1. The Gaseous Phase The three phases of matter, solids, liquids and gases, have different characteristics. A gas expands to fill any container it occupies A liquid has a fixed volume but takes the shape of the volume of the container it occupies A solid has both fixed volume and shape. These characteristics originate from the nature of the interactions between the atoms or molecules

2. On a macroscopic scale, gases are distinguished from solids and liquids by their much smaller values of density. On the microscopic scale, the smaller values of density arise due to the much lower NUMBER DENSITY (number of molecules per cm 3 of the sample) compared with liquids and solids.

3. Understanding the behavior of gases and how reactions occur in the gas phase is of practical importance CH 4 (g) + O 2 (g) --> CO 2 (g) + H 2 O(g) - combustion of fuels N 2 (g) + H 2 (g) --> NH 3 (g) - production of ammonia for fertilizers 2NO(g) + O 2 (g) -> 2NO 2 (g) - responsible for acid rain

4. Properties of Gases - A gas will fill the volume of the container which contains it. - The volume of the gas equals the volume of its container - Gases are highly compressible; when pressure is applied to a gas, its volume readily decreases - Gases form homogenous mixtures with each other regardless of their identity or relative proportions These properties arise because the individual atoms/ molecules are relatively far apart

5. Three properties of gases that are used to describe gases are pressure (P), volume (V) and temperature (T). The volume of a gas is defined by the volume of the container. Typical units for volume of gases is the liter, L.

6. PRESSURE The force exerted by a gas on a unit area of the walls of its container is the pressure exerted by the gas. Pressure = Force Area SI Units for pressure Force is newton, N (=kg m/s 2 ) Area - m 2 Pressure - N/m 2 or pascal (Pa)

7. Atmospheric Pressure Because of gravity, the earth’s atmosphere exerts a downward force and consequently a pressure on the earth’s surface. Atmospheric pressure: pressure exerted by the atmosphere around us A column of air 1m 2 in cross section extending through the atmosphere has a mass or roughly 10,000 kg.

8. The acceleration produced by earth’s gravity is 9.8 m/s 2 force = mass x acceleration Force exerted by this air column is ~ 1 x 10 5 N The pressure exerted by this air column ~ 1 x 10 5 N 1 m 2 ~ 1 x 10 5 Pa More precisely, 1.01325 x 10 5 Pa = 1 atmosphere (atm)

9. A barometer operates on the principle that the height of a liquid in a closed tube depends on the atmospheric pressure Pressure = g h d g ~ 9.8 m/s h is the height of the liquid in the sealed tube d is the density of the liquid

10. a) What is the height of a mercury column in a barometer at atmospheric pressure? b) What is the height of a water column in a barometer at atmospheric pressure? Explains why mercury is used in barometers and not water

11. Units of pressure 1 atm = 760 mm Hg = 760 torr = 1.01325 x 10 5 Pa There are other units of pressure (lbs/in 2, bar) but we will typically deal with atm, mm of Hg or torr and Pa.

12. The Gas Laws Through experimental observations, relationships have been established between the pressure (P), temperature (T) and volume (V) and number of moles (n) of gases. These relationships are called the GAS LAWS. Having defined P, V, T, and n for a gas, this information defines the physical condition or state of a gas. The relationships between P, V, T and n that will be discussed hold for IDEAL gases (or for low pressures; “ideal” conditions)

13. Relationship between Pressure and Volume: Boyle’s Law Boyle noted from the experiments he performed that at a fixed temperature and for a fixed amount of gas, as pressure on a gas increases, the volume occupied by a gas decreases.

14. Boyle’s Law The product of pressure and volume of a sample of gas is a constant, at constant temperature and for a fixed amount of gas. P V = constant At a fixed temperature and for a fixed amount of gas P  1 V P = constant Volume

15. Plot illustrating P-V relationship

16. The conditions of 1.00 atm pressure and 0 o C are called standard temperature and pressure (STP). Under STP conditions, the volume occupied by the gas in the J-tube is 22.4 L. Since, PV = constant P 1 V 1 = P 2 V 2

17. Temperature-Volume Relation - Charles Law The volume of a fixed quantity of gas at constant pressure increases linearly with temperature. V = V 0 +  V 0 t V 0 is the volume of the gas at 0 o C t is the temperature in o C  is the coefficient of thermal expansion

