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Understanding Gases: Kinetic Energy and Molecular Behavior in Chemistry

Chapter 13 delves into the fascinating world of gases, highlighting their unique properties influenced by kinetic energy and attractive forces. As temperature increases, gas molecules move more rapidly, demonstrating higher kinetic energy and lower attractive forces. This chapter explores how these factors affect the behavior of gases, emphasizing the relationship between physical states of matter and molecular dynamics. Understanding these principles is crucial for further studies in chemistry and encompasses both theoretical concepts and practical applications.

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Understanding Gases: Kinetic Energy and Molecular Behavior in Chemistry

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  1. Chemistry B2A Chapter 13 Gases
  2. Gases move faster Kinetic energy ↑ T ↑
  3. Gases Physical sate of matter depends on: Attractive forces Kinetic energy Brings molecules together Keeps molecules apart
  4. Gases Gas High kinetic energy (move fast) Low attractive forces Liquid Medium kinetic energy (move slow) medium attractive forces Solid Low kinetic energy (move slower) High attractive forces
  5. Physical Changes Melting Boiling Change of states
  6. Ideal Gases Kinetic molecular theory: Particles move in straight lines, randomly. Average Kinetic energy of particles depends on temperature. Particles collide and change direction (they may exchange kinetic energies). Their collisions with walls cause the pressure. Gas particles have no volume. No attractive forces (or repulsion) between gas particles. More collision = greater pressure. In reality, no gas is ideal (all gases are real). At low pressure (around 1 atm or lower) and at 0°C or higher, we can consider real gases as ideal gases.
  7. Force (F) Pressure (P) = Area (A) Pressure (P) F: constant A: constant P ↑ P ↑ A ↓ F ↑ Atmosphere (atm) Millimeters of mercury (mm Hg) torr in. Hg Pascal 1.000 atm = 760.0 mm Hg = 760.0 torr = 101,325 pascals = 29.92 in. Hg
  8. Pressure (P) At STP: Standard Temperature & Pressure 1 standard atmosphere = 1.000 atm = 760.0 mm Hg = 760.0 torr = 101,325 pa Pounds per square inch (psi) 1.000 atm = 14.69 psi
  9. Pressure (P) Hg barometer atmospheric pressure manometer pressure of gas in a container
  10. Boyle’s Law m,T: constant Boyle’s law: P 1/αV PV = k (a constant) P1V1 = k (a constant) P2V2 = k (a constant) P1V1 = P2V2 P1V1 P1V1 P2 = V2 = V2 P2
  11. Boyle’s Law
  12. V1 =k (a constant) T1 V2 =k (a constant) T2 V2 V1 = T2 T1 Charles’s Law m,P: constant Charles’s law: V T αV = k (a constant) T V1T2 T1V2 V2 = T2 = T1 V1
  13. Charles’s Law
  14. P1 =k (a constant) T1 P2 =k (a constant) T2 P2 P1 = T2 T1 Gay-Lussac’s Law m,V: constant Gay-Lussac’s law: P P αT = k (a constant) T P1T2 T1P2 P2 = T2 = T1 P1
  15. Gay-Lussac’s Law Pressure (P)
  16. Combined Gas Law Combined gas law: m (or n): constant P1V1 P2V2 PV = = k (a constant) T T1 T2
  17. V1 =k (a constant) n1 V2 =k (a constant) n2 V2 V1 = n2 n1 Avogadro’s Law P,T: constant Avogadro’s law: V V αn = k (a constant) n
  18. Ideal Gas Law Ideal gas law: n: number of moles (mol) R: universal gas constant V: volume (L) P: pressure (atm) T: temperature (K) PV = nRT Standard Temperature and Pressure (STP) T = 0.00°C (273 K) P = 1.000 atm 1 mole → V = 22.4 L (1.000 atm) (22.4 L) PV L.atm R = = = 0.0821 nT mol.K (1 mol) (273 K)
  19. Dalton’s Law Dalton’slaw of partial pressure: PT = P1 + P2 + P3 + …
  20. RT PT= ntotal ( ) V Dalton’s Law P2 P3 P1 n1RT n2RT n3RT P1= P2= P3= V V V PT = P1 + P2 + P3 PT = n1(RT/V) + n2(RT/V) + n3(RT/V) PT = (n1 + n2+ n3) (RT/V)
  21. Dalton’s Law For a mixture of ideal gases, the total number of moles is important. (not the identity of the individual gas particles) 0.75 mol H2 1.00 mol N2 0.75 mol He 0.50 mol O2 V=5L T=20C V=5L T=20C V=5L T=20C 1.75 mol He 0.25 mol Ne 0.25 mol Ar 1.75 mol 1.75 mol PT = 8.4 atm PT = 8.4 atm PT = 8.4 atm Volume of the individual gas particles must not be very important. Forces among the particles must not be very important.
  22. Gas Stoichiometry Only at STP: 1 mole gas = 22.4 L Ex 13.15 (page:433) 2KClO3(s) +  2KCl(s) + 3O2(g) B A 10.5 g KClO3 = ? Volume O2 P = 1.00 atm T = 25.0°C 3 mole O2 1 mole KClO3 ) = 0.128 mole O2 )( 10.5 g KClO3 ( 122.6 g KClO3 2 mole KClO3 1.00V = 0.128  0.0821  298 PV = nRT V = 3.13 L T = 25.0°C + 273 = 298 K
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