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Gas Laws: Physical Characteristics and Mathematical Relationships

This chapter focuses on the physical characteristics of gases, such as their compressibility and ability to mix evenly. It also explores the mathematical relationships between pressure, volume, temperature, and moles of gases, as described by Boyle's Law, Charles' Law, and Avogadro's Law. The chapter includes examples and practice exercises to reinforce these concepts.

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Gas Laws: Physical Characteristics and Mathematical Relationships

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  1. Chapter 5 Gases

  2. Elements that exist as gases at 250C and 1 atmosphere 5.1

  3. 5.1

  4. Physical Characteristics of Gases • 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

  5. Force Area Barometer Pressure = Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 5.2

  6. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm 5.2

  7. h = pressure difference 5.2

  8. V decreases As P (h) increases 5.3

  9. Constant temperature Constant amount of gas Boyle’s Law Pa 1/V P x V = constant P1 x V1 = P2 x V2 5.3

  10. Boyle’s Law : P - V Relationship Pressure is inversely proportional to volume • Or pressure is directly proportional to 1/V • P = or V = or PV=k • If n (number of moles) and T are constant • P1V1 = k P2V2 = k • Then : P1V1 = P2V2 k V k P

  11. 726 mmHg x 946 mL P1 x V1 = 154 mL V2 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg 5.3

  12. T1 V1 T2 V2 = = V1 T1 V2 T2 Charles Law - V - T- Relationship • Volume proportional to T (in K) : V = kT • n and P must be constant • V/T = k or or Temperatures must be expressed in Kelvin .

  13. Charles Law Problem • A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. • V1 = 3.20 L T1 = 125oC = 398 K • V2 = 1.54 L T2 = ? • T2 = 192 K oC = K - 273.15 = 192 - 273 oC = -81oC

  14. Constant temperature Constant pressure Avogadro’s Law Va number of moles (n) V = constant x n Constant same for all gases V1/n1 = V2/n2 5.3

  15. 4NH3 + 5O2 4NO + 6H2O 1 volume NH3 1 volume NO 1 mole NH3 1 mole NO Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of NO are obtained from 10 liters of ammonia at the same temperature and pressure? How many liters of oxygen are required? At constant T and P 5.3

  16. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT 5.4

  17. The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) PV = nRT R = 0.082057 L • atm / (mol • K) 5.4

  18. Problems for Chapter 5 Pages 163-156 1-5, 8, 13-16, 18, 20, 22, 24-30, 32, 34, 36, 38, 40, 42, 44, 48, 50, 52, 60, 62, 65, 66, 72, 76, 80

  19. 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 x 273.15 K V = 1 atm nRT L•atm P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT V = 30.6 L 5.4

  20. What is the pressure exerted by 168 g of argon in a 38.0 L container at 27oC?

  21. P1 = 1.20 atm P2 = ? T1 = 291 K T2 = 358 K nR P = = T V P1 P2 T2 358 K T1 T1 T2 291 K = 1.20 atm x P2 = P1 x Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? n, V and Rare constant PV = nRT = constant = 1.48 atm 5.4

  22. Practice Exercise 5.4 A gas initially at 4.0 L, 1.2 atm, and 66oC undergoes a change so that its temperature and pressure become 42oC and 1.7 atm. What is its final volume?

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

  24. Example 5.5 A chemist has synthesized a greenish-yellow compound of Cl and O and finds its density to be 7.71g/L at 36oC and 2.88 atm. Calculate the molar mass and determine its molecular formula. A 2.10 L vessel contains 4.65 g of gas at 1.00 atm and 27.0oC. What is its density and molar mass?

  25. What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) 6 mol CO2 g C6H12O6 mol C6H12O6 mol CO2V CO2 x 1 mol C6H12O6 1 mol C6H12O6 x 180 g C6H12O6 L•atm mol•K nRT 0.187 mol x 0.0821 x 310.15 K = P 1.00 atm Gas Stoichiometry 5.60 g C6H12O6 = 0.187 mol CO2 V = = 4.76 L 5.5

  26. Ideal Gas Law and Reaction Stoichiometry • Sodium azide (NaN3) is used in some air bags in automobiles. Calculate the volume of nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 . • 2 NaN3 (s) 2 Na (s) + 3 N2 (g)

  27. Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal= P1 + P2 5.6

  28. Dalton’s Law of Partial Pressures • Definition: In a mixture of gases, each gas contributes to the total pressure the amount it would exert if the gas were present in the container by itself. This is called the partialpressure of the gas. • To obtain a total pressure, add all of the partial pressures: Ptotal = p1+p2+p3+...

  29. Dalton’s Law of Partial Pressures • A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of helium at a temperature of 17.0 oC. • What are the partial pressures of each gas, and the total pressure? • T = 17 oC + 273 = 290 K • nCO2 = 3.00 g CO2 x 1 mole CO2 / 44.01 g CO2 • = 0.0682 mole CO2 • pCO2 = nCO2RT/V • pCO2 = • pCO2 = 0.812 atm ( 0.0682 mol CO2) ( 0.08206 L atm/mol K) ( 290 K) (2.00 L)

  30. Dalton’s Law Problem - cont. • nHe = 0.10 g He / 1 mole He / 4.003 g He • = 0.025 mole He • pHe = nHeRT/V • pHe = • pHe = 0.30 atm • PTotal = pCO2 + pHe = 0.812 atm + 0.30 atm • PTotal = 1.11 atm (0.025 mol) ( 0.08206 L atm / mol K) ( 290 K ) ( 2.00 L )

  31. PA = nART nBRT V V PB = nA nB nA + nB nA + nB XA = XB = Consider a case in which two gases, A and B, are in a container of volume V. nA is the number of moles of A nB is the number of moles of B PT = PA + PB PA = XAPT PB = XBPT Pi = XiPT 5.6

  32. 0.116 8.24 + 0.421 + 0.116 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = XiPT PT = 1.37 atm = 0.0132 Xpropane = Ppropane = 0.0132 x 1.37 atm = 0.0181 atm 5.6

  33. Dalton’s Law is used for calculations when a gas is collected over water.Whenever water (or any liquid) is placed in a closed container some water will evaporate and reach a state of equilibrium where the partial pressure of the water remains constant. This pressure is called the equilibrium vapor pressure of the liquid. It depends only on the temperature.

  34. 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2 2 Bottle full of oxygen gas and water vapor 5.6

  35. Relative Humidity pressure of water in air • Rel Hum = x 100% • Example : the partial pressure of water at 15oC is 6.54 mm Hg, what is the relative humidity? • Rel Hum =(6.54 mm Hg/ 12.79 mm Hg )x100% = 51.1 % What if the temperature is increased to 25oC? maximum vapor pressure of water

  36. Collection of Hydrogen Gas over Water - Vapor Pressure • 2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g) • Calculate the mass of hydrogen gas collected over water if 156 ml of gas is collected at 20oC and 769 mm Hg. • PTotal = p H2 + pH2O pH2 = PTotal - pH2O pH2 = 769 mm Hg - 17.4 mm Hg = 752 mm Hg = 0.987 atm • T = 20oC + 273 = 293 K • V = 0.156 L

  37. Kinetic Molecular Theory of Gases • A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. • Gas molecules are in constant motion in random directions. Collisions are perfectly elastic. Pressure is caused by collisions of molecules with the walls of the container. • Gas molecules exert neither attractive nor repulsive forces on one another. • The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 5.7

  38. Kinetic theory of gases explains the gas laws and properties of gases • Compressibility of Gases • Boyle’s Law - Why does a decrease in volume cause an increase in pressure? • Charles’ Law – Why does an increase in temperature cause an increase in pressure? 5.7

  39. Velocity and Energy • Kinetic Energy = 1/2mu2 • All gas molecules have the same average kinetic energy at the same temperature. • Therefore heavier molecules must have slower average speeds. • Consider samples of N2 and He gas at the same temperature. Which has the higher average kinetic energy? Which has the higher average speed?

  40. Molecular Mass and Molecular Speeds Problem: Calculate the molecular speeds of the molecules of hydrogen, methane, and carbon dioxide at 300K! Compare speeds and kinetic energies. Next EH assignment due Mar 12 Section 7 – Properties of Gases

  41. Molecular Mass and Molecular Speeds - III Molecule H2 CH4 CO2 Molecular Mass (g/mol) 2.016 16.04 44.01 Kinetic Energy (J/molecule) 6.213 x 10 - 21 6.0213 x 10 - 21 6.213 x 10 - 21 Velocity (m/s) 1,926 683.8 412.4

  42. Diffusion vs. Effusion • Diffusion - One gas mixing into another gas, or gases, of which the molecules are colliding with each other, and exchanging energy between molecules. • Effusion - A gas escaping from a container into a vacuum. • Graham’s Law

  43. NH3 (g) + HCl(g) = NH4Cl (s) • HCl = 36.46 g/mol NH3 = 17.03 g/mol • RateNH3/ RateHCl = 1.463

  44. NH4Cl Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH3 17 g/mol HCl 36 g/mol 5.7

  45. Apparatus for studying molecular speed distribution 5.7

  46. The distribution of speeds of three different gases at the same temperature 3RT urms = M The distribution of speeds for nitrogen gas molecules at three different temperatures 5.7

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