# TEST4 - Version 078/ABADC TEST4 Tsoi(58020 This print-out...

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

Version 078/ABADC – TEST4 – Tsoi – (58020) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A helium-flled balloon has a volume oF 1 m 3 . As it rises in the Earth’s atmosphere, its vol- ume changes. What is its new volume iF its original tem- perature and pressure are 32 C and 1 . 6 atm and its fnal temperature and pressure are - 26 C and 0 . 4 atm? 1. 0.359202 2. 3.31507 3. 0.753125 4. 0.837191 5. 0.367759 6. 1.25638 7. 0.335075 8. 0.977586 9. 0.911141 10. 3.23934 Correct answer: 3 . 23934 m 3 . Explanation: Given : V i = 1 m 3 , T i = 32 C = 305 K , P i = 1 . 6 atm , T f = - 26 C = 247 K , and P f = 0 . 4 atm . Using ideal gas law P V = nRT we obtain P i V i T i = P f V f T f V f = P i V i T f T i P f = (1 . 6 atm) ( 1 m 3 ) (247 K) (305 K)(0 . 4 atm) = 3 . 23934 m 3 . 002 10.0 points A gold ring has an inner diameter oF 1 . 6214 cm at a temperature oF 21 C. Determine its inner diameter at 129 C. (The coe±cient oF linear expansion is 1 . 42 × 10 5 ( C) 1 .) 1. 2.05116 2. 1.41082 3. 1.6478 4. 2.63208 5. 1.76277 6. 1.2111 7. 1.62389 8. 2.76597 9. 2.67144 10. 2.14839 Correct answer: 1 . 62389 cm. Explanation: Given : L 1 = 1 . 6214 cm , T 1 = 21 C , T 2 = 129 C , and α = 1 . 42 × 10 5 ( C) 1 . The diameter expands by Δ L = αL 1 Δ T = αL 1 ( T 2 - T 1 ) = b 1 . 42 × 10 5 ( C) 1 B × (1 . 6214 cm) (129 C - 21 C) = 0 . 00248658 cm . and L 2 = L 1 + Δ L = 1 . 6214 cm + 0 . 00248658 cm = 1 . 62389 cm . 003 (part 1 oF 2) 10.0 points A vertical cylinder oF cross-sectional area 0 . 2 m 2 is ftted with a tight-ftting, Frictionless piston oF mass 35 kg. The acceleration oF gravity is 9 . 8 m / s 2 .

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

View Full Document
Version 078/ABADC – TEST4 – Tsoi – (58020) 2 h gas m If there are 2 mol of an ideal gas in the cylinder at 363 K, Fnd the pressure inside the cylinder. (Assume that the system is in equilibrium.) 1. 101572.0 2. 106880.0 3. 124520.0 4. 108644.0 5. 137260.0 6. 111045.0 7. 102715.0 8. 121580.0 9. 114720.0 10. 104296.0 Correct answer: 1 . 02715 × 10 5 Pa. Explanation: Given : A = 0 . 2 m 2 , m = 35 kg , R = 8 . 314 J / Kmol , and P atm = 1 . 01 × 10 5 Pa . In equilibrium, P gas A = mg + P atm A P gas = mg A + P atm = (35 kg) ( 9 . 8 m / s 2 ) 0 . 2 m 2 + 1 . 01 × 10 5 Pa = 1 . 02715 × 10 5 Pa . 004 (part 2 of 2) 10.0 points At what height will the piston be in equilib- rium under its own weight? 1. 0.334885
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

TEST4 - Version 078/ABADC TEST4 Tsoi(58020 This print-out...

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

View Full Document
Ask a homework question - tutors are online