H08- Gas Laws-solutions

# H08- Gas Laws-solutions - gonzales(pag757 H08 Gas Laws...

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

gonzales (pag757) – 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 . 8 L at STP and the temperature oF the water remains the same, what is the volume 44 . 72 m below the water’s surFace? Correct answer: 0 . 525943 L. Explanation: P 1 = 1 atm Depth = 44 . 72 m V 1 = 2 . 8 L V 2 = ? 101.325 kPa = 1 atm ±or P 2 : 10.2 m 100 kPa = 44 . 72 m x (10 . 2 m)( x ) = (44 . 72 m)(100 kPa) x = (44 . 72 m)(100 kPa) 10.2 m = 438 . 431 kPa P 2 = 101 kPa + 438 . 431 kPa = 539 . 431 kPa × 1 atm 101.325 kPa = 5 . 32377 atm Applying Boyle’s law, P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = (1 atm) (2 . 8 L) 5 . 32377 atm = 0 . 525943 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. 2400 mm Hg correct 3. 0.042 mm Hg 4. 600 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 10.0 points At standard temperature, a gas has a volume oF 237 mL. The temperature is then increased to 138 C, and the pressure is held constant. What is the new volume? Correct answer: 356 . 802 mL. Explanation: T 1 = 0 C + 273 = 273 K V 1 = 237 mL T 2 = 138 C + 273 = 411 K V 2 = ? V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (237 mL)(411 K) 273 K = 356 . 802 mL 004 10.0 points A sample oF gas in a closed container at a temperature oF 84 C and a pressure oF 8 atm is heated to 370 C. What pressure does the gas exert at the higher temperature? Correct answer: 14 . 409 atm. Explanation: T 1 = 84 C + 273 = 357 K P 1 = 8 atm T 2 = 370 C + 273 = 643 K P 2 = ?

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

View Full Document
gonzales (pag757) – 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 = (8 atm) (643 K) 357 K = 14 . 409 atm 005 10.0 points A gas at 1 . 8 × 10 6 Pa and 11 C occupies a vol- ume of 418 cm 3 . At what temperature would the gas occupy 537 cm 3 at 3 . 33 × 10 6 Pa? Correct answer: 401
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/28/2009 for the course CH 53130 taught by Professor Mccord during the Fall '09 term at University of Texas.

### Page1 / 7

H08- Gas Laws-solutions - gonzales(pag757 H08 Gas Laws...

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

View Full Document
Ask a homework question - tutors are online