Kinetic Molecular Theory Gases. Behavior of gases. Postulate two. Postulate one. Postulate three. Postulate four

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1 Kinetic Molecular Theory Gases Gas particles are so small that their individual volume can be considered to be negligible Gas particles are in constant motion and the collisions of the particles with the walls of the container cause the pressure exerted by the gas Gas particles are assumed to exert no force on each other The average kinetic energy of a collection of gas particles is directly proportional to the Kelvin temperature of the gas Behavior of gases The postulates of the kinetic molecular theory explain the behavior of ideal gases Postulate one Gas particles are so small that their individual volume can be considered to be negligible g Gases have no particular volume Take the volume of their container Postulate two Gas particles are in constant motion and the collisions of the particles with the walls of the container cause the pressure exerted by the gas Gas exerts pressure on the container which holds it. This pressure depends on three things The number of molecules The volume of the gas/container The average kinetic energy of the particles (temperature) Postulate three Gas particles are assumed to exert no force on each other Gas particles are not in any fixed position Spread out and fill the available space Postulate four The average kinetic energy of a collection of gas particles is directly proportional to the Kelvin temperature of the gas A change in the average kinetic energy of the particles result in a significant change in volume 1

2 Influences on the Properties of a gas Pressure Temperature (Kelvin) Volume Number of particles Boyle s Law Robert Boyle If the amount and temperature of a gas remain constant, the pressure varies inversely with the volume. P 1 V 1 = P 2 V 2 Where P 1 is the initial pressure and V 1 is the initial volume and P 2 is the final pressure and V 2 is the final volume Standard pressure 1.0 atm 760 mmhg kpa 76 cmhg Pressures If ml of a gas exerts a pressure of 131 kpa on its container, what will be the pressure when that gas expands to fill a 375 ml container P 1 V 1 = P 2 V 2 (131 kpa)(250.0 ml) = (P 2 )(375 ml) kpaml = (P 2 )(375 ml) 87.3 kpa = P 2 Charles Law Jacques Charles if the amount and pressure of a gas remain constant, the temperature varies directly with the volume and can be calculated using V 1 = V 2 T 1 T 2 Where T 1 is the initial temperature and V 1 is the initial volume and T 2 is the final temperature and V 2 is the final volume Temperatures KMT state that the relationship is to the Kelvin temperature 0.0 C = 273K 2

3 If a gas occupies 425 ml at K, what will the volume be at 275 K V 1 =V 2 T 1 =T ml = V K 275.0K ml K = K V ml = V 2 Gay-Lussac s Law If volume and amount remain constant, pressure and temperature vary directly P 1 = P 2 T 1 T 2 A balloon is filled with helium to a pressure of 1.3 atm at room temperature (27 C). When the balloon vendor stands outside on a hot summer day (98 C) what is the pressure in the balloon? 13atm= 1.3 P 2 27 C 98 C 1.3 atm = P 2 300K 371 K Combined Gas Law Boyle s, Charles, and Gay-Lussac s laws can be used together P 1 V 1 = P 2 V 2 T 1 T K atm = 300K P 2 1.6atm = P 2 If a gas occupies 351ml at 0.8atm and 19 C, what will be its volume at 94.5kPa and 309K? (0.8atm) (351ml) = (94.5kPa) (V 2 ) 19 C 309K (81 kpa) (351ml) = (94.5kPa) (V 2 ) 292 K 309K Molar Volume Just as pressure is unaffected by the identity, so is volume 22.4L of any gas at STP = One mole kpamlk = kpakv ml = V 2 3

