Unformatted text preview: 6 × ln(0.32 × 10 6 ) + 0.68 × 10 6 × ln(0.68 × 10 6 ) ⎤ + k B ( Atotal )ln M ⎣ ⎦ = − kB [13.188 ] + kB ( Atotal )ln M The difference in entropy is given by: ΔS = S2 − S1 = − k B [(13.188 − 13.181) × 10 6 ] = − k B [0.007 × 10 6 ] Since the answer is negative, it means that State ii is less likely than State i (State ii has lower entropy). Now we can calculate the relative value of their multiplicities: Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 6 ⎛W ⎞ ΔS = k B (ln W2 − ln W1 ) = kB ln ⎜ 2 ⎟ ⎝ W1 ⎠ ⇒ ⎛W ⎞ ΔS = ln ⎜ 2 ⎟ = −0.007 × 10 6 = −7000 kB ⎝ W1 ⎠ ⎛W ⎞ ⇒ ⎜ 2 ⎟ = e−7000 ⎝ W1 ⎠ (B) What is the entropy change, in units of J K1, on going from case (i) to case (ii)? ΔS = − k B × 7000 = 1.38 × 10 −23 × 7000 = 9660 × 10 −23 J K 1 6. Consider a system with just two energy levels, 1 and 2. The system has 1 mole (6.0 x 1023) molecules distributed between these two energy levels, E1 and E2. kB = 1.38 x 1023 JK1 E1 = 0.0 J, E2 = 3.0 x 102...
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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 Berkeley.
 Spring '09
 Kuryian

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