Chapter 14 “The Behavior of Gases”

1. Chapter 14“The Behavior of Gases”

2. Section 14.1The Properties of Gases

3. Compressibility • Gases can expand to fill its container, unlike solids or liquids • The reverse is also true: • They are easily compressed, or squeezed into a smaller volume • Compressibility is a measure of how much the volume of matter decreases under pressure

4. Compressibility • This is the idea behind placing “air bags” in automobiles • In an accident, the air compresses more than the steering wheel or dash when you strike it • The impact forces the gas particles closer together, because there is a lot of empty space between them

5. Compressibility • At room temperature, the distance between particles is about 10x the diameter of the particle • Fig. 14.2, page 414 • This empty space makes gases good insulators(example: windows, coats) • How does the volume of the particles in a gas compare to the overall volume of the gas?

6. Variables that describe a Gas • The four variables and their common units: 1. pressure (P) in kilopascals 2. volume (V) in Liters 3. temperature (T) in Kelvin 4. amount (n) in moles • The amount of gas, volume, andtemperature are factors that affect gas pressure.

7. 1. Amount of Gas • When we inflate a balloon, we are adding gas molecules. • Increasing the number of gas particles increases the number of collisions • thus, the pressure increases • If temperature is constant, then doubling the number of particles doubles the pressure

8. Pressure and the number of molecules are directly related • More molecules means more collisions, and… • Fewer molecules means fewer collisions. • Gases naturally move from areas of high pressure to low pressure, because there is empty space to move into – a spray can is example.

9. Common use? • A practical application is Aerosol (spray) cans • gas moves from higher pressure to lower pressure • a propellant forces the product out • whipped cream, hair spray, paint • Fig. 14.5, page 416 • Is the can really ever “empty”?

10. 2. Volume of Gas • In a smaller container, the molecules have less room to move. • The particles hit the sides of the container more often. • As volume decreases, pressure increases. (think of a syringe) • Thus, volume and pressure are inversely related to each other

11. 3. Temperature of Gas • Raising the temperature of a gas increases the pressure, if the volume is held constant. (Temp. and Pres. are directly related) • The molecules hit the walls harder, and more frequently! • Fig. 14.7, page 417 • Should you throw an aerosol can into a fire? What could happen? • When should your automobile tire pressure be checked?

13. Section 14.2The Gas Laws

14. The Gas Laws are mathematical • The gas laws will describe HOW gases behave. • Gas behavior can be predicted by the theory. • The amount of change can be calculated with mathematical equations. • You need to know both of these: the theory, and the math

15. Robert Boyle(1627-1691) • Boyle was born into an aristocratic Irish family • Became interested in medicine and the new science of Galileo and studied chemistry.  • A founder and an influential fellow of the Royal Society of London • Wrote extensively on science, philosophy, and theology.

16. #1. Boyle’s Law - 1662 Gas pressure is inversely proportional to the volume, when temperature is held constant. • Pressure x Volume = a constant • Equation: P1V1 = P2V2 (T = constant)

17. Graph of Boyle’s Law – page 418 Boyle’s Law says the pressure is inverse to the volume. Note that when the volume goes up, the pressure goes down

18. - Page 419

19. Boyle’s Law Practice Ammonia gas occupies a volume of 450. mL at 720. mm Hg. What volume will it occupy at standard pressure? V2 = 426 mL A 3.2-L sample of gas has a pressure of 102 kPa. If the volume is reduced to 0.65 L, what pressure will the gas exert? P2 = 502 kPa

20. Jacques Charles (1746-1823) • French Physicist • Part of a scientific balloon flight on Dec. 1, 1783 – was one of three passengers in the second balloon ascension that carried humans • This is how his interest in gases started • It was a hydrogen filled balloon – good thing they were careful!

21. #2. Charles’s Law - 1787 • The volume of a fixed mass of gas is directly proportional to the Kelvin temperature, when pressure is held constant. • This extrapolates to zero volume at a temperature of zero Kelvin.

