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Gases

Gases. Chapter 10. Elements that exist as gases 25 0 C and 1 atmosphere. 10.1. 10.1. Physical Characteristics of Gases. Gases assume the volume and shape of their containers. Gases are the most compressible state of matter.

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Gases

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  1. Gases Chapter 10

  2. Elements that exist as gases 250C and 1 atmosphere 10.1

  3. 10.1

  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. 10.1

  5. Force Area Barometer Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 bar = 105 Pascals 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 1 atm = 101.325 kPa 10.2

  6. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm 10.2

  7. Manometers Used to Measure Gas Pressures 10.2

  8. Apparatus for Studying the Relationship Between Pressure and Volume of a Gas V decreases As P (h) increases 10.3

  9. Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. 10.3

  10. Constant temperature Constant amount of gas Boyle’s Law Pa 1/V P x V = constant P1 x V1 = P2 x V2 10.3

  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? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg 10.3

  12. As T increases V increases 10.3

  13. Temperature must be in Kelvin Variation of gas volume with temperature at constant pressure. Charles’ & Gay-Lussac’s Law VaT V = constant x T T (K) = t (0C) + 273.15 V1/T1 = V2 /T2 10.3

  14. V T = k • i.e., Charles’s Law • The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. A plot of V versus T will be a straight line. 10.4

  15. 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 10.3

  16. V = kn • Mathematically, this means Avogadro’s Law • The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. 10.3

  17. Constant temperature Constant pressure Avogadro’s Law Va number of moles (n) V = constant x n V1 / n1 = V2 / n2 10.3

  18. 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 volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? At constant T and P 10.3

  19. Final and Initial State Problems • Use Combined Gas Law • List all variables (P,V,n,T) in two columns… one for initial state, one for final state. • Substitute and solve ( make sure units cancel)

  20. Final and Initial State Problems A sample of gas occupies 355 ml and 15oC and 755 torr of pressure. What temperature will the gas have at the same pressure if its volume increases to 453 ml?

  21. Final and Initial State Problems A sample of gas at 30.0o has a pressure of 1.23 bar and a volume of 0.0247 m3 . If the volume of the gas is compressed to 0.00839 m3 at the same temperature, what is its pressure at this volume?

  22. Final and Initial State Problems • A Sample of gas at 27.0 oC has a volume of 2.08 L and a pressure of 750.0 mm Hg. If the gas is in a sealed container, what is its pressure in bar when the temperature (in oC) doubles?

  23. Combining these, we get nT P V Ideal-Gas Equation • So far we’ve seen that V 1/P (Boyle’s law) VT (Charles’s law) Vn (Avogadro’s law) 10.3

  24. 10.3

  25. 10.3

  26. 10.3

  27. nT P nT P V V= R Ideal-Gas Equation The relationship then becomes or PV = nRT 10.4

  28. Ideal-Gas Equation The constant of proportionality is known as R, the ideal gas constant. 10.4

  29. 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 10.4

  30. 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) 10.4

  31. 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 10.4

  32. 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 10.4

  33. Single State Problems • Use Ideal gas Law – you’re given three for the four variables and R - solve for the unknown. • PV = nRT

  34. Single State Problems • How many grams of oxygen gas in a 10.0 L container will exert a pressure of 712 torr at a temperature of 25.0 o C?

  35. Single State Problems • At what temperature will 26.42 grams of sulfur dioxide in a 250.0 ml container exert a pressure of 0.842 atm?

  36. Other calculations involving the Ideal Gas Law • We can often determine the molar mass of a gas by measuring its mass at a given volume, temperature and pressure. • Ideal Gas Law can also be used to determine the density of a gas at given conditions of temperature and pressure.

  37. n V P RT = Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get 10.4

  38. m V P RT = Densities of Gases • We know that • moles  molecular mass = mass n  = m • So multiplying both sides by the molecular mass ( ) gives 10.4

  39. m V P RT d = = Densities of Gases • Mass  volume = density • So, Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas. 10.4

  40. P RT dRT P d =  = Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes 10.4

  41. 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 = 10.4

  42. 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 = 10.4

  43. When 4.93 grams of carbon tetrachloride gas are in a 1.oo liter container at 400.0 K, the gas exerts a pressure of 800.0 torr. What is the molar mass of carbon tetrachloride?

  44. A gaseous compound with a mass of 3.216 grams has a volume of 2236 mL at 27.0oC and 735 mm Hg. Calculate the molar mass of the compound.

  45. Calculate the density of ammonia at 0oC and 1 atm of pressure.

  46. Calculate the molar mass of a compound with a density of 0.630 g/L at 25.0oC and 730.0 torr pressure.

  47. 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 10.5

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

  49. 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

  50. 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

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