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Characteristics of Gases

Passing Gas. Characteristics of Gases. Gases expand to fill a container. Gases form homogeneous mixtures with other gases. Gases can be easily compressed. Readily measured properties of gas are its temperature, volume, and pressure. Kinetic Molecular Theory.

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Characteristics of Gases

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  1. Passing Gas Characteristics of Gases • Gases expand to fill a container • Gases form homogeneous mixtures with other gases • Gases can be easily compressed • Readily measured properties of gas are its temperature, • volume, and pressure

  2. Kinetic Molecular Theory 1. Gases consist if large number of molecules that are in continuous, random motion 2. The volume of all molecules of the gas is negligible compared to the total volume in which the gas is contained. 3. Attractive and repulsive forces between gas molecules are negligible 4. Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature is constant. In other words collisions are perfectly elastic. 5. The average kinetic energy of the molecules is proportional to absolute temperature. At any given temperature all gases have the same average kinetic energy

  3. F F P = P = A A Boy-Am I Under Pressure Pressure may be defined as a force exerted upon any surface: Gravitational force Mass = 1.06 kg Area = 7.85 x 10-3 m2 Since F= mass x acceleration F = 1.06 kg x 9.81 m/sec2 = 10.4 N where N is an SI unit called a Newton with units of kg-m/s2 10.4 N = = 1.32 x 103 N/m2 or pascals (Pa) 7.85 x 10-3 m2

  4. Atmospheric Pressure P=F/A = (10,000 kg)(9.81 m/s2)/1 m2 = 1 x 105 Pa = 101.3 kPa

  5. Atmospheric Pressure Atmospheric pressure can be measured using a barometer 760 mm Hg 760 torr 101.3 kPa 1 atm

  6. Atmospheric Pressure A manometer measures the pressures of encloses gases For a given pressure difference, the difference in heights of the liquid levels in the two arms of the manometer is inversely proportional to the density of the liquid An open-tube manometer measures pressures near atmospheric pressure A closed-tube manometer measures pressures below atmospheric pressure

  7. Atmospheric Pressure A manometer measures the pressures of encloses gases A manometer is filled with dibutylphthalate (D = 1.05 g/ml) If the conditions are such that h = 12.2 cm at 0.964 atm, what is the pressure of the enclosed gas in mm Hg (D =13.6g/ml). 760 mm Hg Patm = 0.964 atm x 733 mm Hg = 1 atm 10 mm 1 ml Ph = 12.2 cm x 1.05 g x x = 9.4 mm Hg 13.6 g 1 cm ml Because the pressure inside the manometer is less than atmospheric pressure, Pgas = P atm + Ph2 Pgas = 733 + 9.4 = 742 mm Hg

  8. Boyle's Law The volume of a fixed quantity of gas maintained at constant temperature is inversely proportional to the pressure

  9. Boyle's Law Boyle’s Law: V  1/P, where (n, T constant) or PV = k

  10. Charles’s Law: V  T, where (n, P constant) or V/T = k Charles' Law The volume of a fixed quantity of gas maintained at constant pressure is inversely proportional to the temperature

  11. Gay-Lussac’s Law • P1 = P2 volume and moles • T1 T2 are constant (n, V)

  12. Examples The pressure of nitrogen gas in a 12.0 L tank at 27°C is 2300 lb/in2. What volume would the gas in the tank have at 1 atm pressure (14.7 lb/in2) if the temperature remains unchanged? Boyle’s Law: V  1/P, where (n, T constant) or PV = k P1V1 = P2V2 (2300 lb/in2)(12.0 l) = (14.7 lb/in2)(V2) V2 = 1880 L The gas pressure in an aerosol can is 1.5 atm at 25° C. Assuming that the gas inside obeys the ideal-gas equation what would be the pressure if the can were heated to 450°C? Gay-Lussac’s Law: P  T, where (n,V constant) or P/T = k P1 P2 = T1 T2 1.5 atm = P2 = 298 K = 723 K P2 = 3.6 atm

  13. Combined Gas Law Boyle’s Law: V  1/P, where (n, T constant) or PV = k P1V1 = P2V2 Charles’s Law: V  T, where (n, P constant) or V/T = k V1 V2 = T1 T2 Gay-Lussac’s Law: P  T, where (n,V constant) or P/T = k P1 P2 = T1 T2 the Combined gas Law is: P1V1 P2V2 = T1 T2

  14. Combined Gas law A quantity of helium gas occupies a volume of 16.5 L at 78°C and 45.6 atm. What is its volume at STP? The Combined gas Law is: P1V1 P2V2 = T1 T2 (45.6 atm)(16.5L) (1 atm)(V2) = 351 K 273 K V2 = 585 L

