Gases amount: n = number of moles; m = mass (g(; MM = molar mass

KMT – Kinetic Molecular Theory of Gases Background for understand physical properties of gases Assumptions: gases are mostly empty space molecules are in constant and chaotic motion collisions are complete elastic pressure caused by molecules colliding with container walls (KE = ½ mv2) Air is mostly molecules (N2 (78%); O2 (21%); Ar (0.09%); CO2 (0.03%); ...) so particles is not generally used amount: n = number of moles; m = mass (g); MM = molar mass

KMT can be used for all three states of matter. Motion: translational rotational vibrational State Gas ✔ Liquid (✔) Solid (explained in detail next slide) amount: n = number of moles; m = mass (g); MM = molar mass

http://www.learnthermo.com/T1-tutorial/ch02/lesson-A/pg04.php

Measurements Volume Amount Temperature Pressure 1 L (1 cm3 = 1 mL) (1 m)3 = (102 cm)3 n = m / MM oC + 273 = K 1 atm 760 mmHg (torr) 101.325 kPa 14.7 psi (1.013 bar) Standard Temperature & Pressure (STP) for Gas Pbs. T = 0oC (273 K) P = 1 atm Standard molar volume (@STP) = 22.4 L Volume: converting vol (L) cubic length (m3): (cubed 1m = cubed 102 cm) amount: n = number of moles; m = mass (g); MM = molar mass AP Reference

Measurements – Measuring Devices Barometer Manometer Sphygmomanometer (atmospheric) (a gas)

Measurements - Manometer Pgas = Patm P gas > Patm Pgas < Patm Pgas is greater than Patm: Pgas is less than Patm (Pgas = Patm + Δh) (Pgas = Patm – Δh) Patm = 760mm Hg Patm = 760 Δh = 20 mm Hg Δh = 20 mm Hg Pgas = 780 mm Hg Pgas = 740 mm Hg Pgas = Patm Pgas = Patm + DHg Pgas = Patm – DHg

Combination of Four Gas Laws Combined Gas Law Combination of Four Gas Laws Boyle (P & V) Charles (V & T) Gay-Lussac (P&T) Avogadro (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 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? determine which law to use (there’s other gas laws). assign values to the variables – organizes: less confusion in initial & final conditions convert T to K; units on both sides of equation are the same rearrange equation substitute values solve; check

Combined Gas Law 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?

PV = nRT IDEAL GAS LAW Units: P = atm V= L n = mol R = 0.0821 Latm/ Kmol T = K Static conditions * (combined = change of conditions) Ideal gas law does depends only on initial & final conditions (State Function)

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) Ideal gas law does depends only on initial & final conditions (State Function) 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? A2(g) + B2(g)  2AB(g) (2) 2AB(g) + B2(g)  2AB2(g) (3) A(g) + B2(g)  AB2(g) (4) 2AB2(g)  A2(g) + 2B2(g)

Vinit & Tinit (from picture) Using Gas Laws to Determine a Balanced Equation Vinit & Tinit (from picture) Pinit = Pinit (from problem) Combined gas law: P1 V1 = P2 V2 n1 T1 n2 T2 From Avogadro’s law, if T doubles, n must be halved. (P & V are constant) number of moles A2(g) + B2(g)  2AB(g) 1 + 1  2 (2) 2AB(g) + B2(g)  2AB2(g) 2 + 1  2 (3) A(g) + B2(g)  AB2(g) 1 + 1  1 (4) 2AB2(g)  A2(g) + 2B2(g) 2  1 + 2

NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l) 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) How many moles of CO2(g) are produced?

NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l) 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) Stoichiometry: 1 CO2 : 1 NaHCO3 Determine mass of 0.234 mole 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: XCO2 = 1.00 / 5.00 = 0.200 partial pressure: 0.200 * 1.26 atm = 0.252 atm (all of the methane is consumed so n = 5)