18. Volume V = V 0 +  V 0 t y = mx + b

19. From a plot of V vs t we can determine V 0 from the y- intercept. From the slope =  V 0,  can be determined Volume

20. Since gases expand by the same relative amount when heated between the same two temperatures (at low pressure) implies that  is the same for all ideal gases. For gases, at low pressure  = 1 ( o C -1 ) 273.15 For liquids and solids  varies from substance to substance

21. Re-writing the expression connecting V and t:  V V0V0 t =t = 273.15 o C [ -1 ] V V0V0 Gas thermometer: By measuring the volume of a gas at 0 o C and measuring the volume change as temperature changes, the temperature can be calculated

22. Absolute temperature - Kelvin Scale V = V 0 [ 1 + ] t 273.15 o C At t = -273.15 o C => volume of gas is zero temperatures negative volume which is physically impossible. Hence 273.15 o C is the lowest temperature that can be physically attained and is the fundamental limit on temperature.

23. This temperature is called ABSOLUTE ZERO and is defined to be the zero point on the Kelvin scale (K) T (Kelvin) = 273.15 + t (Celsius) If we substitute the above expression in V = V 0 [ 1 + ] t 273.15 o C and solve for V V 0 273.15 T V =

24. is a constant V 0 273.15 Hence, V  T Charles’ Law In other words, on an absolute temperature scale, at a constant pressure and for a fixed amount if gas, the volume of the gas is proportional to the temperature Hence, V 1 V2V2 T 1 T2T2 = at a fixed pressure and for a fixed amount of gas Note: T is temperature in Kelvin

25. Quantity-Volume relation - Avogadro’s Law Volume is affected not just by pressure and temperature, but also by the amount of gas. Avogadro’s hypothesis - Equal volumes of gases at the same temperature and pressure contain the same number of molecules.

26. Avogadro’s law: the volume of a gas maintained at constant pressure and temperature is directly proportional to the number of moles of gas. V = constant x n Hence, doubling the moles of gas will cause the volume to double (as long as T and P remain constant)

27. The Ideal-Gas Equation Boyle’s law: V  P -1 (constant n, T) Charles’ law: V  T (constant n, P) Avogadro’s law: V  n(constant P, T) Putting the three laws together: V  n T P

28. P V = n R TIDEAL GAS EQUATION An ideal gas is a gas whose pressure, volume and temperature behavior is completely described by this equation. R is called the universal gas constant since it is the same for all gases. V = R n T P Note: The ideal gas equation is just that - ideal. The equation is valid for the most gases at low pressures. Deviations from “ideal” behavior are observed as pressure increases.

29. The value and units of R depends on the units of P, V, n and T Temperature, T, MUST ALWAYS BE IN KELVIN n is expressed in MOLES P is often in atm and V in liters, but other units can be used. Values of R UnitsNumerical value L-atm/(mol-K)0.08206 cal/(mol-K)1.987 J/(mol-K)8.314 m 3 -Pa/(mol-K)8.314 L-torr/(mol-K)62.36

30. Example: Calcium carbonate, CaCO 3 (s), decomposes upon heating to give CO 2 (g) and CaO(s). A sample of CaCO 3 is decomposed, and the CO 2 collected in a 250. mL flask. After the decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31 o C. How many moles of CO 2 were generated? Given: Volume of CO 2 = 250 mL = 0.250 L Pressure of CO 2 = 1.3 atm temperature of CO 2 = 31 o C First convert temperature to K T = 31 + 273 = 304 K

31. To calculate n: n =P V R T Based on the units given for P and V, use appropriate value for R R = 0.08206 L-atm/(mol K) n = (1.3 atm)(0.250 L) (0.08206 L-atm/(mol-K)) (304K) n = 0.013 mol CO 2

32. Problem The gas pressure in a closed aerosol can is 1.5 atm at 25 o C. Assuming that the gas inside obeys the ideal-gas equation, what would the pressure be if the can was heated to 450 o C? Since, the can is sealed, both V and n stay fixed. P (atm)t ( o C)T(K) Initial1.5 25298 Final?450723 n R V = constant P T = P i T i = P f T f P i T f T i = P f P f = 3.6 atm

33. Molar Mass and Gas Density The ideal gas law, P V = n R T can be used to determine the molar mass of gaseous compounds. The number of moles of a compound = mass of gas sample (m) Molar Mass (M) n = m M

34. Substituting this in the ideal gas equation P V = m M R TM = m R T P V Solving for M, the molar mass P V = m M R T The ideal gas equation can be also be used to determine the density of the gas d = R T P M

35. Gas Stoichiometry If the conditions of pressure and temperature are known, then the ideal gas law can be used to convert between chemical amounts i.e. moles, and gas volume. Hence in dealing with chemical reactions involving gases, we can deal with volumes of gases instead of moles of gases, being that volume is usually an easier quantity to measure.

