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100 K 15kPa 790 dm 3 22.3 mL 2.0 atm If 5.0 ml of a gas at a constant temperature at 1.0 atm , expands to fill 7.5 ml , what is the new pressure ? Identify V 1 , V 2 , and P 1 and P 2 ( 4 pts). temperature pressure volume volume Pressure V 1 = 5.0 mL 7. V 2 = 7.5 mL
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100 K 15kPa 790 dm3 22.3 mL 2.0 atm If 5.0 ml of a gas at a constant temperature at 1.0 atm, expands to fill 7.5 ml, what is the new pressure? Identify V1, V2, and P1 and P2 ( 4 pts) temperature pressure volume volume Pressure V1 = 5.0 mL 7. V2 = 7.5 mL 8. P1 = 1 atm 9. P2 = ? Quiz 5/13/04Identify the quantities from the units: -1 = 89 % -2 = 78% -3 = 67% -4 = 56% -5 = 45 % -6 = 34 % -7 = 23% -8 = 12 % -9 = 1% turn in bonus
Chapter 18: Gases Chapter Homework: HW 17: Definitions page 471 HW 18: Notes p 451-456; # 1-3 page 456; #30 32, 34 page 472 HW 19: Notes page 457-459 #5,6 page 457-460; #26, 3638 p 471 HW 20: notes page 460-461 Do #713 page 461,2 HW 21: notes p 463; Do #15 page 464; #4346 page 473 HW 22: notes page 464469 Do #21,22 page 469 HW 23 Do #48,50 page 473 Turn in all Lab write-ups!
Standard Temperature and Pressure:STP • Standard Temperature = the freezing point of water/ the melting point of water. • 0o C • 273 K • Standard Pressure • = 1 atmosphere • = 101.3 kPa • = 760 mm Hg • = 760 torr • Molar Volume = 22.4 L = 1 mole of any ideal gas AT STP.
1-3 page 456 If 4.41 L of gas are collected at a pressure of 94.2 kPa, what volume will the same gas occupy at standard atmospheric pressure, assuming the temperature remains the same? 101.3 kPa V2 = (4.41 L) (94.2 kPa) (101.3 kPa) V2 = 4.1 L If some oxygen gas at 101 kPa and 25oC is allowed to expand from 5.0 L to 10.0 L without changing the temperature, what pressure will the oxygen gas exert? P2 = 50.5 kPa P2 = (101 kPa) (5.0 L) (10.0L)
P1 = 98.5 kPa V1 = 844 cm3 P2 = 101.3 kPa V2 = ? #3 page 456 Correct the following volumes to standard pressure (101.3 kPa). V2 = 821 cm3 • P2 = 101.3 kPa • V2 = ? • P1 = 59.4 kPa • V1 = 273 cm3 V2 = 160 cm3 • P1 = 90.0 kPa • V1 = 116 m3 • P2 = 101.3 kPa • V2 = ? = 103 m3
#30 Correct the following volumes of gas from the indicated pressures to standard atmospheric pressure. Assume constant temperature. • P1 = 80.8 kPa • V1 = 817 cm3 • P2 = 101.3 kPa • V2 = ? Answers: No credit given unless you show your work! a. 652 cm3; b. 27.2 m3, 11.8 dm3, 589 cm3, 916 cm3 #31. Make the indicated corrections in the following gas volumes. Assume constant temperature. • P1 = 110 kPa • V1 = 0.600 m3 • P2 = 62.4 kPa • V2 = ?
Vocabulary • Correct the volume: figure out the new volume. • Cylinder: A container the shape of a can of soup. • Compressed: Smooshed, particles are packed tightly and at high pressure. • Constant: The beginning and end are the same. • piston: A device used to push a gas or fluid (like in a car engine, or a syringe. • vapor pressure: pressure caused by water molecules which have evaporated.
Boyle’s Law Lab: Page 197 of Lab Book. • Do the lab, follow each instruction one word at a time and answer the questions.
HW 19: Notes page 457-459 #5,6 page 457-460; #26, 3638 p 471 Water Vapor Pressure Graph • Dalton’s Law of Partial Pressures: • Each gas molecule is totally independent of other gas molecules, • to find the total pressure, • we just add all the pressures of each individual gas. • Water evaporates in air to create “vapor pressure” boiling point boiling point
Water Vapor Pressure Water vapor pressure increases as we get closer to the boiling point. Solids usually have a vapor pressure of approximately zero. http://www.chem.purdue.edu/gchelp/liquids/vpress.html link from which moving picture was taken. http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/watvap.html#c1 This is a really nice link – for more information about boiling etc. The graph is from this link. Use this link for extra credit – explain how water vapor forms and how that is related to boiling. http://wine1.sb.fsu.edu/chm1045/notes/Forces/Vapor/Forces05.htm Another good link for more help understanding boiling.