ChemistryS. Martinez – Spring 2009                            1
 KMT – based on idea that particles of matter are always in motion.                                          2
1.   Gases consist of large numbers of tiny     particles that are far apart relative to their     size.2.   Collisions be...
4. There are no forces of attraction or repulsion   between gas particles.5. The average KE of gas particles depends on   ...
 KMTonly applies to ideal gases…ideal gases DO NOT ACTUALLY EXIST but many gases behave nearly ideally if PRESSURE is not...
 Expansion    – gases do not have a definite  shape/volume…completely fill a container  in which they are enclosed and ta...
 Low   Density – Density of a substance in the  gas phase is about 1/1000 the density of the  same substance in the liqui...
8
9
 Diffusion  & Effusion – gases spread out  & mix with another without being stirred. Diffusion is the spontaneous mixing...
 Effusion – is a process by which gas particles pass through a tiny opening. The rates of effusion of different gases are...
 All real gases deviate to some degree from  ideal-gas behavior. Real gas – is a gas that does not behave  completely ac...
 KMT   will hold true for gases whose  particles have little attraction for each  other. Ex: noble gases and nonpolar gas...
 Pressure  is defined as force per unit area  on a surface. Units for pressure:  • Pascal  • Mm Hg  • Torr  • Atm (atmos...
15
16
17
 The gas laws are simple mathematical relationships between the volume, temperature, pressure, and amount of gas.        ...
1.   Boyle’s Law – demonstrates     relationship between volume &     pressure while temp remains constant.     (as one va...
Charles’s Law – demonstrates relationship  between volume & temperature while  pressure remains constant.  - states that t...
3. Gay-Lussac’s Law – demonstrates   relationship between pressure &   temperature while volume remains   constant.   - th...
The Combined Gas Law – calculates  volume, temperature, & pressure  changes.  - expresses the relationship between the  pr...
5. Dalton’s Law of Partial Pressures - used  to calculate partial pressures and total  pressures.  - the pressure of each ...
 Boyle’s Law - P1V1 = P2V2 Charles’s Law - V1/T1 = V2/T2 Gay-Lussac’s Law - P1/T1=P2/T2 Combined Gas Law - P1V1/T1 = P...
A sample of oxygen gas has a volume of 150 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pre...
 P1  = 0.947 atm V1 = 150 mL P2 = 0.987 atm V2 = X mL (0.947)(150) = 0.987x X = 144 mL of O2                        ...
Aballoon filled with helium gashas a volume of 500 mL at apressure of 1 atm. The balloon isreleased and reaches an altitu...
 Answer:   1000 mL He                         28
A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what wi...
 Answer:   3.18 atm                       30
 Diversknow that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at...
 Answer:   0.59 kPa                       32
Asample of neon gas occupies avolume of 752 mL at 25⁰C. Whatvolume will the gas occupy at50⁰C if the pressure remainscons...
 V1/T1= V2/T2 or V1T2/T1 V1 = 752 mL T1 = 25⁰C + 273 = 298K V2 = x mL T2 = 50⁰C + 273 = 323K 752/298 = x/323 or (752...
A helium filled balloon has a volume of 2.75 ml @ 20⁰C. The volume of the balloon decreases to 2.46 ml after it is placed...
 Answer:   262K or -11⁰C                            36
A gas at 65⁰C occupies 4.22L. At what Celsius temperature will the volume be 3.87L, assuming the same pressure? (remember...
 Answer:   37⁰C                   38
 The gas in an aerosol can is at a pressure of 3.00 atm at 25⁰C. Directions on the can warn the user not to keep the can ...
 P1/T1=P2/T2 or P1T2/T1=P2 P1 = 3.00 atm T1 = 25 + 273 = 298K P2 = x atm T2 = 52 + 273 = 325K 3.00/298 = x/325 or (3...
A sample of helium gas has a pressure of 1.20 atm @ 22⁰C. At what celsius temp. will the helium reach a pressure of 2.00 ...
 Answer:   219⁰C or 491.6 K                               42
A helium filled balloon has a volume of 50.0 L @ 25C and 1.08 atm. What volume will it have at 0.855 atm and 10.C? (conve...
 P1  = 1.08 atm V1 = 50.0 L T1 = 25 + 273 = 298K P2 = 0.855 atm V2 = x T2 = 10 + 273 = 283K (1.08)(50.0)/298 = (.85...
Thevolume of a gas is 27.5ml @ 22C and 0.974 atm.What will the volume be at15C and 0.993 atm?                            ...
