hw25a - aldawsari(ma28683 – HW25 – Betancourt –(16856...

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Unformatted text preview: aldawsari (ma28683) – HW25 – Betancourt – (16856) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 20.1-20.2 001 (part 1 of 3) 10.0 points Two moles of a certain diatomic molecular gas are heated at constant pressure from 290 K to 438 K. The heat capacity of this gas under con- stant pressure is 28 . 8 J / mol · K. The univer- sal gas constant is 8 . 31451 J / mol · K. Calculate the heat transferred to the gas. Correct answer: 8524 . 8 J. Explanation: Let : n = 2 mol , T i = 290 K , T f = 438 K , C P = 28 . 8 J / mol · K and R = 8 . 31451 J / mol · K . Note: Do not use the approximation C P = 7 2 R = 7 2 (8 . 31451 J / mol · K) = 29 . 1008 J / mol · K negationslash = 28 . 8 J / mol · K . The amount of heat transferred to the gas in the process with constant pressure is Q = nC P Δ T = nC P ( T f − T i ) = (2 mol) · (28 . 8 J / mol · K) · (438 K − 290 K) = 8524 . 8 J . 002 (part 2 of 3) 10.0 points Calculate the increase in its internal energy. Correct answer: 6063 . 71 J. Explanation: C V = C P − R = 28 . 8 J / mol · K − 8 . 31451 J / mol · K = 20 . 4855 J / mol · K . Note: Do not use the approximation C V = 5 2 R = 5 2 (8 . 31451 J / mol · K) = 20 . 7863 J / mol · K negationslash = 20 . 4855 J / mol · K . The change of internal energy for an ideal diatomic gas is Δ U = nC V Δ T = nC V ( T f − T i ) = (2 mol) · (20 . 4855 J / mol · K) · (438 K − 290 K) = 6063 . 71 J . 003 (part 3 of 3) 10.0 points Calculate the work done by the gas....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.

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hw25a - aldawsari(ma28683 – HW25 – Betancourt –(16856...

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