Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 10 Gases

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1 Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Gases

2

3 A Gas Has neither a definite volume nor shape. Uniformly fills any container. Mixes completely with any other gas Exerts pressure on its surroundings.

4 Earth-like Atmosphere Composition of Earth s Atmosphere Compound %(Volume) Mole Fraction a Nitrogen Oxygen Argon Carbon dioxide Methane 2 x x 10-6 Hydrogen 5 x x 10-7 a. mole fraction = mol component/total mol in mixture.

5 Pressure Pressure is the amount of force applied to an area. P = F A Atmospheric pressure is the weight of air per unit of area.

6 Elevation and Atmospheric Pressure

7 Pressure is equal to force/unit area SI units = Newton/meter 2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa (100,000 Pa = 1 bar) 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr = hpa = psi Meteorologists often report pressure in millibar; 1 mbar =0.001bar =0.1 kpa = 1hPa

8 The barometric pressure reported for Butte by the weather service was P b = 1018 hpa. What is this pressure in Torr, Atm, and Bar?

9 Units of Pressure mm Hg or torr These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. Atmosphere 1 atm = 760 torr = kpa = 14.7 lb/in 2

10 Manometer This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

11 Variables Affecting Gases Pressure (P) Volume (V) Temperature (T) Number of Moles (n)

12 Boyle s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

13 Boyle Law pressure is inversely proportional to volume (at constant T and moles, n).

14 As P and V are inversely proportional A plot of V versus P results in a curve. Since PV = k V = k (1/P) This means a plot of V versus 1/P will be a straight line.

15 Boyle s Law P 1/V (T and n fixed) P V = Constant P 1 V 1 = P 2 V 2

16 Problem. The volume of 1.00 mol of ammonia gas at 1.00 atm of pressure is gradually decreased from 78.0 ml to 39.0 ml. What is the final pressure of ammonia if there is no change in temperature? ans. = 2.00 atm

17 Charles s Law The volume of a gas is directly proportional to Kelvin temperature, and extrapolates to zero at zero Kelvin. V T (P & n are constant) V 1 = V 2 T 1 T 2

18 Problem. A 5.6 L sample of gas is cooled from 78 C to a temperature at which its volume is 4.3 L. What is this new temperature? Assume no change in pressure of the gas. ans. = K

19 The temperature in the troposphere is also proportional to pressure.

20 Avogadro s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V n V 1 = V 2 n 1 n 2

21 Avogadro s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn

22 Ideal-Gas Equation So far we ve seen that V 1/P (Boyle s law) V T (Charles s law) V n (Avogadro s law) Combining these, we get nt V P

23 Combined Gas Law Combining the gas laws the relationship P T(n/V) can be obtained. If n (number of moles) is held constant, then PV/T = constant. P 1 V 1 T 1 = P 2 V 2 T 2

24 Ideal Gas Law PV = nrt R = universal gas constant = L atm K -1 mol -1 P = pressure in atm V = volume in liters n = moles T = temperature in Kelvin

25 Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

26 Ideal-Gas Equation The relationship V nt P then becomes nt V = R P or the classic PV = nrt

27 Problem Calculate the pressure in atmospheres and pascals of a 1.2 mol sample of methane gas in a 3.3 L container at 25 C.

28 Problem If the pressure of ozone, O 3, in the stratosphere is 3.0E-3 atm and the temperature is 250K, how many ozone molecules are in one liter.

29 STP Standard Temperature and Pressure (for gases) P = 1 atmosphere T = 0 C The molar volume of an ideal gas is liters at STP (put 1 mole, 1 atm, R, and 273 K in the ideal gas law and calculate V) Note STP is different for other phases, e.g. solutions or the phases associated with enthalpies of formation.

30 Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get n V = P RT

31 Densities of Gases We know that moles molecular mass = mass n = m So multiplying both sides by the molecular mass ( ) gives m V = P RT

32 Problem Calculate the density of sulfur hexafluoride gas at 707 torr and 21 C.

33 Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: d = P RT Becomes = drt P

34 Problem To identify a diatomic gas (X 2 ), a researcher carried out the following experiment: She weighed an empty 1.00L- bulb, then filled it with the gas at 1.10 atm and 23.0 C and weighed it again. The difference in mass was 1.72 g. Identify the gas.

35

36 Problem Chloroform is a common lab reagent. If the pressure of chloroform in a flask at 25C is 195 mmhg and the density is 1.25 g/l. What is the molar mass of chloroform?

37 Dalton s Law of Partial Pressures For a mixture of gases in a container P Total = P 1 + P 2 + P

38 Consider a mixture of oxygen and nitrogen gases

39 Mole Fraction & Partial Pressure Mole Fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in a mixture. 1 = n 1 = n 1 n TOTAL n 1 + n 2 + n 3 + Mole Fraction in terms of pressure (n = PV/RT) 1 = P 1 (V/RT) P 1 (V/RT) + P 2 (V/RT) + P 3 (V/RT) +

40 Continued 1 = P 1 = P 1 P 1 + P 2 + P 3 + P TOTAL 1 = n 1 = P 1 n TOTAL P TOTAL

41 Mole Fraction Example At 25 0 C, a 1.0 L flask contains moles of nitrogen, mg of oxygen and 4 x molecules of ammonia. A. What is the partial pressure of each gas? B. What is the total pressure in the flask? C. What is the mole fraction of each?

