2. Chapter 13 Outline Gases and their properties

3. Standard #4 The kinetic molecular theory describes the motion of atoms and molecules and explains the properties of gases.

4. *Nitrogen: 78% *Oxygen: 21% *Carbon Dioxide: *Noble Gases trace amounts *Water Vapor (the amount varies) A.Air composed of several kinds of colorless gases Relative Composition of air: 1%

5. B.Some gases have color Chlorine Nitrogen Dioxide is SMOG! C.Any 1 mole of a gas occupies 22.4 L

6. Nature of Gases 1.Gases have Mass

7. 2.Gases are easy to Compress

8. 3.Gases fill their container

10. 4.Gases Diffuse 4.Gases Diffuse Molecules travel from a high concentration to a low concentration

11. I smell something weird!

12. 5.Gases Effuse The escape of gas molecules through a tiny hole into an evacuated space.

13. 5.Gases Effuse Gases Effuse:

14. 6.Exerts Pressure on Wall of container The pressure of the gas depends on the temperature! What do you think will happen to the gas and the amount of pressure if this balloon is placed near a heat source/light?

15. Volume of balloon at room temperature Volume of balloon at 5°C

16. The Kinetic Molecular Theory 1.Gases consist of small particles 2.Gases are separated by large distances 3.Gases are in constant Rapid Motion 4.Collisions are elastic No loss of energy with collisions. Energy is conserved There is a loss of energy with collisions.

17. Measuring Pressure: A. Pressure: Force Area 1. Pressure is high…area is 2. Pressure is low…area is small LARGE

18. Units for measuring PRESSURE 1atm = 760.mmHg101.3kPa14.7lb/in 2 or psi = = End of day 1

19. Barometer Stop here!

20. D.Instrument that measures: A.Pressure: Barometer B.Temperature: Thermometer

21. Units for measuring PRESSURE 1atm = 760.mmHg101.3kPa14.7lb/in 2 or psi = =

22. Boyle’s Law: P and V States that the volume of a gas is inversely proportional to the pressure at constant temperature Robert Boyle 25 January 1627 – 30 December 1691

23. Low pressure= High Volume High pressure= Low Volume Constant temperature and Constant Number of gas particles

24. Boyle’s law animation

25. Jacues Charles In 1808, Charles’ Law was developed. It states that the volume of a gas is directly related to its temperature at constant pressure. Nov. 1746 - Apr. 1823

26. Charles’ Law Visuals

28. John Dalton Dalton's law of partial pressures was stated by in 1801:

29. The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual component gases. The partial pressure is the pressure that each gas would exert if it alone occupied the volume of the mixture at the same temperature

30. Combined Gas Law Problems 500.0 mL of a gas was collected at 20.0 °C and 720.0 mm Hg. What is its volume at STP? 2.00 liters of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated until a volume of 20.0 liters and a pressure of 3.50 atmospheres is reached. What is the new temperature?

31. Ideal Gas Law

32. The Ideal Gas Law PV = nRT

33. Ideal Gases An “ideal” gas exhibits certain theoretical properties. Specifically, an ideal gas … Obeys all of the gas laws under all conditions. Does not condense into a liquid when cooled. Shows perfectly straight lines when its V and T & P and T relationships are plotted on a graph. In reality, there are no gases that fit this definition perfectly. We assume that gases are ideal to simplify our calculations. We have done calculations using several gas laws (Boyle’s Law, Charles’s Law, Combined Gas Law). There is one more to know…

34. The Ideal Gas Law PV = nRT P = Pressure (in kPa)V = Volume (in L) T = Temperature (in K) n = moles R = 8.31 kPa L K mol R is constant. If we are given three of P, V, n, or T, we can solve for the unknown value. Recall, From Boyle’s Law: P 1 V 1 = P 2 V 2 or PV = constant From combined gas law: P 1 V 1 /T 1 = P 2 V 2 /T 2 or PV/T = constant

35. Developing the ideal gas law equation PV/T = constant. What is the constant? At STP: T= 273K, P= 101.3 kPa, V= 22.4 L/mol PV = constant T mol Mol is represented by n, constant by R: PV = R Tn Rearranging, we get: PV = nRT Because V depends on mol, we can change equation to: At STP: (101.3 kPa)(22.4 L) = (1 mol)(R)(273K) R = 8.31 kPa L K mol Note: always use kPa, L, K, and mol in ideal gas law questions (so units cancel)

