All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.

Description

8.4Important LinksDigital Support Online quizzes, study notes, and morewww.edvantagescience.comOrder your own copy of AP Chemistry 1…

Transcript

8.4Important LinksDigital Support Online quizzes, study notes, and morewww.edvantagescience.comOrder your own copy of AP Chemistry 1 www.edvantageinteractive.com8.4 The Kinetic Molecular Theory and Real Gas Behavior Warm Up Assume the box shown here has a volume of 22.4 L and exists under standard temperature and pressure (STP) conditions. Draw a molecular representation of oxygen gas filling the box. List three ways in which your drawing fails to represent the actual situation.A Model for Gases — The Kinetic Molecular TheoryYour first introduction to chemistry in elementary school may have been a study of the kinetic molecular theory of matter (KMT). This theory states that matter is composed of particles such as molecules, atoms, or ions in continuous motion (Figure 8.4.1). In the solid state, the particles are close together and only vibrate. In a liquid, the particles are farther apart and move around. In a gas, the particles are much farther apart and move around freely in a random fashion. A mole of gas occupies 22.4 L, while a mole of most liquids has a volume of only 18 mL. In this section, we will take a closer look at how this theory applies to gases.Figure 8.4.1 The particles of a solid are packed tightly together. Liquid particles are farther apart and canmove freely. A gas is virtually impossible to represent because the particles are no more than tiny points in space with less than 0.01% of the volume occupied by the actual gas molecules. There is a regular distribution of speeds for the molecules. Some are hardly moving while others have far more energy and are moving very quickly.When we use KMT to explain the behavior of gases, we make a number of assumptions about the size of the molecules, the way they behave, and what does and does not occur during particle collisions. Such assumptions are included in the postulates of the kinetic molecular theory.490 Chapter 8 Gases© Edvantage Interactive 2017The Postulates of the Kinetic Molecular Theory 1. The volume of gas particles can be assumed to be negligible relative to the space between them. 2. Gas particles are in constant motion. Gas particles colliding with the walls of their container produce pressure. 3. Gas particles neither attract nor repel one another. 4. The average kinetic energy of a gas sample is directly proportional to the temperature of the gas in Kelvin degrees.Molecular SizeContinuous Motion of Gas Particles The size of individual gas particles is so small compared to the distance between them that at room temperature and standard atmospheric pressure more that 99.99% of the volume of any gas is actually empty space. Because of all this empty space, it is easy to compress a gas so that it occupies a smaller volume just by applying an external force. Individual gas particles are in perpetual, random, straight-line motion. During the course of this motion, they frequently collide with one another and then rebound in a perfectly elastic manner. The continuous random motion of gas particles prevents them from collecting in one area and causes them to be uniformly distributed throughout any container they are placed in. This movement is called “diffusion” and is a property of all gases. Diffusion also allows gases of different types to be mixed together to form solutions of any proportion. Kinetic energy is the energy resulting from movement. The value of the kinetic energy of an objectKinetic Energy of Gas depends on its mass and velocity. This may be expressed mathematically as kinetic energy = ½mv2 where m is the mass and v is the velocity. Particles KE = 1/2mv2 There is a normal distribution of energies in any sample of matter (Figure 8.4.2).Number of particlesNumber of Particles vs. EnergyEnergy Figure 8.4.2 Any gas sample contains a normal distribution of energies.At any given temperature, however, the molecules in all gases have the same average kinetic energy. This average kinetic energy is directly proportional to the Kelvin temperature. KE = 3/2RT© Edvantage Interactive 2017Chapter 8 Gases 491Thus, if we have a sample of hydrogen gas and oxygen gas at the same temperature, both gases will have the same average kinetic energy. It follows that the average