22. Converting Celsius to Kelvin • Gas law problems involving temperature will always require that the temperature be in Kelvin. (Remember that no degree sign is shown with the kelvin scale.) • Reason? There will never be a zero volume, since we have never reached absolute zero. Kelvin = C + 273 °C = Kelvin - 273 and

23. - Page 421

24. Charles's Law Practice Helium occupies 3.8 L at -45°C. What volume will it occupy at 45°C? V2 = 5.3 L At 27°C, fluorine occupies a volume of 0.500 dm3. To what temperature in degrees Celsius should it be lowered to bring the volume to 200. mL? T2 = -153°C (120 K)

25. Joseph Louis Gay-Lussac (1778 – 1850) • French chemist and physicist • Known for his studies on the physical properties of gases. • In 1804 he made balloon ascensions to study magnetic forces and to observe the composition and temperature of the air at different altitudes.

26. #3. Gay-Lussac’s Law - 1802 • The pressure and Kelvin temperature of a gas are directly proportional, provided that the volume remains constant. • How does a pressure cooker affect the time needed to cook food? (Note page 422) • Sample Problem 14.3, page 423

27. Gay-Lussac’s Practice Problems A gas at STP is cooled to -185°C. What pressure in kPa will it have at this temperature (volume remains constant)? P2 = 0.32 atm Chlorine gas has a pressure of 1.05 kPa at 25°C. What pressure will it exert at 75°C? P2 = 1.23 atm

28. #4. The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas. Sample Problem 14.4, page 424

29. The combined gas law contains all the other gas laws! • If the temperature remains constant... P1 V1 P2 x V2 x = T1 T2 Boyle’s Law

30. The combined gas law contains all the other gas laws! • If the pressure remains constant... P1 V1 P2 x V2 x = T1 T2 Charles’s Law

31. The combined gas law contains all the other gas laws! • If the volume remains constant... P1 V1 P2 x V2 x = T1 T2 Gay-Lussac’s Law

32. Combined Gas Law Pracitce A gas occupies 256 mL at 720 kPa and 25°C. What will its volume be at STP? V2 = 220 mL A gas occupies 1.5 L at 850 kPa and 15°C. At what pressure will this gas occupy 2.5 L at 30.0°C? P2 = 540 kPa A gas occupies 125 mL at 125 kPa. After being heated to 75°C and depressurized to 100.0 kPa, it occupies 0.100 L. What was the original temperature of the gas? T1 = 544 K (271°C)

33. Section 14.3Ideal Gases

34. 5. The Ideal Gas Law #1 • Equation: P x V = n x R x T • Pressure times Volume equals the number of moles (n) times the Ideal Gas Constant (R) times the Temperature in Kelvin. • R = 8.31 (L x kPa) / (mol x K) • The other units must match the value of the constant, in order to cancel out. • The value of R could change, if other units of measurement are used for the other values (namely pressure changes)

35. The Ideal Gas Law • We now have a new way to count moles (the amount of matter), by measuring T, P, and V. We aren’t restricted to only STP conditions: P x V R x T n =

36. Ideal Gases • We are going to assume the gases behave “ideally”- in other words, they obey the Gas Laws under all conditions of temperature and pressure • An ideal gas does not really exist, but it makes the math easier and is a close approximation. • Particles have no volume? Wrong! • No attractive forces? Wrong!

37. Ideal Gases • There are no gases for which this is true (acting “ideal”); however, • Real gases behave this way at a) high temperature, and b) low pressure. • Because at these conditions, a gas will stay a gas! • Sample Problem 14.5, page 427

38. #6. Ideal Gas Law 2 • P x V = m x R x T M • Allows LOTS of calculations, and some new items are: • m = mass, in grams • M = molar mass, in g/mol • Molar mass = m R T P V

39. Density • Density is mass divided by volume m V so, m M P V R T D = D = =

40. Ideal Gas Law Practice

41. P1xV1 P2xV2 T1 T2 = (101.6 kPa)x(200 dm3) (98.6 kPa)x(150 dm3) 290 K T2 = Gas Review Problem #1 1)  A quantity of gas has a volume of 200 dm3 at 17oC and 106.6 kPa.  To what temperature (oC) must the gas be cooled for its volume to be reduced to 150 dm3 at a pressure of 98.6 kPa? Write given information: V1  =                                   V2  =  T1  =          T2  =  P1  =                              P2  =  200 dm3 150 dm3 17 oC +  273   =  290 K _______ 106.6 kPa 98.6 kPa Write equation:   Substitute into equation:        Solve for T2:   Recall: oC  +  273  =  K Therefore:  Temperature  =  -71oC T2  =  201 K