  15. Avogadro’s Law: V  n, where (R,T constant) Avogadro's Law H2 Ar N2 Volume 1L Pressure 1 atm Temperature 0°C Mass of gas 1.783 g # of particles 2.66 x 1022 Volume 1L Pressure 1 atm Temperature 0°C Mass of gas 1.250 g # of particles 2.66 x 1022 Volume 1L Pressure 1 atm Temperature 0°C Mass of gas 0.0899 g # of particles 2.66 x 1022 The volume of a fixed quantity of gas maintained at constant pressure and temperature is directly proportional to the number of moles of gas

  16. Charles’s Law: V  T, where (n, P constant) or V/T = k Boyle’s Law: V  1/P, where (n, T constant) or PV = k Avogadro’s Law: V  n, where (R,T constant) or V = kn Ideal Gas Law Since these Are True then: R = 0.0821 L-atm/K mol = 62.36 L mmHg/K mol = 62.36 L torr/K mol = 8.31 L kPa/K mol = 8.31 J/K mol nT V  P nT V = R P PV = nRT

  17. Example • What volume would 3.5g of Methane occupy at STP?

  18. Ideal Gas Law Molar Mass and Gas Densities • The ideal gas law may be used to determine the density or the molecular • mass of a gas P n = Since PV = nRT, then: V RT Let M = the molar mass which is grams per one mole of a substance. n M P M = RT V If we multiply both sides of the expression by M, then the top left-hand side of the expression becomes: or g/L which is density (d) grams moles x mole liter dRT P M or = d = M P RT

  19. Ideal Gas Law Determine the molar mass of a gas if 2.889 g of gas completely fills a 936 ml container. 2.889g (0.0821 l-atm/mol-K) (304K) .936 L dRT = M P 1 atm 735 760 = 79 7 g/mol M

  20. Hey CP!!!! • If you think that this is hard, don’t worry it’s all uphill from here!! =/ • Give up on life now, it’s just not worth it!!!!!!!!!! • Don’t worry there are a few survivors, but you probably won’t be one of them! • Sincerely your AP Chemistry Friends!! • P.S. The Suicide Hotline number is 602-248-8336 (just in case you need it)

  21. Ideal Gas Law Volumes of Gases in Chemical Reactions The industrial synthesis of nitric acid involves the reaction of nitrogen dioxide gas with water: 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g) 1. How grams of nitric acid can be prepared using 540 L of NO2 at a pressure of 5.00 atm and a temperature of 295 K? 2 mol HNO3 46g PV (5.00 atm)( 540 liters) nNO2 = = 3418 g HNO3 = RT 3 mol NO2 mol (0.0821 l-atm/mol-K)(295 K) STOICHIOMETRY

  22. Ideal Gas Law Gas Mixtures and Partial pressure • The ideal gas law may be used to determine the partial pressure of a mixture of gases Since PV = nRT, then: n1RT n2RT n3RT This would be true of a single gas or mixture of gases P1 P2 = P3 = = V V V According to Dalton’s law of partial pressures, the total pressure of a mixture of gas equals the sum of the pressures that each gas would exert if it were present alone. Therefore Pt = (P1 + P2 + P3 +...)V = nRT or = Pt (n1+ n2 + n3+ …) V RT

  23. Ideal Gas Law What pressure, in atm, is exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10 L vessel? 1 mol 2.00 g H2 = 0.990 mol H2 2.02 g 1 mol 8.00 g N2 = 0.286 mol N2 28.0 g Pt (10.0 L) = (0.990 mol + 0.286 mol) (0.0821 l-atm/mol-K)(273 K) = 2.86 atm Pt

  24. If a 5L container with N2 at 3atm is attached to a 3 L container with He at 3.3 atm what

  25. Ideal Gas Law Partial Pressures and Mole Fractions • The mole fraction, X, of any component of a mixture is simple the ratio of the total number of moles in the mixture Moles of component 1 Mole fraction of component 1 = X1 = Total moles of mixture We can relate the mole fractions to partial pressures because n1RT n1 P1 V X1 = = = ntRT nt Pt V P1 = X1Pt Therefore

  26. Ideal Gas Law A study of the effects of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent CO2, 18.0 mol percent O2, and 80.5 mol percent Ar. (a) Calculate the partial pressure of O2 in the mixture if the total pressure of the atmosphere is to be 745 mm Hg. (b) If this atmosphere is to be held in a 120 L space at 295 K, how many moles of O2 are needed? • Converting mole percent to pressure in mm Hg Since P1 = X1Pt P02 = 0.18 (745 mm Hg) = 134 mm Hg • Converting mm Hg pressure atmospheres 1 atm P02 = 134 mm Hg = 0.176 atm 760 mm Hg • Solving for moles of O2 using the ideal gas equation PV (0.176 atm)( 120L) = 0.872 mol nO2 = = RT (0.0821 l-atm/mol-K)(295 K)

