This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Niu (qn269) – Homework 6 – McCord – (91750) 1 This printout should have 32 questions. Multiplechoice 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 = ?...
View
Full
Document
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.
 Summer '07
 Fakhreddine/Lyon

Click to edit the document details