CHEF0314 Topic 2 Gases and Kinetic Molecular Theory PDF

(CHEF0314) TOPIC 2 GASES AND KINETIC-MOLECULAR THEORY (KMT)  An Overview of the Physical States of Matter Distinguishing gases from liquids and solids.  A gas has no definite volume & no fixed shape  Gas volume changes significantly with pressure....

(CHEF0314) TOPIC 2 GASES AND KINETIC-MOLECULAR THEORY (KMT)  An Overview of the Physical States of Matter Distinguishing gases from liquids and solids.  A gas has no definite volume & no fixed shape  Gas volume changes significantly with pressure.  Solid and liquid volumes are not greatly affected by pressure.  Gas volume changes significantly with temperature.  Gases expand when heated and shrink when cooled.  The volume change is 50 to 100 times greater for gases than for liquids and solids.  Particles of gaseous state move randomly, flow very freely & are independent of one another because they have enough energy to overcome the intermolecular forces among them.  Gases have relatively low densities. (Unit is g/L)  Gases form a solution in any proportions.  Gases are freely miscible with each other. 4-2  Figure 5.1 The three states of matter. 4-3  Table 5.1 Common Units of Pressure 1 atm = 760 torr = 760 mmHg = 1.01325 x 105 Pa = 101.325 kPa = 1.01325 x 105 Nm2 4-4  The Gas Laws  The gas laws describe the physical behavior of gases in terms of 4 variables:  pressure (P)  temperature (T)  volume (V)  amount (number of moles, n)  An ideal gas is a gas that exhibits linear relationships among these variables.  No ideal gas actually exists, but most simple gases behave nearly ideally at ordinary temperatures and pressures. 4-5  Boyle’s Law At constant temperature, the volume occupied by a fixed amount of gas is inversely proportional to the external pressure. 1 V or PV = constant P At fixed T and n, P decreases as V increases P increases as V decreases 4-6  Charles’s Law At constant pressure, the volume occupied by a fixed amount of gas is directly proportional to its absolute (Kelvin) temperature. V VT = constant T At fixed P and n, V decreases as T decreases V increases as T increases 4-7  Other Relationship Based on Boyle’s and Charles’s Law At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to its absolute (Kelvin) temperature. P PT = constant T At fixed V and n, P decreases as T decreases P increases as T increases 4-8  Avogadro’s Law At fixed temperature and pressure, the volume occupied by a gas is directly proportional to the amount of gas (mol). V Vn = constant n At fixed P and T, V decreases as n decreases V increases as n increases 4-9  Figure 5.7 The relationship between the volume and amount of a gas. Avogadro’s Law: at fixed temperature and pressure, equal volumes of any ideal gas contain equal numbers of particles (or moles). 4-10  Gas Behavior at Standard Conditions STP or standard temperature and pressure specifies a pressure of 1 atm (760 torr) and a temperature of 0°C ( 273.15 K). The standard molar volume is the volume of 1 mol of an ideal gas at STP. Standard molar volume = 22.4141 L or 22.4 L Meaning here, at STP, 1 mol of any gas occupies the space with a volume of 22.4 L. 4-11  Figure 5.9 Standard molar volume. 4-12  Combined Laws V 1 V  nT P P VT PV  nT Vn Related to Ideal Gas Law PT 4-13  The Ideal Gas Law PV = nRT PV 1 atm x 22.414 L 0.0821 atm·L R= = = nT 1 mol x 273.15 K mol·K R is the universal gas constant; the numerical value of R depends on the units used. Usually we use this gas constant ,R R = 0.0821 L.atm / mol.K 4-14 Figure 5.11 The individual gas laws as special cases of the ideal gas law.  