HW7answers - Cheung, Anthony – Homework 7 – Due: Oct 27...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Cheung, Anthony – Homework 7 – Due: Oct 27 2006, midnight – Inst: McCord 1 This print-out should have 26 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 100kPa with every 10.2m of depth. This means that at 10.2m below the surface, the pressure is 201kPa; at 20.4m below the surface, the pressure is 301kPa; and so forth. If the volume of a bal- loon is 3 . 2 L at STP and the temperature of the water remains the same, what is the vol- ume 54 . 82 m below the water’s surface? Correct answer: 0 . 507854 L. Explanation: P 1 = 1 atm Depth = 54 . 82 m V 1 = 3 . 2 L V 2 = ? 101.325 kPa = 1 atm For P 2 : 10.2m 100kPa = 54 . 82 m x (10 . 2m)( x ) = (54 . 82 m)(100kPa) x = (54 . 82 m)(100kPa) 10.2m = 537 . 451 kPa P 2 = 101kPa + 537 . 451 kPa = 638 . 451 kPa × 1atm 101.325kPa = 6 . 30102 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm)(3 . 2 L) 6 . 30102 atm = 0 . 507854 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. 0.042 mm Hg 3. 2400 mm Hg correct 4. 24 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 A gas at 62 ◦ C occupies 4 . 89 L. At what tem- perature will the volume be 3 . 6 L, assuming the same pressure? Correct answer:- 26 . 3742 ◦ C. Explanation: V 1 = 4 . 89 L V 2 = 3 . 6 L T 1 = 62 ◦ C + 273 = 335 K T 2 = ? V 1 T 1 = V 2 T 2 T 2 = T 1 V 2 V 1 = (335 K)(3 . 6 L) 4 . 89 L = 246 . 626 K ◦ C = K- 273 =- 26 . 3742 ◦ C 004 (part 1 of 1) 10 points At standard temperature, a gas has a volume of 316 mL. The temperature is then increased to 135 ◦ C, and the pressure is held constant. What is the new volume? Correct answer: 472 . 264 mL. Explanation: T 1 = 0 ◦ C + 273 = 273 K V 1 = 316 mL T 2 = 135 ◦ C + 273 = 408 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (316 mL)(408 K) 273 K = 472 . 264 mL Cheung, Anthony – Homework 7 – Due: Oct 27 2006, midnight – Inst: McCord 2 005 (part 1 of 1) 10 points A sample of gas in a closed container at a temperature of 76 ◦ C and a pressure of 7 atm is heated to 383 ◦ C. What pressure does the gas exert at the higher temperature? Correct answer: 13 . 1576 atm. Explanation: T 1 = 76 ◦ C + 273 = 349 K P 1 = 7 atm T 2 = 383 ◦ C + 273 = 656 K P 2 = ?...
View Full Document

This note was uploaded on 11/11/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

Page1 / 8

HW7answers - Cheung, Anthony – Homework 7 – Due: Oct 27...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online