42. P1xV1 P2xV2 T1 T2 P1 P2 T1 T2 = = 98.6 kPa P2 295 K 265 K = (295 K) (295 K) (98.6 kPa)(265) (295) P2 = Gas Review Problem #2 2)  A quantity of gas exerts a pressure of  98.6 kPa at a temperature of 22oC.  If the volume remains unchanged, what pressure will it exert at -8oC? Write given information: V1  =                                V2  =  T1  =         T2  =  P1  =                                P2  =  constant  constant 22 oC +  273   =  295 K -8 oC +  273   =  265 K 98.6 kPa _________ Write equation:   Volume is constant...cancel it out from equation:  Substitute into equation: Solve for P2:   To solve, cross multiply and divide: (P2)(295 K) (98.6 kPa)(265 K) = P2  =  88.6 kPa

43. = mass volume P1xV1P2xV2 Density = (98.7 kPa)x(3.34 L)(101.3 kPa)x(V2) mass 2.85 L = T1 T2 3.17 g/cm3 = 310 K 273K (98.7 kPa)(3.34 dm3) PV n n = = [8.314 (kPa)(dm3)/(mol)(K)](310 K) RT Gas Review Problem #3 What is the mass of 3.34 dm3 sample of chlorine gas if the volume was determined at 37oC and 98.7 kPa?  The density of chlorine gas at STP is 3.17 g/dm3. Write given information: V1  =                                       V2  =  T1  =          T2  =   P1  =                                  P2  =  R  =                    Density  =  n  =  Cl2  =  Two approaches to solve this problem. METHOD 1:  Combined Gas Law & Density Write equation:   Substitute into equation:    Solve for V2: Density  =  3.17 g/dm3 @ STP Recall:    Substitute into equation:    Solve for mass:  __________ 3.34 L 273  K 37 oC +  273   =  310 K 98.7 kPa 101.3 kPa 8.314 kPa L / mol K 3.17 g/dm3 ___________ 71 g/mol 2.85 L V2  =  2.85 L @ STP PV = nRT mass  =  9.1 g chlorine gas

44. METHOD 2:  Ideal Gas Law Write equation:   Solve for moles:    Substitute into equation:    Solve for mole:  n  =  0.128 mol Cl2 Recall molar mass of diatomic chlorine is 71 g/mol Calculate mass of chlorine:  x g Cl2  =  0.128 mol Cl2   =  9.1 g Cl2

45. 1 mol FeS 1 mol H2S 879 g FeS 1 mol FeS (L)(Kpa) (mol)(K) (95.1 kPa)(V) = 1.5 mol H2S 8.314 (303 K) Gas Review Problem #6 Iron (II) sulfide reacts with hydrochloric acid as follows: FeS(s)  +  2 HCl(aq)    FeCl2(aq)  +  H2S(g) What volume of H2S, measured at 30oC and 95.1 kPa, will be produced when 132 g of FeS reacts? Calculate number of moles of H2S... x mole H2S  =  132 g FeS  Write given information: P  =  n  =  R  =  T  =  Equation: Substitute into Equation:  Solve equation for Volume:  132 g X L = 1.50 mol H2S 95.1 kPa 1.5 mole H2S 8.314 L kPa/mol K 30oC +  273  =  303 K PV  =  nRT V  =  39.7 L

46. Gas Review Problem #6 7)  What is the density of nitrogen gas at STP (in g/dm3 and kg/m3)? Write given information: 1 mole N2  =  28 g N2  =  22.4 dm3 @ STP Write equation: Substitute into equation:  Solve for Density:   Density  =  1.35 g/dm3 Recall:  1000 g  =  1 kg     &     1 m3  =  1000 dm3 Convert m3 to dm3:         x dm3  =  1m3                      =  1000 dm3 Convert: Solve: 1.35 kg/m3

47. Real Gases and Ideal Gases

48. Ideal Gases don’t exist, because: • Molecules do take up space • There are attractive forces between particles - otherwise there would be no liquids formed

49. Real Gases behave like Ideal Gases... • When the molecules are far apart. • The molecules do not take up as big a percentage of the space • We can ignore the particle volume. • This is at low pressure

50. Real Gases behave like Ideal Gases… • When molecules are moving fast • This is at high temperature • Collisions are harder and faster. • Molecules are not next to each other very long. • Attractive forces can’t play a role.