 Download Download Presentation Chapter 5

Chapter 5

Télécharger la présentation Chapter 5

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

1. Chapter 5 Gases

2. Properties of a gas 5.1 Pressure • Uniformly fills any container. • Mixes completely with any other gas • Exerts pressure on its surroundings. • Compressible • Gas pressure varies with altitudes and storms

3. Force (N) P (Pa)= Area (m2) Measuring atmospheric pressureTorricellian barometer • Torricelli (1608-1647) studied the problem using mercury rather than H2O. • Mercury is denser than water, so the column wasn’t quite so high. Pascal (SI units) • Gas Pressure • Liquid Pressure = g ·h ·d

4. Pascal and Torricelli Blaise Pascal (1623-1662) Evangelista Torricelli (1608-1647)

5. Barometer 760 mmHg atmospheric pressure P = d·g·h h d - density g - acc. of gravity atmospheric pressure

6. Units of Pressure One atmosphere (1 atm) • Is the average pressure of the atmosphere at sea level • Is the standard atmospheric pressure Standard Atmospheric Pressure: 1 atm = 76 cm Hg = 760 mm Hg = 760 Torr = 101,325 Pa Very small unit, thus it is not commonly used

7. Example A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg 14.7 psi 1.00 atm

8. 5.2 The Gas Laws of Boyle, Charles and Avogadro • Boyle’s Law: PV = const • at constant n, T • Charles’ Law: V/T = const • at constant n, P • Avogadro’s Law: V/n = const • at constant P, T

9. Boyle’s Law V Slope= 1/k PV =k (at constant T and n) P1V1 = P2V2

10. A Plot of PV Versus P for Several Gases at Pressure Below 1 atm Boyle’s holds Only at very Low pressures A gas strictly obeys Boyle’s law is called Ideal gas

11. V/T = b • V = bT • (constant P & n) • V1/T1 = V2/T2 Charles’s Law

12. Plots of V Versus T(Celsius) for Several Gases Volume of a gas Changes by When the temp. Changes by 1oC. I.e., at -273oC , V=0 ???

13. Vn • V = an • (constant P& T) Avogadro’s Law

14. 5.3 The Ideal Gas Law Boyle’s law Charles's law Universal gas constant Avogadro’s law

15. The Ideal Gas Law PV = nRT R = 0.0821 atm L mol-1 K-1 The Ideal Gas Law can be used to derive the gas laws as needed!

16. Molar Volume At STP 4.0 g He 16.0 g CH4 44.0 g CO2 1 mole 1 mole 1mole (STP) (STP) (STP) V = 22.4 L V = 22.4 L V = 22.4 L

17. The value of R • What is R for 1.00 mol of an ideal gas at STP (25 oC and 1.00 atm)?Given that V of 1 mol of gas at STOP= 22.4L

18. Example • A reaction produces enough CO2(g) to fill a 500 mL flask to a pressure of 1.45 atm at a temperature of 22 oC. How many moles of CO2(g) are produced? PV = nRT

19. The Ideal Gas Law: Final and initial state problems

20. Ideal Gas Law • The ideal gas law is an equation of state. • Independent of how you end up where you are at. Does not depend on the path. • The state of the gas is described by: P, V, T and n • Given 3 you can determine the fourth. • Ideal gas equation is an empirical equation - based on experimental evidence.

21. Example A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

22. Data Table Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL T2 = ?? ??

23. Solution Solve for T2 Enter data T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C

24. Calculation Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K - 273 = 331 °C

25. Example A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

26. Solution T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL = 178 K - 273 = - 95°C

27. 5.4 Gas Stoichiometry • Reactions happen in moles • At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occuppies 22.4 L. • If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

28. Example A.What is the volume at STP of 4.00 g of CH4? 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He

29. Example A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder. P = nRT V PV = nRT = (3.74 mol)(0.082L•atm)(297.5K) (12.25 L) mol•K ? P = V = n = R = T = 12.25 L mol = 1.009 atm 75.5 g = mol 3.74 20.18 g = 5670 torr 0.082 L•atm mol•K 24.5 + 273 = 297.5 K

30. Example • 30.2 mL of 1.00 M HCl are reacted with excess FeS. What volume of gas is generated at STP? STP means standard temperature and pressure . . . 0 oC and 1 atm. 2 HCl + FeS  FeCl2 + H2S HCl + FeS  FeCl2 + H2S Now . . . Go for moles,

31. 2 HCl + FeS  FeCl2 + H2S PV = nRT V = nRT/P

32. Example The decomposition of sodium azide, NaN3, at high temperatures produces N2(g). What volume of N2(g), measured at 735 mm Hg and 26°C, is produced when 70.0 g NaN3 is decomposed. 2 NaN3(s) → 2 Na(l) + 3 N2(g)

33. nRT (1.62 mol)(0.08206 L atm mol-1 K-1)(299 K) V= = P 1.00 atm (735 mm Hg) 760 mm Hg 2 NaN3(s) → 2 Na(l) + 3 N2(g) Determine moles of N2: 1 mol NaN3 3 mol N2 = 1.62 mol N2 nN2 = 70 g NaN3 X X 65.01 g N3/mol N3 2 mol NaN3 Determine volume of N2: = 41.1 L

34. Molar mass of a gas P x V = n x R x T • P x V = m x R x T M • m = mass, in grams • M = molar mass, in g/mol • Molar mass = m R T P V

35. Density • Density (d) is mass divided by volume • P x V = m x R x T M • P = m x R x T V x M • d = m V • P = d x R x T M 

36. Example A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25°C (d=0.9970 g cm-3) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0°C. What is the molar mass of polypropylene?  Volume of the vessel

37. Vflask = = 98.41 cm3 = 0.09841 L mgas = mfilled- mempty = (40.2959 g – 40.1305 g) = 0.1654 g

38. m RT PV = M (0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K) M = (0.9741 atm)(0.09841 L) m RT M = PV PV = nRT M = 42.08 g/mol

39. Example Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L d = PXM R x T

40. The density of O2 gas at STP is 1.43 grams per liter

41. m RT M = PV Example • 2.00 g sample of SX6(g) has a volume of 329.5 Cm3 at 1.00 atm and 20oC. Identify the element X. Name the compound • P= 1.00 atm • T = 273+20 = 293K

42. = 146 g SX6 /mol f Molar mass of(X6 )= 146- 32 = 114 g/mol Molar mass of X = (114 g/mol X6)/6 = 19 X = with a molar mass of 19 = F The compound isSF6

43. 5.5 Dalton’s Law of Partial Pressures • For a mixture of gases in a container, the total pressure is the sum of the pressure each gas would exert if it were alone in the container. • The total pressure is the sum of the partial pressures. • PTotal = P1 + P2 + P3 + P4 + P5 ... • For each gas:

44. Partial pressure • Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone • PTotal = n1RT + n2RT + n3RT +... V V V • In the same container R, T and V are the same. • PTotal = (n1+ n2 + n3+...)RT V Thus,

45. A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. The total pressure is due to the partial pressures of each of these gases. so: For He: 1.00 x 10-3 g He mol _____________________ = mol He 2.50 x 10-4 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = mol H2 9.92 x 10-4 2.016 g

46. A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. so: 1.00 x 10-3 g He mol _____________________ = mol He For He: 2.50 x 10-4 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = mol H2 9.92 x 10-4 2.016 g And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V) = (0.001242 mol)(0.0821 L•atm)(25 + 273)K mol•K (0.2500 L) Ptotal= 0.1216 atm