Gas Laws - McCord - vu (tv2894) H09: Gas Laws mccord...

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vu (tv2894) – H09: Gas Laws – mccord – (50970) 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 3 . 6 L at STP and the temperature oF the water remains the same, what is the volume 32 . 77 m below the water’s surFace? Correct answer: 0 . 863822 L. Explanation: P 1 = 1 atm Depth = 32 . 77 m V 1 = 3 . 6 L V 2 = ? 101.325 kPa = 1 atm ±or P 2 : 10.2 m 100 kPa = 32 . 77 m x (10 . 2 m)( x ) = (32 . 77 m)(100 kPa) x = (32 . 77 m)(100 kPa) 10.2 m = 321 . 275 kPa P 2 = 101 kPa + 321 . 275 kPa = 422 . 275 kPa × 1 atm 101.325 kPa = 4 . 16753 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (3 . 6 L) 4 . 16753 atm = 0 . 863822 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. 2400 mm Hg correct 2. 0.042 mm Hg 3. 600 mm Hg 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 278 mL. The temperature is then increased to 135 C, and the pressure is held constant. What is the new volume? Correct answer: 415 . 473 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 278 mL T 2 = 135 C + 273 = 408 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (278 mL)(408 K) 273 K = 415 . 473 mL 004 10.0 points A sample oF gas in a closed container at a temperature oF 67 C and a pressure oF 8 atm is heated to 286 C. What pressure does the gas exert at the higher temperature? Correct answer: 13 . 1529 atm. Explanation: T 1 = 67 C + 273 = 340 K P 1 = 8 atm T 2 = 286 C + 273 = 559 K P 2 = ?
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vu (tv2894) – 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 = (8 atm) (559 K) 340 K = 13 . 1529 atm 005 10.0 points A gas at 1 . 69 × 10 6 Pa and 29 C occu- pies a volume of 370 cm 3 . At what tem- perature would the gas occupy 542 cm 3 at 2 . 61 × 10 6 Pa? Correct answer: 410
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Gas Laws - McCord - vu (tv2894) H09: Gas Laws mccord...

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