Solving Gas Law Problems Generally the gas law problems can be grouped in two types: (1) There is a changing condition – means one (or more) of the four variables, which causes a change in another variable while other variables remain constant. (2) One variable is unknown, but the other three variables are known So look at the given data and what the question asked whether it is a changing condition or not. For changing condition, the ideal gas law can also be expressed/arranged by the combined equation depends on given data. PV = nRT where R is constant, so we have: P1V1 = n1RV1 and P2V2 = n2RV2 P1V1 = R and P2V2 = R n1T1 n2T2 P1V1 = P2V2 So n1T1 n2T2 Be careful in rearranging the variables, depends on the given data.  P1V1 = P2V2 n1T1 n2T2 For example, one or two variables remain constant where others are changing: P1V1 P2V2 = where n remains constant T1 T2 P1V1 = P2V2 where T and n remain constant V1 V2 = where P and T remain constant n1 n2  Applying the Volume-Pressure Relationship PROBLEM: A 3.50 L sample of neon gas has a pressure of 924 torr. What pressure (in atm) is required to compress the gas into a tank with a volume of 1350 cm3 at constant temperature? SOLUTION: P1 = 924 torr (convert to atm) P2 = unknown (in atm) n and T V1 = 3.50 L V2 = 1350 cm3 (convert to L) are constant 924 torr x 1 atm = 1.22 atm 760 torr 1350 cm3 x 1 mL x 1L = 1.35 L 1 cm3 103 mL P1V1 P2V2 = P1V1 = P2V2 n1T1 n2T2 P2 = P1 x V1 = 1.22 atm x 3.50 L = 3.16 atm V2 1.35 L PROBLEM: A balloon is filled with 1.95 L of air at 25°C and then placed in a car in the sun. What is the volume of the balloon when the temperature in the car reaches 90°C?  SOLUTION: V1 = 1.95 L V2 = unknown T1 = 25ºC (convert to K) T2 = 90ºC (convert to K) P and n remain constant T1 (K) = 25°C + 273.15 = 298.15 K T2 (K) = 90°C + 273.15 = 363.15 K P1V1 PV V V = 2 2 or 1 = 2 n1T1 n2T2 T1 T2 T2 363.15 K V2=V1× =1.95L× = 2.38L T1 298.15 K PROBLEM: A helium-filled balloon has a volume of 15.8 L at a pressure of 0.980 atm and 22°C. What is its volume at the summit of Mt. Hood, Oregon’s highest mountain, where the atmospheric pressure is 532 mmHg and the temperature is 0°C?  Solution: V1 = 15.8 L V2 = unknown T1 = 22°C (convert to K) T2 = 0°C (convert to K) P1 = 0.980 atm P2 = 532 mmHg (convert to atm) n remains constant T1 (K) = 22°C + 273.15 = 295.15 K T2 (K) = 0°C + 273.15 = 273.15 K 1 atm P2 (atm) = 532 mmHg x = 0.700 𝑎𝑡𝑚 760 mmHg P1V1 PV PV PV = 2 2 or 1 1 = 2 2 n 1T 1 n 2T 2 T1 T2 P1T2 (0.980 atm) (273.15 K) V2 = V1 x = 15.8 L x = 𝟐𝟎. 𝟓 𝐋 P2T1 (0.700 atm) (295.15 K) Sample Problem 5.6 Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 21oC. PLAN: We are given V, T and mass, which can be converted to moles (n). Use the ideal gas law to find P. SOLUTION: V = 438 L T = 21°C = 294 K n = 0.885 kg O2 (convert to mol) P is unknown 3 Mol O2 = 0.885 kg O2 x 10 g x 1 mol O2 = 27.7 mol O2 1kg 32.00 g O2 atm·L 27.7 mol x 0.0821 x 294.15 K nRT mol·K P= = = 1.53 atm V 438 L Further Applications of The Ideal Gas Law Equation The density of a gas is - directly proportional to its molar mass and - inversely proportional to its temperature. - directly proportional to the pressure PV = nRT m Where; moles = M m m PV = RT density, d = M V m M xP =d= V RT  Sample Problem 5.8 Calculating Gas Density PROBLEM: Find the density (in g/L) of CO2 (g) (a) at STP (b) at room conditions (20.°C and 1.00 atm). SOLUTION: (a) At STP, or 273 K and 1.00 atm: M xP 44.01 g/mol x 1.00 atm d= = = 1.96 g/L RT 0.0821 atm·L x 273 K mol·K (b) At 20.°C and 1.00 atm: T = 20.