velocity of an oxygen molecule must be less than that of a hydrogen molecule since the mass of the oxygen is 16 times greater. The relatively high velocities of molecules of hydrogen and atoms of helium account for the fact that Earth’s atmosphere contains very little of these two gases. They escaped from Earth’s gravitational attraction many millions of years ago. In any sample of gas, the particles are moving very quickly. Gas particles at room temperature orThe Elasticity of Gas 25°C move at speeds in the order of 200 to 2000 m/s, which convert to 450 to 4500 miles per hour. As explained on the previous page, the speed is related to the mass of the gas particles (Figure 8.4.3). Particles As a result of this tremendous speed, gas particles collide with one another very frequently. One of the assumptions of the KMT is that no energy is lost during these collisions. In other words, the collisions are perfectly elastic. If this were not the case, the particles would gradually lose their kinetic energy. In fact, particles do transfer kinetic energy from one to another during collisions, but the average kinetic energy of all the particles present in a sample remains constant. Collisions between the gas particles and the walls of their container produce pressure (Figure 8.4.3).Low temperatureHigh temperatureFigure 8.4.3 The average velocity of gas particles is directly related to the Kelvin temperature of the sample.There is a normal distribution of energies in any gas sample with some particles barely moving while others move extremely quickly. Collisions between the particles are perfectly elastic and collisions with the walls of the container produce pressure. The representations above show the molecules much larger than they actually are. In reality, less than 0.01% of the volume of the container is occupied by gas particles.Quick Check Use the postulates of the kinetic molecular theory to explain each of the following gas laws: (a) Boyle’s law(b) Charles’s law(c) Gay-Lussac’s law(d) Dalton’s law of partial pressures492 Chapter 8 Gases© Edvantage Interactive 2017The Kelvin temperature is a measure of the average kinetic energy (KE) of any sample of gas particles.Root Mean Square Velocity For a mole of any gas:KEavg =3 RT 2 (I)where R = 8.31 J/K·mol Also, KEavg = NA(½mμ2) (II) where NA = the number of particles in a mole (Avogadro’s number) m = the mass of the gas particle μ2 = average of the squares of the particle velocities Thus, the average kinetic energy of one gas particle in a sample is equal to ½mμ2. The average velocity of a gas particle is called the root mean square velocity and is written as: Vrms = √ μ2 Rearranging equations (I) and (II) from above gives:3 RT = ½mμ2 2μrms =3RT mIt is necessary to insert the molar mass in kg to allow proper unit analysis. One kg•m2 joule is equivalent to one . The molar mass in kg cancels, leaving the square root s2 of m2/s2, which gives an appropriate unit for velocity of m/s.© Edvantage Interactive 2017Chapter 8 Gases 493Sample Problem — Root Mean Square Velocity and Average Kinetic Energy Calculate the root mean square velocity and the kinetic energy of carbon monoxide gas at 25°C.What To Think AboutHow To Do It1. The root mean square velocity equals the square root of 3RT/M, where R is 8.31 J/mol·K, T is in Kelvin degrees and M is the molar mass in kg.µrms = 2. Be sure to divide before taking the square root.µrms = kg•m2 for s2 J and insertion of the molar mass in kg/mol.3. Unit analysis requires the substitution of 4. The kinetic energy is simply 3/2RT where R = 8.31 J/ mol·K and the temperature is in Kelvin degrees.µrms = √ √3RT M√kg•m2/s2 3(8.31 )(298 K) mol•K 0.028 kg/mol3(8.31 J/mol•K)(298 K) 0.0028 kg = 515 m/s KEavg = 3/2RT = 3/2 x 8.31 J/mol K x 298 K = 3710 JPractice Problems — Root Mean Square Velocity & Average Kinetic Energy 1. Calculate the root mean square velocity and the average kinetic energy of atoms of Ar gas at 25°C.2. Calculate the root mean square velocity and the average kinetic energy of ozone molecules, O3, high in the stratosphere where the temperature is –83°C.494 Chapter 8 Gases© Edvantage Interactive 2017In a real gas, there are large numbers of collisions between gas particles. When the lid is removedRange of Velocities in from a perfume bottle, it takes some time for the odor to move through the air. This is due to collisions occurring between the perfume vapor and the oxygen and