1. Gas Laws IB Chemistry 1

2. Phases of Matter Solid – tightly packed, vibrate in one position, definite shape and definite volume. Liquid – packed close together but now able to slide past each other, definite volume but not definite shape. Gas – rapid random, far apart molecules with no definite shape or volume.

3. Phase Diagram

4. Heating Curve

5. Gases Variable volume and shape Expand to occupy volume available Volume, Pressure, Temperature, and the number of moles present are interrelated Can be easily compressed Exert pressure on whatever surrounds them Easily diffuse into one another 5

6. Mercury Barometer Used to define and measure atmospheric pressure On the average at sea level the column of mercury rises to a height of about 760 mm. This quantity is equal to 1 atmosphere It is also known as standard atmospheric pressure 6

7. Pressure Units & Conversions The above represent some of the more common units for measuring pressure. The standard SI unit is the Pascal or kilopascal. Pounds per square inch or PSI is widely used in the United States. Most other countries use only the metric system. 7 1 atm = 760 Torr = 760 mmHg = 100 kPa = 1 x 10 5 Pa = 14.7 psi

8. Boyle’s Law According to Boyle’s Law the pressure and volume of a gas are inversely proportional at constant pressure. P α 1/V. P 1 V 1 = P 2 V 2 8

9. Boyle’s Law A graph of pressure and volume gives an inverse function A graph of pressure and the reciprocal of volume gives a straight line 9

10. = 340 kPa If the pressure of helium gas in a balloon has a volume of 4.00 dm 3 at 210 kPa, what will the pressure be at 2.50 dm 3 ? P 1 V 1 = P 2 V 2 (210 kPa) (4.00 dm 3 ) = P 2 (2.50 dm 3 ) P 2 = (210 kPa) (4.00 dm 3 ) (2.50 dm 3 ) Sample Problem 1: 10

11. Charles’ Law According to Charles’ Law the volume of a gas is proportional to the Kelvin temperature as long as the pressure is constant V α T V 1 = T 1 V2 T2V2 T2 Note: The temperature for gas laws must always be expressed in Kelvin where Kelvin = o C +273.15 (or 273 to 3 significant digits) 11

12. Charles’ Law A graph of temperature and volume yields a straight line. Where this line crosses the x axis (x intercept) is defined as absolute zero 12

13. Sample Problem 2 A gas sample at 40 o C occupies a volume of 2.32 dm 3. If the temperature is increased to 75 o C, what will be the final volume? 2.58 dm 3 13 V 1 = V 2 T 1 T 2 Convert temperatures to Kelvin. 40 o C = 313K 75 o C = 348K 2.32 dm 3 = V 2 313 K 349K (313K)( V 2 ) = (2.32 dm 3 )(348K) V 2 =

14. Gay-Lussac’s Law Gay-Lussac’s Law defines the relationship between pressure and temperature of a gas. The pressure and temperature of a gas are directly proportional P 1 = P 2 T1T1 T2T2 14 P α T

15. Sample Problem 3: The pressure of a gas in a tank is 3.20 atm at 22 o C. If the temperature rises to 60 o C, what will be the pressure in the tank? 3.6 atm 15 P 1 = P 2 T 1 T 2 Convert temperatures to Kelvin. 22 o C = 295K 60 o C = 333K 3.20 atm = P 2 295 K 333K (295K)( P 2 ) = (3.20 atm)(333K) V 2 =

16. The Combined Gas Law 1. If the amount of the gas is constant, then Boyle’s Charles’ and Gay-Lussac’s Laws can be combined into one relationship 2. P 1 V 1 = P 2 V 2 T2T2 T1T1 16

17. Sample Problem 4: A gas at 110 kPa and 30 o C fills a container at 2.0 dm 3. If the temperature rises to 80 o C and the pressure increases to 440 kPa, what is the new volume? V 2 = 0.58 dm 3 17 P 1 V 1 = P 2 V 2 T 1 T 2 Convert temperatures to Kelvin. 30 o C = 303K 80 o C = 353K V 2 = V 1 P 1 T 2 P 2 T 1 = (2.0 dm 3 ) (110 kPa ) (353K) (440 kPa ) (303 K)