°C + 273.15 = 293.15 K M xP 44.01 g/mol x 1.00 atm d= = = 1.83 g/L RT 0.0821 atm·L x 293 K mol·K Molar Mass from the Ideal Gas Law m PV n= = M RT mRT dRT M= OR M= PV P Sample Problem 5.9 Finding the Molar Mass of a Volatile Liquid PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a preweighed flask and puts the flask in boiling water, causing the liquid to vaporize and fill the flask with gas. She closes the flask and reweighs it. She obtains the following data: Volume (V) of flask = 213 mL T = 100.0°C P = 754 torr mass of flask + gas = 78.416 g mass of flask = 77.834 g Calculate the molar mass of the liquid. PLAN: The variables V, T and P are given. We find the mass of the gas by subtracting the mass of the flask from the mass of the flask with the gas in it, and use this information to calculate M. Sample Problem 5.9 SOLUTION: m of gas = (78.416 - 77.834) = 0.582 g V = 213 mL x 1 L = 0.213 L T = 100.0°C + 273.15 = 373.15 K 103 mL P = 754 torr x 1 atm = 0.992 atm 760 torr atm·L 0.582 g x 0.0821 x 373 K mRT mol·K M= = = 84.4 g/mol PV 0.213 L x 0.992 atm The Partial Pressure of Each Gas in a Mixtures of Gases  Gases mix homogeneously in any proportions.  Each gas in a mixture behaves as if it were the only gas present. (assuming no chemical interactions)  The pressure exerted by each gas in a mixture is called its partial pressure.  Dalton’s Law of partial pressures states that in a mixture of unreacting gases, the total pressure in a mixture is the sum of the partial pressures of the component gases. Ptotal = PA + PB + PC + ………. For example: Assume V and T are constant, each gas, N2 and H2 behaves independently. nN2 RT nH2 RT PN2 = PH2 = V V Ptotal = PN2 + PH2 X is mole fraction where the mol (nA) of each gas is divided with the total mol (ntotal) of gas. So; nN2 RT nH2 RT = + nN2 nN2 V V XN2 = = ntotal nN2 + nH2 = (nN2 + nH2)RT V So; ntotal RT Ptotal = V (from here relate with XN2 to determine partial pressure of N2)  For N2 Where: ntotal RT nN2 Ptotal = XN2 = V ntotal nN2 RT nN2 Ptotal = ntotal = VXN2 XN2 nN2 RT Given before: PN2 = V Same steps applied for H2 PN2 So: Ptotal = XN2 That is how to get: PN2 = XN2 x Ptotal Generally: PA = XA x PT  Applying Dalton’s Law of Partial Pressures  PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mole % N2, 17 mole % 16O2, and 4.0 mole % 18O2. (The isotope 18O will be measured to determine the O uptake.) 2 The total pressure of the mixture is 0.75 atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.  SOLUTION: 4.0 mol %18O2 X18O 2= = 0.040 100 P18O2 = X18O2 × PTotal = 0.040 x 0.75 atm = 0.030 atm Applying Dalton’s Law of partial pressure: Collecting a Gas over Water  Whenever a gas is in contact with water, some of the water vaporizes into the gas.  The water vapor that mixes with the gas contributes the vapor pressure, a portion of the total pressure that depends only on the water temperature.  (refer Table 5.2) Table 5.2 Vapor Pressure of Water (P ) at Different T H2O T(0C) P (torr) T(0C) P (torr) H2O H2O 0 4.6 40 55.3 5 6.5 45 71.9 10 9.2 50 92.5 12 10.5 55 118.0 14 12.0 60 149.4 16 13.6 65 187.5 18 15.5 70 233.7 20 17.5 75 289.1 22 19.8 80 355.1 24 22.4 85 433.6 26 25.2 90 525.8 28 28.3 95 633.9 30 31.8 100 760.0 35 42.2 Figure 5.12 Collecting a water-insoluble gaseous product and determining its pressure. Ptotal = Pgas + PH2O Sample Problem 5.11 Calculating the Amount of Gas Collected over Water PROBLEM: Acetylene (C2H2) is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(aq) A collected sample of acetylene has a total gas pressure of 