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Understanding Gas Laws and States of Matter | Lecture Summary

Learn about the kinetic molecular theory of gases, gas laws (Boyle's, Charles's, Avogadro's), and ideal gas behavior. Practice calculations and interpret graphs. Discover how the states of matter are interconvertible and explore the gaseous state's parameters.

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Understanding Gas Laws and States of Matter | Lecture Summary

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  1. CHAPTER 5 STATES OF MATTER 7 HOURS 5.1 GAS 5.2 LIQUID 5.3 SOLID 5.4 PHASE DIAGRAM 1

  2. LECTURE 1 GAS Learning Outcomes: (a) Explain qualitatively the basic assumptions of the kinetic molecular theory of gases for an ideal gas. (b) Define gas laws: i.Boyle’s law ii.Charles’s law iii.Avogadro’s law (c) Sketch and interpret the graphs of Boyle’s and Charles’s Laws. (d) Perform calculations involving gas laws and ideal gas equation 2

  3. In principle, the 3 states of matter are interconvertible 3

  4. There are four parameters describing the gaseous state: ParameterUnit (SI) • Quantity, n moles • Volume, V litres • Temperature, T kelvin • Pressure, P pascal/ Nm-2 4

  5. Relationship between units: 1 atm = 101325 Pa = 101.325 kPa = 101325 Nm-2 = 760 mmHg = 760 torr 1 Pa = 1 Nm-2 1 mmHg = 1 torr 0C = 273.15 K Example: Convert 749 mmHg to atm Solution: 1 atm = 760 mmHg 0.986 atm= 749 mmHg 5

  6. Kinetic Molecular Theory of Gases 4 basic assumption (postulate) : • The gas consists of tiny particles of negligible volume. 2. Intermolecular forces attraction do not exist between gas particles. 3. The molecules of a gas are in continuous random motion. The gaseous particles are perfectly elastic. 4. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature. 6

  7. Ideal gas behaviour • The gas consists of tiny particles of negligible volume. • Intermolecular forces attraction do not exist between gas particles. 7

  8. BOYLE’S LAW GAS LAW CHARLES’S LAW AVOGADRO’S LAW KMP/ChemUnit/SK017-09/10 8

  9. BOYLE’S LAW ( The pressure –volume relationship) • For a fixed amount of gas at a constant temperature, gas volume is inversely proportional to gas pressure. • As the pressure (P) increases, the volume (V) decreases. • V= 1/P at constant T and n • P1V1 = k1 (a constant) P1V1 = P2V2 9

  10. 10

  11. Graph A : Pressure against Volume Illustrate that : pressure is inversely proportional to volume 11

  12. Graph B : Pressure against 1/Volume Illustrate that : P is directly proportional to 1/ V KMP/ChemUnit/SK017-09/10 12

  13. Graph C: PV against Pressure, P Illustrate that: PV = constant KMP/ChemUnit/SK017-09/10 13

  14. Example:- An inflated balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? Note : n & T = kept constant. ~ so we used Boyle’s Law 14

  15. Exercise 1 • A sample of chlorine gas occupies a volume of 750 ml at pressure of 5.0 atm. Calculate the pressure of the gas if the volume is reduced at constant temperature to 200 ml. • At 46 °C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-tenth (0.10) of the original value at the same temperature? Answer: 19 atm Answer: 53 atm 15

  16. CHARLES’S LAW( The Temperature – Volume relationship) • The volume of a fixed amount of gas at constant pressure is directly proportional to the absolute temperature V  Tat constant P and n V = constant T T in Kelvin (K) !!! T(K) = TC + 273.15 V1= V2 T1 T2 16

  17. Variation of gas volume with temperature at constant temperature P1 P2 T 17

  18. Reminder !!!! Absolute zero ( 0 K or -273.15 ºC) is the temperature at which an ideal gas would have zero volume. Example: 452-mL sample of fluorine gas is heated from 22C to 187C at constant pressure. What is its final volume? Answer: Remember to convert oC to K when solving gas-law problems. KMP/ChemUnit/SK017-09/10 18

  19. Exercise 2 • A sample of carbon monoxide gas occupies 3.20 L at 125 oC. Calculate the temperature at which the gas will occupy 1.54 L if the pressure remains constant. • A balloon filled with air has a volume of 2.00 L at 27oC. What will be its volume at -23oC if the pressure remains constant? ( Answer: 192 K @ -81oC ) ( Answer: 1.67 L 19

