ProbSet7_solutions(2)

What is the multiplicity e3 200 molecules e2 1100

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Unformatted text preview: system? What is the difference in energy between the two levels? Answer: i) Initial entropy = -Ntkb∑pilnpi = -100000J/K x (2x0.5 ln 0.5) = 693 kJ/K Final entropy = -100000J/K x (0.65 ln 0.65+0.35 ln 0.35) = 647 kJ/K The change in entropy (∆S) is -46 kJ/K ii) ∆S = qrev/T ; qrev = ∆SxT = -46kJ/K x 293K = -13478 kJ As a check, we can verify that heat is removed from the system because the entropy of the system decreases during this process. iii) From Boltzmann distribution: − (U 2 −U1 ) P2 0. 5 = = e RT P1 0. 5 ΔU = − RT ln( = +1.53 kJ 0. 5 ) = −2.5 kJ * ln(0.54 ) 0. 5 20. What is the entropy of the following energy distribution? What is the multiplicity? E3 = 200 molecules E2 = 1100 molecules E1 = 10500 molecules Answer: We first compute the entropy using the probabilistic definition Initial entropy = -Ntkb∑pilnpi = -11800 J/K x (0.02 ln 0.02 + 0.09 ln 0.09 + 0.89 ln 0.89) = -11800 J/K x (-0.40) = 1220 J/K Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 13 S = -kblnW; W = eS/kb = e 1220 ~ 1060 . 21. A system has a partition function of 1.3 at 293K. At that temperature, an energy level is populated by 10% of the molecules. What is the energy of that level? Answer: e-u/kbT = p x Q; u = -kb x T x ln (p x Q) = -1J/K x 293K x ln (0.1x1.3) = -598J 22. Multiple Choice/TrueFalse 1) For a system converting from state 1 to state 2: (kBlnW2−kBlnW1)= qrev/T True/False 2) If the second lowest ene...
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