Kinetic Theory of Matter and Gas Law | Jamb Chemistry

Jamb(UTME) key points on Phenomena to support the kinetic theory of matter

Phenomena Supporting the Kinetic Theory of Matter

Melting

1. Definition: Melting is the process where a solid turns into a liquid when heated.
2. Molecular Motion in Melting: Heat energy increases the kinetic energy of solid particles, causing them to vibrate more vigorously.
3. Intermolecular Forces: Increased energy weakens the strong bonds holding particles in place.
4. Phase Transition: Particles break free from their fixed positions, entering a more fluid state.
5. Example: Ice melting into water.
6. Kinetic Theory Explanation: Particles gain enough energy to overcome the rigid lattice structure.
7. Latent Heat: Energy absorbed during melting is used to break bonds, not raise temperature.
8. Uniform Temperature: Melting occurs at a constant temperature for pure substances.
9. Endothermic Process: Energy is absorbed from the surroundings.
10. Real-World Application: Melting metals for casting demonstrates the principle.
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Vaporization and Boiling

11. Definition: Vaporization is the change of a liquid to gas, including evaporation and boiling.
12. Evaporation: Occurs at the surface of a liquid below its boiling point as particles with sufficient energy escape.
13. Boiling: A rapid form of vaporization occurring throughout the liquid at the boiling point.
14. Molecular Motion: Particles gain enough kinetic energy to overcome intermolecular attractions and escape as vapor.
15. Temperature Dependence: Boiling occurs when vapor pressure equals atmospheric pressure.
16. Kinetic Energy Distribution: Not all particles have the same energy, so only the highest-energy ones evaporate.
17. Brownian Motion: Molecules in the gas phase exhibit random, constant motion.
18. Latent Heat of Vaporization: Energy absorbed during boiling without a temperature rise breaks intermolecular bonds.
19. Example: Water boiling at 100°C under normal atmospheric pressure.
20. Real-Life Example: Steam engines harness energy from boiling water.
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Freezing

21. Definition: Freezing is the process where a liquid becomes a solid as it loses heat.
22. Molecular Motion in Freezing: Kinetic energy decreases, causing particles to move slower.
23. Intermolecular Forces: Strong forces cause particles to lock into a fixed lattice structure.
24. Exothermic Process: Energy is released to the surroundings.
25. Example: Water freezing into ice at 0°C.
26. Uniform Temperature: Freezing occurs at a constant temperature for pure substances.
27. Latent Heat of Fusion: Energy released during freezing strengthens intermolecular bonds.
28. Significance: Freezing is used in food preservation (e.g., freezing vegetables).
29. Phase Transition: Indicates equilibrium between liquid and solid phases.
30. Real-Life Application: Frost formation on cold surfaces.
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Condensation

31. Definition: The transition from gas to liquid upon cooling.
32. Molecular Motion: Particles lose kinetic energy, slowing down and moving closer together.
33. Intermolecular Forces: Increased attraction causes gas particles to form a liquid.
34. Exothermic Process: Heat is released during condensation.
35. Example: Water vapor condensing on a cold windowpane.
36. Uniform Temperature: Occurs at the dew point temperature under specific pressure.
37. Role in Nature: Formation of dew and clouds relies on condensation.
38. Industrial Use: Condensers in power plants use this principle.
39. Latent Heat of Condensation: Energy released during the transition.
40. Significance: Condensation helps in water cycle processes like precipitation.
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Brownian Motion

41. Definition: The random movement of particles suspended in a fluid due to collisions with fast-moving molecules.
42. Observation: First observed in pollen grains by Robert Brown.
43. Kinetic Theory Insight: Molecules in liquids and gases are in constant motion, colliding with suspended particles.
44. Evidence: Demonstrates the existence and movement of molecules.
45. Temperature Effect: Higher temperatures increase the motion of molecules, intensifying Brownian motion.
46. Applications: Used in determining particle sizes in colloidal systems.
47. Example: Dust particles moving randomly in a beam of light.
48. Gas Behavior: Supports the idea of molecular motion in gases.
49. Real-Life Application: Helps explain diffusion processes.
50. Significance: Reinforces the kinetic molecular theory.
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Distinguishing Between Solids, Liquids, and Gases

