HW12SOLUTIONS - Niu(qn269 – HW12 – Tsoi –(58020 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Niu (qn269) – HW12 – Tsoi – (58020) 1 This print-out should have 19 questions. Multiple-choice 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....
View Full Document

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.

Page1 / 6

HW12SOLUTIONS - Niu(qn269 – HW12 – Tsoi –(58020 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online