  20. AVOGADRO’S LAW( The Volume-amount relationship) At constant pressure and temperature, the volume of a gas is directly proportional to number of moles of the gas present. V  n ( P and T remain constant ) …………….. 4 20

  21. Avogadro’s “equal volumes-equal numbers” hypothesis can be stated in either of two ways: • At the same temperature and pressure, equal volume of different gases contain the same number of molecules (or atoms if the gas is monatomic). • Equal numbers of molecules of different gases compared at the same temperature and pressure occupy equal volumes. • At constant P and T, if VCO2 = VN2 = VAr  nCO2 = nN2 = nAr • At STP: T = 273.15K (0C), P = 1.0 atm; • 1 mol of gas = (6.023x1023 particles) = 22.4 dm3. • The volume of 22.4 dm3 gases at STP is called molar volume of a gas at STP 21

  22. Example: Calculate the volume (in litres) at STP occupied by • 0.20 mol of CO2 gas • 6.0 g of Cl2 gas 1 mol of any gas occupies 22.4 L at STP Solution: 22

  23. Exercise 3 • The density of a gas at S.T.P is 1.78 dm-3. Calculate the relative molecular mass of the gas. • Calculate the volume (at s.t.p) occupied by the following gasses. • 0.25 mol of oxygen gas • 106.5 g of chlorine gas Answer: 39.87 Answer: 5.6 dm-3 Answer: 33.6 dm-3 23

  24. COMBINED GAS LAW P1V1 = P2V2 T1 T2 ………….. 3 Example: A gas initially at 4.0 L, 600 torr and 66°C undergoes a change so that its final volume and temperature become 1.7 L and 42°C. What is the final pressure? Assume the number of moles remains unchanged? Therefore : KMP/ChemUnit/SK017-09/10 24

  25. Exercise 4 1. A volume of gas, 60 ml, is collected at 60°C and 1.05 x 105 Nm-2. What volume would the gas occupy at STP? 2. A meteorogical balloon has a volume of 6.15 m3 when filled with Helium at 14°C and 762 mmHg. What will its temperature be if the volume expands to 6.37 m3 and the pressure falls to 749 mmHg? Answer: 51 ml Answer: 19 °C 25

  26. LECTURE 2 Learning Outcomes: (e) Define and perform calculation using Dalton’s law 26

  27. IDEAL GAS » V n.T P » V = R. n.T P Where R = gas constant 27

  28. The Ideal Equation PV = nRT ……………............... (6) R= gas constant = 0.08206 L atm K-1 mol-1 = 8.314 N m K-1 mol -1 P = cRT KMP/ChemUnit/SK017-09/10 28

  29. Gas constant L mmHg mol -1 K-1 * SI unit 29

  30. Example: 1. Calculate the pressure (in atm) exerted by 2.12 moles of nitric oxide (NO) in a 9.0 L container at 76 C. Answer: V = 9.0 L n = 2.12 mol T = 76C + 273 = 349 K R = 0.08206 atm L mol1 K1 P V = n R T  P = P = = 6.75 atm 30

  31. 2. Calculate the density of ammonia (NH3) in grams per litre (g/L) at 752 mmHg and55 C. Answer: 31

  32. Exercise 5 : • 6.32 g of gas occupies 2,200 cm3 at 100°C and 101 kPa. Calculate the relative molecular mass of the gas. • The density of a gas is 1.25 g dm-3 at 25°C and 101 kPa. Calculate the relative molecular mass of the gas. • Sketch a graph for a constant number of moles of an ideal gas for i. PV against T ii. PV/T against T iii. V against T Explain the graphs. Answer: 88.15 g/mol Answer: 30.6 32

  33. Dalton’s Law? The total pressure of a mixture of gases is the sum of the partial pressures of all the components in the mixture. PT = PA + PB + PC + ……. Where PA, PB, PC = partial pressure of A,B and C PT = total pressure of the mixture KMP/ChemUnit/SK017-09/10 33

  34. Definition: Dalton’s Law states that the total pressure of a mixture of gases is the sum of the partial pressures of all the components in the mixture 34

  35. Partial pressures is the pressures of individual gas components in the mixture. Partial pressure of gas A in the mixture is given as: PA = XA . Ptotal XA = mole fraction of gas A in the mixture = nA/ntotal The total mole fraction of all gasses in the mixture is equal to 1 XA + XB + XC = 1 35

  36. Volume and temperature are constant Combining the gases B C PT = PA + PB + PC A In a mixture of gases A, B and C, the total pressure PT is the result of collisions of 3 types of molecules, A, B and C, with the wall of container. Thus, according to the Dalton’s law, PTotal = PA + PB + PC = nART + nBRT + nCRT V V V = RT ( nA + nB + nC ) V = nTotalRT V nTotal = nA + nB + nC 36

  37. The ratio number of moles of gas A to the total number of moles, XA = nA = PA( because PA = nART/V ) ntotal Ptotal PT nTRT/V Mole fraction of gas A PA = XA PT Where : XA + XB + XC + ….. = 1 37

  38. Example • A mixture of gases contains 0.1 mol of CO2 and 0.3 mol of O2. Calculate the partial pressures of the gases if the total pressure is 760 mmHg at a certain temperature. Answer: KMP/ChemUnit/SK017-09/10 38

  39. Example • What volume of H2 (in cm3) at 30 C and 90 kPa must be admitted to a 500cm3 flask that contains gas N2 at 20 C and 100 kPa in order to increase the total pressure of the mixture to 110 kPa at 20 C? Answer: KMP/ChemUnit/SK017-09/10 39

  40. Exercise 6: • A 5.8 dm3 container contains 0.04 mol nitrogen, 0.12 mol carbon monoxide and 0.08 mol hydrogen at 25°C. Calculate : • The partial pressure of each gas. • The total pressure of the mixture • A cylinder of capacity 40 dm3 contains 350 g CO2, 800 g O2 and 5000 g nitrogen at 28°C. Calculate the total pressure of the mixture. Answer: PN2= 1.708 x 104 Pa PH2 = 3.416 x 104 Pa PCO = 5.124 x 104 Pa PT = 1.025 x 105 Pa Answer: PT = 1.323 x 107 Pa KMP/ChemUnit/SK017-09/10 40

  41. Water vapour in a gas mixture KMP/ChemUnit/SK017-09/10 41

  42. The total pressure in the collection of a gas over does not only result from the pressure of that particular gas but also from the water vapour. PT = P atm = P gas + P H2O 42

  43. Exercise 7 Ammonium nitrite decomposes upon heating. NH4NO2 N2 + 2H2O When a sample of NH4NO2 is decomposed in a test tube 511mL of gas is collected over water at 26oC and 745 torr total pressure. How many grams of NH4NO2 were decomposed ? (vapour pressure of water at 26oC = 25 torr) Answer: 1.26 g 43

  44. LECTURE 3 Learning Outcomes: (f) Explain the ideal and non-ideal behaviors of gases in terms of intermolecular forces and molecular volume. (g) Explain the conditions at which real gases approach the ideal behavior. 44

  45. Real Gases- Deviation From Ideality • Boyle’s Law and Charles Law have led to the derivation of an ideal gas equation, PV=nRT. • Ideal gas obey the equation and fits the assumption of the Kinetic Molecular Theory. • However, real gases showed deviation from ideal behavior. • There are two main reason: • Volume of molecules • Intermolecular forces 45

  46. Volume of molecules • Ideal gas assume that gases consist of tiny molecules that does not occupy any space. • However, for a real gas, the molecules have a certain volume. • When a gas is compressed, the volume of the gas is decreased. • Thus, the volume of molecules begins to occupy a sizable portion of the container. KMP/ChemUnit/SK017-09/10 46

  47. Intermolecular forces • Ideal gas assume that, there are no attractive force or repulsive force between gaseous particles. • However, • When the volume of a container becomes smaller (by increasing the pressure) • The distance between molecules decrease • The force of attraction between the molecules increase. • These caused the behavior of real gases to deviate from ideal behavior. KMP/ChemUnit/SK017-09/10 47

  48. PV RT 1.0 Figure A (at constant temperature)

  49. For 1 mole of ideal gas, the value of PV/RT is equal to 1. • The lines plotted display the deviation of real gases from the ideal behavior. • However, all the lines converge to 1.0 when P is near zero. • Thus, real gas behave ideally at very low pressure. 49

  50. When gases are compressed, the molecules are closed enough to experience the attractive force among them.(curve below 1.0, PV<RT) • NH3 (polar molecule) shows the largest deviation because it has strongest attractive force. • At higher pressure, the molecules are pushed too close to each other that cause the repulsive forces among them. • This repulsion makes them less compressible, hence the line above 1.0, PV>RT. 50

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