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Unformatted text preview: Iype, Shelby – Homework 8 – Due: Oct 31 2007, midnight – Inst: Vandenbout 1 This printout should have 33 questions. Multiplechoice 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 100kPa with every 10.2m of depth. This means that at 10.2m below the surface, the pressure is 201kPa; at 20.4m below the surface, the pressure is 301kPa; and so forth. If the volume of a bal loon is 3 . 4 L at STP and the temperature of the water remains the same, what is the vol ume 39 . 15 m below the water’s surface? Correct answer: 0 . 710578 L. Explanation: P 1 = 1 atm Depth = 39 . 15 m V 1 = 3 . 4 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2m 100kPa = 39 . 15 m x (10 . 2m)( x ) = (39 . 15 m)(100kPa) x = (39 . 15 m)(100kPa) 10.2m = 383 . 824 kPa P 2 = 101kPa + 383 . 824 kPa = 484 . 824 kPa × 1atm 101.325kPa = 4 . 78484 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm)(3 . 4 L) 4 . 78484 atm = 0 . 710578 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. 24 mm Hg 3. 2400 mm Hg correct 4. 0.042 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 At standard temperature, a gas has a volume of 336 mL. The temperature is then increased to 119 ◦ C, and the pressure is held constant. What is the new volume? Correct answer: 482 . 462 mL. Explanation: T 1 = 0 ◦ C + 273 = 273 K V 1 = 336 mL T 2 = 119 ◦ C + 273 = 392 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (336 mL)(392 K) 273 K = 482 . 462 mL 004 (part 1 of 1) 10 points A sample of gas in a closed container at a temperature of 85 ◦ C and a pressure of 8 atm is heated to 267 ◦ C. What pressure does the gas exert at the higher temperature? Correct answer: 12 . 067 atm. Explanation: T 1 = 85 ◦ C + 273 = 358 K P 1 = 8 atm T 2 = 267 ◦ C + 273 = 540 K P 2 = ? Applying the GayLussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (8 atm)(540 K) 358 K = 12 . 067 atm Iype, Shelby – Homework 8 – Due: Oct 31 2007, midnight – Inst: Vandenbout 2 005 (part 1 of 1) 10 points A gas at 1 . 84 × 10 6 Pa and 14 ◦ C occu pies a volume of 380 cm 3 . At what tem perature would the gas occupy 530 cm 3 at 3 . 11 × 10 6 Pa? Correct answer: 403 . 576 ◦ C. Explanation: P 1 = 1 . 84 × 10 6 Pa P 2 = 3 . 11 × 10 6 Pa V 1 = 380 cm 3 T 1 = 14 ◦ C + 273 = 287 K V 2 = 530 cm 3 T 2 = ?...
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This note was uploaded on 03/30/2009 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas.
 Spring '07
 Fakhreddine/Lyon
 Chemistry

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