18. Advogadro’s Law Equal volumes of a gas under the same temperature and pressure contain the same number of particles. If the temperature and pressure are constant the volume of a gas is proportional to the number of moles of gas present V α n where n is the number of moles of gas V/n = constant V 1 /n 1 = constant = V 2 /n 2 V 1 /n 1 = V 2 /n 2 18

19. Universal Gas Equation Based on the previous laws there are four factors that define the quantity of gas: Volume, Pressure, Kevin Temperature, and the number of moles of gas present (n). Putting these all together: PV nT = Constant = R The proportionality constant R is known as the universal gas constant 19

20. Universal Gas Equation The Universal gas equation is usually written as PV = nRT Where P = pressure V = volume T = Kelvin Temperature n = number of moles The numerical value of R depends on the pressure unit (and perhaps the energy unit) Some common values of R include: R = 62.36 dm 3 torr mol -1 K -1 = 0.08206 dm 3 atm mol -1 K -1 = 8.314 dm 3 kPa mol -1 K-1 20

21. Standard Temperature and Pressure (STP) The volume of a gas varies with temperature and pressure. Therefore it is helpful to have a convenient reference point at which to compare gases. For this purpose standard temperature and pressure are defined as: Temperature = 0 o C 273 K Pressure = 1 atmosphere = 760 torr = 100 kPa This point is often called STP 21

22. Sample Problem 5 Example: What volume will 25.0 g O 2 occupy at 20 o C and a pressure of 0.880 atmospheres? : V = (0.781 mol)(0.08206 dm -3 atm mol -1 K -1 )(293K) 0.880 atm V = 21.3 dm 3 (25.0 g) n = ----------------- = 0.781 mol (32.0 g mol -1 ) V =? P = 0.880 atm; T = (20 + 273)K = 293K R = 0.08206 dm -3 atm mol -1 K -1 PV = nRT so V = nRT/P Data Formula Calculation Answer 22

23. Density (d) Calculations d = m V = PMrPMr RT m is the mass of the gas in g M r is the molar mass of the gas Molar Mass ( M r ) of a Gaseous Substance dRT P M r = d is the density of the gas in g/L Universal Gas Equation –Alternate Forms 23

24. A 2.10 dm 3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0 o C. What is the molar mass of the gas? Sample Problem 6 24

25. A 2.10 dm 3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0 o C. What is the molar mass of the gas? dRT P M r = d = m V 4.65 g 2.10 dm 3 = = 2.21 g dm 3 M r = 2.21 g dm 3 1 atm x 0.08206 x 300.15 K dm 3 atm molK Mr =Mr = 54.6 g/mol Sample Problem 6 Solution 25

26. Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases (partial pressures). P T = P 1 + P 2 + P 3 + P 4 +.... whereP T = total pressure P 1 = partial pressure of gas 1 P 2 = partial pressure of gas 2 P 3 = partial pressure of gas 3 P 4 = partial pressure of gas 4 26

27. Dalton’s Law of Partial Pressures Applies to a mixture of gases Very useful correction when collecting gases over water since they inevitably contain some water vapor. 27

28. Sample Problem 7 Henrietta generates Hydrogen gas and collected it over water. If the volume of the gas is 250 cm 3 and the barometric pressure is 765.0 torr at 25 o C, what is the pressure of the “dry” hydrogen gas at STP? (P H2O = 23.8 torr at 25 o C) 28

29. Sample Problem 8 -- Solution Henrietta Minkelspurg generates Hydrogen gas and collected it over water. If the volume of the gas is 250 cm 3 and the barometric pressure is 765.0 torr at 25 o C, what is the pressure of the “dry” hydrogen gas at STP? (P H2O = 23.8 torr at 25 o C) 29

30. Sample Problem 9 Henrietta generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm 3 and the barometric pressure is 765 torr at 25 o C, what is the volume of the “dry” oxygen gas at STP? 30

31. Sample Problem 9 -- Solution Henrietta generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm 3 and the barometric pressure is 765 torr at 25 o C, what is the volume of the “dry” oxygen gas at STP? From the previous calculation the adjusted pressure is 742.2 torr V 2 = (250 cm3)(742.2 torr)(273K) (298K)(760.torr) V 2 = 223.7 cm 3 P 1 = P H2 = 742.2 torr; P 2 = Std Pressure = 760 torr V 1 = 250 cm 3 ; T 1 = 298K; T 2 = 273K; V 2 = ? (V 1 P 1 /T 1 ) = (V 2 P 2 /T 2 ) therefore V 2 = (V 1 P 1 T 2 )/(T 1 P 2 ) 31

