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hw29b - aljabr(faa335 – Hw29 – Ross –(89251 1 This...

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Unformatted text preview: aljabr (faa335) – Hw29 – Ross – (89251) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Three moles of a certain diatomic molecu- lar gas are heated at constant pressure from 300 K to 442 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: 12268 . 8 J. Explanation: Let : n = 3 mol , T i = 300 K , T f = 442 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 ) = (3 mol) · (28 . 8 J / mol · K) · (442 K − 300 K) = 12268 . 8 J . 002 (part 2 of 3) 10.0 points Calculate the increase in its internal energy. Correct answer: 8726 . 82 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 ) = (3 mol) · (20 . 4855 J / mol · K) · (442 K − 300 K) = 8726 . 82 J . 003 (part 3 of 3) 10.0 points Calculate the work done by the gas....
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hw29b - aljabr(faa335 – Hw29 – Ross –(89251 1 This...

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