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Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of Illinois

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2 CHEMISTRY 100 Dr. Jimmy Hwang Sections 5 and 7 Tu & Th 9:30a.m.-10:45a.m. Textbook: Zumdahl, Introductory Chemistry, 5 th Edition Office Hours: by appointment 619-421-6700 Ext. 3399 (voicemail) Email: jhwang@swccd.edu

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3 Gases Chapter 12

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4 Overview In Chapter 12, our goals are for the students to: 1. Learn about atmospheric pressure and how barometers work. 2. Learn the various units of pressure. 3. Understand the law that relates the pressure and volume of a gas and use it in calculations. 4. Learn about absolute zero. 5. Learn about the law relating the volume and temperature of a sample of gas at a constant number of moles and pressure and use it in calculations. 6. Understand the law relating the volume and number of moles of a a sample of gas at constant temperature and pressure and use it in calculations. 7. Understand the ideal gas law and use it in calculations.

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5 Overview In Chapter 12, our goals are for the students to: 8.Understand the relationship between the partial and total pressures of a gas mixture and use it in calculations. 9.Understand the relationship between laws and models (theories). 10.Understand the basic postulates of the kinetic molecular theory. 11.Understand the term temperature. 12.Learn how the kinetic molecular theory explains the gas laws. 13.Understand the molar volume of an ideal gas. 14.Learn the definition of STP. 15.Do stoichiometric calculations involving gases.

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6 Gases Chapter 12 Become familiar with the definition and measurement of gas pressure Learn the gas laws: Boyle’s law, Charle’s law, and Avogadro’s law The ideal gas law Learn Dalton’s law of partial pressures Review of Laws and Models Study the kinetic molecular theory of gases Study the implications of the kinetic theory of gases Look at gas stoichiometry

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7 Properties of Gases Expand to completely fill their container Take the Shape of their container Low Density –much less than solid or liquid state Compressible Mixtures of gases are always homogeneous Fluid

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8 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. Exerts pressure on its surroundings. Physical Characteristics of Gases

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9 The pressure exerted by the gases in the atmosphere can be demonstrated by boiling water in a large metal can (a) and then turning off the heat and sealing the can.

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10 As the can cools, the water vapor condenses, lowering the gas pressure inside the can. This causes the can to crumple.

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11 Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 torr = 1 mm Hg 1 atm 760 torr 1 atm 101,325 Pa 760 torr 101,325 Pa Pressure = Force Area Pressure can be measured by a Barometer →

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12 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

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13

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14 Boyle’s Law

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15 As P (h) increasesV decreases

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16 P 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Boyle’s Law Constant temperature Constant amount of gas

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17 Pressure and Volume: Boyle’s Law Pressure is inversely proportional to Volume –constant T and amount of gas –graph P vs V is curve –graph P vs 1/V is straight line as P increases, V decreases by the same factor P x V = constant P 1 x V 1 = P 2 x V 2

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18 Boyle’s Law * Pressure Volume = Constant (T = constant) P 1 V 1 = P 2 V 2 (T = constant) V 1/P (T = constant) ( * Holds precisely only at very low pressures.)

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19 As pressure increases, the volume of SO 2 decrease (at const. T) P 1 V 1 P 2 V 2 =

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20 Gas Pressure Pressure = total force applied to a certain area –larger force = larger pressure –smaller area = larger pressure Gas pressure caused by gas molecules colliding with container or surface More forceful collisions or more frequent collisions mean higher gas pressure

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21 Air Pressure Constantly present when air present Decreases with altitude –less air Varies with weather conditions Measured using a barometer –Column of mercury supported by air pressure –Longer mercury column supported = higher pressure –Force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury

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22 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

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23 Measuring Pressure of a Trapped Gas Use a manometer Open-end manometer –if gas end lower than open end, P gas = P air + diff. in height of Hg –if gas end higher than open end, P gas = P air – diff. in height of Hg