32. Kinetic Molecular Theory Matter consists of particles (atoms or molecules) that are in continuous, random, rapid motion The Volume occupied by the particles has a negligibly small effect on their behavior Collisions between particles are elastic Attractive forces between particles have a negligible effect on their behavior Gases have no fixed volume or shape, but take the volume and shape of the container The average kinetic energy of the particles is proportional to their Kelvin temperature 32

33. Maxwell-Boltzman Distribution Molecules are in constant motion Not all particles have the same energy The average kinetic energy is related to the temperature An increase in temperature spreads out the distribution and the mean speed is shifted upward 33

34. The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature u rms = 3RT MrMr  Velocity of a Gas 34

35. Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH 3 17.0 g/mol HCl 36.5 g/mol NH 4 Cl Diffusion 35

36. DIFFUSION AND EFFUSION Diffusion is the gradual mixing of molecules of different gases. Diffusion is the gradual mixing of molecules of different gases. Effusion is the movement of molecules through a small hole into an empty container. Effusion is the movement of molecules through a small hole into an empty container. 36

37. Graham’s Law Graham’s law governs effusion and diffusion of gas molecules. Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv 2 Thomas Graham, 1805-1869. Professor in Glasgow and London. The rate of effusion is inversely proportional to its molar mass. 37

38. Sample Problem 10 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 o C according to the equation: 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O(g) Using Graham's Law, what is the ratio of the effusion rates of NH 3 (g) to O 2 (g)? 38

39. Sample Problem 10 Solution 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 o C according to the equation: 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O(g) Using Graham's Law, what is the ratio of the effusion rates of NH 3 (g) to O 2 (g)? 39

40. Sample Problem 11 What is the rate of effusion for H 2 if 15.00 cm 3 of CO 2 takes 4.55 sec to effuse out of a container? 40

41. Sample Problem 11 Solution What is the rate of effusion for H 2 if 15.00 cm 3 of CO 2 takes 4.55 sec to effuse out of a container? Rate for CO 2 = 15.00 cm 3 /4.55 s = 3.30 cm 3 /s 41

42. Sample Problem 12 What is the molar mass of gas X if it effuses 0.876 times as rapidly as N 2 (g)? 42

43. Sample Problem 12 Solution What is the molar mass of gas X if it effuses 0.876 times as rapidly as N 2 (g)? 43

44. Ideal Gases v Real Gases Ideal gases are gases that obey the Kinetic Molecular Theory perfectly. The gas laws apply to ideal gases, but in reality there is no perfectly ideal gas. Under normal conditions of temperature and pressure many real gases approximate ideal gases. Under more extreme conditions more polar gases show deviations from ideal behavior. 44

45. In an Ideal Gas --- The particles (atoms or molecules) in continuous, random, rapid motion. The particles collide with no loss of momentum The volume occupied by the particles is essentially zero when compared to the volume of the container The particles are neither attracted to each other nor repelled The average kinetic energy of the particles is proportional to their Kelvin temperature At normal temperatures and pressures gases closely approximate idea behavior 45

46. Real Gases These deviations occur because Real gases do not actually have zero volume Polar gas particles do attract if compressed For ideal gases the product of pressure and volume is constant. Real gases deviate somewhat as shown by the graph pressure vs. the ratio of observed volume to ideal volume below. 46

47. van der Waals Equation (P + n 2 a/V 2 )(V - nb) = nRT The van der Waals equation shown below includes corrections added to the universal gas law to account for these deviations from ideal behavior wherea => attractive forces between molecules b => residual volume of molecules The van der Waals constants for some elements are shown below Substance a (dm 6 atm mol -2 ) b (dm 3 mol -1 ) He0.03410.02370 CH 4 2.250.0428 H 2 O5.460.0305 CO 2 3.590.0437 47

48. Sample Problem 13 What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) 48

49. Sample Problem 13 Solution What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.08206 x 310.15 K dm 3 atm molK 1.00 atm = = 4.76 dm 3 49