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II. Ideal Gas Law

II. Ideal Gas Law. Ch. 14 - Gases. Ideal Gas Law and Gas Stoichiometry. Part 1. Ideal Gas Law. A. Avogadro’s Principle. Molar Volume at STP. S tandard T emperature & P ressure 0°C and 1 atm. 1 mol of a gas=22.4 L at STP. A. Avogadro’s Principle. V. n.

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II. Ideal Gas Law

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  1. II. Ideal Gas Law Ch. 14 - Gases Ideal Gas Law and Gas Stoichiometry

  2. Part 1 Ideal Gas Law

  3. A. Avogadro’s Principle Molar Volume at STP StandardTemperature&Pressure 0°C and 1 atm 1 mol of a gas=22.4 L at STP

  4. A. Avogadro’s Principle V n • Equal volumes of gases contain equal numbers of moles • at constant temp & pressure • true for any gas • n = number of moles

  5. B. Ideal Gas Law = k V n Merge the Combined Gas Law with Avogadro’s Principle: PV T PV nT = R UNIVERSAL GAS CONSTANT R=0.08206 Latm/molK R=8.315 dm3kPa/molK

  6. B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.08206 Latm/molK R=8.315 dm3kPa/molK

  7. C. Ideal Gas Law Problems • Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.08206Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

  8. C. Ideal Gas Law Problems WORK: 85 g 1 mol O2 = 2.7 mol 32.00 g O2 • Find the volume of 85 g of O2 at 25°C and 104.5 kPa. GIVEN: V=? n=85 g T=25°C = 298 K P=104.5 kPa R=8.315dm3kPa/molK = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molKK V = 64 dm3

  9. D. Applications of Ideal Gas Law • Can be used to calculate the molar mass of a gas from the density • Substitute this into ideal gas law • And m/V = d in g/L, so

  10. D. Applications of Ideal Gas Law • The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas. GIVEN: P = 1.50 atm T = 27°C = 300. K d = 1.95 g/L R = 0.08206Latm/molK MM = ? WORK: MM = dRT/P MM=(1.95)(0.08206)(300.)/1.50 g/L Latm/molK K atm MM = 32.0 g/mol

  11. D. Applications of Ideal Gas Law • Calculate the density of carbon dioxide gas at 25°C and 750. torr. WORK: 750 torr 1 atm = .987 atm 760 torr GIVEN: d = ? g/L CO2 T = 25°C = 298 K P = 750. torr R = 0.08206Latm/molK MM = 44.01 g/mol MM = dRT/P →d = MM P/RT d=(44.01 g/mol)(.987 atm) (0.08206 Latm/molK )(298K) d = 1.78 g/L CO2 = .987 atm

  12. Part 2 Gas Stoichiometry

  13. * Stoichiometry Steps Review * 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. • Mole ratio - moles  moles • Molar mass - moles  grams • Molarity - moles  liters soln • Molar volume - moles  liters gas • Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

  14. A. Molar Volume at STP StandardTemperature&Pressure 0°C and 1 atm 1 mol of a gas=22.4 L at STP

  15. A. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.02  1023 particles/mol

  16. B. Gas Stoichiometry • Liters of one Gas  Liters of another Gas: • Avogadro’s Principle • Coefficients give mole ratios and volume ratios • Moles (or grams) of A  Liters of B: • STP – use 22.4 L/mol • Non-STP – use ideal gas law & stoich • Non-STP • Given liters of gas? • start with ideal gas law • Looking for liters of gas? • start with stoichiometry conv

  17. C. Gas Stoichiometry Problem – STP • How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3 2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L O2 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

  18. D. Gas Stoichiometry Problem – Non-STP • What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? CaCO3 CaO + CO2 5.25 g ? Lnon-STP Looking for liters: Start with stoich and calculate moles of CO2. P = 103 kPa V = ? n = ? R = 8.315 dm3kPa/molK T = 298K 5.25 g CaCO3 1 mol CaCO3 100.09g CaCO3 1 mol CO2 1 mol CaCO3 NEXT  = 0.0525 mol CO2 Plug this into the Ideal Gas Law for n to find liters

  19. D. Gas Stoichiometry Problem – Non-STP • What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: P=103 kPa V = ? n=0.0525 mol T=25°C = 298 K R=8.315dm3kPa/molK WORK: PV = nRT (103 kPa)V=(0.0525mol)(8.315dm3kPa/molK) (298K) V = 1.26 dm3 CO2

  20. B. Gas Stoichiometry Problem • How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2 2 Al2O3 15.0 L non-STP ? g GIVEN: P=97.3 kPa V = 15.0 L n=? T=21°C = 294 K R=8.315dm3kPa/molK WORK: PV = nRT (97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K) n = 0.597 mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT 

  21. B. Gas Stoichiometry Problem • How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2 2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3 0.597 mol O2 2 mol Al2O3 3 mol O2 101.96 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

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