H08- Gas Laws-solutions

# H08- Gas Laws-solutions - marquez(ecm739 H08 Gas Laws...

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marquez (ecm739) – H08: Gas Laws – McCord – (53110) 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 . 1 L at STP and the temperature oF the water remains the same, what is the volume 58 . 43 m below the water’s surFace? Correct answer: 0 . 315775 L. Explanation: P 1 = 1 atm Depth = 58 . 43 m V 1 = 2 . 1 L V 2 = ? 101.325 kPa = 1 atm ±or P 2 : 10.2 m 100 kPa = 58 . 43 m x (10 . 2 m)( x ) = (58 . 43 m)(100 kPa) x = (58 . 43 m)(100 kPa) 10.2 m = 572 . 843 kPa P 2 = 101 kPa + 572 . 843 kPa = 673 . 843 kPa × 1 atm 101.325 kPa = 6 . 65031 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (2 . 1 L) 6 . 65031 atm = 0 . 315775 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. 24 mm Hg 3. 2400 mm Hg correct 4. 0.042 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 233 mL. The temperature is then increased to 130 C, and the pressure is held constant. What is the new volume? Correct answer: 343 . 952 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 233 mL T 2 = 130 C + 273 = 403 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (233 mL)(403 K) 273 K = 343 . 952 mL 004 10.0 points A sample oF gas in a closed container at a temperature oF 70 C and a pressure oF 2 atm is heated to 400 C. What pressure does the gas exert at the higher temperature? Correct answer: 3 . 9242 atm. Explanation: T 1 = 70 C + 273 = 343 K P 1 = 2 atm T 2 = 400 C + 273 = 673 K P 2 = ?

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marquez (ecm739) – H08: Gas Laws – McCord – (53110) 2 Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (2 atm) (673 K) 343 K = 3 . 9242 atm 005 10.0 points A gas at 1 . 69 × 10 6 Pa and 28 C occu- pies a volume of 391 cm 3 . At what tem- perature would the gas occupy 534 cm 3 at 3 . 48 × 10 6 Pa? Correct answer: 573
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H08- Gas Laws-solutions - marquez(ecm739 H08 Gas Laws...

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