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Ch.12: Study Guide (answers)

Ch.12: Study Guide (answers). small. insignificant. compressibility. forces. expand. random. elastic. Kinetic energy. Kelvin. AT. AT. NT. NT. AT. c. d. e. a. b.

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Ch.12: Study Guide (answers)

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  1. Ch.12: Study Guide(answers)

  2. small insignificant compressibility forces expand random elastic Kinetic energy Kelvin

  3. AT AT NT NT AT

  4. c d e a b

  5. 1) Gas volume is insignificant compared to the space between the particles. This explains why all gases take up the same space at a given P and T. 2) There is no significant attractive or repulsive forces between gas particles. This explains why gases spread out to fill their container. 3) Gases move constantly in random straight lines and collide with each other and the walls without losing energy. This explains why gases don’t slow down and become a liquid or solid.

  6. doubles half decreasing doubles temperature doubles

  7. AT NT AT NT NT

  8. b a b a

  9. Kelvin temperature is directly related to pressure. This means that if the Kelvin temperature increases the pressure also increases or if Kelvin temperature decrease so does the pressure. The driver can increase the pressure or decrease temperature (refrigerated trailer) so they can get more gas in each load.

  10. inversely increases Boyle’s amount Kelvin Charles Gay-Lussac’s directly Combined none

  11. NT AT ST NT NT AT

  12. c b e a d

  13. P2 P1 (473) (55) = P2 = T2 T1 173 P1 = 55 kPa P2 (473) 55 (473) = T1 = -100.0 °C 473 173 + 273 Change Temp to KELVIN!! 173 K P2 = _____ kPa = 150 kPa T2 = 200 °C + 273 = 473 K

  14. V2 P2 P1 V1 = T2 T1 91 x 75 x 273 V1 = (303 x 101.3) V1 = _____ mL V1 (101.3) (91) (75) = T1 = 0 °C = 273K 303 273 P1 = 101.3 kPa V2 = 75.0 mL = 60.7 mL T2 = 30 °C + 273 = 303 K P2 = 91 kPa

  15. number of moles PV = nRT n ideal gas constant kPa·L mol·K 8.31 ideal no forces volume

  16. AT AT NT NT AT

  17. d c b a

  18. V P n = T R PV = n kPa·L mol·K RT R = 8.31 n = _____ mol O2 (25,325) (12.5) V = 12.5 L = (8.31) (295) (8.31 x 295) P = 25,325 kPa T = 22 °C + 273 = 295 K n = 129 mol O2

  19. V P n = T R Moles or grams indicates that you are using the Ideal gas law. You can convert moles to grams for any formula PV = n kPa·L mol·K RT R = 8.31 n = _____ mol NO2 (240) (0.275) V = 275 mL 1 L 1000 mL = 0.275 L x = (8.31) (301) (8.31 x 301) P = 240.0 kPa T = 28 °C + 273 = 301 K n = 46.01 g NO2 1 mol NO2 0.0264 mol NO2 = x 1.21 g NO2

  20. Avogadro’s Temperature pressure 1 mole 22.4 L total sum inversely molar mass Graham’s law of effusion

  21. NT AT NT AT AT

  22. d a b c

  23. Kinetic Energy of Gas Particles At the same conditions of temperature, all gases have the same average kinetic energy. m = mass v = velocity

  24. KE = ½ mv2 B Both objects at the same temperature. A Same temp. means same KEave KEA=KEB 2• ½ mAvA2=½ mBvB2 2• multiply both sides by 2 mAvA2=mBvB2 vA2mB = mAvB2 mAvB2 vB2 mA Take the square root of both sides. divide both sides by mA and VB2

  25. g i j f d e c a h b

  26. 83.73g - 83.32g = 0.41g 83.82g - 83.39g = 0.43g 1 L 267 mL x = 0.267 L 1000 mL 99.0 + 273 = 372.0 K 99.0 + 273 = 372.0 K 773.5mm Hg x 101.3 kPa = 103.1 kPa 760 mm Hg 812.0mm Hg x 101.3 kPa = 108.2 kPa 760 mm Hg

  27. 1 L 267 mL x = 0.267 L 1000 mL 99.0 + 273 = 372.0 K 99.0 + 273 = 372.0 K 773.5mm Hg x 101.3 kPa = 103.1 kPa 760 mm Hg 812.0mm Hg x 101.3 kPa = 108.2 kPa 760 mm Hg PV = nRT PV RT (103.1)(0.267) (8.31) (372.0) n = = = 0.00890 moles (108.2)(0.267) (8.31) (372.0) n = = 0.00935 moles

  28. 83.73g - 83.32g = 0.41g 83.82g - 83.39g = 0.43g grams mole Molar mass = 1 L 267 mL x = 0.267 L 1000 mL 99.0 + 273 = 372.0 K 0.41 g 0.00890 moles = 46.1 g/mol 99.0 + 273 = 372.0 K 0.43 g 0.00935 moles 773.5mm Hg x 101.3 kPa = 46.0 g/mol = 103.1 kPa 760 mm Hg 812.0mm Hg x 101.3 kPa = 108.2 kPa 760 mm Hg PV RT (103.1)(0.267) (8.31) (372.0) n = = = 0.00890 moles (108.2)(0.267) (8.31) (372.0) n = PV = nRT = 0.00935 moles

  29. H H O-H H H O-H H H H H C C C grams mole 0.41 g 0.00890 moles Molar mass = = 46.1 g/mol 0.43 g 0.00935 moles = 46.0 g/mol (46.1 + 46.0)/2 = 46.05 g/mol C = 12.01 Ethanol H = 1.01 O = 16.00 16 1.01 1.01 1.01 12.01 12.01 12.01 1.01 1.01 1.01 1.01 16 1.01 1.01 1.01 methanol ethanol =32.05 g/mol = 46.08 g/mol Which one has a molar mass of 46 g/mol?

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