Gas Laws

# Gas Laws - solis(nrs545 H08 Gas Laws McCord(53130 This...

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

solis (nrs545) – H08: Gas Laws – McCord – (53130) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 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 balloon is 2 . 6 L at STP and the temperature oF the water remains the same, what is the volume 47 . 49 m below the water’s surFace? Correct answer: 0 . 464967 L. Explanation: P 1 = 1 atm Depth = 47 . 49 m V 1 = 2 . 6 L V 2 = ? 101.325 kPa = 1 atm ±or P 2 : 10.2 m 100 kPa = 47 . 49 m x (10 . 2 m)( x ) = (47 . 49 m)(100 kPa) x = (47 . 49 m)(100 kPa) 10.2 m = 465 . 588 kPa P 2 = 101 kPa + 465 . 588 kPa = 566 . 588 kPa × 1 atm 101.325 kPa = 5 . 59179 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (2 . 6 L) 5 . 59179 atm = 0 . 464967 L 002 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. 24 mm Hg 2. 600 mm Hg 3. 0.042 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 003 10.0 points At standard temperature, a gas has a volume oF 304 mL. The temperature is then increased to 112 C, and the pressure is held constant. What is the new volume? Correct answer: 428 . 718 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 304 mL T 2 = 112 C + 273 = 385 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (304 mL)(385 K) 273 K = 428 . 718 mL 004 10.0 points A sample oF gas in a closed container at a temperature oF 70 C and a pressure oF 7 atm is heated to 316 C. What pressure does the gas exert at the higher temperature? Correct answer: 12 . 0204 atm. Explanation: T 1 = 70 C + 273 = 343 K P 1 = 7 atm T 2 = 316 C + 273 = 589 K P 2 = ?

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

View Full Document
solis (nrs545) – H08: Gas Laws – McCord – (53130) 2 Applying the Gay-Lussac law, P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (7 atm) (589 K) 343 K = 12 . 0204 atm 005 10.0 points A gas at 1 . 9 × 10 6 Pa and 28 C occupies a vol- ume of 430 cm 3 . At what temperature would the gas occupy 580 cm 3 at 3 . 04 × 10 6 Pa? Correct answer: 376
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/24/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas.

### Page1 / 7

Gas Laws - solis(nrs545 H08 Gas Laws McCord(53130 This...

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

View Full Document
Ask a homework question - tutors are online