Lecture 1 the kinetic theory of gases
- 1. Designed By: Dr. Sagir
- 2. Gases are one of the most pervasive aspects of our environment on the Earth. We continually exist with constant exposure to gases of all forms. The steam formed in the air during a hot shower is a gas. The Helium used to fill a birthday balloon is a gas. The oxygen in the air is an essential gas for life.
- 3. A windy day or a still day is a result of the difference in pressure of gases in two different locations. A fresh breeze on a mountain peak is a study in basic gas laws.
- 4. Important Characteristics of Gases 1) Gases are highly compressible An external force compresses the gas sample and decreases its volume, removing the external force allows the gas volume to increase. 2) Gases are thermally expandable When a gas sample is heated, its volume increases, and when it is cooled its volume decreases. 3) Gases have high viscosity Gases flow much easier than liquids or solids. 4) Most Gases have low densities Gas densities are on the order of grams per liter whereas liquids and solids are grams per cubic cm, 1000 times greater. 5) Gases are infinitely miscible Gases mix in any proportion such as in air, a mixture of many gases.
- 5. Helium He 4.0 Neon Ne 20.2 Argon Ar 39.9 Hydrogen H2 2.0 Nitrogen N2 28.0 Nitrogen Monoxide NO 30.0 Oxygen O2 32.0 Hydrogen Chloride HCl 36.5 Ozone O3 48.0 Ammonia NH3 17.0 Methane CH4 16.0 Substances That Are Gases under Normal Conditions Substance Formula MM(g/mol)
- 6. To fully understand the world around us requires that we have a good understanding of the behavior of gases. The description of gases and their behavior can be approached from several perspectives. The Gas Laws are a mathematical interpretation of the behavior of gases. However, before understanding the mathematics of gases, a chemist must have an understanding of the conceptual description of gases. That is the purpose of the Kinetic Molecular Theory.
- 7. The Kinetic Molecular Theory is a single set of descriptive characteristics of a substance known as the Ideal Gas. All real gases require their own unique sets of descriptive characteristics. Considering the large number of known gases in the World, the task of trying to describe each one of them individually would be an awesome task. In order to simplify this task, the scientific community has decided to create an imaginary gas that approximates the behavior of all real gases. In other words, the Ideal Gas is a substance that does not exist. The Kinetic Molecular Theory describes that gas. While the use of the Ideal Gas in describing all real gases means that the descriptions of all real gases will be wrong, the reality is that the descriptions of real gases will be close enough to correct that any errors can be overlooked.
- 8. Three basic assumptions of the kinetic theory as it applies to gases: 1. Gas is composed of particles- usually molecules or atoms ◦Small, hard spheres ◦Insignificant volume; relatively far apart from each other ◦No attraction or repulsion between particles
- 9. 2. Particles in a gas move rapidly in constant random motion ◦Move in straight paths, changing direction only when colliding with one another or other objects ◦Average speed of O2 in air at 20 o C is an amazing 1660 km/h! (1.6km=1mile)
- 10. 3. Collisions are perfectly elastic- meaning kinetic energy is transferred without loss from one particle to another- the total kinetic energy remains constant Newtonian Cradle- Where the collisions between the balls elastic? Yes, because kinetic energy was transferred with each collision
- 11. Why did the balls eventually stop swinging? The collisions were not perfectly elastic, some kinetic energy was lost as heat during each collision. At constant temperatures and low to moderate pressures, collisions between gas particles are perfectly elastic
- 12. Gas consists of large number of particles (atoms or molecules) Particles make elastic collisions with each other and with walls of container There exist no external forces (density constant) Particles, on average, separated by distances large compared to their diameters No forces between particles except when they collide Remember the assumptions
- 13. The potential energy of the ball Which is converted to kinetic energy in the ball Which is converted to potential energy in the ball Is converted to kinetic energy in the ball Which is converted into the potential energy of the ball………….. …..but in reality the ball loses height and eventually stops bouncing Why does this happen?
- 14. Through friction with the air (air resistance) Through sound when it hits the floor Through deformation of the ball Through heat energy in the bounce
- 16. The gas consists of objects with a defined m The gas particles travel randomly in straight- All collisions involving gas particles are elast The gas particles do not interact with each o The gas phase system will have an average k
- 19. Kinetic Molecular Theory (KMT) for an ideal gas states that all gas particles: are in random, constant, straight-line motion. are separated by great distances relative to their size; the volume of the gas particles is considered negligible. have no attractive forces between them. have collisions that may result in the transfer of energy between gas particles, but the total energy of the system remains constant.