4 What would be the volume of 56.2 g of carbon dioxide gas at STP? What would be the mass of 122 L of oxygen gas at STP? 56.2g CO 2 /1 mole CO 2 / 22.4 L CO 2 1 / 44.0 g CO 2 / 1 mol CO L CO L O 2 / 1 mol O 2 / 32 g O 2 1 / 22.4 L O 2 / 1 mol O g O 2 What would be the volume of 11.7g of chlorine gas at 1.2atm and 300K? 11.7g Cl 2 / 1 mol Cl 2 / 22.4LCl 2 1 / 71.0g Cl 2 / 1 mol Cl g Cl 2 / 1 mol Cl 2 / 22.4LCl 2 /1.0atm/300K g / 71.0g Cl 2 / 1 mol Cl 2 / 273K/1.2 atm Ideal Gas Equation Combines all the postulates of the KMT PV = nrt P = pressure V= volume N = number of moles R = molar gas constant T = temperature Molar gas constant Derived from the molar volume of a gas at STP P = standard pressure 1 atm; 760 mmhg; kpa V = 22.4 L N = one mole T 273K What pressure is exerted by moles of gas in a 25.9 L container at 282 K? P(25 9 L) ( l)( L/ lk ) (282K) P(25.9 L)=( mol)(0.0821atml/molk ) (282K) P (25.9L) = atm L P = atm 4

5 What volume is occupied by mol of a gas at 743 mmhg and 282 K? (743 mmhg)v = mol (62.4 mmhg L/molK )(282K ) (743 mmhg)v = mmhg L V = 16.2 L What is the density of carbon dioxide gas at STP? Use STP to find the mols in one liter, then use the molar mass of carbon dioxide to find the density So use PV = nrt (1.0 atm)(1.0 l) = n (0.0821atm l/molk)(273k) 1.0 atm l = n (22.41atml/mol) 10atml 1.0 = n 22.41atml/mol mol = n molar mass of CO2 is 44 g/mol mol / 44 g CO 2 1 l / 1 mol CO 2 Density = 1.96 g/l Dalton s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone At constant temperature and volume, the total pressure of a gas sample is determined by the total number of moles of gas present, whether this represents a single substance, or a mixture Therefore the pressure is not affected by the identity of the gas (postulates 1 and 3) In a scuba tank with a total pressure of 113 kpa, the mixture of gases is 42.0L of nitrogen and 26.0L of oxygen, at a temperature of 25 C. What is the partial pressure of the oxygen in the tank? 42.0L (N 2 ) / 68.0L (total) t = ( 113 kpa) = P N 69.8 kpa = P N 26.0L /68.0L = (113 kpa) = P O 43.2 kpa= P O Graham s Law When does the identity of the gas matter? When it spreads from one container to another Diffusion Describes the mixing of gases Effusion Describes the passage of a gas through an opening» Measures the speed with which the gas is transferred into the new container Graham s Law of diffusion The rate at which gases diffuse is inversely proportional to the square root of the density Since volumes of gases contain the same number of particles, the number of moles per liter at the same T and P remains constant; thus the density is directly proportional to the molar mass 5

6 Graham s Law of Effusion The rate of effusion is inversely proportional to the square root of either the density or the molar mass of the gas Graham s Law and KMT Since average kinetic energy depends only on temperature, gases at the same temperature have the same AKE Therefore, relative rates of effusion of two gases are inversely proportional to the square root of the masses of gas particles Rate of effusion of gas 1 = sq root mol. mass of gas 2 Rate of effusion of gas 2 sq. root mol. mass of gas 1 Rates of diffusion are different in that the particles are moving through air and must mix with the air molecules However, the relative rates of diffusion may be calculated using Grahams law, since gases travelling through the same mixture are affected in the same manner If we know that the ratio of the rates of diffusion of the basic gas to the acidic gas in a neutralization reaction is 1.47, and that the base is ammonia (NH 3 ), is the vaporous acid most likely hydrobromic acid or hydrochloric acid? Rate of effusion = square root of mol mass A square root of mol mass B 1.47 = sq rt X sq rt 17 (1.47) 2 = X = X Real Gases The differences between real gases and ideal gases is most obvious at high pressures and low temperatures Demonstrated by the fact that at high pressures and/or low temperatures gases can be compressed into liquids This is because real gases have intermolecular This is because real gases have intermolecular forces These intermolecular forces account for a compression factor Z In ideal gases Z = 1 In real gases» If Z>1 the repulsions are dominant; Z<1 attractions are dominant 6