How much heat was lost answer change in entropy

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Unformatted text preview: Would this equation be true for such a system? Explain your answer. For the following questions, assume that the value of the Boltzmann constant, kB is 1.0. 12. A system starts with a multiplicity of 2000. 2kJ of heat are transferred into the system reversibly at 298K. What is the multiplicity now? Answer: Initial entropy = kB ln W = 1*ln2000 = 7.6 J/K Change in entropy = 2000J/298K = 6.7 J/K Final entropy = 14.3 J/K Final multiplicity = eS/kB = 1 623 346 13. A system starts with a multiplicity of 1099. Heat is transferred out of the system reversibly at 298K such that the final multiplicity is 105. How much heat was lost? Answer: Change in entropy = kBlnW2- kBlnW1 = 1*ln(105) – 1*ln(1099) = -216.4 J/K qrev = ∆SxT = -216.4J/K*298K = -64 487.2 J = 64.5 kJ was lost from the system 14. A system of non-interacting atoms has a positional multiplicity of 120 and an energetic multiplicity of 5. What is the total multiplicity? What are the positional, energetic and total entropy? Answer: Total multiplicity = positional multiplicity x energetic multiplicity = 120 x 5 = 600 positional entropy = kb l...
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This note was uploaded on 01/27/2011 for the course MCB 100A taught by Professor Kuryian during the Spring '09 term at University of California, Berkeley.

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