exam2cor - October 24, 2006 Corrections to Midterm # 2...

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October 24, 2006 Corrections to Midterm # 2 solution: Q1B.2: There was a numerical error in the solution…… 1, 2, 3, 1, 3, () (0 1 3) 4 (1 4 1) 6 7.42 AAA BBB uuu pA e e e pB e e e −++ −+ + == = = Q3B.3: There was a numerical error in the solution…… 10.5 7.5 3 2 2 181 2 AB BC L L Q4A: W left = 8! 2!6! = 28 W right = 4! 1!3! = 4 W total = W right x W left = 28 x 4 = 112 If one move piston one to right, W total,right = 36 x 3 = 108 If one move piston one to left, W total,left = 21 x 5 = 105 The system is at equilibrium since entropy is maximized.
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October 24, 2006 Q4B: There is an alternative way to solve this problem besides the solution provided and it yields the same answer. Compute entropy of both the left side and right side of the system A and B. Compute total entropy of systems A and B by adding the entropy of the left side and right side for each respective system. To calculate the change in entropy, simply subtract the total
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exam2cor - October 24, 2006 Corrections to Midterm # 2...

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