HW-8 - Rahman, Tarique – Homework 8 – Due: Oct 31 2007,...

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Unformatted text preview: 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 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 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 : 10.2m 100kPa = 57 . 19 m x (10 . 2m)( x ) = (57 . 19 m)(100kPa) x = (57 . 19 m)(100kPa) 10.2m = 560 . 686 kPa P 2 = 101kPa + 560 . 686 kPa = 661 . 686 kPa × 1atm 101.325kPa = 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 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 = ?...
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This note was uploaded on 02/04/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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HW-8 - Rahman, Tarique – Homework 8 – Due: Oct 31 2007,...

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