HW 02 Gas Laws-solutions

HW 02 Gas Laws-solutions - nanni (arn437) HW 02 Gas Laws...

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nanni (arn437) – HW 02 Gas Laws – vanden bout – (50985) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. Questions 2, 3 and 7 are not multiple choice. ±or questions 2 and 3, your answer should have fve decimal places. ±or question 7, your answer should have Four decimal places. 001 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. 600 mm Hg 3. 24 mm Hg 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 002 10.0 points A sample oF gas in a closed container at a temperature oF 85 C and a pressure oF 8 atm is heated to 267 C. What pressure does the gas exert at the higher temperature? Correct answer: 12 . 067 atm. Explanation: T 1 = 85 C + 273 = 358 K P 1 = 8 atm T 2 = 267 C + 273 = 540 K P 2 = ? Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (8 atm) (540 K) 358 K = 12 . 067 atm 003 10.0 points A ²ask containing 170 cm 3 oF hydrogen was collected under a pressure oF 20 . 7 kPa. What pressure would have been required For the vol- ume oF the gas to have been 72 cm 3 , assuming the same temperature? Correct answer: 48 . 875 kPa. Explanation: V 1 = 170 cm 3 V 2 = 72 cm 3 P 1 = 20 . 7 kPa P 2 = ? Applying Boyle’s law, P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 V 2 = (20 . 7 kPa)(170 cm 3 ) 72 cm 3 = 48 . 875 kPa 004 10.0 points A 5.00-L sample oF a gas exerts a pressure oF 1040 torr at 50.0 C. In what volume would the same sample exert a pressure oF 1.00 atm at 50.0 C? 1. 0.581 L 2. 10.5 L 3. 3.33 L 4. 6.84 L correct 5. 1.95 L Explanation: P 1 = (1040 torr) p 1 atm 760 torr P = 1 . 36842 atm P 2 = 1 atm V 1 = 5 L Since the number oF moles and the temper- ature oF the gas sample are held constant, the change in the volume is determined by Boyle’s Law: P 1 V 1 = P 2 V 2
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nanni (arn437) – HW 02 Gas Laws – vanden bout – (50985) 2 V 2 = P 1 V 1 P 2 = (1 . 36842 atm)(5 L) 1 atm = 6 . 84211 L 005 10.0 points Consider 2 Al + 6 HCl 2 AlCl 3 + 3 H 2 , the reaction of Al with HCl to produce hy- drogen gas. This reaction has a yield of 82.5 percent. How much HCl are needed to pro- duce 14.0 L of H 2 at 351 K and 1.11 atm? 1. 1.08 mol 2. 1.31 mol correct 3. 0.655 mol 4. 0.890 mol 5. 0.540 mol 6. 0.223 mol 7. 0.446 mol Explanation: V = 14 L T = 351 K P = 1 . 11 atm p = 82 . 5% P V = n R T n = P V R T = (1 . 11 atm)(14 L) (0 . 0821 L · atm mol · K )(351 K) = 0 . 539263 mol H 2 . For 82.5% yield,
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This note was uploaded on 09/19/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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HW 02 Gas Laws-solutions - nanni (arn437) HW 02 Gas Laws...

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