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Unformatted text preview: Niu (qn269) – HW12 – Tsoi – (58020) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points At 15 ◦ C, an aluminum ring has an inner di ameter of 7 cm and a brass rod has a diameter of 7 . 06 cm. If the temperature coefficient of expansion for brass is α b = 1 . 9 × 10 − 5 ( ◦ C) − 1 and the temperature coefficient of expansion for alu minum is α a = 2 . 4 × 10 − 5 ( ◦ C) − 1 , to what temperature must the ring be heated so that it will just slip over the rod? Correct answer: 372 . 143 ◦ C. Explanation: Given : T = 15 ◦ C , d a = 7 cm , d b = 7 . 06 cm , α b = 1 . 9 × 10 − 5 ( ◦ C) − 1 , and α a = 2 . 4 × 10 − 5 ( ◦ C) − 1 . The new length will be d b = d a (1 + α Δ T ) d b = d a + d a α Δ T T 1 = T + Δ T = T + d b d a d a α a = 15 ◦ C + 7 . 06 cm 7 cm (7 cm)(2 . 4 × 10 − 5 ( ◦ C) − 1 ) = 372 . 143 ◦ C . 002 (part 2 of 2) 10.0 points To what temperature must both be heated so that the ring just slips over the rod? Correct answer: 1787 ◦ C. Explanation: We need L Al = L brass for some Δ T . Using the same law, L Al (1 + α a Δ T ) = L brass (1 + α b Δ T ) d a + d a α a Δ T = d b + d b α b Δ T Δ T = T + d b d a α a d a α b d b α a d a α b d b = bracketleftbig 2 . 4 × 10 − 5 ( ◦ C) − 1 bracketrightbig (7 cm) bracketleftbig 1 . 9 × 10 − 5 ( ◦ C) − 1 bracketrightbig (7 . 06 cm) = 3 . 386 × 10 − 5 m / ◦ C so Δ T = T 2 T T 2 = T + Δ T = T + d b d a α a d a α b d b = 15 ◦ C + 7 . 06 cm 7 cm 3 . 386 × 10 − 5 m / ◦ C = 1787 ◦ C . 003 (part 1 of 2) 10.0 points An ideal gas occupies a volume of 366 cm 3 at 98 ◦ C and a pressure of 594 Pa. Determine the number of moles of gas in the container. Correct answer: 7 . 04785 × 10 − 5 mol. Explanation: Given : V = 366 cm 3 = 0 . 000366 m 3 , T = 98 ◦ C = 371 K , P = 594 Pa , R = 8 . 31451 J / Kmol , and N A = 6 . 02 × 10 23 mol − 1 . Using the equation P V = nRT , the number of moles of gas in the container is n = P V RT = (594 Pa) ( . 000366 m 3 ) (8 . 31451 J / Kmol) (371 K) = 7 . 04785 × 10 − 5 mol . Niu (qn269) – HW12 – Tsoi – (58020) 2 004 (part 2 of 2) 10.0 points Calculate the number of molecules in the con tainer....
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This note was uploaded on 09/01/2009 for the course PHY M taught by Professor Staff during the Spring '09 term at University of Texas.
 Spring '09
 Staff
 Physics

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