2. Quiz 5/13/04 Identify the quantities from the units: 1.100 K 2.15kPa 3.790 dm 3 4.22.3 mL 5.2.0 atm 6.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) 1.temperature 2.pressure 3.volume 4.volume 5.Pressure 6.V 1 = 5.0 mL 7.V 2 = 7.5 mL 7.V 2 = 7.5 mL 8.P 1 = 1 atm 8.P 1 = 1 atm 9.P 2 = ? 9.P 2 = ? -1 = 89 % -2 = 78% -3 = 67% -4 = 56% -5 = 45 % -6 = 34 % -7 = 23% -8 = 12 % -9 = 1% turn in bonus

3. 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, 36  38 p 471 HW 20: notes page 460-461 Do #7  13 page 461,2 HW 21: notes p 463; Do #15 page 464; #43  46 page 473 HW 22: notes page 464  469 Do #21,22 page 469 HW 23 Do #48,50 page 473 Turn in all Lab write-ups!

4. Standard Temperature and Pressure: STP  Standard Temperature = the freezing point of water/ the melting point of water.  0 o 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.

5. 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 V 2 = (4.41 L) (94.2 kPa) (101.3 kPa) V 2 = 4.1 L If some oxygen gas at 101 kPa and 25 o C is allowed to expand from 5.0 L to 10.0 L without changing the temperature, what pressure will the oxygen gas exert? P 2 = (101 kPa) (5.0 L) (10.0L) P 2 = 50.5 kPa

6. #3 page 456 Correct the following volumes to standard pressure (101.3 kPa).  P 1 = 98.5 kPa  V 1 = 844 cm 3  P 2 = 101.3 kPa  V 2 = ?  P 1 = 59.4 kPa  V 1 = 273 cm 3  P 2 = 101.3 kPa  V 2 = ?  P 1 = 90.0 kPa  V 1 = 116 m 3  P 2 = 101.3 kPa  V 2 = ? V 2 = 821 cm 3 V 2 = 160 cm 3 = 103 m 3

7. #30 Correct the following volumes of gas from the indicated pressures to standard atmospheric pressure. Assume constant temperature.  P 1 = 80.8 kPa  V 1 = 817 cm 3  P 2 = 101.3 kPa  V 2 = ? Answers: No credit given unless you show your work! a. 652 cm 3 ; b. 27.2 m 3, 11.8 dm 3, 589 cm 3, 916 cm 3 #31. Make the indicated corrections in the following gas volumes. Assume constant temperature.  P 1 = 110 kPa  V 1 = 0.600 m 3  P 2 = 62.4 kPa  V 2 = ?

8. 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.

9. Boyle’s Law Lab: Page 197 of Lab Book.  Do the lab, follow each instruction one word at a time and answer the questions.

10. HW 19: Notes page 457-459 #5,6 page 457-460; #26, 36  38 p 471  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 Water Vapor Pressure Graph

11. 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://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. http://www.chem.purdue.edu/gchelp/liquids/vpress.html link from which moving picture was taken.