36. Sample problems How many moles of H 2 is in a 3.1 L sample of H 2 measured at 300 kPa and 20°C? PV = nRT (300 kPa)(3.1 L) = n (8.31 kPaL/Kmol)(293 K) (8.31 kPaL/Kmol)(293 K) (300 kPa)(3.1 L) = n = 0.38 mol How many grams of O 2 are in a 315 mL container that has a pressure of 12 atm at 25°C? P = 300 kPa, V = 3.1 L, T = 293 K PV = nRT (8.31 kPaL/Kmol)(298 K) (1215.9 kPa)(0.315 L) = n = 0.1547 mol P= 1215.9 kPa, V= 0.315 L, T= 298 K 0.1547 mol x 32 g/mol = 4.95 g

37. Ideal Gas Law Questions 1.How many moles of CO 2 (g) is in a 5.6 L sample of CO 2 measured at STP? 2.a) Calculate the volume of 4.50 mol of SO 2 (g) measured at STP. b) What volume would this occupy at 25°C and 150 kPa? (solve this 2 ways) 3.How many grams of Cl 2 (g) can be stored in a 10.0 L container at 1000 kPa and 30°C? 4.At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate its molar mass. 5.98 mL of an unknown gas weighs 0.087 g at SATP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas?

38. P=101.325 kPa, V=5.6 L, T=273 K PV = nRT (101.3 kPa)(5.6 L) = n (8.31 kPaL/Kmol)(273 K) 1.Moles of CO 2 is in a 5.6 L at STP? (8.31 kPaL/Kmol)(273 K) (101.325 kPa)(5.6 L) = n = 0.25 mol 2.a) Volume of 4.50 mol of SO 2 at STP. P= 101.3 kPa, n= 4.50 mol, T= 273 K PV=nRT (101.3 kPa)(V)=(4.5 mol)(8.31 kPaL/Kmol)(273 K) (101.3 kPa) (4.50 mol)(8.31 kPaL/Kmol)(273 K) V == 100.8 L

39. 2.b) Volume at 25°C and 150 kPa (two ways)? Given: P = 150 kPa, n = 4.50 mol, T = 298 K (150 kPa) (4.50 mol)(8.31 kPaL/Kmol)(298 K) V == 74.3 L From a): P = 101.3 kPa, V = 100.8 L, T = 273 K Now P = 150 kPa, V = ?, T = 298 K P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (101.3 kPa)(100 L) (273 K) = (150 kPa)(V 2 ) (298 K) (101.3 kPa)(100.8 L)(298 K) (273 K)(150 kPa) =(V 2 ) =74.3 L

40. 3.How many grams of Cl 2 (g) can be stored in a 10.0 L container at 1000 kPa and 30°C? PV = nRT (8.31 kPaL/Kmol)(303 K) (1000 kPa)(10.0 L) = n = 3.97 mol P= 1000 kPa, V= 10.0 L, T= 303 K 3.97 mol x 70.9 g/mol = 282 g 4.At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate molar mass. PV = nRT (8.31 kPaL/Kmol)(423 K) (100 kPa)(1.00 L) = n = 0.02845 mol P= 100 kPa, V= 1.00 L, T= 423 K g/mol = 2.506 g / 0.02845 mol = 88.1 g/mol

41. 5.98 mL of an unknown gas weighs 0.081 g at SATP. Calculate the molar mass. PV = nRT (8.31 kPaL/Kmol)(298 K) (100 kPa)(0.098 L) = n = 0.00396 mol P= 100 kPa, V= 0.098 L, T= 298 K g/mol = 0.081 g / 0.00396 mol = 20.47 g/mol It’s probably neon (neon has a molar mass of 20.18 g/mol)

42. Determining the molar mass of butane Using a butane lighter, balance, and graduated cylinder determine the molar mass of butane. Determine the mass of butane used by weighing the lighter before and after use. The biggest source of error is the mass of H 2 O remaining on the lighter. As a precaution, dunk the lighter & dry well before measuring initial mass. After use, dry well before taking final mass. (Be careful not to lose mass when drying). When you collect the gas, ensure no gas escapes & that the volume is 90 – 100 mL. Place used butane directly into fume hood. Submit values for mass, volume, & g/mol.