H09- Gas Laws-solutions

# H09- Gas Laws-solutions - goldstein(ddg625 H09 Gas Laws...

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goldstein (ddg625) – 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 . 7 L at STP and the temperature oF the water remains the same, what is the volume 49 . 65 m below the water’s surFace? Correct answer: 0 . 637845 L. Explanation: P 1 = 1 atm Depth = 49 . 65 m V 1 = 3 . 7 L V 2 = ? 101.325 kPa = 1 atm ±or P 2 : 10.2 m 100 kPa = 49 . 65 m x (10 . 2 m)( x ) = (49 . 65 m)(100 kPa) x = (49 . 65 m)(100 kPa) 10.2 m = 486 . 765 kPa P 2 = 101 kPa + 486 . 765 kPa = 587 . 765 kPa × 1 atm 101.325 kPa = 5 . 80079 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (3 . 7 L) 5 . 80079 atm = 0 . 637845 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. 24 mm Hg 4. 2400 mm Hg correct 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 344 mL. The temperature is then increased to 117 C, and the pressure is held constant. What is the new volume? Correct answer: 491 . 429 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 344 mL T 2 = 117 C + 273 = 390 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (344 mL)(390 K) 273 K = 491 . 429 mL 004 10.0 points A sample oF gas in a closed container at a temperature oF 92 C and a pressure oF 6 atm is heated to 327 C. What pressure does the gas exert at the higher temperature? Correct answer: 9 . 86301 atm. Explanation: T 1 = 92 C + 273 = 365 K P 1 = 6 atm T 2 = 327 C + 273 = 600 K P 2 = ?

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goldstein (ddg625) – 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 = (6 atm) (600 K) 365 K = 9 . 86301 atm 005 10.0 points A gas at 1 . 81 × 10 6 Pa and 15 C occu- pies a volume of 357 cm 3 . At what tem- perature would the gas occupy 503 cm 3 at 2 . 69 × 10 6 Pa? Correct answer: 330
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## This note was uploaded on 11/11/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

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H09- Gas Laws-solutions - goldstein(ddg625 H09 Gas Laws...

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