 Answer:   26.3 ml                      46
A700.0 ml gas sample atSTP is compressed to avolume of 200.0 ml, and thetemp is increased to 30.0C.What is the new pressu...
 Answer:   3.94 x 105 Pa or 394 kPa                                       48
Helium   gas is collected over water @ 25*C. What is the partial pressure of the helium, given that the barometric pressu...
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Chapter 10 – Physical Characteristics of Gases

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Overview of the Kinetic Molecular Theory and how it relates to gases along with their characteristics.

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Chapter 10 – Physical Characteristics of Gases

  1. 1. ChemistryS. Martinez – Spring 2009 1
  2. 2.  KMT – based on idea that particles of matter are always in motion. 2
  3. 3. 1. Gases consist of large numbers of tiny particles that are far apart relative to their size.2. Collisions between gas particles and between particles and container walls are elastic collisions(one in which there is no net loss of KE)3. Gas particles are in continuous, rapid, random motion…therefore possess KE, which is energy of motion. 3
  4. 4. 4. There are no forces of attraction or repulsion between gas particles.5. The average KE of gas particles depends on the temp of the gas. [KE = 1/2mv2](the average speeds and KEs of gas particles increase with an increase in temp and decrease with a decrease in temp…all gases at the same temp have the same KE but might have different speeds due to their mass) 4
  5. 5.  KMTonly applies to ideal gases…ideal gases DO NOT ACTUALLY EXIST but many gases behave nearly ideally if PRESSURE is not very high or TEMPERATURE is not very low. 5
  6. 6.  Expansion – gases do not have a definite shape/volume…completely fill a container in which they are enclosed and take its’ shape. Fluidity – attractive forces between gas particles are insignificant, gas particles slide past one another…..this ability allows them to behave similarly to liquids…gases are referred to as fluids. 6
  7. 7.  Low Density – Density of a substance in the gas phase is about 1/1000 the density of the same substance in the liquid or solid state….due to gas particles being spread farther apart. Compressibility – gas particles are able to crowd close together…when the pressure in a container increases the volume of gas particles may increase by 100 times compared to the same container that is not pressurized. 7
  8. 8. 8
  9. 9. 9
  10. 10.  Diffusion & Effusion – gases spread out & mix with another without being stirred. Diffusion is the spontaneous mixing of the particles of 2 substances caused by their random motion. The rate of diffusion depends upon the particles speeds, diameters, and attractive forces between them. 10
  11. 11.  Effusion – is a process by which gas particles pass through a tiny opening. The rates of effusion of different gases are directly proportional to the velocities of their particles. 11
  12. 12.  All real gases deviate to some degree from ideal-gas behavior. Real gas – is a gas that does not behave completely according to the assumptions of the KMT. Real gases occupy space and exert attractive forces on each other…at high pressure and low temps, deviations may be considerable. 12
  13. 13.  KMT will hold true for gases whose particles have little attraction for each other. Ex: noble gases and nonpolar gases such as Ne, H2, & O2. The more polar a gas’s molecules are the greater the attractive forces between them and the more the gas will deviate from ideal gas behavior, ex: NH3 or H2O. 13
  14. 14.  Pressure is defined as force per unit area on a surface. Units for pressure: • Pascal • Mm Hg • Torr • Atm (atmospheres) STP (standard temp and pressure): 1 atm & 0*C Temp: Kelvins…0*C = 273 K 14
  15. 15. 15
  16. 16. 16
  17. 17. 17
  18. 18.  The gas laws are simple mathematical relationships between the volume, temperature, pressure, and amount of gas. 18
  19. 19. 1. Boyle’s Law – demonstrates relationship between volume & pressure while temp remains constant. (as one variable increases the other decreases) - states that the volume of a fixed mass of gas varies INVERSELY with the pressure at constant temperature. P1V1 = P2V2 19