Solids

51. Particle Arrangement: Fixed, closely packed in a regular lattice.
52. Particle Motion: Limited to vibrations about fixed positions due to low kinetic energy.
53. Properties: Definite shape, definite volume, incompressible.
54. Example: Ice, metals, wood.
55. Intermolecular Forces: Very strong, holding particles in fixed positions.
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Liquids

56. Particle Arrangement: Closely packed but not fixed, allowing flow.
57. Particle Motion: Particles slide past one another with moderate kinetic energy.
58. Properties: No definite shape but definite volume; slightly compressible.
59. Example: Water, oil, alcohol.
60. Intermolecular Forces: Weaker than solids but stronger than gases.
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Gases

61. Particle Arrangement: Particles are far apart and in random motion.
62. Particle Motion: Rapid, random motion with high kinetic energy.
63. Properties: No definite shape or volume; highly compressible.
64. Example: Air, oxygen, carbon dioxide.
65. Intermolecular Forces: Extremely weak or negligible.
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Inferences from Molecular Motion

66. Reason for State Changes: Changes in kinetic energy alter the balance between molecular motion and intermolecular forces.
67. Heat Addition: Increases molecular motion, leading to melting, boiling, or evaporation.
68. Heat Removal: Decreases molecular motion, causing freezing or condensation.
69. Brownian Motion Evidence: Shows that particles in a fluid are constantly moving, confirming the kinetic molecular theory.
70. Practical Implication: Understanding molecular motion aids in designing refrigeration systems, industrial heating processes, and climate control mechanisms.
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Jamb(UTME) Key points on The laws of Boyle, Charles, Graham and Dalton (law of partial pressure) etc

Boyle's Law

1. Definition: Boyle's Law states that the pressure (P) of a gas is inversely proportional to its volume (V) at constant temperature.
   Formula: $P_1V_1 = P_2V_2$.
2. Key Concept: As volume decreases, pressure increases if the temperature remains constant.
3. Graphical Representation: A hyperbolic curve when plotting P against V.
4. Applications: Used in syringes, vacuum pumps, and breathing mechanisms.
5. Example: Compressing air in a bicycle pump demonstrates Boyle’s Law.
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Charles’ Law

6. Definition: Charles’ Law states that the volume (V) of a gas is directly proportional to its temperature (T) in Kelvin, at constant pressure.
   Formula: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
7. Key Concept: Gases expand when heated and contract when cooled.
8. Graphical Representation: A straight line when plotting V against T in Kelvin.
9. Applications: Hot air balloons and temperature compensation in gas flow meters.
10. Example: Inflating a balloon in a warm environment causes it to expand.
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Graham’s Law

11. Definition: Graham’s Law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass.
    Formula: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
12. Key Concept: Lighter gases diffuse faster than heavier gases.
13. Diffusion: Spreading of gas particles from high concentration to low concentration.
14. Effusion: Movement of gas through a small hole without collisions between particles.
15. Applications: Separating isotopes, gas leak detection, and air purification.
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Dalton’s Law of Partial Pressure

16. Definition: Dalton’s Law states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas.
    Formula: $P_{\text{total}} = P_1 + P_2 + P_3 + \dots$.
17. Key Concept: Each gas in a mixture behaves independently.
18. Applications: Scuba diving gas mixtures and respiratory physiology.
19. Example: Atmospheric pressure is the sum of partial pressures of nitrogen, oxygen, carbon dioxide, and other gases.
20. Real-Life Use: In gas analysis and medical respiratory equipment.
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Combined Gas Law

21. Definition: The Combined Gas Law combines Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law, showing the relationship between pressure, volume, and temperature.
    Formula: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.
22. Key Concept: Useful when all three variables change simultaneously.
23. Applications: Weather balloons and gas storage systems.
24. Graphical Representation: Complex relationships depending on which variables are held constant.
25. Example: Calculating the volume of gas at varying pressures and temperatures.
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Molar Volume and Atomicity of Gases

26. Molar Volume: At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.
27. Atomicity: Refers to the number of atoms in a molecule of a gas (e.g., O₂ has atomicity of 2, He has atomicity of 1).
28. Importance: Helps calculate the amount of gas using the mole concept.
29. Applications: Used in stoichiometry and gas reaction predictions.
30. Example: 1 mole of CO₂ occupies 22.4 L at STP.
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The Ideal Gas Equation

31. Definition: The Ideal Gas Equation relates pressure (P), volume (V), number of moles (n), temperature (T), and the universal gas constant (R).
    Formula: $PV = nRT$.
32. Gas Constant (R): $R = 0.0821 \, \text{L·atm/(mol·K)}$.
33. Key Assumptions: Gases consist of tiny particles in random motion with negligible volume and no intermolecular forces.
34. Applications: Predicts gas behavior under various conditions.
35. Example: Finding the number of moles of a gas using $n = \frac{PV}{RT}$.
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Relationship Between Vapour Density and Relative Molecular Mass

36. Vapour Density (VD): Defined as the mass of a given volume of gas compared to the mass of the same volume of hydrogen under identical conditions.
    Formula: $\text{VD} = \frac{\text{Molar Mass of Gas}}{\text{2}}$.
37. Relative Molecular Mass (M): $M = 2 \times \text{VD}$.
38. Key Concept: Vapour density helps determine the molecular mass of unknown gases.
39. Applications: Identifying gases in laboratory and industrial settings.
40. Example: A gas with a vapour density of 16 has a molecular mass of $16 \times 2 = 32 \, \text{g/mol}$.
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Advanced Applications and Key Inferences

Practical Applications

41. Boyle's Law: Design of pressure regulators and scuba diving equipment.
42. Charles' Law: Predicting gas volume changes in meteorology.
43. Graham's Law: Isotope separation in nuclear science.
44. Dalton's Law: Calculating oxygen requirements in medical ventilators.
45. Ideal Gas Law: Designing airbags and predicting explosion risks in chemical plants.
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Key Inferences

46. Kinetic Theory Basis: All gas laws are derived from the kinetic theory of gases.
47. STP Conditions: Standardized for consistent gas calculations.
48. Molar Ratios: Essential for chemical reaction stoichiometry involving gases.
49. Gas Behavior Deviations: Real gases deviate under high pressure and low temperature from ideal behavior.
50. Significance of Gas Laws: Fundamental for understanding atmospheric phenomena, industrial processes, and biological respiration.
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Jamb(UTME) Simple calculations based on these laws, equations and relationships

Boyle's Law

1. Q: A gas occupies 5 L at 2 atm. What is its volume at 4 atm?
   A: $P_1V_1 = P_2V_2$. $2 \times 5 = 4 \times V_2$.
   $V_2 = \frac{10}{4} = 2.5L$.
2. Q: If 8 L of a gas at 1 atm is compressed to 4 L, what is the new pressure?
   A: $P_1V_1 = P_2V_2$. $1 \times 8 = P_2 \times 4$.
   $P_2 = \frac{8}{4} = 2 atm$.
3. Q: A gas at 0.5 atm occupies 12 L. What volume will it occupy at 3 atm?
   A: $P_1V_1 = P_2V_2$. $0.5 \times 12 = 3 \times V_2$.
   $V_2 = \frac{6}{3} = 2L$.
4. Q: The volume of a gas decreases from 10 L to 2 L. If the initial pressure was 1 atm, what is the final pressure?
   A: $P_1V_1 = P_2V_2$. $1 \times 10 = P_2 \times 2$.
   $P_2 = \frac{10}{2} = 5atm$.
5. Q: A gas has a pressure of 3 atm and volume of 6 L. What is its pressure when the volume is 9 L?
   A: $P_1V_1 = P_2V_2$. $3 \times 6 = P_2 \times 9$.
   $P_2 = \frac{18}{9} = 2atm$.

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Charles' Law

6. Q: A gas occupies 3 L at 300 K. What is its volume at 600 K?
   A: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. $\frac{3}{300} = \frac{V_2}{600}$.
   $V_2 = \frac{3 \times 600}{300} = 6L$.
7. Q: A gas at 200 K occupies 4 L. What is its volume at 400 K?
   A: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. $\frac{4}{200} = \frac{V_2}{400}$.
   $V_2 = \frac{4 \times 400}{200} = 8L$.
8. Q: A gas occupies 10 L at 273 K. What is its volume at 546 K?
   A: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. $\frac{10}{273} = \frac{V_2}{546}$.
   $V_2 = \frac{10 \times 546}{273} = 20L$.
9. Q: A gas at 400 K has a volume of 8 L. What is its volume at 300 K?
   A: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. $\frac{8}{400} = \frac{V_2}{300}$.
   $V_2 = \frac{8 \times 300}{400} = 6L$.
10. Q: A gas at 150 K occupies 2 L. What is its volume at 450 K?
    A: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. $\frac{2}{150} = \frac{V_2}{450}$.
    $V_2 = \frac{2 \times 450}{150} = 6L$.

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Graham's Law

11. Q: Find the relative rate of diffusion of oxygen (O₂) and hydrogen (H₂).
    A: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
    $\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{32}{2}} = 4$.
    Hydrogen diffuses 4 times faster than oxygen.
12. Q: Compare the rates of diffusion of CO₂ and He.
    A: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
    $\frac{r_He}{r_{CO_2}} = \sqrt{\frac{44}{4}} = 3.32$.
    Helium diffuses 3.32 times faster than carbon dioxide.
13. Q: Calculate the relative rate of effusion of nitrogen (N₂) and chlorine (Cl₂).
    A: $\frac{r_{N_2}}{r_{Cl_2}} = \sqrt{\frac{71}{28}} = 1.59$.
    Nitrogen effuses 1.59 times faster than chlorine.
14. Q: If a gas diffuses 9 times faster than another gas, find the ratio of their molar masses.
    A: $\frac{r_1}{r_2} = 9 = \sqrt{\frac{M_2}{M_1}}$.
    $M_2/M_1 = 81$.
15. Q: Determine the diffusion rate ratio of NH₃ and HCl.
    A: $\frac{r_{NH_3}}{r_{HCl}} = \sqrt{\frac{36.5}{17}} = 1.46$.
    Ammonia diffuses 1.46 times faster than hydrogen chloride.

Dalton’s Law

16. Q: A mixture contains O₂ (3 atm) and N₂ (2 atm). Find the total pressure.
    A: $P_{total} = P_{O_2} + P_{N_2}$.
    $P_{total} = 3 + 2 = 5atm$.
17. Q: A gas mixture has H₂ (1 atm), O₂ (2 atm), and N₂ (3 atm). Find the total pressure.
    A: $P_{total} = 1 + 2 + 3 = 6atm$.
18. Q: Find the partial pressure of CO₂ if the total pressure is 4 atm and it constitutes 25% of the mixture.
    A: $P_{CO_2} = 0.25 \times 4 = 1atm$.
19. Q: A gas mixture has a total pressure of 10 atm. O₂ constitutes 40%. Find its partial pressure.
    A: $P_{O_2} = 0.4 \times 10 = 4atm$.
20. Q: If the partial pressures of two gases are 2 atm and 3 atm, what is the total pressure?
    A: $P_{total} = 2 + 3 = 5atm$.

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Combined Gas Law

21. Q: A gas at 1 atm, 10 L, and 300 K changes to 2 atm and 500 K. Find its volume.
    A: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.
    $\frac{1 \times 10}{300} = \frac{2 \times V_2}{500}$.
    $V_2 = 8.33L$.
22. Q: A gas at 5 atm, 2 L, and 400 K changes to 1 atm and 200 K. Find its volume.
    A: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.
    $\frac{5 \times 2}{400} = \frac{1 \times V_2}{200}$.
    $V_2 = 5L$.

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