# HW9 - Lin(cgl286 Homework 9 Lyon(52370 1 This print-out...

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Unformatted text preview: Lin (cgl286) Homework 9 Lyon (52370) 1 This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This HW assignment is due Tuesday, April 14, by 11PM. 001 10.0 points If the volume of a sample of a gas is in- creased 5 times the original volume while the temperature is held constant, what happens to the pressure of the gas? 1. It increases by 1 5 . 2. It increases 5 times the original pressure. 3. It stays the same. 4. It decreases by 1 5 . correct Explanation: Boyles Law describes an inverse relation- ship between volume and pressure when the moles of gas and temperature are held con- stant. P 1 V 1 = P 2 V 2 V 1 V 2 P 1 = P 2 V 1 5 ( V 1 ) P 1 = P 2 1 5 P 1 = P 2 parenleftbigg the pressure decreases by 1 5 parenrightbigg 002 10.0 points A certain quantity of a gas occupies 61.3 mL at 68 C. If the pressure remains constant, what would the volume of the gas be at 17 C? 1. 32 mL 2. 52 mL correct 3. 72 mL 4. 92 mL Explanation: T 1 = 68 C + 273 = 341 K V 1 = 61 . 3 mL T 2 = 17 C + 273 = 290 K Charles law relates the volume and abso- lute (Kelvin) temperature of a sample of gas. V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (61 . 3 mL)(290 K) 341 K = 52 mL 003 10.0 points A balloon is inflated outdoors on a cold day in North Dakota at a temperature of- 40 C to a volume of 2.00 L. The pressure remains constant. What is the volume of the balloon indoors at a temperature of 25 C? 1. 2.0 L 2. 1.6 L 3. 2.6 L correct 4.- 3 . 2 L Explanation: T 1 =- 40 C + 273 = 233 K V 1 = 9 . 0 L T 2 = 25 C + 273 = 298 K Charless law relates the volume and ab- solute (Kelvin) temperature of a sample of gas: V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (2 . 00 L)(298 K) 233 K = 2 . 6 L 004 10.0 points A sample of gas with a volume of 750 mL exerts a pressure of 98 kPa at 30 C. What pressure will the sample exert when it is compressed to 250 mL and cooled to- 25 C? 1. 241 kPa correct Lin (cgl286) Homework 9 Lyon (52370) 2 2. 353 kPa 3. 39.9 kPa 4. 359 kPa 5. 26.7 kPa Explanation: V 1 = 750 mL T 1 = 30 C + 273 . 15 = 303 . 15 K V 2 = 250 mL T 2 =- 25 C + 273 . 15 = 248 . 15 K P 1 = 98 kPa We can use the combined gas law and solve for P 2 : P 1 V 1 T 1 = P 2 V 2 T 2 P 2 = P 1 V 1 T 2 T 1 V 2 = (98 kPa) (750 mL) (248 . 15 K) (303 . 15 K) (250 mL) = 240 . 66 kPa 005 10.0 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....
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## This note was uploaded on 10/26/2010 for the course CH 52370 taught by Professor Lyon during the Spring '09 term at University of Texas.

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HW9 - Lin(cgl286 Homework 9 Lyon(52370 1 This print-out...

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