  27. Ideal Gas Law Collecting Gases Over Water P total = P gas + PH2O

  28. Ideal Gas Law Collecting Gases Over Water Suppose the 2.00 L of oxygen is collected over water. The temperature of the water and gas is 26°C and the pressure is 750 mm Hg. How many moles of O2 are collected? What volume would the O2 gas collected occupy when dry, at the same temperature and pressure? (a) Keep in mind that water vapor may be dealt with in Dalton’s partial pressure law just like any other gas. P total = P gas + PH2O PO2 = P total - PH2O PO2 = 750- 25 = 725 mm Hg PV (725 mm Hg/760 Hg)(0.200 L) nO2 = = = 7.77 x 10-3 moles O2 RT (0.0821 l-atm/mol-K)(299 K) (b) P1V1 = P2V2 (725 mm Hg)(0.200 L)1 = (750 mm Hg)(V2) V2= 0.193 L Note: ( the vapor pressure of water at 26 °C is 25 mm Hg., Appendix B)

  29. Kinetic Molecular Theory How does KMT explain PV=k, P/T = k At higher temperatures a greater fraction of molecules is moving at greater speeds. This results in a higher average kinetics energy Root-mean-square (rms) speed Indicates the speed of a particle possessing average kinetic energy ( )such that  = ½ m2

  30. Kinetic Molecular Theory • = √3RT M If two molecules which differ in mass have the same kinetic energy, then because Ek = ½ m  2 is true, the velocities of the particles must be different; lighter particles must move with a higher average speed.

  31. Kinetic Molecular Theory Molecules Effusion and Diffusion Calculate the rms speed () of a N2 molecules at 25 °C Keep in mind that R needs to fit the specific circumstances you are dealing with. Since we are dealing with particles possessing mass and traveling at some velocity, R = 8.31 J/K-mol and J = kg-m2/s2, therefore R = 8.31 kg-m2/s2-K-mol 3(8.31 kg-m2/s2-K-mol)( 298K)  = 1 kg 28 g 1000g  = 5.15 x 102 m/s

  32. Kinetic Molecular Theory Molecules Effusion and Diffusion Diffusion Effusion

  33. Kinetic Molecular Theory Graham’s Law of Effusion The effusion rate of a gas is inversely proportional to the square root of its molar mass 1 r1 M2 3RT/ M1 = = = 2 r2 M1 3RT/ M2 r1 M2 = M1 r2 or T1 = √M1 T2 M2

  34. Kinetic Molecular Theory Graham’s Law of Effusion If an unknown gas effuses at a rate that is only 0.468 times that of O2 at the same temperature, what is the molar mass of the unknown gas? rx M2 = rO2 M1 32 g/mol 0.468 = M1 = 146 g/mol M1

  35. Kinetic Molecular Theory Diffusion and Mean Free Path Diffusion speeds are slower than molecular speeds The average distance traveled by a molecule between collisions is called the mean free path. The higher the density of the gas the smaller the mean free path (the shorter the distance between collisions

  36. Deviations From Ideal Behavior n = 10 atm PV=nRT deviates at high pressure

  37. Deviations From Ideal Behavior Note: as the temperature increases, the PV/RT behavior of gases approaches ideal behavior Ideal Gas PV=nRT deviates at low temperature

  38. Deviations From Ideal Behavior The Van der Waals Equation n2a (V-nb) = nRT P + V2 (V-nb) is a correction for the finite volume of the gas molecules The Van der Waal constant, b, varies with each gas and is the actual volume occupied by the gas molecules and increases with an increase in the mass or complexity of the molecule The correction for pressure, n2a/V2, takes into account the intermolecular attractions between molecules.

  39. Deviations From Ideal Behavior The Van der Waals Equation If 1.000 mol of an ideal gas were confined to 22.41 L at 0.0°C, it would exert a pressure of 1.000 atm. Use the Van der Waals equation and the constants in Table 10.3 to estimate the pressure exert by 1.000 mol of Cl2(g) in 22.4 L at 0.0 °C n2a (V-nb) = nRT P + V2 nRT n2a Solving for pressure: P = - V-nb V2 (1.000 mol)(0.0821 l-atm/mol-K)(273.2 K) (1.000 mol)(6.49 L2-atm)/mol2 - P = (22.4 L)2 22.4 L -(1.000 mol)(0.0562 l/mol) P = 0.990 atm

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