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Gases KMT – Kinetic Molecular Theory of Gases • Background for understand physical properties of gases • Assumptions: o gases are mostly empty space o molecules are in constant and chaotic motion o collisions are complete elastic o pressure caused by molecules colliding with container walls (KE = ½ mv2) • KMT can be used for all three states of matter. Motion: State Gas Liquid Solid translational rotational vibrational ✔ (✔) ✔ ✔ ✔ ✔ ✔ Measurements Volume Amount Temperature Pressure 1L n = m / MM oC + 273 = K 1 atm 760 mmHg (torr) 101.325 kPa 14.7 psi (1.013 bar) (1 cm3 = 1 mL) (1 m)3 = (102 cm)3 Standard Temperature & Pressure (STP) for Gas Pbs. • T = 0oC (273 K) P = 1 atm Standard molar volume (@STP) = 22.4 L Measurements – Measuring Devices Barometer (atmospheric) Manometer (a gas) Sphygmomanometer Measurements - Manometer Pgas = Patm Pgas = Patm P gas > Patm Pgas = Patm + DHg Pgas < Patm Pgas = Patm – DHg Combined Gas Law Combination of Four Gas Laws • Boyle (P & V) • Charles (V & T) • Gay-Lussac (P&T) • Avogadro PV = k1 V = k2 T P1V1 = P2V2 V1 V2 = T1 T2 P = k3 T V = k4 n P1 P2 = T1 T2 V1 V2 = n1 n2 (at a fixed T & P, V is directly related to number of moles) Graphically (Ideal Gases) Boyle’s Law Charles’s Law Gay-Lussac’s Law Combined Gas Law P1V1 P2V2 = n1T1 n2T2 A gas-filled balloon contains 1.1 mol in a 26.5 L at 25oC and 101 kPa of pressure. How many moles of gas are added or removed to produce 27.0 L at 52oC and 1050 mmHg? 101kPa 1atm P2V2 n1T1 P1 = x = 1.00atm n2 = 101.325kPa T2 P1V1 V1 = 26.5L n1 = 1.1mol (1.38atm)(27.0L)(1.1mol)(298K) n2 = T1 = 25 + 273 = 298K (325K)(1.0atm)(26.1L) 1050mmHg 1atm P2 = x = 1.38atm 760mmHg n2 =1.4mol V2 = 27.0L Dn = +0.3mol n2 = ? T2 = 52 + 273 = 325K Combined Gas Law 1. A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22oC. What would the pressure be if the can was heated to 100.oC? P1 = 3.80atm V1 = V2 n1 = n2 T1 = 22 + 273 = 295K P2 = ? T2 = 100 + 273 = 373K P1V1 P2V2 = n1T1 n2T2 P1 T2 P2 = T1 (3.8atm)(373K) P2 = (295) P2 = 4.8atm IDEAL GAS LAW PV = nRT Units: P = atm V= L n = mol R = 0.0821 Latm/ Kmol T=K PV = nRT • A 47.3-L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature? P = 1.85 atm V = 47.3 L n = 1.62 mol R = 0.0821 L-atm/K-mol T=? T = (1.85 atm)(47.3 L) (1.62 mol)(0.0821 L-atm/K-mol) T = 658K (=385oC) Ideal Gas Law – MOLAR MASS rearrange for n: n = PV RT Let M = molar mass (g/mol) (m=mass of sample) (n=number of moles) M = m/n => M = mRT PV Ideal Gas Law – DENSITY M = mRT PV M = m RT V P M = dRT P IDEAL GAS LAW conventional form: PV = nRT molar mass MM = mRT PV density MM = dRT P Using Gas Laws to Determine a Balanced Equation A piston expands after a gaseous reaction that is carried out under constant pressure & volume. (Assume isolated system) Which of the following balanced chemical equations describes the reaction? (1) A2(g) + B2(g) 2AB(g) (3) A(g) + B2(g) AB2(g) (2) 2AB(g) + B2(g) 2AB2(g) (4) 2AB2(g) A2(g) + 2B2(g) Using Gas Laws to Determine a Balanced Equation 1. Vinit & Tinit (from picture) 2. Pinit = Pinit (from problem) 3. Combined gas law: P1 V1 = P2 V2 n1 T1 n2 T2 4. From Avogadro’s law, if T doubles, n must be halved. (P & V are constant) number of moles (1) A2(g) + B2(g) 2AB(g) (g) + B (g) 2AB (g) (2) 2AB 2 2 1 + 1 2 2 + 1 2 (3) A(g) + B2(g) AB2(g) 1 + 1 1 (4) 2AB2(g) A2(g) + 2B2(g) 2 1 + 2 Problem Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25oC and 2.00 atm. NaHCO3(s) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l) 1. How many moles of CO2(g) are produced? PV PV = nRT Þ n = RT (2.00atm)(2.97L) n= = 0.234 mol CO2 (0.0821L.atm / K.mol)(298K) Problem Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25oC and 2.00 atm. NaHCO3(s) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l) n = 0.234 mol CO2 2. Stoichiometry: 1 CO2 : 1 NaHCO3 3. Determine mass of 0.234 mole NaHCO3 0.234 mol NaHCO3 84.01 g NaHCO3 x = 19.66 g mol NaHCO3 1 mol NaHCO3 Gas Mixtures (Dalton’s Law of Partial Pressures) PT = Pi + Pii ... mole fraction (Xi) = Pi/PT so, Pi = Xi PT When 1 mole of methane (CH4) is heated with 4 moles of O2, 1 mol CO2 and 2 mol H2O are formed. What are the partial pressures if PT = 1.26 atm. 1CO2 + 2H2O 1CO2 + 2H2O CO2: mole fraction: partial pressure: XCO2 = 1.00 / 5.00 0.200 * 1.26 atm (all of the methane is consumed so n = 5) = 0.200 = 0.252 atm