36. Problem Dinitrogen monoxide, N 2 O, better known as nitrous oxide or laughing gas, is shipped in steel cylinders as a liquid at pressures of 10 MPa. It is produced as a gas in aluminium trays by the decomposition of ammonium nitrate at 200 o C. NH 4 NO 3 (s) --> N 2 O(g) + 2H 2 O(g) What volume of N 2 O(g) at 1.00 atm would be produced from 100.0g of NH 4 NO 3 (s) after separating out the H 2 O and cooling the N 2 O gas to 273K. Assume a 100% yield in the production of N 2 O(g). Assuming a 100% yield => all the NH 4 NO 3 (s) is converted to N 2 O(g) Moles of NH 4 NO 3 (s) decomposed = 100.0 g/80.04g/mol = 1.249 mol

37. Hence, 1.249 mol of N 2 O(g) formed. To calculate the volume of N 2 O(g) produced, use the ideal gas equation. V = n R T/ P V =[ (1.249 mol) (0.08206 L-atm/(mol-K)) (273)]/(1.00 atm) = 28.0 L N 2 O(g)

38. Mixtures of Gases Can we use the ideal gas equation to determine the properties of gases in a mixture? Dalton observed that the total pressure of a mixture of gases equals the sum of the pressures that each would exert if each were present alone. The partial pressure of a gas in a mixture of gases is defined as the pressure it exerts if it were present alone in the container.

39. Dalton’s law states that the total pressure is the sum of the partial pressures of each gas in the mixture. For example, consider a mixture of two gases A and B in a closed container Assuming that the pressure is low enough, A and B obey the ideal gas equation. The fact that A and B behave as ideal gases implies that A and B do not interact with each other.

40. P A = n A RT V The pressure exerted by A, P A is then and that exerted by B is: P B = n B RT V

41. From Dalton’s law: P total = P A + P B n A RT V n B RT V =+ = (n A + n B ) RT V = n total RT V Where n total = n A + n B is the total number of moles

42. Mole Fraction What is the fraction of the number of moles of A in the mixture? To find this out, we need to divide the number of moles of A by the total number of moles of gases in the mixture nAnA n total The quantity is called the MOLE FRACTION of A, X A Note: mole fraction is unitless since it is a fraction of two quantities with the same unit. Also, sum of mole fractions of the components in a mixture =1

43. For the component A in the mixture, we can write its pressure as P A = n A RT V = n total RT V The total pressure is P total Dividing P A /P total PAPA P total nAnA n total == X A Hence, P A = X A P total

44. A study of the effect of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent of CO 2, 18.0 mol percent of O 2 and 80.5 mol percent Ar. a) Calculate the partial pressure of O 2 in the mixture if the total pressure of the atmosphere is to be 745 torr? b) If this atmosphere is to be held in a 120-L space at 295 K, how many moles of O 2 are needed? a) P O 2 = X O 2 P total = (0.180) (745 torr) = 134 torr b) PV = n R T (134 torr) (1 atm) (120 L) = n (0.08206 L-atm/molK)(295 K) (760 torr) n = 0.872 mol

45. Kinetic Theory of Gases The ideal gas equation describes how gases behave. In the 19th century, scientists applied Newton’s laws of motion to develop a model to explain the behavior of gases. This model, called the kinetic theory of gases, assumes that the atoms or molecules in a gas behave like billiard balls. In the gas phase, atoms and molecules behave like hard spheres and do not interact with each other.

46. Assumptions of Kinetic Theory of Gases 1) A gas consists of a large number of particles that are so small compared to the average distance separating them, that their own size can be considered negligible. 2) The particles of an ideal gas behave totally independent, neither attracting nor repelling each other. 3) Gas particles are in constant, rapid, straight-line motion, incessantly colliding with each other and with the walls of the container. All collisions between particles are elastic. 4) A collection of gas particles can be characterized by its average kinetic energy, which is proportional to the temperature on the absolute scale.