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25 Units of Gas Pressure atmosphere (atm) height of a column of mercury ( mm Hg, in Hg ) torr Pascal (Pa) pounds per square inch (psi, lbs./in 2 ) 1.000 atm = 760.0 mm Hg = 29.92 in Hg = 760.0 torr = 101,325 Pa = 101.325 kPa = 14.69 psi

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26 Summary: Boyle’s Law Pressure is inversely proportional to Volume –constant T and amount of gas –graph P vs V is curve –graph P vs 1/V is straight line as P increases, V decreases by the same factor P x V = constant P 1 x V 1 = P 2 x V 2

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27 P 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Boyle’s Law Constant temperature Constant amount of gas

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28 ¬Write down the given amounts P 1 = 56 torrP 2 = 150 torr V 1 = 1.5 L.V 2 = ? L Convert values of like quantities to the same units both Pressure already in torr value of V 2 will come out in L Example 12.2 What is the new volume if a 1.5 L sample of freon-12 at 56 torr is compressed to 150 torr?

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29 ®Choose the correct Gas Law Since we are looking at the relationship between pressure and volume we use Boyle’s Law P 1 x V 1 = P 2 x V 2 ¯Solve the equation for the unknown variable Example 12.2 (continued) What is the new volume if a 1.5 L sample of freon-12 at 56 torr is compressed to 150 torr?

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30 °Plug in the known values and calculate the unknown P 1 = 56 torrP 2 = 150 torr V 1 = 1.5 L.V 2 = ? L Example 12.2 (continued) What is the new volume if a 1.5 L sample of freon-12 at 56 torr is compressed to 150 torr?

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31 Absolute Zero Theoretical temperature at which a gas would have zero volume and no pressure –calculated by extrapolation 0 K = -273.15 °C = -459 °F Kelvin T = Celsius T + 273.15 Never attainable –though we’ve gotten real close! All gas law problems use Kelvin temperature scale!

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32 As T increasesV increases Charle’s Law: Volume-Temperature Relationship (n and P const.)

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33 Figure 12.7: Plots of V (L) versus T (°C) for several gases.

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34 Figure 12.8: Plots of V versus T as in Figure 12.7, except that here the Kelvin scale is used for temperature.

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35 Volume and Temperature: Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. V = bT (n and P = constant) b = a proportionality constant

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36 Charles’s Law

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37 Summary: Charles’ Law Volume is directly proportional to Temperature –constant P and amount of gas –graph of V vs T is straight line as T increases, V also increases V = constant x T –if T measured in Kelvin V 1 = V 2 T 1 T 2

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38 (Calculating Volume using Charle’s Law). A 2.0-L sample of air is collected at 298 K and then cooled to 278 K. The pressure is held constant at 1.0 atm. Calculate the volume of the air at 278 K. Does the volume increase or decrease? V 1 = 2.0 L T 1 = 298 K V 2 = ? T 2 = 278 K V 2 = T 2 x V 1 T1T1 278 K x 2.0 L 278 K = = 1.9 L The volume gets smaller when the temperature decreases.

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39 (Calculating Volume using Charle’s Law). A sample of gas at 15 C (at 1 atm) has a volume of 2.58 L. The temperature is then raised to 38 C (at 1 atm). Calculate the new volume. Does the volume of the gas increase or decrease? V 1 = 2.58 L T 1 = 15 °C = 15+273 = 288 K V 2 = ? T 2 = 38 °C = 38+273 = 311 K V 2 = T 2 x V 1 T1T1 311 K x 2.58 L 288 K = = 2.79 L V 1 /T 1 = V 2 /T 2 The volume gets smaller when the temperature decreases.

