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Learning Log

Learning Log. Why are you advised to open windows slightly if a tornado approaches?. Ch. 10 & 11 - Gases. Dalton’s Law Ideal Gas Law (p. 322-325, 340-346). Dalton’s Law of Partial Pressures.

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Learning Log

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  1. Learning Log • Why are you advised to open windows slightly if a tornado approaches?

  2. Ch. 10 & 11 - Gases Dalton’s LawIdeal Gas Law(p. 322-325, 340-346)

  3. Dalton’s Law of Partial Pressures • Total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. • In the absence of a chemical reaction PT = P1+ P2+ P3+…

  4. Practice Problem A mixture of four gases exerts a total pressure of 1200 mm Hg. Gases A and B each exert 420 mm Hg. Gas C exerts 280 mm Hg. What pressure is exerted by gas D? PT = P1 + P2 + P3 + P4 1200 = 420 mm Hg + 420 mm Hg + 280 mm Hg + P4 1200 = 1120 mm Hg + P4 P4 = 80 mm Hg

  5. B. Vapor pressure of water • Gases are often collected in lab by water displacement and are mixed with water vapor • Patm = Pgas + PH2O • To determine the pressure of the gas collected – subtract the vapor pressure of the water at that temperature from the current atmospheric pressure

  6. C. Ideal Gas Law • The mathematical relationship among pressure, volume, temperature and the number of moles of a gas. • Derived by combining the gas laws. PV=nRT

  7. D. Ideal Gas Constant = k V n Merge the Combined Gas Law with Avogadro’s Principle: PV T PV nT = R IDEAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK

  8. E. 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 = 0.0821L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

  9. E. Ideal Gas Law Problems WORK: 85 g 1 mol = 2.7 mol 32.00 g • 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

  10. Finding Molar Mass from the Ideal Gas Law n = mass n = m Molar mass M PV = mRTOR M PV=nRT M =mRT PV

  11. Finding Density from the Ideal Gas Law D = mass or D = m volume V M = DRT P M =mRT PV D =MP RT

  12. PRACTICE PROBLEMS • At 28°C and 0.974 at, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas? • P = 0.974 atm V = 1.00 L • T = 28°C = 273 = 301 K m = 5.16 g • M = mRT PV = (5.16 g) (0.0821 L∙atm/mol∙K) (301K) (0.974 atm)(1.00 L) = 131g/mol

  13. PRACTICE PROBLEMS • What is the density of a sample of ammonia gas, NH3, if the pressure is 0.928 atm and the temperature is 63.0°C? • P = 0.928 atm T = 63.0°C + 273 = 336 K M = 17.034 g/mol R = 0.0821 L∙atm/mol∙K • D = MP = (17.034 g/mol)(0.928 atm) RT (0.0821L∙atm/mol∙K)((336 K) = 0.572 g/L NH3

  14. Homework Assignment • Workbook. • Complete problems #1 and 2 on pp. 173-174, 1-2 p. 178, 1-2 p. 180

  15. Practice Test Part 2 • P. 181 – 182 #1-7, 12, 18 - 21

  16. Practice test Part 2 • Workbook p. 182 #1-7

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