H09- Gas Laws-solutions - phan (dtp377) H09: Gas Laws...

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phan (dtp377) – H09: Gas Laws – mccord – (51600) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd 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 . 6 L at STP and the temperature oF the water remains the same, what is the volume 41 . 68 m below the water’s surFace? Correct answer: 0 . 516936 L. Explanation: P 1 = 1 atm Depth = 41 . 68 m V 1 = 2 . 6 L V 2 = ? 101.325 kPa = 1 atm ±or P 2 : 10.2 m 100 kPa = 41 . 68 m x (10 . 2 m)( x ) = (41 . 68 m)(100 kPa) x = (41 . 68 m)(100 kPa) 10.2 m = 408 . 627 kPa P 2 = 101 kPa + 408 . 627 kPa = 509 . 627 kPa × 1 atm 101.325 kPa = 5 . 02963 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (2 . 6 L) 5 . 02963 atm = 0 . 516936 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. 600 mm Hg 2. 0.042 mm Hg 3. 2400 mm Hg correct 4. 24 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 302 mL. The temperature is then increased to 112 C, and the pressure is held constant. What is the new volume? Correct answer: 425 . 897 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 302 mL T 2 = 112 C + 273 = 385 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (302 mL)(385 K) 273 K = 425 . 897 mL 004 10.0 points A sample oF gas in a closed container at a temperature oF 84 C and a pressure oF 6 atm is heated to 392 C. What pressure does the gas exert at the higher temperature? Correct answer: 11 . 1765 atm. Explanation: T 1 = 84 C + 273 = 357 K P 1 = 6 atm T 2 = 392 C + 273 = 665 K P 2 = ?
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phan (dtp377) – H09: Gas Laws – mccord – (51600) 2 Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (6 atm) (665 K) 357 K = 11 . 1765 atm 005 10.0 points A gas at 1 . 74 × 10 6 Pa and 16 C occu- pies a volume of 429 cm 3 . At what tem- perature would the gas occupy 552 cm 3 at 2 . 63 × 10 6 Pa? Correct answer: 289
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H09- Gas Laws-solutions - phan (dtp377) H09: Gas Laws...

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