11. What are the basic assumptions of KMT? 1. Gases consist of tiny (submicroscopic) particles. The distance between particles is large compared with the size of the particles themselves. The volume occupied by a gas consists mostly of empty space. 3. Gas particles have no attraction for one another. Gas particles move in a straight line in all directions, colliding frequently with one another and with the walls of the container. No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic. 6. The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature.

15. What are the characteristics of an ideal gas? An “ideal gas” behaves according to KMT. No ideal gases exist, but under certain conditions of temperature and pressure, real gases approach ideal behavior, or at least show only small deviation from it. At very high pressure, low temperature real gases deviate greatly from ideal behavior.

The barometer reads 715 mm Hg. Calculate the corresponding pressure: a) 715 mm Hg x 1 atm = 0.941 atm 760 mm Hg b) 715 mm Hg x 29.9 in Hg = 28.1 in Hg 760 mm Hg c) 715 mm Hg x 14.7 psi = 13.8 psi 760 mm Hg

27. Express the following pressures in atm: a) 28 mm Hg x 1 atm = 0.037 atm 760 mm Hg b) 6000. cm Hg x 1 atm = 78.95 atm 76 cm Hg c) 795 torr x 1 atm = 1.05 atm 760 torr d) 5.00 kPa x 1 atm = 0.0494 atm 101.3 kPa

P1V1 = P2V2 or V2 = P1V1 P2 (500. mm Hg)(400.mL) 760 mm Hg 29. A gas occupies a volume of 400. mL at 500. mm Hg pressure. What will be its volume, at constant temperature, if pressure is changed to (a) 760 mm Hg? (b) 250 torr? P1V1 = P2V2 or V2 = P1V1 P2 (500. mm Hg)(400.mL) 760 mm Hg 250 torr V2 = 2.6 x 102 mL V2 = 8.0 x 102 mL

31. P1 V1 = P2 V2 P2 = P1 V1 = (640. mm Hg)(500. mL) V2 855 mL 31. A 5oo. –mL sample of a gas is at a pressure of 640. mm Hg. What must be the pressure, at constant temperature, if volume is changed to 855 mL? P2 = P1 V1 = (640. mm Hg)(500. mL) V2 855 mL P2 = 374 mm Hg 31. P1 V1 = P2 V2

33. Given 6.00 L of N2 gas at -25C, what volume will the nitrogen occupy at (a) 0.0C (b) 100. K? (Assume constant pressure) V1 = V2 or V2 = V1T2 T1 T2 T1 (a) (6.00 L)(273 K) = 6.60 L 248 K (b) (6.00 L)( 100. K) = 2.42 L

Use the combined gas law P1V1 = P2V2 V2 = P1V1 T2 T1 T2 P2T1 35. A gas occupies a volume of 410 mL at 17C and 740 mm Hg pressure. Calculate the volume the gas would occupy at STP. Use the combined gas law P1V1 = P2V2 V2 = P1V1 T2 T1 T2 P2T1 V2 = (740 mm Hg)(410 ml)(273 K) (760 mm Hg) (300.K) V2 = 360 mL

A sample of propane gas, C3H8, was collected over water at 22.5 oC and 745 torr. The volume of the wet gas is 1.25 L. What will be the volume of the dry propane at standard pressure? P Total = PC3H8 + PH2O @ 22.5 oC PH2O = 20.5 torr PC3H8 = 745 – 20.5 = 725 torr

To calculate the volume of the dry propane, note that the temperature is constant, so … P1V1 = P2V2 P1V1 = V2 P2 (725 torr)(1.25 L) = 1.19 L C3H8 760 torr

Given the equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O How many moles of NH3 are required to produce 5.5 mol of NO? 4 NH3 = X mol NH3 4 NO 5.5 mol NO X = 5.5 mol NH3

Given the equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O b) How many liters of NO can be made from 12 L of O2 and 10 L of NH3 at STP? 4 NH3 = 10 L NH3 4 NO X NO X = 10 L NO 5 O2 = 12 L O2 4 NO X NO X = 9.6 L NO

Given the equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O c) At constant temperature and pressure, what is the maximum volume, in Liters, of NO can be made from 3.0 L of NH3 and 3.0 L of O2? 4 NH3 = 3.0 L NH3 4 NO X NO X = 3.0 L NO 5 O2 = 3.0 L O2 4 NO X NO X = 2.4 L NO

63) Given the equation: 4 FeS (s) + 7 O2 (g) → 2 Fe2O3 (s) + 4 SO2 (g) How many liters of O2, measured at STP, will react with 0.600 kg of FeS? 600. g FeS 1 mol = 6.83 mol FeS 87.9 g 6.83 mol FeS 7 O2 = 12.0 mol O2 4 FeS 12.0 mol O2 22.4 L = 269 L O2 1 mol

Sketch a graph to show each of the following relationships: V P V T P V T n

At 27 0C and 750 torr pressure, what will be the volume of 2.3 mol Ne? V = nRT P PV = nRT 27 0C + 273 = 300. K 750 torr 1atm = 0.99 atm 760 torr V = (2.3)(0.0821)(300.) = 57 L Ne 0.99

91a) Using the ideal gas equation, calculate the volume of 0.510 mol of H2 at 47 0C and 1.6 atm pressure PV = nRT V = nRT P 47 0C + 273 = 320. K V = (0.510)(0.0821)(320.) = 8.4 L H2 1.6

What is the Kelvin temperature of a system in which 4.50 mol of a gas occupy 0.250 L at 4.15 atm? PV = nRT T = PV nR T = (4.15)(0.250) = 2.81 K (4.50)(0.0821)

How many moles of N2 gas occupy 5.20 L at 250 K and 0.500 atm? PV = nRT n = PV RT n = (0.500 atm)(5.20 L) = 0.13 mol N2 (0.0821)(250 K)