examII07 - NAME and Section No. Chemistry 391 Exam II KEY...

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––––––––––––––––––––––––––––––– NAME and Section No. 1 Chemistry 391 Fall 2007 Exam II KEY 1. (30 Points) ***Do 5 out of 7 ***(If 6 or 7 are done only the first 5 will be graded)*** a). How does the efficiency of a reversible engine compare with that of an irreversible engine? How can you improve the efficiency of an irreversible engine? The efficiency of an irreversible engine must be less than that of a reversible, with equality only when both are reversible. The only way to improve the efficiency of an irreversible engine is to make it operate closer to reversibly, e.g., by making it operate more slowly. b). A refrigerator is a Carnot cycle run “backwards”. That is, heat is now withdrawn from the cold reservoir at cold T and is deposited in the hot reservoir at hot T . The coefficient of performance of a refrigerator η is defined as cold cycle q w . Express in terms of cold T and hot T . Since 0 cycle cold hot dU w q q == + + v , and since cold hot cold hot qq TT =− () cold cold cold cycle cold hot hot cold T wq q T T = −+ c). You are told that Δ S sys = 0 for a process in which the system is coupled to its surroundings. Can you conclude that the process is reversible? Justify your answer. No. The criterion for reversibility is Δ S sys + Δ S surroundings = 0. To decide if this criterion is satisfied, Δ S surroundings must be known.
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––––––––––––––––––––––––––––––– NAME and Section No.
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examII07 - NAME and Section No. Chemistry 391 Exam II KEY...

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