738 torr and a volume of 523 mL. At the temperature of the gas (23oC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? P C2H2 = (738 - 21) torr = 717 torr 1 atm P C H = 717 torr x = 0.943 atm 2 2 760 torr V = 523 mL x 1 L = 0.523 L 103 mL T = 23°C + 273.15 K = 296.15 K  Sample Problem 5.11 SOLUTION: n PV = 0.943 atm x 0.523 L C2H2 = atm·L = 0.0203 mol RT 0.0821 x 296 K mol·K 26.04 g C2H2 mass of C2H2 = 0.0203 mol x = 0.529 g C2H2 1 mol C2H2 The Ideal Gas Law and Stoichiometry P, V, T P, V, T of gas A of gas B Amount (mol) Amount (mol) of gas A of gas B Figure 5.13 The relationships among the amount (mol, n) of gaseous reactant (or product) and the gas pressure (P), volume (V), and temperature (T). Sample Problem 5.13 Using Gas Variables to Find Amounts of Reactants and Products PROBLEM: What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of potassium metal? PLAN: First we must write a balanced equation. Since the quantities of both reactants are given, we must next determine which reactant is limiting. We will use the ideal gas law to calculate the moles of Cl2 present. SOLUTION: The balanced equation is: Cl2(g) + 2K(s) → 2KCl(s) For Cl2: P = 0.950 atm V = 5.25 L T = 293 K n = unknown  Sample Problem 5.13 n PV 0.950 atm x 5.253 L Cl2 = = atm·L = 0.207 mol Cl2 RT 0.0821 x 293 K mol·K For Cl2: Mol KCl = 0.207 mol Cl2 x 2 mol KCl = 0.414 mol KCl 1 mol Cl2 For K: Mol KCl = 17.0 g K x 1 mol K x 2 mol KCl = 0.435 mol KCl 39.10 g K 2 mol K Cl2 is the limiting reactant because it yields less mol of KCl Mass KCl = 0.414 mol KCl x 74.55 g KCl = 30.9 g KCl 1 mol KCl The Kinetic-Molecular Theory(KMT): A Model for Gas Behavior A gas that behaves exactly as outlined by the KMT is known as an ideal gas or perfect gas. The KMT is based on the motion of particles. KMT is based on these 5 assumptions: (1) Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls. (2) Gas particles are tiny with large spaces between them. The volume of each particle is negligible (means can be neglected) compared to the total volume of the container. (3) Attractive forces between gas molecules is also negligible. Meaning assume that there is no attractive forces or repulsive forces between the gas molecules. (4) Collisions are perfectly elastic, meaning that colliding particles exchange energy but do not lose any energy due to friction. (5) The average kinetic energy of the gas molecules is proportional to absolute temperature. The average kinetic energy of gas molecules is constant at constant temperature. The main point to explain about ideal gas is that we assume the volume of gas molecules is negligible, same goes there is no attractive forces between gas molecules Use KMT to explain about ideal gas The pressure of a gas is caused by the collisions with the wall of container. The more frequently they collide the wall, the greater the pressure is. 1 V For example: to explain Boyle’s law P - When V decreases, the distance between gas molecules decreases, there are more collisions of gas molecules with wall of container and pressure of gas also increases. (Assuming there is no attractions between gas molecules) - When V increases, the distance between gas molecules increases, there are fewer collisions of gas molecules with wall of container and pressure of gas also decreases. At constant volume, pressure is directly proportional with absolute temperature - When T increases, the average kinetic energy of gas molecules also increases, there are more collisions of gas molecules with wall of container and pressure of gas also increases.