- 22. Kinetic Theory Assumptions ◦ Point Mass ◦ No Forces Between Molecules ◦ Molecules Exert Pressure Via Elastic Collisions With Walls xx (courtesy F. Remer)
- 23. Non-Ideal Gas ◦ Violates Assumptions Volume of molecules Attractive forces of molecules (courtesy F. Remer)
- 24. A real gas is most like an ideal gas when the real gas is at low pressure and high temperature. At high pressures gas particles are close therefore the volume of the gas particles is considered. At low temperatures gas particles have low kinetic energy therefore particles have some attractive force Example Dry ice, liquid oxygen and nitrogen
- 25. Behave as described by the ideal gas equation; no real gas is actually ideal Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less In real gases, particles attract each other reducing the pressure Real gases behave more like ideal gases
- 26. Weight of column of air above your head. We can measure the density of the atmosphere by measuring the pressure it exerts.
- 27. Effect of Atmospheric Pressure on Objects at the Earth’s Surface
- 29. Pressure = Force per Unit Area Atmospheric Pressure is the weight of the column of air above a unit area. For example, the atmospheric pressure felt by a man is the weight of the column of air above his body divided by the area the air is resting on P = (Weight of column)/(Area of base) Standard Atmospheric Pressure: 1 atmosphere (atm) 14.7 lbs/in2 (psi) 760 Torr (mm Hg) 1013.25 KiloPascals or Millibars (kPa = N/m2 )
- 30. Torricelli determined from this experiment that the pressure of the atmosphere is approximately 30 inches or 76 centimeters (one centimeter of mercury is equal to 13.3 millibars. He also noticed that height of the mercury varied with changes in outside weather conditions. For climatological and meteorological purposes, standard sea-level pressure is said to be 76.0 cm or 29.92 inches or 1013 millibars
- 31. Atmospheric pressure results from the collisions of air molecules with objects ◦Decreases as you climb a mountain because the air layer thins out as elevation increases Barometer is the measuring instrument for atmospheric pressure; dependent upon weather
- 32. Common Units of Pressure Unit Atmospheric Pressure Scientific Field pascal (Pa); 1.01325 x 105 Pa SI unit; physics, kilopascal(kPa) 101.325 kPa chemistry atmosphere (atm) 1 atm* Chemistry millimeters of mercury 760 mmHg* Chemistry, medicine, ( mm Hg ) biology torr 760 torr* Chemistry pounds per square inch 14.7 lb/in2 Engineering ( psi or lb/in2 ) bar 1.01325 bar Meteorology, chemistry, physics
- 33. Converting Units of Pressure Problem: A chemist collects a sample of carbon dioxide from the decomposition of limestone (CaCO3) in a closed end manometer, the height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure in torr, atmospheres, and kilopascals. Plan: The pressure is in mmHg, so we use the conversion factors from Table 5.2(p.178) to find the pressure in the other units. Solution: PCO2 (torr) = 341.6 mm Hg x = 341.6 torr 1 torr 1 mm Hg converting from mmHg to torr: converting from torr to atm: PCO2( atm) = 341.6 torr x = 0.4495 atm 1 atm 760 torr converting from atm to kPa: PCO2(kPa) = 0.4495 atm x = 45.54 kPa101.325 kPa 1 atm
- 34. Change in average atmospheric pressure with altitude.
- 35. Gas Pressure – defined as the force exerted by a gas per unit surface area of an object ◦Due to: a) force of collisions, and b) number of collisions ◦No particles present? Then there cannot be any collisions, and thus no pressure – called a vacuum
- 37. Rules of thumb: When evaluating, start from the known pressure end and work towards the unknown end At equal elevations, pressure is constant in the SAME fluid When moving down a manometer, pressure increases When moving up a manometer, pressure decreases Only include atmospheric pressure on open ends Manometers measure a pressure difference by balancing the weight of a fluid column between the two pressures of interest
- 40. Find the pressure at point A in this open u- tube manometer with an atmospheric pressure Po PD = γH2O x hE-D + Po Pc = PD PB = PC - γHg x hC-B PA = PB Example 2 P = γ x h + PO
- 41. What would Polly Parcel look like if she had no gas molecules inside? zero molecules = zero pressure inside zero pressure inside = zero force on the inside
- 42. In order to describe gases, mathematically, it is essential to be familiar with the variables that are used. There are four commonly accepted gas law variables Temperature Pressure Volume Moles
- 43. The temperature variable is always symbolized as T. It is critical to remember that all temperature values used for describing gases must be in terms of absolute kinetic energy content for the system. Consequently, T values must be converted to the Kelvin Scale. To do so when having temperatures given in the Celsius Scale remember the conversion factor Kelvin = Celsius + 273 According to the Kinetic Molecular Theory, every particle in a gas phase system can have its own kinetic energy. Therefore, when measuring the temperature of the system, the average kinetic energy of all the particles in
- 44. The pressure variable is represented by the symbol P. The pressure variable refers to the pressure that the gas phase system produces on the walls of the container that it occupies. If the gas is not in a container, then the pressure variable refers to the pressure it could produce on the walls of a container if it were in one. The phenomenon of pressure is really a force applied over a surface area. It can best be expressed by the equation
- 45. Consider the Pressure equation and the impact of variables on it. The force that is exerted is dependent upon the kinetic energy of the particles in the system. If the kinetic energy of the particles increases, for example, then the force of the collisions with a given surface area will increase. This would cause the pressure to increase. Since the kinetic energy of the particles is increased by raising the temperature, then an increase in temperature will cause an increase in pressure. If the walls of the container were reduced in total surface area, there would be a change in the pressure of the system. By allowing a given quantity of gas to occupy a container with a smaller surface area, the pressure of the system would increase.
- 46. As this container of gas is heated, the temperature increases. As a result, the average kinetic energy of the particles in the system increases. With the increase in kinetic energy, the force on the available amount of surface area increases. As a result, the pressure of the system increases. Eventually,...................... ....Ka-Boom
- 47. The Volume variable is represented by the symbol V. It seems like this variable should either be very easy to work with or nonexistent. Remember, according to the Kinetic Molecular Theory, the volume of the gas particles is set at zero. Therefore, the volume term V seems like it should be zero. In this case, that is not true. The volume being referred to here is the volume of the container, not the volume of the gas particles. The actual variable used to describe a gas should be the amount of volume available
- 48. Since the Kinetic Molecular Theory states that the volume of the gas particles is zero, then the equation simplifies. As a result, the amount of available space for the gas particles to move around in is approximately equal to the size of the container. Thus, as stated before, the variable V is the volume of the container.
- 49. The final gas law variable is the quantity of gas. This is always expressed in terms of moles. The symbol that represents the moles of gas is n. Notice that, unlike the other variables, it is in lower case. Under most circumstances in chemistry, the quantity of a substance is usually expressed in grams or some other unit of mass. The mass units will not work in gas law mathematics. Experience has shown that the number of objects in a system is more descriptive than the mass of the objects. Since each different gas will have its own unique mass for the gas particles, this would create major difficulties when working with gas law mathematics. The whole concept of the Ideal Gas says that all gases can be approximated has being the same. Considering the large difference in mass of the many different gases available, using mass as a measurement of quantity would cause major errors in the Kinetic Molecular Theory. Therefore, the mole will standardize the mathematics for all gases and minimize the chances for errors.
- 50. There are four variables used mathematically for describing a gas phase system. While the units used for the variables may differ from problem to problem, the conceptual aspects of the variables remain unchanged. 1. T, or Temperature, is a measure of the average kinetic energy of the particles in the system and MUST be expressed in the Kelvin Scale. 2. P, or Pressure, is the measure of the amount of force per unit of surface area. If the gas is not in a container, then P represents the pressure it could exert if it were in a container. 3. V, or Volume, is a measure of the volume of the container that the gas could occupy. It represents the amount of space available for the gas particles to move around in. 4. n, or Moles, is the measure of the quantity of gas. This expresses the number of objects in the system and does not directly indicate their masses.