47. Gas particles are constantly colliding with each other and the walls of the container. It is the collisions between the gas particles and the walls of the container that define the pressure of the gas. Every time a gas particle collides with the wall of the container, the gas particle imparts its momentum to the wall momentum = mass x velocity

48. The pressure exerted by the gas is proportional to the momentum of the particle and the number of collisions per unit time, the collision frequency. Pressure  (momentum of particle) x (rate of collisions with the wall) The rate of collision is proportional to the number of particles per unit volume (N/V) and the speed of the particle (u). P  (m x u) x [ N x u] V

49. P  N m u 2 V P V  N m u 2 The speed, u, is the average speed of the particles, since not all the particles move with the same speed. P V  N m u 2 Replace u 2 with the mean-square speed, u 2

50. Particles are moving in a 3-D space: P V  N m u 2 3 1 Comparing this equation with the ideal gas equation P V = n R T n R T  N m u 2 3 1 This equation relates the speed of the gas particles with the temperature of the gas

51. n N o = Nwhere N o is Avogadro’s number n R T  n N o m u 2 3 1 R T  N o m u 2 3 1 (m N o ) is the molar mass of the gas, M R T  M u 2 3 1 M 3 R T u2u2 = The mean square speed depends on T and M

52. The kinetic energy of a particle is = 1 m u 2 2 The average kinetic energy of a mole of particles is = 1 N o m u 2 2 Temperature and Kinetic Energy

53. R T  N o m u 2 3 1 From the equation: 2 1 N o m u 2 Average kinetic energy per mole of particles = 3 R T = N o m u 2 Average kinetic energy per mole of particles = RT 3 2 Hence the average kinetic energy of the molecules in a gas depends only on the temperature of the gas and is independent of the mass or density of the gas.

54. The kinetic theory of gases explains the observed behavior of ideal gases; i.e. it explains, the three gas laws Boyle’s Law: P V  N m u 2 3 1 The pressure exerted by a the gas is due to the collisions between the particles and the walls of the container. If the temperature stays the same, then the average speed of the particles is the same The pressure will depend on the number of collisions per unit area of the wall per unit time. Reducing the volume of the container, will result in more frequent collisions and hence a higher pressure.

55. Charles’ Law: P V  N m u 2 3 1 u 2 depends on T. As T increases, u 2 increases. With increasing temperature, the number of collisions with the walls of the container must increase as u 2 increases. If the P and n are kept constant, the volume of the container must increase with increasing T.

56. Avogadro’s Law: P V  N m u 2 3 1 At fixed pressure and temperature, the volume of the gas is proportional to the number of particles. This is explicit in the equation above, for fixed P and T.

57. Distribution of Molecular Speeds Atoms or molecules in a gas do not all travel at the same speed. There is a distribution of speeds, with some travelling slow, others fast, and the majority peaked about a value called the most probable speed. A plot of the number of molecules travelling at a given speed versus speed, at a fixed temperature is called the MAXWELL- BOLTZMANN distribution.

59. u mp u u rms

60. The root mean square speed, is the square root of the square of the average speed Root mean square speed : sum the squares of the speeds, divide by the number of particles and then square root the resulting number. Average speed: divide the sum of the speed of all particles by the number of particles. 1, 2, 3, 4, 5, 6 Average = (1+2+3+4+5+6)/6 = 3.5 RMS value = ((1+4+9+16+25+36)/6) 1/2 = 3.89 The rms value is always slightly higher than the average value.

61. For the same molecule, as temperature increases the most probable speed shifts to higher values; the distribution also broadens. For two gases at the same temperature, but different masses: the rms speed for the lighter gas is higher than that for the heavier gas particles. The average kinetic energy of all gases at the same temperature is the same, regardless of mass.

62. Temperature describes a system of gaseous molecules only when their speed distribution is represented by the Maxwell- Boltzmann distribution. A collection of gas molecules whose speed distribution can be represented by a Maxwell-Boltzmann distribution is said to be at thermal equilibrium.

63. Motion of Gas Molecules- Diffusion & Effusion Gas molecules do not travel in a straight line, but undergo a more random type motion. Each time a gas molecule collides with another its direction changes. The average distance covered by a gas molecule between two collisions is the mean free path. Lower the gas pressure, longer is the mean free path.

64. If the pathway of a gas molecule from point A to B is tracked, its path would look like this:

65. This type of irregular motion is called DIFFUSION and is responsible for gases mixing; like an the odor filling up a room, The rate of diffusion depends inversely on the mass of the molecule; heavier molecules diffuse more slowly than lighter molecules. rate of diffusion of A rate of diffusion of B = √MB√MB √MA√MA

66. Effusion Effusion is the motion of gas molecules through a small hole. Within the container, each gas molecule undergoes the random motion, colliding with other gas molecules. During this process, if the gas molecule encounters the hole in the container it will emerge out of the container.