  20. 20. Charles’s Law – demonstrates relationship between volume & temperature while pressure remains constant. - states that the volume of a fixed mass of gas at a constant pressure varies DIRECTLY with the Kelvin temperature. K = 273.15 + ____ ⁰C V1/T1 = V2/T2 20
  21. 21. 3. Gay-Lussac’s Law – demonstrates relationship between pressure & temperature while volume remains constant. - the pressure of a fixed mass of gas at constant volume varies DIRECTLY with the Kelvin temperature. P1/T1=P2/T2 21
  22. 22. The Combined Gas Law – calculates volume, temperature, & pressure changes. - expresses the relationship between the pressure, volume, and temperature of a fixed amount of gas. P1V1/T1 = P2V2/T2 22
  23. 23. 5. Dalton’s Law of Partial Pressures - used to calculate partial pressures and total pressures. - the pressure of each gas in a mixture is called the partial pressure of that gas. - states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. - PT = P1 + P2 + P3 +…. 23
  24. 24.  Boyle’s Law - P1V1 = P2V2 Charles’s Law - V1/T1 = V2/T2 Gay-Lussac’s Law - P1/T1=P2/T2 Combined Gas Law - P1V1/T1 = P2V2/T2 Dalton’s Law of Partial Pressures - PT = P1 + P2 + P3 +…. Conversion from Celsius to Kelvin - K = 273.15 + ⁰C 24
  25. 25. A sample of oxygen gas has a volume of 150 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? 25
  26. 26.  P1 = 0.947 atm V1 = 150 mL P2 = 0.987 atm V2 = X mL (0.947)(150) = 0.987x X = 144 mL of O2 26
  27. 27. Aballoon filled with helium gashas a volume of 500 mL at apressure of 1 atm. The balloon isreleased and reaches an altitudeof 6.5 km, where the pressure is0.5 atm. Assuming that thetemperature has remained thesame, what volume does the gasoccupy at this height? 27
  28. 28.  Answer: 1000 mL He 28
  29. 29. A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be, assuming constant temperature? 29
  30. 30.  Answer: 3.18 atm 30
  31. 31.  Diversknow 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, the pressure is 301 kPa; and so forth. Given that the volume of a balloon is 3.5 L at STP and that the temperature of the water remains constant, what is the volume 51 m below the water’s surface? 31
  32. 32.  Answer: 0.59 kPa 32
  33. 33. Asample of neon gas occupies avolume of 752 mL at 25⁰C. Whatvolume will the gas occupy at50⁰C if the pressure remainsconstant? 33
  34. 34.  V1/T1= V2/T2 or V1T2/T1 V1 = 752 mL T1 = 25⁰C + 273 = 298K V2 = x mL T2 = 50⁰C + 273 = 323K 752/298 = x/323 or (752)(323)/298 X = 815 mL 34
  35. 35. A helium filled balloon has a volume of 2.75 ml @ 20⁰C. The volume of the balloon decreases to 2.46 ml after it is placed outside on a cold day. What is the outside temperature in K? in ⁰C? 35
  36. 36.  Answer: 262K or -11⁰C 36
  37. 37. A gas at 65⁰C occupies 4.22L. At what Celsius temperature will the volume be 3.87L, assuming the same pressure? (remember temp must be in Kelvin in the formula…so you will have to convert to Kelvin and then from Kelvin to get the answer) 37
  38. 38.  Answer: 37⁰C 38
  39. 39.  The gas in an aerosol can is at a pressure of 3.00 atm at 25⁰C. Directions on the can warn the user not to keep the can in a place where temp exceeds 52⁰C. What would the gas pressure in the can be at 52⁰C? 39
  40. 40.  P1/T1=P2/T2 or P1T2/T1=P2 P1 = 3.00 atm T1 = 25 + 273 = 298K P2 = x atm T2 = 52 + 273 = 325K 3.00/298 = x/325 or (3.00)(325)/298 X = 3.27 atm 40
  41. 41. A sample of helium gas has a pressure of 1.20 atm @ 22⁰C. At what celsius temp. will the helium reach a pressure of 2.00 atm?(hint: you have to convert to Kelvin and then from Kelvin) 41
  42. 42.  Answer: 219⁰C or 491.6 K 42
  43. 43. A helium filled balloon has a volume of 50.0 L @ 25C and 1.08 atm. What volume will it have at 0.855 atm and 10.C? (convert celcius to kelvins first) 43
  44. 44.  P1 = 1.08 atm V1 = 50.0 L T1 = 25 + 273 = 298K P2 = 0.855 atm V2 = x T2 = 10 + 273 = 283K (1.08)(50.0)/298 = (.855)x/283 X= 60.0 L 44
  45. 45. Thevolume of a gas is 27.5ml @ 22C and 0.974 atm.What will the volume be at15C and 0.993 atm? 45
  46. 46.  Answer: 26.3 ml 46
  47. 47. A700.0 ml gas sample atSTP is compressed to avolume of 200.0 ml, and thetemp is increased to 30.0C.What is the new pressure ofthe gas in Pa? (hint: 1 atm =1.013 25 x 10 5 Pa) 47
  48. 48.  Answer: 3.94 x 105 Pa or 394 kPa 48
  49. 49. Helium gas is collected over water @ 25*C. What is the partial pressure of the helium, given that the barometric pressure is 750.0 mm Hg? (hint: use table A-8) 49

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