nitrogen particles that make up a Gas SampleRelative number of moleculesmost of the air. The average distance a particle travels between collisions in a sample of gas is called the mean free path. This distance is typically very small and in the range of 60 nanometers. The many collisions result in a wide range of velocities as particles collide and exchange kinetic energy. Figure 8.4.4 and Figure 8.4.6 illustrate the velocity distributions for various gases at STP conditions and a particular gas under changing conditions of temperature. The peak of each curve indicates the most common particle velocity.O2 CO2 N2 H2 SpeedFigure 8.4.4 The velocity distributions reflect the kinetic energy distribution for all the particles in samples ofRelative number of moleculesCO2, O2, N2, and H2 gas. While the average velocity of an oxygen molecule is 500 m/s, many O2 particles have other velocities. As the mass of a particle increases, its average velocity decreases.Hot Room Temp. Cold VelocityFigure 8.4.5 Increasing the temperature of a gas sample broadens the entire distribution of velocities, causingthe average velocity of a particle to increase.Effusion and Diffusion © Edvantage Interactive 2017Two important terms for describing gases are diffusion and effusion. The odor emitted from a perfume bottle takes several moments to permeate an entire room as the vapor bounces off oxygen and nitrogen molecules, zigzagging its way through the air in random straight-line motion. Diffusion is the term used to describe the mixing of one gas through another. The rate of diffusion is the rate of this mixing. The rate of diffusion is calculated by directly relating the relative distance travelled by two types of molecules in the same period of time to their relative velocities. Both of these ratios are inversely related to the square root of the ratio of the molecules’ masses.Chapter 8 Gases 495distance travelled by gas distance travelled by1gas 1 µrms v forforgas gas11 m2 = = = rm s = distance travelled by gas distance travelled by 2 gas 2 µrms vrm sfor forgas gas22 m1Effusion is the passage of a gas through a tiny orifice into an evacuated chamber. The rate of effusion measures the speed at which the gas is transferred into the chamber.This might take a while...No problem!NeHeFigure 8.4.6 Lighter gases move more quickly and effuse more rapidly than heavier ones, as predicted by theequation, KE = 1/2mv2.Scottish chemist Thomas Graham developed Graham’s law of effusion. It states the rate of effusion is inversely proportional to the square root of the mass of its particles. PinholeHelium (He)Evacuated chamberEthylene oxide (C2H4O)Evacuated chamberFigure 8.4.7 The rate of effusion of a light gas such as helium is much faster than that of a more massive gaslike ethylene oxide.Since the effusion rate for a gas depends directly on the average velocity of its particles, the rates of effusion of two gases at the same temperature may be compared by an equation similar to the one used for diffusion above.496 Chapter 8 Gases© Edvantage Interactive 2017Sample Problem — Graham’s Law Carbon monoxide effuses at a rate that is 2.165 times faster than that of an unknown noble gas. Calculate the molar mass of the unknown gas and determine its identity.What to Think AboutHow To Do It1. Graham’s law of effusion is required to solve this problem. 2. Identify the lighter gas as gas 1 so the effusion rate ratio is >1. This makes the algebra a little easier.effusion Rate for gas 1 effusion Rate for gas 2=M2 M1effusion Rate for gas 1 effusion Rate for gas 2= M of X M of CO3. Substitute 2.165 for the effusion rate ratio. The molar mass of CO may be substituted as 28.0 g/mol. It is not necessary to convert the molar mass to kg as the kilo prefix will cancel in the ratio form of the equation.2.165= Mof X 28.0 g / mol4. Determine the identity of the noble gas by comparison of known molar masses of the noble gas family with the calculated molar mass.∴ M of X=2.165( 28.0 g / mol) = 11.4561 g / mol∴ M of X = 11.45612=131 g / mol The molar mass matches that of xenon gas.Practice Problems — Graham’s Law 1. Calculate the ratio of the effusion rates of H2 and UF6. UF6 is a gas used to produce fuel for nuclear reactors.2. A popular experiment involves the determination of the distances travelled by NH3(g) and HCl(g) from opposite ends of a glass tube. When the two gases, meet they form a visible white ring of NH4Cl. NH3 17 g/molNH4ClHCl 36.5 g/moldistance 100 cm(a) Which end of the tube will be closer to the white ring of ammonium chloride? Justify your answer with an explanation.Continued© Edvantage Interactive 2017Chapter 8 Gases 497Practice Problems — Continued 2. (b) Calculate the ratio of the distance traveled by NH3(g) to distance traveled by HCl(g).