2. 1 IB Topic 1: Quantitative Chemistry 1.4: Mass Relationships in Chemical Reactions  1.4.4 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.  1.4.5 Solve problems using the ideal gas equation, PV = nRT  1.4.6 Apply Avogadro’s law to calculate reacting volumes of gases  1.4.7 Apply the concept of molar volume at standard temperature and pressure in calculations  1.4.8 Analyze graphs relating to the ideal gas equation

3. 2 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

4. 3 Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

5. 4

6. 5 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.

7. 6 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

8. 7 And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure 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

9. 8 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Boyle’s Law Relates Pressure-Volume  As pressure increases, volume decreases if temperature and amount remain constant.  Spaces between particles so particles can move close closer together  P 1 x V 1 = P 2 x V 2

10. 9 Boyle’s Law 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 Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

11. 10 Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

12. 11 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Charles’s Law Relates Temperature-Volume  As temperature increases, volume increases if pressure and amount remain constant  Particles gain kinetic energy, move farther apart  V 1 = V 2 T 1 T 2  T has to be in Kelvin ; K= o C + 273

13. 12 Charles’s original balloon Modern long-distance balloon

14. 13 Charles’s Law

15. 14 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! Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

16. 15 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Gay-Lussac’s Law Relates Temperature-Pressure  As temperature increases, pressure increases if volume and amount remain constant.  Particles gain kinetic energy so they move faster and have more collisions  T has to be in Kelvin ; K = o C + 273

17. 16 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)

18. 17 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 No, it’s not related to R2D2

19. 18 Combined Gas Law You must go through Kelvin in order to get the right answer!!!!!

20. 19 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 = ??

21. 20 Calculation P 1 = 0.800 atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ?? P 1 V 1 P 2 V 2 = P 1 V 1 T 2 = P 2 V 2 T 1 T 1 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T 2 = 604 K - 273 = 331 °C = 604 K

22. 21 Learning Check 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?

23. 22 Solution T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = 0.315 L = 315 mL P 1 = 0.850 atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL = 179 K - 273 = - 94°C

24. 23 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?

25. 24 Solution Complete the following setup: Initial conditionsFinal conditions V 1 = 785 mLV 2 = ? T 1 = 21°C = 294 KT 2 = 0°C = 273 K Since P is constant, P cancels out of the equation. V 1 V 2 V 1 T 2 = V 1 T 2 = T 1 V 2 = V 2 T 1 T 2 T 1 = 729 mL Check your answer: If temperature decreases, V should decrease.

26. 25 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases Avogadro’s Hypothesis: Equal volumes of gases at the same temperature and pressure contain equal numbers of particles (moles). At STP (standard temperature & pressure: 273 K and 101.3 kPa) 1 mole of any gas occupies a volume of 22.4 dm 3 (L).

27. 26 And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure 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

28. 27 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases Assuming STP, how many dm 3 of oxygen are needed to produce 19.8 dm 3 SO 3 according to: 2SO 2 (g) + O 2 (g)  2SO 3 (g) Since equal volumes of gases contain the same number of moles, we can use the equation coefficients with the volumes. 19.8 dm 3 SO 3 x 1 O 2 = 9.90 dm 3 O 2 2 SO 3

29. 28 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm 3 of nitrogen dioxide are produced when 3.4 cm 3 of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.

30. 29 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm 3 of nitrogen dioxide are produced when 3.4 cm 3 of oxygen reacts with and excess of nitrogen monoxide? Assume conditions of STP. 2NO + O 2  2NO 2 3.4 cm 3 O 2 x 2 NO 2 = 6.8 cm 3 O 2 1 O 2

31. 30 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? b) How many grams of Sn are needed to react with 20.0 dm 3 HF at STP? c) What volume of H 2 is produced from 37.4 g Sn?

32. 31 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

33. 32 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

34. 33 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

35. 34 1.4.7 Solve problems using the ideal gas equation, PV = nRT Gas Laws: Ideal Gas Law Relates Temperature-Pressure-Volume-Amount PV = nRT  P = pressure in kPa  V = volume in dm 3 or L  n = moles  R = Gas constant (8.31 dm 3 kPa mol -1 K -1 )  T = temperature in K Ideal gas: particles have no volume and are not attracted to each other

36. 35 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (101.3 kPa)(22.414dm 3 ) (1 mol)(273.15 K) R = 8.31242 dm 3 kPa / (mol K) = 8.31 dm 3 kPa mol -1 K -1 ) 5.4 Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 dm 3.

37. 36 Applying STP  Calculate the mass of 40.0 dm 3 of NO gas at standard conditions.

38. 37 Applying STP  Calculate the mass of 40.0 dm 3 of NO gas at standard conditions. Using PV = nRT… 101.3kPa(40.0 dm 3 ) = n (8.31)(273) n = 1.786 mol NO x 30.01 g NO= 53.6 g NO 1 mol NO

39. 38 Applying STP  Calculate the mass of 40.0 dm 3 of NO gas at standard conditions. Using 1 mole @ STP = 22.4 dm 3 40.0 dm 3 NO x 1 mol NO = 1.786 mol NO 22.4 dm 3 NO n = 1.786 mol NO x 30.01 g NO= 53.6 g NO 1 mol NO

40. 39 More Practice  1.0 g of Mg is dropped into a solution of hydrochloric acid. a) Write the balanced chemical equation for the reaction. b) What volume will the evolved hydrogen occupy at STP?

41. 40 More Practice  1.0 g of Mg is dropped into a solution of hydrochloric acid. a) Write the balanced chemical equation for the reaction. Mg + 2 HCl  MgCl 2 + H 2 b) What volume will the evolved hydrogen occupy at STP? 1.0 g Mg x 1 mol Mg x 1 mol H 2 x 22.4 dm 3 H 2 24.31 g Mg 1 mol Mg 1 mol H 2 = 0.92 dm 3 H 2

42. 41 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L

43. 42 1.4.8 Analyze graphs relating to the ideal gas equation

44. 43 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

45. 44

46. 45 Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

47. 46 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0 C. What is the molar mass of the gas? 5.3 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x 0.0821 x 300.15 K Latm molK M = 54.6 g/mol