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mcbC100A-fa08-mt3-Kuriyan_Krantz-soln

# mcbC100A-fa08-mt3-Kuriyan_Krantz-soln - UC BERKELEY CHEM...

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UC BERKELEY, CHEM C130/MCB 100A, FALL 2008, MID-TERM EXAM 3. YOUR NAME_____________________________ Page 2 of 9 QUESTION 1. (25 pts.) A system has three energy levels ( E1 , E2 and E3 ) given as multiples of the value of k B . Throughout the question particles are indicated as solid filled circles. (Q1A) 15 pts. Consider how 21 particles will distribute among the 3 energy levels when the system reaches thermal equilibrium with its surroundings at 27 ºC. (Q1A.i) 2 pts. What type of distribution will the system resemble at equilibrium? Fill in the blank. Boltzmann distribution (Q1A.ii) 8 pts. Calculate the precise number of particles in each energy level upon reaching equilibrium. Whole number rounding is appropriate, but the final total must be 21. Show your work. N i = number of particles in i th level N = total number of particles p i = probability of being in the i th level p i = N i /N = exp(-E i /(k B T))/Q N i = N/Q exp(-E i /(k B T)) Q = Σ exp(-E i /(k B T)) Energy levels are in multiples of k B T. Q = exp(0) + exp(-1) + exp(-2) = 1 + 0.367 + 0.135 = 1.5 N = N 1 + N 2 + N 3 = 21 N 1 = 21/1.5 × exp(0) = 14 N 2 = 21/1.5 × exp(-1) = 5.2 ≈ 5 N 3 = 21/1.5 × exp(-2) = 1.9 ≈ 2 (Q1A.iii) 5 pts. What is the Helmholtz free energy, A , of the distribution calculated in Q1A.i ? The temperature is 27 ºC. Express the free energy in units of k B . Show your work. A = U – TS T = 300 K U = Σ N i E i = 14(0) + 5(300) k B + 2(600) k B = 2700 k B S = k B ln W W = 21!/(14!5!2!) = 2.44 × 10 6 S = 14.7 k B A = 2700 k B – 300 × 14.7 k B A = -1712 k B E1 = 0 k B E2 = 300 k B E3 = 600 k B
UC BERKELEY, CHEM C130/MCB 100A, FALL 2008, MID-TERM EXAM 3. YOUR NAME_____________________________ Page 3 of 9 (Q1B) 4 pts. Rank these four possible 21-particle distributions in terms of the Helmholtz free energy, A , at 300 K. Place a number, 1, 2, 3 or 4, under each given depiction in the blank provided. ‘1’ corresponds to the smallest (or most negative), and ‘4’ corresponds to the largest (most positive) free energy, A . Compare these situations first before calculating anything. 3

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mcbC100A-fa08-mt3-Kuriyan_Krantz-soln - UC BERKELEY CHEM...

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