hw 8 answers - Iype Shelby Homework 8 Due midnight Inst...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Iype, Shelby – Homework 8 – Due: Oct 31 2007, midnight – Inst: Vandenbout 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 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 . 4 L at STP and the temperature of the water remains the same, what is the vol- ume 39 . 15 m below the water’s surface? Correct answer: 0 . 710578 L. Explanation: P 1 = 1 atm Depth = 39 . 15 m V 1 = 3 . 4 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2 m 100 kPa = 39 . 15 m x (10 . 2 m)( x ) = (39 . 15 m)(100 kPa) x = (39 . 15 m)(100 kPa) 10.2 m = 383 . 824 kPa P 2 = 101 kPa + 383 . 824 kPa = 484 . 824 kPa × 1 atm 101.325 kPa = 4 . 78484 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (3 . 4 L) 4 . 78484 atm = 0 . 710578 L 002 (part 1 of 1) 10 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. 600 mm Hg 2. 24 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 (part 1 of 1) 10 points At standard temperature, a gas has a volume of 336 mL. The temperature is then increased to 119 C, and the pressure is held constant. What is the new volume? Correct answer: 482 . 462 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 336 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 = (336 mL)(392 K) 273 K = 482 . 462 mL 004 (part 1 of 1) 10 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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Iype, Shelby – Homework 8 – Due: Oct 31 2007, midnight – Inst: Vandenbout 2 005 (part 1 of 1) 10 points A gas at 1 . 84 × 10 6 Pa and 14 C occu- pies a volume of 380 cm 3 . At what tem- perature would the gas occupy 530 cm 3 at 3 . 11 × 10 6 Pa? Correct answer: 403 . 576 C. Explanation: P 1 = 1 . 84 × 10 6 Pa P 2 = 3 . 11 × 10 6 Pa V 1 = 380 cm 3 T 1 = 14 C + 273 = 287 K V 2 = 530 cm 3 T 2 = ? P 1 V 1 T 1 = P 2 V 2 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 = (3 . 11 × 10 6 Pa) (530 cm 3 ) (287 K) (1 . 84 × 10 6 Pa) (380 cm 3 ) = 676 . 576 K = 403 . 576 C 006 (part 1 of 1) 10 points A sample of ideal gas occupies 250 mL at 25 C and 740 torr. What is its volume at STP? 1. 266 mL 2. 235 mL 3. 250 mL 4. 223 mL correct 5. 280 mL 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 007 (part 1 of 1) 10 points What is the percent yield if 4.51 moles of CH 4 produces 16 L of CO 2 at STP?
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern