Kinetic Theory of Gases PDF

Kinetic Theory of Gases Table 1. shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states.  Gas is a substance that is characterized by widely separated molecules in rapid motion. Gases assume the volume and sh...

Kinetic Theory of Gases Table 1. shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states.  Gas is a substance that is characterized by widely separated molecules in rapid motion. Gases assume the volume and shape of their containers Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.  Gases are normally in the gaseous state at 25°C and 1 atm pressure Vapor is the gaseous form of any substance that is liquid or solid at normal temps. or pressure.  Pressure is the forces exerted by gas on the walls of the container Force Pressure = Area SI unit is the Pascal = N/m2 = (kg m s-2)/m2 = 1 kg / ms2 The pressure of a gas is considered as the measurements of a gas pressure relative to the atmospheric pressure  The atmospheric pressure is the change in gas pressure with distance from earth's surface The gas pressure is measured by Barometer that is invented by Torricelli. The unit of the barometer measurement is 1 Torr = 1mm Hg 1 atm pressure = 760mm Hg = 760 Torr 1 atm is equivalent to 101325 Pa =101.325 KPa  The Volume occupied by the gas is inversely proportional to the pressure at constant temperature and number of moles P a 1/V P *V = constant P1 * V1 = P2 * V2 A weather balloon has a volume of 400L at atmospheric pressure (1.0atm). The balloon is released to a height of 6.5 km where the pressure is 0.4 atm. What is the new volume of the balloon? Answer: Using Boyle's Law P1 V1 = P2 V2 V2 = V1 * P1/P2 V2 = 400L * 1.0atm/0.40atm = 1000 L  At constant pressure, the volume of a fixed quantity of the gas is directly dependent on its absolute temperature. VaT V = constant *T V1/T1 = V2 /T2 Temperature must be in Kelvin T (K) = t (0C) + 273.15 As T increases V increases A sample of CO2 occupies 7.50 L at 150°C. If the pressure remains constant, calculate the temperature at which the CO2 will occupy 3.76 L. Answer: V1 = 7.60L V2 = 3.76L T1 = (273.15 + 150) K = 423.2 K T2 = ? T2 = V2 / V1 x 423.2 K = 3.76 L / 7.50 L x 423.2 K = 212 K.  The pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. P P1 P2 PaT T = a con stant or T1 = T2 A car tire in the summer has a pressure of 220 kPa at a temperature of 25°C.In the winter, the tire pressure is measured at 176 kPa. Calculate the temperature of the air inside the tire. Answer: P1 = 220 kPa P2 = 176 kPa T1 = (25 + 273) K = 298K T2 = ? The volume of a gas at constant T and P is directly proportional to the number of moles of gas. V a number of moles (n) V = constant x n V1 / n1 = V2 / n2 Example 1 mole CO2 occupies 22.414 L at 273.15K, and P = 1.000 atm; 2 moles of CO2 gas would occupy 44.86 L. V a 1/P (at constant n and T) Boyle’s Law V a T (at constant n and P ) Charles’ and Gay Lussac's Law Van (at constant P and T ) Avogadro’s Law PaT ( at constant V ) Amonton’s Law  Va nT P V = constant x nT = R nT P P PV = nRT  Calculate the volume occupied by 32.06 g of Ne gas at 5.0°C and 630 mm Hg. Answer: P = 630 mm Hg x 1 atm/760 mm Hg P = 0.8289 atm T = (5 °C + 273 °C) 1 ºC / 1 K = 278 K n Ne = 32.06 g Ne x 1 mol Ne / 20.18 g = 1.589 moles P V = n RT V = n RT / P V= 1.589 moles x 0.08206 L atm / (K mol) x 278. K / 0.829 atm = 43.7L  PV = nRT m is the mass of the gas in g m M is the molar mass of the gas PV=M RT m d = V = PM RT Molar Mass (M ) of a Gaseous Substance dRT M= P d is the density of the gas in g/L A 2.10 L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas? M = dRT P m 4.65 g g d=V = = 2.21 2.10 L L g 2.21 L x 0.0821 L atmx300.15 K mol K M= 1 atm M = 54.5 g/mol The total pressure, Ptotal, of a mixture of gases is the sum of their individual gas partial pressures V and T are constant P11 P2 Ptotal = P1+ P2 Consider a case in which two gases, 1 and 2, are in a container of volume V. P1 / Pt = [n1 RT/V] / [(n1 + n2)RT/V] = n1 / (n1 + n2) = X1 where X1 is defined as the mole fraction of gas 1 The mole fraction is a dimensionless quantity; it gives the ratio of the number of moles of gas 1 to the total number of moles of gases present in the mixture  Partial pressure of gas 1 (P1) = X1 * PT  Partial pressure of gas 2 (P2) = X2 * PT In general, in gaseous mixtures, the partial pressure of the ith component is related to the total pressure by Pi = Xi PT where Xi is the mole fraction of gas i. A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 = 0.0132 Xpropane = 8.24 + 0.421 + 0.116 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm (A) If 4 moles of gas are added to a container that already holds 1 mole of gas, how will the pressure change inside the container? a. The pressure will be five times higher. b. The pressure will double. c. The pressure will be four times higher. d. The pressure will not change (B)At a certain temperature and pressure, 0.20 mol of carbon dioxide has a volume of 3.1 L. A 3.1-L sample of hydrogen at the same temperature and pressure ____. a. has the same mass b. contains the same number of atoms c. has a higher density d. contains the same number of molecules Under constant T, P, the diffusion rates for gaseous substances are inversely proportional to the squares roots of the molar masses. Diffusion process is the gradual mixing of gas molecules caused by kinetic properties. 𝑟1 𝑀2 = 𝑟2 𝑀1 Effusion - the process by which a gas under pressure goes (escapes) from one compartment of a container to another by passing through a small opening.  𝑡2 𝑀2 = 𝑡1 𝑀1 where t1 and t2 are the effusion times of gases 1 and 2 and M1 and M2 again represent the molar masses of gas 1 and gas 2, respectively. How quickly the molecules diffuse depends on their molecular speed and the mean free path In an effusion apparatus, pure O2 required 55.0 min to reach the detector. Under the same conditions, an unknown gas needed 85.0 min. Calculate the molecular mass of the unknown gas. t unk Munk 85.0 Munk    t O2 MO 2 55.0 32.00g / mole 2  85.0  2  Munk  Munk    1545   2  .  55.0   32.00 g / mole  32.00 g / mole Munk 2.388 =  Munk  76.4 g / mole 32.00 g / mole  Real gases do not obey ideal gas equation under all conditions. They nearly obey ideal gas equation at higher temperatures and very low pressures. However, they show deviations from ideality at low temperatures and high pressures. The deviations from ideal gas behavior can be illustrated as follows: The isotherms obtained by plotting pressure, P against volume, V for real gases do not coincide with that of ideal gas, as shown below. It is clear from above graphs that the volume of real gas is more than or less than expected in certain cases. The deviation from ideal gas behavior can also be expressed by compressibility factor, Z.  The compressibility factor is the ratio of PV to nRT from ideal gas equation: PVperfect = nRT Therefore Note: For ideal or perfect gases, the compressibility factor, Z = 1. But for real gases, Z ≠1. Case-I : If Z>1 Vreal > Videal The repulsion forces become more significant than the attractive forces. The gas cannot be compressed easily. Usually, the Z > 1 for so called permanent gases like He, H2. Case-II: If Z < 1 Vreal < Videal. The attractive forces are more significant than the repulsive forces. The gas can be liquefied easily. Usually, the Z < 1 for gases like