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Homework 10 Solution

# Homework 10 Solution - Version PREVIEW Homework 10...

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Version PREVIEW – Homework 10 – Savrasov – (39825) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Carnot Efficiency 001 10.0 points A heat engine operates between two reservoirs at 18 . 9 C and 168 C. What is the maximum efficiency possible for this engine? Correct answer: 0 . 338095. Explanation: Given : T h = 168 C = 441 K and T c = 18 . 9 C = 291 . 9 K . Efficiency e = 1 - T c T h = 1 - 291 . 9 K 441 K = 0 . 338095 . Entropy in Expansion 002 10.0 points 58 . 1 mol of an ideal gas is allowed to undergo a free expansion. If the initial volume is 20 cm 3 and the final volume is 100 cm 3 , find the change in entropy. Correct answer: 777 . 054 J / K. Explanation: Given : n = 58 . 1 mol , V 1 = 20 cm 3 , and V 2 = 100 cm 3 . The free expansion is not quasi-static; we have to use a quasi-static process connecting the initial and the final equilibrium states to calculate the change of entropy. Since the two states have the same temperature, we connect them with an isothermal process which gives Δ S = n R ln parenleftbigg V 2 V 1 parenrightbigg = (58 . 1 mol)(8 . 31 J / mol · K) ln parenleftbigg 100 cm 3 20 cm 3 parenrightbigg = 777 . 054 J / K .

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Homework 10 Solution - Version PREVIEW Homework 10...

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