67. Although each particle traces its own unique path to the hole, the faster the molecules move, the more quickly will they emerge from the hole. The rate of effusion is proportional to u rms. For a mixture of two different gases, A and B, in the same container, and hence at the same temperature and pressure, the rate of effusion for each depends on u rms of each. rate of effusion of A rate of effusion of B = u rms (A) u rms (B) (3 R T/M A ) 1/2 (3 R T/M B ) 1/2 = rate of effusion of A rate of effusion of B = √MB√MB √MA√MA

68. Isotope separation by Diffusion Rate of diffusion of a gas is inversely proportional to the square root of the mass Light atoms diffuse through a porous barrier faster than heavier atoms. 235 U is fissile, not 238 U

69. Natural U is about 99.28% 238 U and 0.72% 235 U For a nuclear reactor ~ 10% 235 U For a weapons-grade ~ 90% 235 U To enrich U in 235 U one of the ways is to take advantage of the different diffusion rates of 235 U vs 238 U Need a gas; use gaseous UF 6 The enrichment factor, is theoretically 0.43%, but in practice only about 0.14% To produce 99% uranium-235 from natural uranium ~ 4000 stages are required. The process requires the use of thousands of miles of pipe, thousands of pumps and motors, and intricate control mechanisms.

70. The biggest obstacle was finding a suitable material for the "porous barrier" that was able to withstand the corrosive properties of the uranium gas - one of the contributions of the Manhattan Project at Columbia Note: other methods of enriching U with 235 U were also used

71. 4 stories high and almost a half mile long; enclosed ~2 million square feet of space, making it the largest building in the world at the time. The eventual cost of the K-25 complex > $500 million. K-25, Oak Ridge National Lab

72. Real Gases: Deviations from Ideal Behavior The fact that gases can liquefy at low temperatures or high pressures, indicate that gases do not behave “ideally” over all ranges of temperature and pressure. Gases liquefy because of interactions between molecules become important as the molecules come closer together. For Boyle’s law to hold a gas must never liquefy; it must remain a gas at all pressures. This means that there must be no interactions between gas molecule.

73. Temperature at which He(g) condenses to He(l) ~ 4 K Ar(g) condenses to Ar(l) ~ 87 K

74. When a gas is compressed by an increase in pressure and corresponding decrease in volume, gas molecules are forced closer together. As the pressure increases, the amount by which the gas can be compressed decreases because of the finite volume occupied by each gas molecule. For Boyle’s law to be valid over all ranges of pressure means that gas molecules must have zero volume

75. The ideal gas equation is important in determining limiting values of pressure, volume, and is a useful description of the behavior of gases at low pressures and high temperatures. Deviations from ideal behavior can be quantified by a compressibility factor, Z: P V n R T = Z If Z = 1, the gas behaves as an ideal gas The further the compressibility factor is from 1, the greater the deviation of the gas from an ideal gas

76. Nitrogen

77. Equation of State for Real Gases: van der Waals Equation Real gases: 1) Particles of a real gas occupy space 2) Attractive and repulsive forces do exist between gas molecules. The van der Waals equation of state accounts for the real behavior of gases. The ideal gas equation: P V = n R T must be modified to account for the non-zero volume of each gas molecule, and the interactions between gas molecules.

78. Accounting for Volume Because of the non-zero volume of each gas molecule, the volume available to a gas molecules is less than the volume of the container by V- nb b is the volume occupied by 1 mole of gas molecules (L/mol) n is the number of moles of gas (mol) V is the volume of the container. Accounting for Pressure Since real gas molecules interact with each other, the observed pressure is lower than the ideal gas pressure. P + n2an2a V2V2 The term accounts for the effect of interactions between molecules on the pressure of the gas

79. Accounting for real gas behavior results in van der Waal’s equation of state (P + ) n2an2a V2V2 (V - n b) = n R T The van der Waal’s constants, a and b, are different for different gases.

80. The constant b is related to the size of the gas particle. Larger the value of b, larger is the particle. Atom/Moleculeb (L/mol) Ar0.03219 Cl 2 0.05622 He0.02370 H 2 0.02661

81. The magnitude of the constant a is a measure of the attractive forces between molecules. Gases with larger a values liquefy or solidify more easily than gases with smaller a values since the attractive forces between molecules are strong. Atoma (L 2 atm/(mol 2 ))boiling pt (K) Ar1.34587.3 He0.03412 4.2 In general atoms or molecules like He and H 2 which have small a and b values exhibit behavior fairly close to ideal.