HW6solutions

# HW6solutions - Niu (qn269) – Homework 6 – McCord –...

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Unformatted text preview: Niu (qn269) – Homework 6 – McCord – (91750) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 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. 0.042 mm Hg 2. 24 mm Hg 3. 600 mm Hg 4. 2400 mm Hg correct 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 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 bal- loon is 3 . 5 L at STP and the temperature of the water remains the same, what is the vol- ume 56 . 53 m below the water’s surface? Correct answer: 0 . 541253 L. Explanation: P 1 = 1 atm Depth = 56 . 53 m V 1 = 3 . 5 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2 m 100 kPa = 56 . 53 m x (10 . 2 m)( x ) = (56 . 53 m)(100 kPa) x = (56 . 53 m)(100 kPa) 10.2 m = 554 . 216 kPa P 2 = 101 kPa + 554 . 216 kPa = 655 . 216 kPa × 1 atm 101.325 kPa = 6 . 46648 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (3 . 5 L) 6 . 46648 atm = 0 . 541253 L 003 10.0 points At standard temperature, a gas has a volume of 350 mL. The temperature is then increased to 137 ◦ C, and the pressure is held constant. What is the new volume? Correct answer: 525 . 641 mL. Explanation: T 1 = 0 ◦ C + 273 = 273 K V 1 = 350 mL T 2 = 137 ◦ C + 273 = 410 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (350 mL)(410 K) 273 K = 525 . 641 mL 004 10.0 points A sample of ideal gas occupies 250 mL at 25 ◦ C and 740 torr. What is its volume at STP? 1. 280 mL 2. 235 mL 3. 223 mL correct 4. 250 mL 5. 266 mL Niu (qn269) – Homework 6 – McCord – (91750) 2 Explanation: P 1 = 740 torr P 2 = 760 torr V 1 = 250 mL T 2 = 273.15 K T 1 = 25 ◦ C + 273.15 = 298.15 K Using the Combined Gas Law, P 1 V 1 T 1 = P 2 V 2 T 2 and recalling that STP implies standard tem- perature (273.15 K) and pressure (1 atm or 760 torr), we have V 2 = P 1 V 1 T 2 P 2 T 1 = (740 torr) (250 mL) (273 . 15 K) (760 torr) (298 . 15 K) = 223 . 01 mL 005 10.0 points A gas at 1 . 96 × 10 6 Pa and 14 ◦ C occu- pies a volume of 354 cm 3 . At what tem- perature would the gas occupy 516 cm 3 at 2 . 7 × 10 6 Pa? Correct answer: 303 . 283 ◦ C. Explanation: P 1 = 1 . 96 × 10 6 Pa P 2 = 2 . 7 × 10 6 Pa V 1 = 354 cm 3 T 1 = 14 ◦ C + 273 = 287 K V 2 = 516 cm 3 T 2 = ?...
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## This note was uploaded on 07/30/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Summer '07 term at University of Texas.

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HW6solutions - Niu (qn269) – Homework 6 – McCord –...

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