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40 (Calculating Temperature using Charle’s Law). A balloon is filled with helium, and its volume is 2.1 L at 298 K. The balloon will burst if its volume exceeds 2.3 L. At what temperature would you expect the balloon to burst? V 1 = 2.1 L T 1 = 298 K V 2 = 2.3 L T 2 = ? T 2 = V 2 x T 1 V1V1 2.3 L x 298 K 2.1 L = = 362 K V 1 /T 1 = V 2 /T 2

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41 Volume and Moles: Avogadro’s Law Volume directly proportional to the number of gas molecules –V = constant x n (moles) –Constant P and T –More gas molecules = larger volume Count number of gas molecules by moles One mole of any ideal gas occupies 22.414 L at standard conditions - molar volume Equal volumes of gases contain equal numbers of molecules –It doesn’t matter what the gas is!

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42 Avogadro’s Law V number of moles (n) V = constant x n V 1 /n 1 = V 2 /n 2 Constant temperature Constant pressure

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43 Figure 12.9: The relationship between volume V and number of moles n (at constant T and P). As the number of moles is increased from 1 to 2 (a to b), the volume doubles. When the number of moles is tripled (c), the volume is also tripled.

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44 Ideal Gas Law By combing the proportionality constants from the gas laws we can write a general equation R is called the gas constant The value of R depends on the units of P and V –Generally use R = 0.08206 when P in atm and V in L Use the ideal gas law when have gas at one condition Most gases obey this law when pressure is low (at or below 1 atm) and temperature is high (above 0°C) If a gas changes some conditions, the unchanging conditions drop out of the equation PV = nRT

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45 Ideal Gas Law Charles’ law: V T (at constant n and P) Avogadro’s law: V n (at constant P and T) Boyle’s law: V (at constant n and T) 1 P V V nT P V = constant x = R nT P P R is the gas constant, can be calculated from expt. PV = nRT A single equation has been derived which relates to amount, temperature, and pressure. Relation between variables:

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46 Ideal Gas Law PV = nRT R = proportionality constant = 0.08206 L atm mol P = pressure in atm; 1 atm = 101.325 kPa; 1 atm = 760 mmHg V = volume in liters n = amount in moles T = temperature in Kelvin (T K = tºC + 273) Holds closely at P < 1 atm

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47 Combined Gas Law

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48 What is the volume (in liters) occupied by 49.8 g of HCl at 1.00 atm? PV = nRT V = nRT P T = 0 °C = 273 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.08206 x 273 K Latm molK V = 30.6 L

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49 Example 12.8 Using the Ideal Gas Law in Calculations Question: A sample of hydrogen gas, H2, has a volume of 8.56 L at a temperature of 0 °C and a pressure of 1.5 atm. Calculate the number of moles of H 2 present in this gas sample. (Assume that the gas behaves ideally.) Answer: 0.57 mol

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50 Example 12.8 Using the Ideal Gas Law Calculations Involving Conversion of Units Question: What volume is occupied by 0.250 mol of carbon dioxide at 25 ºC and 371 torr ? Answer: 12.5 L

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51 Example 12.11. Calculating Volume Changes Using the Ideal Gas Law Question: A sample of diborane gas, B 2 H 6, a substance that bursts into flames when exposed to air, has a pressure of 0.454 atm at a temperature of –15 ºC and a volume of 3.48 L. If conditions are changed so that the temperature is 36 ºC and the pressure is 0.616 atm, what will be the new volume of the sample? Hint: Calculate V 2 from the combined gas law: Answer: 3.07L

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52 Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures each gas would exert independently –Partial pressures is the pressure a gas in a mixture would exert if it were alone in the container –P total = P gas A + P gas B + … Particularly useful for determining the pressure a dry gas would have after it is collected over water –P air = P wet gas = P dry gas + P water vapor –P water vapor depends on the temperature, look up in table

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53 Partial Pressures The partial pressure of each gas in a mixture can be calculated using the Ideal Gas Law

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54 Figure 12.10: When two gases are present, the total pressure is the sum of the partial pressures of the gases.

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55 Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +...

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56 Figure 12.11: The total pressure of a mixture of gases depends on the number of moles of gas particles (atoms or molecules) present, not on the identities of the particles.

