2. 1.4 Moles and Gaseous Volume Relationships in Chemistry

3. Assessment Objectives 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases. 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations. 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. 1.4.7 Solve problems using the ideal gas equation, PV = nRT 1.4.8 Analyse graphs relating to the ideal gas equation.

4. Kinetic Molecular Theory 1) All substances are composed of particles and these particles are in continuous motion. 2) Collisions between particles are perfectly elastic. 3) Average energy of a particle is directly proportional to temperature Temperature causes particles to move faster (with liquids and gases) or vibrate more. At 0 K, there is no motion in particles

5. 4 Gaseous Volume Relationships in Chemical Reactions Kinetic Theory: Tiny particles in all forms of matter are in constant motion Application to Gases 1) A gas is composed of particles that are considered to be small, hard spheres that have insignificant volume and are relatively far apart from one another. Between the particles there is empty space. No attractive or repulsive forces exist between the particles. 2) The particles in a gas move rapidly in constant random motion. They travel in straight paths and move independently of each other. They change direction only after a collision with one another or other objects. 3) All collisions are perfectly elastic. Total kinetic energy remains constant.

6. Demo 1: Diffusion of KMnO 4 Observations: KMnO 4 (potassium permanganate) diffuses faster in hot water. Explanation: KMnO 4 dissolves in water. As the KMnO 4 dissolves, it collides with H 2 O molecules and spreads out. Because molecules in hot water are moving faster, the solid dissolves faster and diffuses faster. Cold water Hot water

7. Demo 2: thermal expansion of a liquid Observations: coloured water moves slowly up the tube Explanation: heat increases the kinetic energy of liquid particles. The particles move faster (greater vibrational, rotational, and translational energy). This greater movement increases the distance between molecules. Thus, the volume expands. Glass tube Rubber stopper Coloured water in florence flask Hot plate

8. Demo 4: Kill da wabbit Question: what will happen when the air pressure surrounding a balloon is decreased? Why? Explanation: A balloon normally stays the same size because the pressure inside the balloon is equal to the pressure outside the balloon (rate of collisions with the wall of the balloon is equal). Reducing the pressure on the outside eliminates opposing collisions, allowing the balloon to expand. For more lessons, visit www.chalkbored.com www.chalkbored.com

9. 8 Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. 5.1 Physical Characteristics of Gases

10. 9 Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

11. 10 5.1

12. 11 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Variables That Describe a Gas Pressure (P) measured in kPa, mm Hg, atm 101.3 kPa = 760 mm Hg = 1.00 atm = 101,300 Pa Volume (V) measured in dm 3 or L 1 dm 3 = 1000 cm 3 = 1 L = 1000 mL Temperature (T) measured in K (Kelvin) K = o C + 273 Amount of matter (n) measured in moles

13. 12 Standard Pressure and Temperature (STP) Standard Pressure = 1 atm (or an equivalent, such as 101.3 kPa) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

14. Reactions with Gases When we have reactions with gases, it is often difficult to have exact amounts (moles) of reactants to achieve an amount of product. It is often common to have one of the gas reactants in EXCESS to make sure that the required amount of product is achieved.

15. Assessment Objectives 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases.

16. Avogadro’s Hypothesis The same amount (moles) of different gases occupy the same volume (at the same pressure and temperature). This means that the volumes of reactants and products are in the same RATIO as the coefficients in balanced equations. e.g 2CO + O 2  2CO 2 2moles 1mole 2moles 2 Vol 1 Vol 2 Vol

17. Question 10cm 3 of ethyne (C 2 H 2 ) is reacted with 50cm 3 hydrogen to produce ethane (C 2 H 6 ). Calculate the total volume and composition of the remaining gas mixture (assuming temperature and pressure remain constant). C 2 H 2 + 2H 2  C 2 H 6 1 Vol 2 Vol 1 Vol 10cm 3 20cm 3 10cm 3

18. Therefore Hydrogen is in excess Volume of hydrogen remaining = 50 cm 3 – 20 cm 3 = 30 cm 3 Therefore the total gas mixture remaining is 40 cm 3 (10 cm 3 of ethane and 40 cm 3 of hydrogen)

19. Assessment Objectives 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations.

20. Molar Gas Volume 1 mole of gas occupies 22.4 dm 3 Amount of gas (moles) = Volume of gas Molar Volume(22.4) (At standard temperature and pressure 101.3 kPa and 273K) Worksheet Questions

