Chemistry HW 8 - Kim Eugene – Homework 8 – Due midnight...

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Unformatted text preview: Kim, Eugene – Homework 8 – Due: Oct 31 2007, midnight – Inst: McCord 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 4 L at STP and the temperature of the water remains the same, what is the volume 41 . 57 m below the water’s surface? Correct answer: 0 . 796973 L. Explanation: P 1 = 1 atm Depth = 41 . 57 m V 1 = 4 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2m 100kPa = 41 . 57 m x (10 . 2m)( x ) = (41 . 57 m)(100kPa) x = (41 . 57 m)(100kPa) 10.2m = 407 . 549 kPa P 2 = 101kPa + 407 . 549 kPa = 508 . 549 kPa × 1atm 101.325kPa = 5 . 01899 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm)(4 L) 5 . 01899 atm = 0 . 796973 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. 0.042 mm Hg 2. 2400 mm Hg correct 3. 24 mm Hg 4. 600 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 397 mL. The temperature is then increased to 121 ◦ C, and the pressure is held constant. What is the new volume? Correct answer: 572 . 96 mL. Explanation: T 1 = 0 ◦ C + 273 = 273 K V 1 = 397 mL T 2 = 121 ◦ C + 273 = 394 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (397 mL)(394 K) 273 K = 572 . 96 mL 004 (part 1 of 1) 10 points A sample of gas in a closed container at a temperature of 91 ◦ C and a pressure of 4 atm is heated to 345 ◦ C. What pressure does the gas exert at the higher temperature? Correct answer: 6 . 79121 atm. Explanation: T 1 = 91 ◦ C + 273 = 364 K P 1 = 4 atm T 2 = 345 ◦ C + 273 = 618 K P 2 = ? Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (4 atm)(618 K) 364 K = 6 . 79121 atm Kim, Eugene – Homework 8 – Due: Oct 31 2007, midnight – Inst: McCord 2 005 (part 1 of 1) 10 points A gas at 1 . 57 × 10 6 Pa and 18 ◦ C occu- pies a volume of 364 cm 3 . At what tem- perature would the gas occupy 539 cm 3 at 3 . 14 × 10 6 Pa? Correct answer: 588 . 808 ◦ C. Explanation: P 1 = 1 . 57 × 10 6 Pa P 2 = 3 . 14 × 10 6 Pa V 1 = 364 cm 3 T 1 = 18 ◦ C + 273 = 291 K V 2 = 539 cm 3 T 2 = ?...
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Chemistry HW 8 - Kim Eugene – Homework 8 – Due midnight...

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