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57 2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O 22 Experimental setup for Example 12.13

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58 Using Dalton’s Law of Partial Pressures using Fig. 12.12 Question: A sample of solid potassium chlorate, KClO 3, was heated in a test tube (see Figure 12.12) and decomposed according to the reaction KClO 3 (s) 2 KCl (s) + 3 O 2 (g) The oxygen produced was collected by displacement of water at 22 ºC. The resulting mixture of O 2 and H 2 O vapor had a total pressure of 754 torr and a volume of 0.650 L. Calculate the partial pressure of O 2 in the gas collected and the number of moles of O 2 present. The vapor pressure of water at 22 ºC is 21 torr. Ans: P(O 2) = 0.964 atm and n(O 2) = 2.59 10 -2 mol

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59 Example using Dalton’s Law of Partial Pressure Not all pollution is due to human activity. Natural sources, including volcanoes, also contribute to air pollution. A scientist tries to generate a mixture of gases similar to those found in a volcano by introducing 15.0 g of water vapor, 3.5 g of SO 2 and 1.0 g of CO 2 into a 40.0L vessel held at 120 ºC. Calculate the partial pressure of each gas and the total pressure. Ans.: P H 2 O = 0.672 atm, P SO 2 = 0.044 atm, P CO 2 = 0.019 atm; P tot = 0.735 atm.

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60 Laws and Models: A Review Aim: To understand the relationship between laws and models (theories). The laws are the ideal gas laws, and the model (theory) we want to look at is the Kinetic Molecular Theory of Gases.

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61 Real Gases Deviate at least slightly from the ideal gas law because of two factors: 1. Gas molecules attract one another 2. Gas molecules occupy a finite volume. Both of these factors are neglected in the ideal gas law. Both increase in importance when the molecules are close together (high P, low T). This means that at high pressures and/or low temperatures, the properties of gases deviate significantly from the predictions of the ideal gas equation. An ideal gas is really a hypothetical substance. At low pressures and/or high temperatures, real gases approach the behavior expected for an ideal gas.

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62 Real Gases Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

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63 Kinetic - Molecular Theory The properties of solids, liquids and gases can be explained based on the speed of the molecules and the attractive forces between molecules In solids, the molecules have no translational freedom, they are held in place by strong attractive forces –May only vibrate

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64 Kinetic - Molecular Theory In liquids, the molecules have some translational freedom, but not enough to escape their attraction for neighboring molecules –They can slide past one another, rotate as well as vibrate In gases, the molecules have “complete” freedom from each other, they have enough energy to overcome “all” attractive forces Kinetic energy depends only on the temperature

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65 Describing a Gas using KM theory Gases are composed of tiny particles The particles are small compared to the average space between them –Assume the molecules do not have volume Molecules constantly and rapidly moving in a straight line until they bump into each other or the wall –Average kinetic energy proportional to the temperature –Results in gas pressure Assumed that the gas molecules attraction for each other is negligible

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66 Postulates of the Kinetic - Molecular Theory of Gases Gases consist of tiny particles (atoms or molecules). These particles are so small, compared with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero). The particles are in constant random motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas. The particles are assumed not to attract or to repel each other. The average kinetic energy of the gas particle is directly proportional to the Kelvin temperature of the gas.

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67 Basic equation of the kinetic- molecular theory for the P gas This equation can be rearranged to give, for N A molecules, The Kelvin temperature (T) of a gas is directly proportional to the average translational energy of its molecules.

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68 The Meaning of Temperature per mole per molecule The Kelvin temperature (T) of a gas is directly proportional to the average translational energy of its molecules. New meaning of temperature: The zero temperature is the temperature at which translational molecular motion should cease.

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69 Gas Properties Explained Gases have indefinite shape and volume because the freedom of the molecules allows them to move and fill the container they’re in Gases are compressible and have low density because of the large spaces between the molecules

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70 The Meaning of Temperature Temperature is a measure of the average kinetic energy of the molecules in a sample –Not all molecules have same kinetic energy Kinetic energy is directly proportional to the Kelvin Temperature –average speed of molecules increases as the temperature increase (actually as T). The Kelvin temperature (T) of a gas is directly proportional to the average translational energy of its molecules.