(c) Give two factors that must be carefully controlled to ensure the expected result.3. If the average velocity of a N2 molecule is 475 m/s at 25°C, what is the average velocity of a He atom at 25°C?The Effects of Volume and Intermolecular Forces on Real GasesMeasurement of the volume of one mole of most gases at STP results in a number that is near, but seldom exactly equal to 22.4 L. As pressure increases and temperature decreases, the volume of one mole of gas deviates even further from this value. In a previous section it was stated that gases behave most ideally when pressure is low and temperature is high. The reason for this is two-fold: 1. Real gases do have some volume, while ideal gases are assumed to have no volume and to exist as tiny points in space. 2. Real gases exert some attractive forces on their neighboring particles, while ideal gases are assumed to be completely independent of the gas particles that surround them. In 1873, Johannes van der Waals modified the ideal gas law to fit the behavior of real gases. Van der Waals inserted two small correction factors. The first accounts for the volume of the gas particles themselves and the second corrects for the IMFs existing between gas particles. The Van der Waal’s equation is really just a more complex form of the ideal gas law.n P+a V{( )2ideal gas pressureactual volume ×(V – nb) = nRT{actual pressureideal gas volumeThe “correction factors” include an a factor and a b factor, which account for the attractive forces and the volume that real gases demonstrate. Every gas has its own unique values for a and b. Note that the units of a and b allow proper cancellation to produce units of pressure and volume respectively when substituted into the van der Waals equation.498 Chapter 8 Gases© Edvantage Interactive 2017Table 8.4.1 Van der Waals Constants for Common Gasesa atm liters2 / mole2b liters/moleN21.390.0391O21.360.0318CO1.490.0399CO23.590.0427CCl420.40.138HCl3.670.408H20.2440.0266H2O5.640.0237CH42.250.0428He0.0340.0237Ne0.2110.0171Ar1.350.0322Kr2.320.0398Xe4.190.0511The effect of IMF’s and gas volume on real gas behavior becomes quite obvious when you examine a graph of PV/nT versus P for real gases (Figure 8.4.8). Since PV = nRT, PV/nT should equal R.Departure from Ideal Gas Behaviour H28.4PV (J/mol K) nTIdeal gas R = 8.3145 8.3N2 CO8.2 8.1 8.0 7.9 0O2 5101520P (atmospheres) Figure 8.4.8 As pressure increases, deviation from ideal behavior becomes more pronounced. The deviationis most noticeable for larger gases. This makes sense as larger gases have a greater volume and exert larger London dispersion forces on each other.The effect is further demonstrated with examination of a graph of PV/RT versus P for one mole of a gas at three different temperatures (Figure 8.4.9). Since PV = nRT, PV/RT should equal 1 for one mole of gas.© Edvantage Interactive 2017Chapter 8 Gases 4993200 K 500 K2 PV RT1000 K Ideal gas10 0300600900P (atmospheres) Figure 8.4.9â€‚ As temperature decreases, deviation from ideal behavior becomes more pronounced. Gasesbehave most ideally when the temperature is high.Examination of the data in Table 8.4.1 and the graphs in Figure 8.4.8 and Figure 8.4.9 provides support for the statement that gases behave most ideally when pressure is low and temperature is high. Low pressure and high temperature make real gas volume and real gas attractive forces less noticeable.500â€ƒ Chapter 8 GasesÂŠ Edvantage Interactive 2017Quick Check 1. Examine the graphs in Figure 8.4.8 and Figure 8.4.9. (a) Which particle deviates the most from ideal behavior? Give two reasons for this behavior.(b) What temperature provides the most ideal behavior? Explain.2. Explain the general pattern for the values of a and b moving down the noble gas family. (A simple statement of the trend is not an adequate answer.)3. Explain the same pattern moving from N2 to CO2 to CCl4.4. Explain the same pattern moving from N2 to CO to HCl to H2O.ÂŠ Edvantage Interactive 2017Chapter 8 Gasesâ€ƒ 501

Related Search

We Need Your Support

Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks