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Chapter 10 Gases

Lecture Presentation. Chapter 10 Gases. Unlike liquids and solids, gases Expand to fill their containers. Are highly compressible. Have extremely low densities. Characteristics of Gases. F A. P =. Pressure. Pressure is the amount of force applied to an area:.

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Chapter 10 Gases

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  1. Lecture Presentation Chapter 10Gases

  2. Unlike liquids and solids, gases Expand to fill their containers. Are highly compressible. Have extremely low densities. Characteristics of Gases

  3. F A P = Pressure • Pressure is the amount of force applied to an area:

  4. pascal(Pa); kilopascal(kPa) 1.01325x105Pa; 101.325 kPa SI unit; physics, chemistry atmosphere(atm) 1 atm* chemistry millimeters of mercury(Hg) 760 mm Hg* chemistry, medicine, biology torr 760 torr* chemistry pounds per square inch (psi or lb/in2) 14.7lb/in2 engineering bar 1.01325 bar meteorology, chemistry, physics Common Units of Pressure Unit Atmospheric Pressure Scientific Field *This is an exact quantity; in calculations, we use as many significant figures as necessary.

  5. PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Dh = 291.4 mm Hg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. 1torr 1 mmHg 1 atm 760 torr 101.325 kPa 1 atm Sample Problem 5.1 Converting Units of Pressure SOLUTION: 291.4 mmHg = 291.4 torr 291.4 torr = 0.3834 atm 0.3834 atm = 38.85 kPa

  6. Normal atmospheric pressure at sea level is referred to as standard pressure. Standard Pressure • It is equal to • 1.00 atm • 760 torr (760 mmHg) • 101.325 kPa

  7. The Gas Laws • Gas experiments revealed that four variables affect the state of a gas: • Temperature, T • Volume, V • Pressure, P • Quantity of gas present, n (moles) • These variables are related through equations know as the gas laws.

  8. The volume occupied by a gas is inversely related to its pressure Boyle’s Law 1662 Boyle’s Law: P1 x V1 = P2 x V2

  9. 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? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg 5.3

  10. Charles’ Law 1787 Charles’s Law: • At constant pressure, the volume occupied by a fixed amount of gas directly proportional to its absolute temperature • First conceived by Guillaume Amontons in 1702 and published by Joseph Gay-Lussac 1802. V1 /T1 = V2 /T2

  11. 1.54 L x 398.15 K V2 x T1 = 3.20 L V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K T2 = = 192 K 5.3

  12. Avogadro’s Law Avogadro’s Law: • At fixed temperature and pressure, equal volumes of any ideal gas contain equal number of particles (or moles) V1 /n1= V2 /n2

  13. Standard molar volume. Figure 5.7

  14. V 1/P (Boyle’s law) VT (Charles’s law) Vn (Avogadro’s law) nT P V Ideal-Gas Equation • So far we’ve seen that • Combining these, we get

  15. Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

  16. The relationship nT P V nT P V= R Ideal-Gas Equation then becomes or PV = nRT

  17. We can manipulate the density equation to enable us to find the molecular mass of a gas: P RT d = dRT P  = Molecular Mass Becomes  molar mass of a gaseous substance

  18. d = = m M = V 4.65 g 2.10 L g 2.21 L x 0.0821 x 300.15 K 54.6 g/mol M = 1 atm g = 2.21 L dRT L•atm mol•K P A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas? M = 5.4

  19. 1)What is the volume (in liters) occupied by 49.8 g of HCl at STP? 2) 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)? 5.4

  20. 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

  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. Formulated in 1801 The total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it was alone. Dalton’s Law of Partial Pressures

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

  24. PA = nART nBRT V V PB = nA nB nA + nB nA + nB XB = XA = ni mole fraction (Xi) = nT 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

  25. 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

  26. In Groups

  27. What is the volume of CO2 produced at 37 0C 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

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