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71 Root Mean Square Velocity Derivation: The root mean square velocity can be derived by equating the kinetic energy of the molecule to be equal to the thermal energy, i.e. ½ m u 2 = 3/2 kT.

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72 Pressure and Temperature As the temperature of a gas increases, the average speed of the molecules increases the molecules hit the sides of the container with more force (on average) the molecules hit the sides of the container more frequently the net result is an increase in pressure Gay-Lussac’s Law states that for a fixed amount of gas (fixed number of moles) at a fixed volume, the pressure is proportional to the temperature. Gay-Lussac’s Law:

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73 Volume and Temperature In a rigid container, according to KM theory, raising the temperature increases the pressure For a cylinder with a piston, the pressure outside and inside stay the same To keep the pressure from rising, the piston moves out increasing the volume of the cylinder –as volume increases, pressure decreases Therefore KM theory predicts that the volume of a gas will increase as we raise its temperature at a constant pressure. This agrees with experimental observation (as summarized by Charle’s law).

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74 Kinetic Molecular Theory of Gases 1.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 (V zero). 2.Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the absolute temperature of the gas in K. Any two gases at the same temperature will have the same average kinetic energy.

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75 Kinetic Molecular Theory of Gases 1.Gases consist of atoms or molecules in continuous, random motion. 2.Collisions between gas particles are elastic. 3.V gas particles << V container 4.Attractive forces between particles 0). 5.The average kinetic energy of the molecules is proportional to the absolute temperature; E trans = c T 6.At a given temperature, all gases have the same average translational kinetic energy. In other words, c is a universal constant (for all gases). E trans = ½ m u 2 = c T where m = mass molecule, u = avg. speed, T = temp. in K, and c is a constant which has the same value for all gases.

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76 Kinetic theory of gases and … Compressibility of Gases : Boyle’s Law P collision rate with wall Collision rate number density (Average collision rate with the walls.htm)Average collision rate with the walls.htm Number density 1/V P 1/V Charles’ Law P collision rate with wall Collision rate average kinetic energy of gas molecules Average kinetic energy T P T

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77 Kinetic theory of gases and … Avogadro’s Law P collision rate with wall Collision rate number density Number density n P nP n 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 P total = P i

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78 Gas Stoichiometry Use the general algorithms discussed previously to convert masses or solution amounts to moles Use gas laws to convert amounts of gas to moles –or vice versa

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79 Stoichiometry of Reactions Involving Gases

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80 Gas Stoichiometry What is the volume of CO 2 produced at 37 º C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.08206 x 310.15 K Latm molK 1.00 atm = = 4.76 L

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81 The conditions 0 ºC and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.42L) (1 mol)(273.15 K) R = 0.08206 L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.42 L.

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82 What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L

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83 Example 12.16. A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N 2 are present? PV = nRT n = PV RT STP: T = 0 0 C = 273 K,P = 1 atm n = X 273K 1.00 atm x 1.75 L L atm mol K V = 7.81 x 10 -2 mol 0.08206 At STP, 1 mole of an ideal gas occupies 22.42 L. Using Avogadro’s Law, V/n = const Two methods to solve this problem: 1. By using Ideal Gas Law or 2. By using Avogadro’s Law.

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84 Example 12.17. Gas Stoichiometry: Reactions Involving Gases at STP Answer: This is a gas stoichiometry problem. First calculate the number of moles of CO 2 from the balanced chemical equation. Then use Avogadro’s Law to calculate the volume of CO 2 gas produced at STP. Remember 1.000 mol = 22.4 L (STP) Quicklime, CaO, is produced by heating calcium carbonate, CaCO 3. Calculate the volume of CO 2 produced at STP from the decomposition of 152 g of CaCO 3 according to the reaction CaCO 3 (s) CaO (s) + CO 2 (g)

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