NH3, CO2, SO2. The isotherms for one mole of different gases, plotted between the Z value and pressure, P at 0oC  For gases like He, H2 the Z value increases with increase in pressure (positive deviation). It is because, the repulsive forces become more significant and the attractive forces become less dominant. Hence these gases are difficult to be condensed. For gases like CH4, CO2, NH3 etc., the Z value decreases initially (negative deviation) but increases at higher pressures. It is because: at low pressures, the attraction forces are more dominant over the repulsion forces, whereas at higher pressures the repulsion forces become significant as the molecules approach closer to each other.  But for all the gases, the Z value approaches one at very low pressures, indicating the ideal behavior. The graphs of Z vs P for a particular gas, N2 at different temperatures. In above graph, the curves are approaching the horizontal line with increase in the temperature i.e., the gases approach ideal behavior at higher temperatures. The deviations from ideal gas behavior can be illustrated by kinetic theory of gases. The real volume of the gas molecules is negligible when compared to the volume of the container. There are no forces of attraction or repulsion between the gas molecules. Hence van der Waals suggested the following corrections:  The gas molecules possess finite volume and hence should not be neglected. It is especially true at high pressures and low temperatures and should be accounted for. In case of real gases, both the forces of attraction as well as repulsion operate between gas molecules. Note: If the gases obey the kinetic theory of gases, then they cannot be compressed since the attractions between the gas molecules is negligible. Also the following corrections are applied by van der Waals to the ideal gas equation:  Volume correction The volume available for the gas molecules is less than the volume of the container, V. The available volume is obtained by subtracting excluded volume of ‘n’ moles of gas, nb from the volume of the container. Available volume = V - nb where 'b' is a constant characteristic of a gas. The ideal gas equation can be written after correcting for this as: P(V-nb) = nRT Pressure correction The pressure of the real gas is less than the expected pressure due to attractions between the molecules and this results in: reduction of frequency of collisions over the walls reduction in the force with which the molecules strike the walls. Hence Pideal = Preal + p Where p = reduction in pressure. However, the reduction in pressure is proportional to the square of molar concentration, n/V.  Because the molecules are attracted to each other, the pressure on the container will be less than ideal by an amount a(n/v)2 “Attractive forces”. Pressure depends on the number of molecules per liter. Attraction forces increases as the number of molecules increases in the container (n/V ) Molecular attraction decreases the number of collisions on the walls of the container ( n/V ) The reduction in pressure is proportional to the square of molar concentration, n/V P a (n/v)2 The reduction in pressure will be P = a (n/v)2 where ‘a’ is a proportionality constant characteristic of a gas, Therefore: Pideal = Preal + p = P + a (n/v)2 The New ideal gas equation can be modified by introducing the pressure and volume corrections as: [P + a (n/v)2] (V-nb) = nRT Note that: P is the observed pressure of the real gas [P + a (n/v)2] is the corrected pressure Characteristics of Vander weal constants ‘a’ is called van der Waals constant of attraction. Higher values of ‘a’ indicate greater attraction between gas molecules. The easily compressible gases like ammonia, HCl possess higher ‘a’ values. Greater the value of ‘a’ for a gas easier is the liquefaction. ‘b’ is called excluded volume of the gas The density () of steam at 100.0oC and 1.01325 bar is 0.5974 kg m-3 Calculate: i. the compressibility factor. ii. the deviation of the volume from the ideal behavior. Solution i- Submitting n = m/M and  = m/v (m = mass of the gas) into the equation 𝐏𝐕 𝐏𝐌𝐕 𝐏𝐌 𝐙= = = 𝐧𝐑𝐓 𝐦𝐑𝐓 𝛒𝐑𝐓 (𝟏. 