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Unformatted text preview: 1 eV = 0.160 x 1018 J n/N = exp ( 0.68x0.160x1018 /13.8x1024 x 933) = 2.1x104 . 1. A crystal has an enthalpy of formation for vacancies of 1.5 eV, establishing a certain equilibrium concentration of vacancies at the temperature 1200 K. By how much does the temperature have to be raised to increase the vacancy concentration by a factor of 10? n/N = exp (1.5 x0.160x1018 /13.8x1024 x1200) = exp(14.49) = 5.094x107 10n/N = 5.094x106 = exp [1.5 x0.160x1018 /13.8x1024 x (1200+t)] = exp [ 17391/(1200+t)]12.187 =  17391/(1200+t); 1200+t = 1427 t = 227 ° C 2. The figure on the next page represents a (100) section through the structure of CaO, which has the NaCl (rock salt) structure type. Five defects are shown in the structure. Write the defect symbols using the common convention, e.g. V Na ’ V O .. A B ’. V ca ”, V O .. , F O . , Fe Ca . , Li Ca ’...
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 Spring '08
 Tomozawa
 Thermodynamics, Enthalpy, Sodium chloride, black balls

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