Hw 9 - baum(awb464 H09 Gas Laws mccord(50970 This print-out should have 22 questions Multiple-choice questions may continue on the next column or

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baum (awb464) – 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 38 . 5 m below the water’s surFace? Correct answer: 0 . 762398 L. Explanation: P 1 = 1 atm Depth = 38 . 5 m V 1 = 3 . 6 L V 2 = ? 101.325 kPa = 1 atm ±or P 2 : 10.2 m 100 kPa = 38 . 5 m x (10 . 2 m)( x ) = (38 . 5 m)(100 kPa) x = (38 . 5 m)(100 kPa) 10.2 m = 377 . 451 kPa P 2 = 101 kPa + 377 . 451 kPa = 478 . 451 kPa × 1 atm 101.325 kPa = 4 . 72194 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 . 72194 atm = 0 . 762398 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. 24 mm Hg 2. 600 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 215 mL. The temperature is then increased to 119 C, and the pressure is held constant. What is the new volume? Correct answer: 308 . 718 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 215 mL T 2 = 119 C + 273 = 392 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (215 mL)(392 K) 273 K = 308 . 718 mL 004 10.0 points A sample oF gas in a closed container at a temperature oF 90 C and a pressure oF 5 atm is heated to 277 C. What pressure does the gas exert at the higher temperature? Correct answer: 7 . 57576 atm. Explanation: T 1 = 90 C + 273 = 363 K P 1 = 5 atm T 2 = 277 C + 273 = 550 K P 2 = ?

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baum (awb464) – 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 = (5 atm) (550 K) 363 K = 7 . 57576 atm 005 10.0 points A gas at 1 . 59 × 10 6 Pa and 26 C occu- pies a volume of 442 cm 3 . At what tem- perature would the gas occupy 507 cm 3 at 3 . 48 × 10 6 Pa? Correct answer: 477
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This note was uploaded on 11/21/2010 for the course CHEM 50970 taught by Professor Dr.mccord during the Fall '10 term at University of Texas at Austin.

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Hw 9 - baum(awb464 H09 Gas Laws mccord(50970 This print-out should have 22 questions Multiple-choice questions may continue on the next column or

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