42

43

44 Problem A sample of KClO 3 is heated and decomposes to produce O 2 gas. The gas is collected by water displacement at 25 C. The total volume of the collected gas is 370 ml at a pressure of 754 torr. How many moles of oxygen are formed? Hint: The gas collected is a mixture so use Dalton s Law to calculate the pressure of oxygen then the ideal gas law to find the number of moles oxygen. P T = P O 2 + P H 2O

45 Collecting a Gas Over Water

46 P H2O (25 C) = 23.8 torr

47 Problem A sample of 4.10 ml of diethylether (d = g/mL) is introduced into a 5.10 L vessel that already contains a mixture of N 2 and O 2, whose partial pressures are atm and atm, respectively. The temperature is held at 35.0 C, and the diethylether totally evaporates. a. Calculate the partial pressure of the diethylether. b. Calculate the total pressure in the container.

48 a. Calculate the partial pressure of the diethylether.

49 b. Calculate the total pressure in the container.

50 Kinetic Molecular Theory 1. The volume of the gas molecules is negligible compared with the container s volume. 2. Gas molecules move randomly and constantly. 3. The motion of these molecules is associated with their average kinetic energy that is proportional to the absolute temperature of the gas.

51 Kinetic Molecular Theory 4 Gas molecules continuously and elastically collide with one another and container walls. 5 Each molecule acts independently of all the other molecules in the sample. There are no forces of attraction (or repulsion) between molecules.

52 The temperature of an ideal monatomic gas is a measure related to the average kinetic energy of its atoms as they move. In this animation, the size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure. These room-temperature atoms have a certain, average speed (slowed down here two trillion fold).

53 Main Tenets of Kinetic- Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

54 The path of individual particles is a Random Walk The Mean Free Path is the average distance traveled between collisions.

55 Kinetic Molecular Theory K.E. = 1/2mu 2 rms U rms - the root-meansquared speed of the molecules At Constant Temp U rms = 3RT M

56 u rms 3RT M root mean square velocity Calculate the root mean square velocity of O 2 at 60 C (333 K). [3 (8.314 kg m 2 s -2 mol -1 K -1 ) (333 K)/(0.032 kg mol -1 )] 1/2 = 509 m s -1

57 Problem. Calculate the velocity (in mi/hr) for water molecules at room temperature.

58 Diffusion and Effusion Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Effusion: describes the escape of gas through a tiny hole into a space of lower pressure.

59 The difference in the rates of effusion for helium and nitrogen, for example, explains why a helium balloon deflates faster. Effusion

60

61 At a given temperature the u rms velocity of a gas particle is dependent on temperature and mass of the particle. (from Kinetic Molecular Theory) The u rms equation can be used to derive Graham s Law of effusion (or diffusion). rate of rate of effusion effusion gas gas 1 2 M( gas 2) M( gas1)

62 Problem List the following gases, which are at the same temperature, in the order of increasing rates of diffusion. O 2, He, & NO

63 What will be the relative rate of effusion of hydrogen gas as compared to oxygen? r H 2 M O r O 2 M H

64 A sample of helium diffuses 4.58 times faster than an unknown gas. What is the likely identity of the unknown gas? (use Graham s Law) r r He unk M M unk He M He M unk M unk 83.9 [krypton(83.8)]

65 Real Gases For real gasses correct ideal gas behavior assumption when at high pressure (smaller volume) and low temperature (attractive forces become important).

66 Deviations from Ideal Behavior For 1 mole of an ideal gas under standard conditions we can define a compressibility factor, Z = 1 deviations of Z from 1 reflect non-ideal behavior.

67 In general, deviations from ideal behavior become more significant the closer a gas is to a phase change, the lower the temperature or the larger the pressure.

68 The value of Z generally increases with pressure and decreases with temperature.

69 The value of Z generally increases with pressure and decreases with temperature. Briefly At high pressures molecules are colliding more often, and at low temperatures they are moving less rapidly. This allows attractive forces between molecules to have a notable effect, making the volume of the real gas (V real ) less than the volume of an ideal gas (V ideal ) which causes Z to drop below one When pressures are lower or temperatures higher, the molecules are more free to move. In this case repulsive forces dominate, making Z > 1. The closer the gas is to its critical point or its boiling point, the more Z deviates from the ideal case.

70 Real Gases van der Waals Equation [ P + a ( n / V) ] x ( V - nb ) = nrt obs 2 corrected pressure corrected volume P ideal V ideal

71 Table 8.4 Van der Waal Constants for Selected Gases Substance a (L 2 atm/mol 2 ) b (L 2 /mol) He H CH CO SO P n 2 a V 2 V n b n R T

72 The van der Waals Equation (P + n 2 a V 2 ) (V nb) = nrt

73 At high pressures, real gases do not behave ideally. Use the van der Waals equation and data in the text to calculate the pressure exerted by mol H 2 at 25 C in a 1.00 L container. P P P nrt V nb L atm 17.55mol 298K mol K 1.00L 17.55mol L mol 729atm 2 n a 2 V 2 2 L atm 17.55mol mol L Repeat the calculation assuming that the gas behaves like an ideal gas. (check yourself, ans. 429 atm)

74 (a) Calculate the pressure exerted by 1.00 mol of CO 2 in a 1.00 L vessel at 300 K, assuming that the gas behaves ideally. (b) Repeat the calculation by using the van der Waals equation.

75 (b) Repeat the calculation by using the van der Waals equation.co2; a: 3.59 b: n 2 a P V n b n V 2 R T

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