Gas Density. Standard T & P (STP) 10/29/2011. At STP, 1 mol of any ideal gas occupies 22.4 L. T = 273 K (0 o C) P = 1 atm = kpa = 1.

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1 Standard T & P (STP) T = 73 K (0 o C) P = 1 atm = kpa = bar At STP, 1 mol of any ideal gas occupies.4 L.4 L Gas Density We can use PV = nrt to determine the density of gases. What are the units of density? He balloons mass/volume What does this suggest about gas density? It will depend strongly on T, P, and the mass of the gas molecules. Hot-air balloon Contrast with liquids and solids, whose densities depend somewhat on T, but far less on P. 1

2 What is the density of O gas in g/l at 5 o C and atm? Calculate # moles in 1 L, use MM of O to get g, divide by V. P = atm V = 1 L n =? R = L atm/mol K T = 5 C + 73 = 98 K n = Gas Density Example PV = nrt atm1 L L atm 98 K mol K PV n = RT = mol O? g/l O = mol O 3.00 g O 1 1 mol O 1 L Note: A value for gas density is meaningful only if accompanied by the T and P at which it was measured. = 1.11 g/l O 5 C, atm Gas Density and Molar Mass We can develop an expression relating density and MM using PV = nrt. Substituting: m PV = RT M PV = nrt Density!! m PM = RT V m n = M PM = drt Therefore, the density of an ideal gas depends on T, P, V, and Molar Mass We can use the density of a gas to determine its molar mass.

3 PM = drt Example A.30-g sample of an unknown solid was vaporized in a 345-mL vessel. If the vapor has a pressure of 985 mmhg at 148 o C, what is the molecular weight of the solid? P = 985 mmhg M =? d =.30 g L T = 148 o C = 41 K = 1.96 atm = g/l R = L atm/mol K V = 345 ml mass of solid =.30 g Heat applied mass of gas =.30 g M = drt = P g/l L atm 41 K 1.96 atm mol K = g/mol Chemical Equations and Calculations Atoms (Molecules) Avogadro s Number Reactants 6.0 x 10 3 mol -1 Moles Molar Mass Mass g/mol Products Molarity moles / L PV = nrt Solutions Gases 3

4 Gas Stoichiometry Example What volume of N (g) is produced when 70.0 g NaN 3 is decomposed? P = 735 mmhg, T = 6 o C NaN 3 (s) Na(l) + 3 N (g) 70.0 g? L g/mol 735 mmhg 6 o C 8.0 g/mol 1. The Stoichiometry Part: mass NaN 3 mol NaN 3 mol N. The Gas Law Part: mol N, P, T V of N (g) Example: The Stoichiometry Part What volume of N (g) is produced when 70.0 g NaN 3 is decomposed? P = 735 mmhg, T = 6 o C NaN 3 (s) Na(l) + 3 N (g) 1 mol NaN? mol N 3 3 mol N = 70.0 g NaN g NaN 3 mol NaN 3 = mol N 4

5 What volume of N (g) is produced when 70.0 g NaN 3 is decomposed? (P = 735 mmhg, T = 6 o C) P = 735 mmhg 1 atm 760 mmhg V =? L n = mol N R = L.atm/mol.K T = 6 o C + 73 = 99 K V = Example: The Gas Law Part = atm PV = nrt V = nrt P mol N L atm 99 K V = L = 41.0 L Yay!!! atm mol K Dalton s Law of Partial Pressures P He = 00 torr P Ar = 500 torr P = 700 torr Recall that according to the ideal gas law, gas molecules are non-interacting point particles. Increasing the number of point particles increases the pressure by an amount that is proportional to the number of particles. For a mixture of ideal gases in a container: pressure = the sum of the individual gas pressures. 5

6 Dalton s Law of Partial Pressures Say we have a container with some amount of three different gases inside, at a certain T and P. n = n 1 + n + n3 Dalton s Law says that the pressure exerted by the three gases is the sum of the individual pressures. P = P 1 + P + P 3 P = P 1 + P + P 3 1 P = + + P V = n1rt + nrt + n3rt P V = n RT n RT n RT n3rt V V V P V = n + n + n RT 1 3 Partial Pressures Example Mixtures of He and O are used in scuba tanks to help prevent the bends. For a particular dive, 1 L of O at 5 o C and 1.0 atm was pumped along with 46 L of He at 5 o C and 1.0 atm into a 5.0-L tank. What is the partial pressure of each gas? What is the pressure? 1. Find the number of moles of each gas that were delivered to the tank.. Find the partial pressure of each gas in the tank. 3. Add them up! 6

7 Mixtures of He and O are used in scuba tanks to help prevent the bends. For a particular dive, 1 L of O at 5 o C and 1.0 atm was pumped along with 46 L of He at 5 o C and 1.0 atm into a 5.0-L tank. What is the partial pressure of each gas? What is the pressure? O Data: P = 1.0 atm V = 1 L n =? mol R = L atm/mol K T = 5 o C = 98 K He Data: P = 1.0 atm V = 46 L n =? mol R = L atm/mol K T = 5 o C = 98 K PV n = RT n = O n = He 1.0 atm1 L L atm 98 K mol K 1.0 atm46 L L atm 98 K mol K = 0.49 mol O = 1.9 mol He Using the moles of each gas, the temperature, and volume of the tank we can now calculate the partial pressure of each gas, then add them to get the pressure. P = O P = He 0.49 mol O L atm 98 K 5.0 L mol K 1.9 mol He L atm 98 K 5.0 L mol K P = =.4 atm = 9.3 atm P = 11.7 atm Partial pressures of O and He in the 5.0 L scuba tank nrt V 7

