homework29 - Rodriguez, Brandon Homework 29 Due: Dec 5...

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Unformatted text preview: Rodriguez, Brandon Homework 29 Due: Dec 5 2006, 11:00 am Inst: Andrew J Rader 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA- TIONS TO AT LEAST THREE SIGNIFI- CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 2) 4 points Air in the cylinder of a diesel engine at 47 . 8 C is compressed from an initial pres- sure of 1 . 232 atm and of volume of 820 cm 3 to a volume of 32 . 5 cm 3 . Assuming that air behaves as an ideal gas ( = 1 . 40) and that the compression is adia- batic and reversible, find the final pressure. Correct answer: 113 . 061 atm. Explanation: Let : P i = 1 . 232 atm , V i = 820 cm 3 , and V f = 32 . 5 cm 3 . Using the relation P i V i = P f V f , we find that P f = P i V i V f = (1 . 232 atm) 820 cm 3 32 . 5 cm 3 1 . 40 = 113 . 061 atm 002 (part 2 of 2) 4 points Find the final temperature under the same assumptions as above. Correct answer: 893 . 827 C. Explanation: Let : T i = 47 . 8 C = 320 . 8 K . Since P V = n R T is always valid during the process and since no gas escapes from the cylinder, P i V i T i = P f V f T f = n R Therefore, T f = P f V f P i V i T i = (113 . 061 atm)(32 . 5 cm 3 ) (1 . 232 atm)(820 cm 3 ) (320 . 8 K) = 1166 . 83 K = 893 . 827 C keywords: 003 (part 1 of 2) 4 points...
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.

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homework29 - Rodriguez, Brandon Homework 29 Due: Dec 5...

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