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Unformatted text preview: Killough, Ian Homework 5 Due: Oct 19 2006, 11:00 pm Inst: Donna C Lyon 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. This HW assignment is due Thursday, Oc tober 19, by 11PM. 001 (part 1 of 1) 10 points The pressure at 20,000 feet above sea level is about 400 mmHg. To what does this corre spond? 1. 53.3 kPa correct 2. 0.526 Torr 3. 53 . 3 g m 1 s 2 4. 526 Pa 5. 0.526 kPa Explanation: 002 (part 1 of 1) 10 points A 6.5 L sample of nitrogen at 25 C and 0.76 atm is allowed to expand to 13.0 L. The tem perature remains constant. What is the final pressure? 1. 0.12 atm 2. 0.063 atm 3. 0.75 atm 4. 0.38 atm correct 5. 3.0 atm Explanation: P 1 = 0 . 76 atm P 2 = ? V 1 = 6 . 5 L V 2 = 13 . 0 L Boyles law, P 1 V 1 = P 2 V 2 , relates the vol ume and pressure of a sample of gas. P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 V 2 = (0.76 atm)(6.5 L) 13 L = 0.38 atm 003 (part 1 of 1) 10 points A certain quantity of a gas occupies 61.3 mL at 68 C. If the pressure remains constant, what would be the volume of the gas at 128 C? 1. 72 mL correct 2. 92 mL 3. 32 mL 4. 52 mL Explanation: V 1 = 61 . 3 mL T 1 = 68 C = 341 K T 2 = 128 C = 401 K Charless law V 1 T 1 = V 2 T 2 relates the volume and absolute (Kelvin) temperature of a sam ple of gas: V 2 = V 1 T 2 T 1 = (61 . 3mL)(401 K) 341 K = 72 L 004 (part 1 of 1) 10 points A sample of N 2 gas occupies a volume of 746 mL at STP. What volume would N 2 gas occupy at 155 C at a pressure of 368 torr? 1. 3295 mL 2. 1792 mL 3. 312 mL 4. 983 mL 5. 323 mL 6. 588 mL 7. 566 mL 8. 2415 mL correct Killough, Ian Homework 5 Due: Oct 19 2006, 11:00 pm Inst: Donna C Lyon 2 Explanation: V 1 = 746 mL T 1 = 273 K P 1 = 1atm = 760torr T 2 = 155 C = 428 K P 2 = 368torr Combined Gas Law: P 1 V 1 T 1 = P 2 V 2 T 2 V 2 = P 1 V 1 T 2 T 1 P 2 = (760torr)(746mL)(428K) (273 K)(368torr) = 2415 . 3814mL 005 (part 1 of 1) 10 points The normal respiratory rate for a human be ing is 15.0 breaths per minute. The average volume of air for each breath is 505 cm 3 at 20 C and 9 . 95 10 4 Pa. What is the volume of air at STP that an individual breathes in one day? Give your answers in cubic meters. 1. None of these 2. 9 . 98m 3 / day correct 3. 100m 3 / day 4. 505m 3 / day 5. 9 . 95cm 3 / day 6. 505cm 3 / day Explanation: P 1 = 9 . 95 10 4 Pa P 2 = 1 . 01325 10 5 Pa T 1 = 20 C + 273 = 293 K V 1 = 505 cm 3 T 2 = 0 C + 273 = 273 K V 2 = ? P 1 V 1 T 1 = P 2 V 2 T 2 V 2 = P 1 V 1 T 2 P 2 T 1 = (9 . 95 10 4 Pa)(505 cm 3 )(273 K) (1 . 01325 10 5 Pa)(293 K) = 0 . 000462054 m 3 15 breaths min 60 min hour 24 hours day = 21600 breaths / day 21600 breaths day . 000462054 m 3 breath = 9 . 98037 m 3 / day 006 (part 1 of 1) 10 points What is the density of hydrogen gas at STP?...
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This note was uploaded on 04/13/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Fakhreddine/Lyon

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