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HW7answers - Cheung Anthony Homework 7 Due midnight Inst...

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Cheung, Anthony – Homework 7 – Due: Oct 27 2006, midnight – Inst: McCord 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Divers know that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa; at 20.4 m below the surface, the pressure is 301 kPa; and so forth. If the volume of a bal- loon is 3 . 2 L at STP and the temperature of the water remains the same, what is the vol- ume 54 . 82 m below the water’s surface? Correct answer: 0 . 507854 L. Explanation: P 1 = 1 atm Depth = 54 . 82 m V 1 = 3 . 2 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2 m 100 kPa = 54 . 82 m x (10 . 2 m)( x ) = (54 . 82 m)(100 kPa) x = (54 . 82 m)(100 kPa) 10.2 m = 537 . 451 kPa P 2 = 101 kPa + 537 . 451 kPa = 638 . 451 kPa × 1 atm 101.325 kPa = 6 . 30102 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (3 . 2 L) 6 . 30102 atm = 0 . 507854 L 002 (part 1 of 1) 10 points A gas is enclosed in a 10.0 L tank at 1200 mm Hg pressure. Which of the following is a reasonable value for the pressure when the gas is pumped into a 5.00 L vessel? 1. 600 mm Hg 2. 0.042 mm Hg 3. 2400 mm Hg correct 4. 24 mm Hg Explanation: V 1 = 10.0 L V 2 = 5.0 L P 1 = 1200 mm Hg Boyle’s law relates the volume and pressure of a sample of gas: P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 V 2 = (1200 mm Hg)(10 . 0 L) 5 L = 2400 mm Hg 003 (part 1 of 1) 10 points A gas at 62 C occupies 4 . 89 L. At what tem- perature will the volume be 3 . 6 L, assuming the same pressure? Correct answer: - 26 . 3742 C. Explanation: V 1 = 4 . 89 L V 2 = 3 . 6 L T 1 = 62 C + 273 = 335 K T 2 = ? V 1 T 1 = V 2 T 2 T 2 = T 1 V 2 V 1 = (335 K) (3 . 6 L) 4 . 89 L = 246 . 626 K C = K - 273 = - 26 . 3742 C 004 (part 1 of 1) 10 points At standard temperature, a gas has a volume of 316 mL. The temperature is then increased to 135 C, and the pressure is held constant. What is the new volume? Correct answer: 472 . 264 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 316 mL T 2 = 135 C + 273 = 408 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (316 mL)(408 K) 273 K = 472 . 264 mL
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Cheung, Anthony – Homework 7 – Due: Oct 27 2006, midnight – Inst: McCord 2 005 (part 1 of 1) 10 points A sample of gas in a closed container at a temperature of 76 C and a pressure of 7 atm is heated to 383 C. What pressure does the gas exert at the higher temperature? Correct answer: 13 . 1576 atm. Explanation: T 1 = 76 C + 273 = 349 K P 1 = 7 atm T 2 = 383 C + 273 = 656 K P 2 = ? Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (7 atm) (656 K) 349 K = 13 . 1576 atm 006 (part 1 of 1) 10 points A gas at 658 × 10 4 Pa and 13 C occupies a vol- ume of 754 cm 3 . At what temperature would the gas occupy 957 cm 3 at 9 . 03 × 10 4 Pa? Correct answer: - 268 . 018 C. Explanation: P 1 = 6 . 58 × 10 6 Pa P 2 = 9 . 03 × 10 4 Pa V 1 = 754 cm 3 T 1 = 13 C + 273 = 286 K V 2 = 957 cm 3 T 2 = ? P 1 V 1 T 1 = P 2 V 2 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 = (9 . 03 × 10 4 Pa) (957 cm 3 ) (286 K) (6 . 58 × 10 6 Pa) (754 cm 3 ) = 4 . 9816 K = - 268 . 018 C 007 (part 1 of 1) 10 points A sample of ideal gas occupies 250 mL at 25 C and 740 torr.
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