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5.27X104. Suppose the molecules of a gas at 600. K can exist in two possible energy states: E = 2.49x100, 2 Calculate the fra
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states: Given : The temperatue o gos = 600k Two possible energy sta E = 2.49x10 ²2 J E = 5.27X 10 22 J Solution : - I Fractioof stato, i. Ei is the the so energy se 6 -BER Degeneracy of Energy ?=1 Degeneracy of Energy 2 = 1 BEI kot (1:38065x10-235/K)Function is het Ni = le - (1. 2072x102 ) (2.49x1022 ) 1.9088 -0.03006 se 1.9088 6.9704 1.9088 y = 0.5084 Faaction in e BE -(1

The fraction of molecule in E1 using the Boltzmann distribution equation = 0.5084

The fraction of molecule in E2 using the Boltzmann distribution equation = 0.4916

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