HW06-key

# HW06-key - Chemistry 111 Homework #6 - KEY Due: October 12,...

This preview shows pages 1–4. Sign up to view the full content.

Chemistry 111 Homework #6 - KEY Due: October 12, 2005 (Wednesday) 1. The volume of a gas under normal conditions is observed to be . directly proportional to the temperature and . inversely proportional to the pressure. Your answer should be either directly or inversely in either case and should be entered as two words separated by a comma and a space in Blackboard. P V = n R T Æ if the temperature increases, the volume increases too Æ if the pressure increases, the volume decreases 2. A plot of pressure versus inverse volume (1/V) at constant temperature (isothermal = same temperature) is… a) a straight line with a positive slope at 45 o . b) a straight line with a negative slope. c) a straight line with a positive slope going through the point (0,0). d) a hyperbolic curve e) a logarithmic curve. P V = constant = c T,n (at constant T and n) Æ P = c T,n x 1/V this is the equation for a straight line with a slope of c T,n and intercept of (0,0) 3. True or False . A plot of pressure versus temperature for a given number of moles of gas at a given volume (isochoric = same volume) gives exactly the same for all gases (assume ideal gas behavior). P / T = n R / V = c n,V (at constant n and V) this equation works for ALL gases (that behave ideally) 4. A plot of volume versus 1/T for a given number of moles of gas at constant pressure (isobaric = same pressure) is a) a logarithmic curve. b) a straight line with a negative slope. d) a straight line with a positive slope going through the point (0,0). d) a hyperbolic curve e) none of these.

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

View Full Document
Chemistry 111 2 Homework #6 / Fall 2005 In this case, V = T (n R / P) Æ more logarithmic in shape than any of the other choices, though it also looks a bit hyperbolic. But technically, it's not any of those shapes. So we accepted (a), (d) and (e). 5. Convert 54.2 kPa into torr. a) 305 torr b) 232 torr c) 407 torr d) 760 torr e) 229 torr f) None of these. We need the conversion factor between kPa and atm. 54.2 kPa × 1a tm 101.3 kPa × 760 torr 1 atm = 407 torr . 6. At the peak of Mt. Everest, the pressure is approximately 270 mm Hg. Determine the pressure in atmospheres at that height. a) 0.36 atm b) 0.52 atm c) 0.27 atm d) 0.40 atm e) None of the above. Remember – a mm of Hg is the same as a torr. Therefore, 270 torr × 1atm 760 torr = 0.36 atm . V 1/T
Chemistry 111 3 Homework #6 / Fall 2005 For questions 7 – 12 answer True or False. Bozo the Clown climbs Mt. Everest carrying a big red helium balloon (that doesn't leak). At the top, the pressure and temperature (in K) are about 1/3 and 5/6 times their values when the balloon was filled in his warm, sea-level studio. When Bozo finally gets to the top, 7. The balloon is bigger than when he started.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/08/2010 for the course CHEM 111 taught by Professor Kenney during the Spring '08 term at Case Western.

### Page1 / 9

HW06-key - Chemistry 111 Homework #6 - KEY Due: October 12,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online