𝟎𝟏𝟑𝟐𝟓𝒙𝟏𝟎𝟓 𝑷𝒂)(𝟏𝟖. 𝟎𝟏𝟓𝟐𝒙𝟏𝟎−𝟑 𝒌𝒈) 𝒁= = 𝟎. 𝟗𝟖𝟒𝟖 (𝟎. 𝟓𝟗𝟕𝟓𝒌𝒈 m−3)(𝟖. 𝟑𝟏𝟒𝒎𝟑 𝑷𝒂𝒌−𝟏 𝒎𝒐𝒍−𝟏 )(𝟑𝟕𝟑. 𝟏𝟓𝒌)  __ 𝑹𝑻 𝟖. 𝟑𝟏𝟒𝒙𝟑𝟕𝟑. 𝟏𝟓 𝑽𝒊𝒅𝒆𝒂𝒍 = = = 𝟎. 𝟎𝟑𝟎𝟔𝟏𝟖𝒎𝟑 = 𝟑𝟎. 𝟏𝟓𝟑 L 𝑷 𝟏𝟎𝟏𝟑𝟐𝟓 𝑉𝑟𝑒𝑎𝑙 = 𝑍𝑉𝑖𝑑𝑒𝑎𝑙 = 0.9848 x 30.618 = 30.152 L 𝑽𝒊𝒅𝒆𝒂𝒍 −𝑽𝒓𝒆𝒂𝒍 𝟑𝟎𝟔𝟏𝟖 − 30152.6 Deviation = 𝒙𝟏𝟎𝟎 = 𝒙𝟏𝟎𝟎 = 𝟏. 𝟓𝟐% 𝑽𝒊𝒅𝒆𝒂𝒍 𝟑𝟎𝟔𝟏𝟖 So this gas shows 1.52% deviation from ideal behavior under these conditions.  Calculate the pressure needed to confine 1 mol of carbon dioxide gas to a volume of 0.5 L at 298 K using van der Waals equation. a = 3.592 L2 atm mol-2 and b = 0.0427 L mol-1. nRT an2 P= V−nb − V2 Preal = 48.87 atm – 7.18 atm = 41.69 atm. Pideal = nRT / V = 1 x 0.082 x 298 / 0.5 = 48.87 atm The pressure is appreciably less 7 atm than that would be needed if carbon dioxide behaved ideally.  Application of Van der Waals equation The Vander Waal's equation holds good for real gases up to moderately high pressures. It explains the isotherms of PV/RT vs P for various gases. From this equation it is possible to obtain expressions for Boyle's temperature, critical constants and inversion temperature in terms of the Vander Waal's constants 'a' and 'b'. Liquefaction of gases Liquefaction of gases is physical conversion of a gas into a liquid state. Many gases can be put into a liquid state at normal atmospheric pressure by simple cooling; a few, such as carbon dioxide, require pressurization At higher temperatures, say 50 oC, the isotherms show ideal behavior. As the temperature is lowered, the isotherms show deviation from ideal behavior. At 30.98 oC, carbon dioxide remains as gas up to 73 atm. But liquid appears for the first time at 73 atm (represented by point ‘O’). Hence 30.98 oC is called critical temperature for CO2. And above 73 atm. there is a steep rise in the pressure. This steep portion of the curve represents the isotherm of liquid state for which small decrease in volume results in steep rise in the pressure. At even lower temperature, 20 oC, the liquid appears at point ‘A’. Further compression does not change the pressure up to point ‘B’. After point, B the curve again becomes steep representing the isotherm for liquid CO2.  TC, PC and VC are collectively known as ‘critical constants’. Definitions: 1. Critical Temperature: (TC) Is the temperature at which it just becomes possible to liquefy a gas under compression. Above this temperature the gas cannot be liquefied however large the applied pressure may be. 2. Critical pressure: (PC) It is the pressure necessary to liquefy a gas at critical temperature. 3. Critical Volume: (VC) It is the volume which unit mass (or one mole of a gas) of a gas occupies at the critical temperature and pressure.  𝟏 𝒃 = 𝑽𝑪 𝟑 𝒂 = 𝟑𝑷𝑪 𝑽𝟐𝑪 𝟖𝑷𝑪 𝑽𝑪 𝑹= 𝟑𝑻𝑪 These equations can be rearranged to get the compressibility factor at the critical point, 𝐏𝐂 𝐕𝐂 𝟑 𝐙= = 𝐑𝐓𝐂 𝟖 By eliminating the VC term, by using VC = 3b, from the other expression of Eq.(2.15). The resulting expressions for a and b are 27 R2 𝑇𝐶2 𝑅𝑇𝐶 𝑎= 𝑎𝑛𝑑 𝑏 = 64 PC 8𝑃𝐶 Evaluate the van der Waals constants for O2 using TC = -118.40C and PC = 50.1 atm. Solution 𝟐𝟕𝑹𝟐 𝑻𝟐𝑪 𝟐𝟕(𝟎. 𝟎𝟖𝟐) L atm K−1 𝒎𝒐𝒍−𝟏 )𝟐 (𝟏𝟓𝟒. 𝟖 K)𝟐 𝒂= = 𝟔𝟒𝑷𝑪 𝟔𝟒(𝟓𝟎. 𝟏 atm) = 𝟏. 𝟑𝟔𝟎𝑳𝟐 𝒂𝒕𝒎 mol−2 𝑹𝑻𝑪 (𝟎. 𝟎𝟖𝟐 L atm K−1 mol−1 )𝟐 (𝟏𝟓𝟒. 𝟖 K)𝟐 𝒃= = 𝟖𝑷𝑪 𝟖(𝟓𝟎. 𝟏 atm) = 𝟎. 𝟎𝟑𝟏𝟕 L𝟐 𝒂𝒕𝒎 mol−2