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HW-8 - Rahman Tarique Homework 8 Due midnight Inst...

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Rahman, Tarique – Homework 8 – Due: Oct 31 2007, midnight – Inst: Vandenbout 1 This print-out should have 33 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 2 . 9 L at STP and the temperature of the water remains the same, what is the vol- ume 57 . 19 m below the water’s surface? Correct answer: 0 . 444081 L. Explanation: P 1 = 1 atm Depth = 57 . 19 m V 1 = 2 . 9 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2 m 100 kPa = 57 . 19 m x (10 . 2 m)( x ) = (57 . 19 m)(100 kPa) x = (57 . 19 m)(100 kPa) 10.2 m = 560 . 686 kPa P 2 = 101 kPa + 560 . 686 kPa = 661 . 686 kPa × 1 atm 101.325 kPa = 6 . 53034 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (2 . 9 L) 6 . 53034 atm = 0 . 444081 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. 2400 mm Hg correct 3. 24 mm Hg 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 290 mL. The temperature is then increased to 138 C, and the pressure is held constant. What is the new volume? Correct answer: 436 . 593 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 290 mL T 2 = 138 C + 273 = 411 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (290 mL)(411 K) 273 K = 436 . 593 mL 004 (part 1 of 1) 10 points A sample of gas in a closed container at a temperature of 67 C and a pressure of 6 atm is heated to 314 C. What pressure does the gas exert at the higher temperature? Correct answer: 10 . 3588 atm. Explanation: T 1 = 67 C + 273 = 340 K P 1 = 6 atm T 2 = 314 C + 273 = 587 K P 2 = ? Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (6 atm) (587 K) 340 K = 10 . 3588 atm
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Rahman, Tarique – Homework 8 – Due: Oct 31 2007, midnight – Inst: Vandenbout 2 005 (part 1 of 1) 10 points A gas at 1 . 62 × 10 6 Pa and 26 C occu- pies a volume of 399 cm 3 . At what tem- perature would the gas occupy 542 cm 3 at 3 . 38 × 10 6 Pa? Correct answer: 574 . 421 C. Explanation: P 1 = 1 . 62 × 10 6 Pa P 2 = 3 . 38 × 10 6 Pa V 1 = 399 cm 3 T 1 = 26 C + 273 = 299 K V 2 = 542 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 = (3 . 38 × 10 6 Pa) (542 cm 3 ) (299 K) (1 . 62 × 10 6 Pa) (399 cm 3 ) = 847 . 421 K = 574 . 421 C 006 (part 1 of 1) 10 points A sample of ideal gas occupies 250 mL at 25 C and 740 torr. What is its volume at STP? 1. 223 mL correct 2. 280 mL 3. 250 mL 4. 266 mL 5. 235 mL Explanation: P 1 = 740 torr P 2 = 760 torr V 1 = 250 mL T 2 = 273.15 K T 1 = 25 C + 273.15 = 298.15 K Using the Combined Gas Law, P 1 V 1 T 1 = P 2 V 2 T 2 and recalling that STP implies standard tem- perature (273.15 K) and pressure (1 atm or 760 torr), we have V 2 = P 1 V 1 T 2 P 2 T 1 = (740 torr) (250 mL) (273 . 15 K) (760 torr) (298 . 15 K) = 223 . 01 mL 007 (part 1 of 1) 10 points What is the percent yield if 4.51 moles of CH 4 produces 16 L of CO 2 at STP?
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