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H09- Gas Laws-solutions

H09- Gas Laws-solutions - truitt(wtt255 H09 Gas Laws...

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truitt (wtt255) – H09: Gas Laws – mccord – (50970) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 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 balloon is 2 . 3 L at STP and the temperature of the water remains the same, what is the volume 57 . 17 m below the water’s surface? Correct answer: 0 . 352307 L. Explanation: P 1 = 1 atm Depth = 57 . 17 m V 1 = 2 . 3 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2 m 100 kPa = 57 . 17 m x (10 . 2 m)( x ) = (57 . 17 m)(100 kPa) x = (57 . 17 m)(100 kPa) 10.2 m = 560 . 49 kPa P 2 = 101 kPa + 560 . 49 kPa = 661 . 49 kPa × 1 atm 101.325 kPa = 6 . 5284 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (2 . 3 L) 6 . 5284 atm = 0 . 352307 L 002 10.0 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 10.0 points At standard temperature, a gas has a volume of 235 mL. The temperature is then increased to 110 C, and the pressure is held constant. What is the new volume? Correct answer: 329 . 689 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 235 mL T 2 = 110 C + 273 = 383 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (235 mL)(383 K) 273 K = 329 . 689 mL 004 10.0 points A sample of gas in a closed container at a temperature of 68 C and a pressure of 3 atm is heated to 391 C. What pressure does the gas exert at the higher temperature? Correct answer: 5 . 84164 atm. Explanation: T 1 = 68 C + 273 = 341 K P 1 = 3 atm T 2 = 391 C + 273 = 664 K P 2 = ?
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truitt (wtt255) – H09: Gas Laws – mccord – (50970) 2 Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (3 atm) (664 K) 341 K = 5 . 84164 atm 005 10.0 points A gas at 1 . 56 × 10 6 Pa and 21 C occu- pies a volume of 429 cm 3 . At what tem- perature would the gas occupy 521 cm 3 at 2 . 62 × 10 6 Pa? Correct answer: 326 . 659 C. Explanation: P 1 = 1 . 56 × 10 6 Pa P 2 = 2 . 62 × 10 6 Pa V 1 = 429 cm 3 T 1 = 21 C + 273 = 294 K V 2 = 521 cm 3 T 2 = ?
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