- 51. (1) When temperature is held constant, the density of a gas is proportional to pressure, and volume is inversely proportional to pressure. Accordingly, an increase in pressure will cause an increase in density of the gas and a decrease in its volume. – Boyles’s Law (2) If volume is kept constant, the pressure of a unit mass of gas is proportional to temperature. If temperature increase so will pressure, assuming no change in the volume of the gas. (3) Holding pressure constant, causes the temperature of a gas to be proportional to volume, and inversely proportional to density. Thus, increasing temperature of a unit mass of
- 53. Hyperbolic Relation Between Pressure and Volume p V p – V Diagramp – V Diagram isotherms T1 T2 T3 T3 >T2>T1 (courtesy F. Remer)
- 55. Linear Relation Between Temperature and Pressure P T (K) 0 100 200 300 P – T DiagramP – T Diagram isochorsisochorsV1 V2 V3 V1 <V2 <V3 (courtesy F. Remer)
- 56. Real data must be obtained above liquefaction temperature. Experimental curves for different gasses, different masses, different pressures all extrapolate to a common zero.
- 60. What would Polly Parcel look like if she had a temperature of absolute zero inside? absolute zero = no molecular motion no molecular motion = zero force on the inside
- 61. The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written PV = nRT R = ideal gas constant
- 62. R is known as the universal gas constant Using STP conditions P V R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K) n T = 0.0821 L-atm mol-K
- 63. What is the value of R when the STP value for P is 760 mmHg?
- 64. What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1mol) (273K) = 62.4 L-mm Hg mol-K
- 65. Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?
- 66. Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L 20.0 L T = 23°C + 273 296 K n = 2.86 mol 2.86 mol P = ? ?
- 67. Rearrange ideal gas law for unknown P P = nRT V Substitute values of n, R, T and V and solve for P P = (2.86 mol)(62.4L-mmHg)(296 K) (20.0 L) (K-mol)
- 68. A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?
- 69. Solve ideal gas equation for n (moles) n = PV RT = (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K) = 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2
- 70. What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? n = PV = (0.813 atm) (0.215 L) = 0.00703 mol RT (0.0821 L-atm/molK) (303K) Molar mass = g = 0.250 g = 35.6 g/mol
- 71. Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L PV = nRT PV = nRT P = n RTV RTV RT
- 72. Substitute (1.00 atm ) mol-K = 0.0446 mol O2/L (0.0821 L-atm) (273 K) Change moles/L to g/L 0.0446 mol O2 x 32.0 g O2 = 1.43 g/L 1 L 1 mol O2 Therefore the density of O2 gas at STP is
- 73. A gas has a % composition by mass of 85.7% carbon and 14.3% hydrogen. At STP the density of the gas is 2.50 g/L. What is the molecular formula of the gas?
- 74. Calculate Empirical formula 85.7 g C x 1 mol C = 7.14 mol C/7.14 = 1 C 12.0 g C 14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H 1.0 g H Empirical formula = CH2
- 75. Using STP and density ( 1 L = 2.50 g) 2.50 g x 22.4 L = 56.0 g/mol 1 L 1 mol n = EF/ mol = 56.0 g/mol = 4 14.0 g/EF molecular formula CH2 x 4 = C4H8
- 76. On December 1, 1783, Charles used 1.00 x 103 lb of iron filings to make the first ascent in a balloon filled with hydrogen Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) At STP, how many liters of hydrogen gas were generated?
- 77. lb Fe → g Fe → mol Fe → mol H2 → L H2 1.00 x 103 lb x 453.6 g x 1 mol Fe x 1 mol H2 1 lb 55.9 g 1 mol Fe x 22.4 L H2 = 1.82 x 105 L H2 1 mol H2
- 78. How many L of O2 are need to react 28.0 g NH3 at 24°C and 0.950 atm? 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
- 79. Find mole of O2 28.0 g NH3 x 1 mol NH3 x 5 mol O2 17.0 g NH3 4 mol NH3 = 2.06 mol O2 V = nRT = (2.06 mol)(0.0821)(297K) = 52.9 L
- 82. A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?
- 83. A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?
- 84. Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.
- 85. The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. STP P = 1.00 atm P = 1.00 atm 1.0 mol He 0.50 mol O2 + 0.20 mol He + 0.30 mol N2
- 86. When a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood,
- 87. A 5.00 L scuba tank contains 1.05 mole of O2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?
- 88. P = nRT PT = PO + PHe V 2 PT = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L (K mol) = 7.19 atm
Editor's Notes
- #40: Difficult to place a barometer inside a gas to be measured. Manometers compare gas pressure and barometric pressure.