Jones