21. 20 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) a) How many dm 3 of HF are needed to produce 9.40 dm 3 H 2 at STP? 9.40 dm 3 H 2 x 2 dm 3 HF 1 dm 3 H 2 = 18.8 dm 3 HF

22. 21 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) b) How many grams of Sn are needed to react with 20.0 dm 3 of HF at STP? Remember 1 mol = 22.4 dm 3 20.0 dm 3 HF x 1 mol HF x 1 mol Sn x 118.69 g Sn 22.4 dm 3 HF 2 mol HF 1 mol Sn = 53.0 g

23. 22 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) c) What volume of H 2 at STP is produced from 37.4 g Sn? 37.4 g Sn x 1 mol Sn x 1 mol H 2 x 22.4 dm 3 118.69 g Sn 1 mol Sn 1 mol H 2 = 7.06 dm 3

24. Assessment Objective 1.4.7 Solve problems using the ideal gas equation, PV = nRT

25. Ideal Gas Equation An ideal gas is one where the particles have neglible mass, there is no attractive force between molecules anf the kinetic energy is directly proportional to the temperature. This is a good approximation for most gases except at low temperatures and high pressures.

26. PV = nRT R is the ideal gas rate constant 8.314 J K -1 mol -1 P = pressure kPa V = volume dm 3 n = moles The Ideal Gas Equation R = 8.314 J/K mol 82.06 cm3 atm/mol K 0.08206 L atm/mol K 62.4 L torr/mol K

27. How many moles of H 2 is in a 3.1 L sample of H 2 measured at 300 kPa and 20°C?

28. 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 STP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas?

29. Assessment Objective 1.4.8 Analyse graphs relating to the ideal gas equation.

30. 29 1.4.8 Analyze graphs relating to the ideal gas equation

31. 30 1.4.8 Analyze graphs relating to the ideal gas equation Real gases deviate from ideal behavior at low and high pressures and temperatures. Gas molecules do have some attraction for each other Gas molecules have a volume

32. If we keep the amount of gas constant, there are essentially 3 variables in the ideal gas equation: Pressure Volume Temperature Lets look at 2 variables at a time, to examine their relationship.

33. Boyle-Mariotte Law Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

34. Boyle-Mariotte Law P 1 V 1 = P 2 V 2 A gas with volume 1 m 3 at 2 MPa is compressed to a final volume of 0.5 m 3. If the temperature is kept constant the final pressure can be calculated as: P 2 = P 1 V 1 / V 2 = (2 MPa) (1 m 3 ) / (0.5 m 3 ) = 4 MPa

35. Worksheet Questions P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P 1 V 1 = P 2 V 2

36. Kinetic Molecular Theory Can you describe why the pressure would increase when we decrease the volume using the Kinetic molecular theory?

37. Charles Law T has to be in Kelvin K= o C + 273

38. Temperature vs. Volume Graph ) 5 10 15 20 25 30 Volume (mL) Temperature (  C) 0 100 – 273 25 mL at 22C 31.6 mL, 23.1 mL Y=0.0847x + 23.137

39. 38 Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V 1 V 2 = T 1 T 2 If one temperature goes up, the volume goes up! If one temperature goes up, the volume goes up! Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

40. 39 Charles’s original balloon Modern long-distance balloon

41. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K?

42. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

43. Kinetic Molecular Theory Can you describe why the volume would increase when we increase the temperature using the Kinetic molecular theory?

44. Pressure (Gay Lussac’s) Law P 1 /T 1 =P 2 /T 2 Worksheet Questions

45. 44 Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P 1 P 2 = T 1 T 2 If one temperature goes up, the pressure goes up! Joseph Louis Gay- Lussac (1778-1850)

46. Kinetic Molecular Theory Can you describe why the pressure would increase when we increase the temperature using the Kinetic molecular theory?

47. Combining the gas laws Jacques CharlesRobert Boyle P1V1P1V1 =P2V2P2V2 V1V1 T1T1 = V2V2 T2T2 These are all subsets of a more encompassing law: the combined gas law P1P1 T1T1 = P2P2 T2T2 P 1 V 1 P 2 V 2 T 1 T 2 = Joseph Louis Gay-Lussac

48. 47 Combined Gas Law The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION ! P 1 V 1 P 2 V 2 = T 1 T 2

49. 48 Combined Gas Law Problem 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? Set up Data Table P 1 = 0.800 atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ??

50. 49 One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?