8 Mole Fraction and Partial Pressure In the last example, we determined the pressure by adding the partial pressures. We could have also added the moles of each gas, and determined a pressure. These two approaches suggest that a relationship exists between the moles of each gas and the pressure. Mole Fraction and Partial Pressure Mole Fraction (): ratio of the number of moles of a component in a mixture to the number of moles in the mixture. n1 n1 1 n n n n ni i n 1 3 V Pi RT V P RT Pi P P P i i Dalton s Law The fraction of moles of a certain gas in a mixture is equal to the ratio of its partial pressure to the pressure of the mixture. 8

9 Mixtures of He and O are used in scuba tanks to help prevent the bends. For a particular dive, 1 L of O at 5 o C and 1.0 atm was pumped along with 46 L of He at 5 o C and 1.0 atm into a 5.0-L tank. What is the partial pressure of each gas? What is the pressure? n O = 0.49 mol n He = 1.9 mol 0.49 mol O O =.39 mol gas n =.39 mol P = 11.7 atm = 0.05 He 1.9 mol He =.39 mol gas = P = atm O P = atm He =.39 atm = 9.8 atm.4 atm 9.3 atm Previous results. Partial pressures of O and He in the 5.0 L scuba tank Saturation

10 Production of O (g) by thermal decomposition of KCIO 3 O is mixed with water vapor in the collection vessel. Collection of Hydrogen Gas Over Water Vapor pressure - I HCl (aq) + Zn (s) ZnCl (aq) + H (g) Calculate the mass of hydrogen gas collected over water if 156 ml of gas is collected at 0 o C and 769 mm Hg. P Total = P H + P H O P H = P Total - P H O FROM TABLE P H = 769 mm Hg mm Hg = 75 mm Hg T = 0 o C + 73 = 93 K P H = (75 mm Hg )(1 atm/ 760 mm Hg) = atm V = L 10

11 Collection of Hydrogen Gas Over Water Vapor Pressure - II P H V = n H RT n H = P H V / RT n H = (0.989 atm) (0.156 L) (0.081 L atm/mol K) (93 K) n H = mol Mass H = mol x (.016 g H / mol H ) = g H Kinetic Molecular Theory (KMT) The gas laws of Boyle, Charles, and Avogadro are empirical, meaning they are based on observation of a macroscopic property. These laws offer a general description of behavior based on many experiments. The empirical gas laws can tell you what happens to an ideal gas under certain conditions, but not why it happens. KMT is a theoretical, molecular-level model of ideal gases, which can be used to predict the macroscopic behavior of a gaseous system. KMT Simulation: 11

12 Postulates of KMT Gas particles are so small that their volume is negligible. Gas particles are in constant, random motion. This motion is associated with an average kinetic energy that is directly proportional to the Kelvin temperature of the gas. Gas molecules constantly collide with each other and with the container walls. The collisions of the particles with the container walls are the cause of the pressure exerted by the gas. Collisions are elastic. The particles are assumed to exert no forces on each other; they neither attract or repel their neighbors. KMT: Central Points The main ideas you should take from KMT are that we can describe temperature and pressure from a molecular perspective. Pressure: arises from molecules banging into the container walls. Temperature: Is directly related to the kinetic energy of the gas molecules. The more KE they have, the greater their temperature. 1

13 KMT Let s consider the average KE per molecule and see how it determines molecular speed. Total KE in 1 mol of gas: 3 RT From KMT. Note: T is measured in Kelvin Average KE per molecule: 1 mu Where u is an average of molecular velocity, and m is the mass of one molecule mu RT N A We are apportioning the KE in the mole of gas among all the molecules in an average fashion. KMT (cont.) mu RT N A We are apportioning the KE in the mole of gas among all the molecules in an average fashion. 3RT u Note that mn A = M. mn A u rms 3RT M u rms is the speed of a molecule that has the average KE. u rms gives us a formal connection between average gas speed, T, and M. 13

14 Distribution of Molecular Speeds Maxwell-Boltzmann curve (a statistical distribution) This plot represents the fraction of gas molecules in a sample that are traveling at a given velocity. u m most probable speed u av average speed u rms the speed of a molecule with the average molecular kinetic energy (m/s) 3RT u rms M Increased T increased average KE increased u rms. Increased M decreased u rms Increased T increased average KE increased u rms Maximum of curve shifts to higher u, and distribution spreads out. Distribution of speeds will be shorter and fatter at higher temperatures. Increased M decreased u rms Heavier molecules have lower average speed than lighter molecules at a given temperature. Distribution of speeds for heavier gases will be taller and skinnier than for lighter molecules. 3RT u rms M NOTE: There are always some molecules with low velocity in a Boltzmann distribution!! 14