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Gases. Chapter 5 Problems: 14, 19, 20, 21, 26, 32, 34, 36, 40, 42, 43, 44, 47, 48, 50, 52, 54, 59, 60, 63, 66, 67, 76, 84, 88, 102, Graham’s law problems. Elements that exist as gases at 25 0 C and 1 atmosphere. 5.1. 5.1. Physical Characteristics of Gases.

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  1. Gases Chapter 5 Problems: 14, 19, 20, 21, 26, 32, 34, 36, 40, 42, 43, 44, 47, 48, 50, 52, 54, 59, 60, 63, 66, 67, 76, 84, 88, 102,Graham’s lawproblems

  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 = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mm Hg = 760 torr = 101,325 Pa = 14.7 psi = 29.92 in. Hg 5.2

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

  7. Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P1V1 = P2 V2 Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

  8. Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V1 V2 = T1 T2 • If one temperature goes up, the volume goes up! Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  9. Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P1 P2 = T1 T2 If one temperature goes up, the pressure goes up! Joseph Louis Gay-Lussac (1778-1850)

  10. Combined Gas Law • The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P1 V1 P2 V2 = T1 T2 No, it’s not related to R2D2

  11. And now, we pause for this commercial message from STP • 1 mol of any gas at STP has a volume of 22.4 L Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

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

  13. 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

  14. 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

  15. 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 Solve for n, which = mass/molar mass Or, M = d is the density of the gas in g/L 5.4

  16. 54.6 g/mol M = L•atm mol•K A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas? PV = nRT 2.10 L X 1 atm n = 0.0821 x 300.15 K n = 0.0852 mol g n = M 5.3

  17. 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

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

  19. 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

  20. 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

  21. 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2 2 Bottle full of oxygen gas and water vapor 5.6

  22. 5.6

  23. P V Chemistry in Action: Scuba Diving and the Gas Laws 5.6

  24. 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 among molecules are perfectly elastic. • 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

  25. Kinetic theory of gases and … • Compressibility of Gases • Boyle’s Law • Pa collision rate with wall • Collision rate a number density • Number density a 1/V • Pa 1/V • Charles’ Law • Pa collision rate with wall • Collision rate a average kinetic energy of gas molecules • Average kinetic energy aT • PaT 5.7

  26. Kinetic theory of gases and … • Avogadro’s Law • Pa collision rate with wall • Collision rate a number density • Number density an • Pan • Dalton’s Law of Partial Pressures • Molecules do not attract or repel one another • P exerted by one type of molecule is unaffected by the presence of another gas • Ptotal = SPi 5.7

  27. n = = 1.0 PV RT Deviations from Ideal Behavior 1 mole of ideal gas Repulsive Forces PV = nRT Attractive Forces 5.8

  28. Effect of intermolecular forces on the pressure exerted by a gas. 5.8

  29. ( ) P + (V – nb) = nRT } } corrected pressure corrected volume an2 V2 Van der Waals equation nonideal gas 5.8

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

  31. 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

  32. Gas Diffusionrelation of mass to rate of diffusion • HCl and NH3 diffuse from opposite ends of tube. • Gases meet to form NH4Cl • HCl heavier than NH3 • Therefore, NH4Cl forms closer to HCl end of tube.

  33. diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container. GAS DIFFUSION AND EFFUSION

  34. GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv2 Rate of effusion is inversely proportional to its molar mass. Thomas Graham, 1805-1869. Professor in Glasgow and London.

  35. GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is • proportional to T • inversely proportional to M. Therefore, He effuses more rapidly than O2 at same T. He

  36. Graham’s Law Problem 1 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 degrees celsius according to the equation: 4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g) Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?

  37. Graham’s Law Problem 2 What is the rate of effusion for H2 if 15.00 ml of CO2 takes 4.55 sec to effuse out of a container?

  38. Graham’s Law Problem 3 What is the molar mass of gas X if it effuses 0.876 times as rapidly as N2(g)?

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