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Chapter 1 The Nature of Physical Chemistry and the Kinetic Theory of Gases

Chapter 1 The Nature of Physical Chemistry and the Kinetic Theory of Gases . 1.1 The Nature of Physical Chemistry. ~Physical chemistry is the application of the methods of physics to chemical problems .

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Chapter 1 The Nature of Physical Chemistry and the Kinetic Theory of Gases

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  1. Chapter 1 The Nature of Physical Chemistry and the Kinetic Theory of Gases 1.1 The Nature of Physical Chemistry ~Physical chemistry is the application of the methods of physics to chemical problems. ~Physical chemistryincludes the qualitative and quantitative study, both experimental and theoretical, of the general principles determining the behavior of matter, particularly the transformation of one substance into another. ~Although physical chemists use many of the methods of the physicist, they apply the methods to chemical structures and chemical processes. ~Physical chemistryis not so much concerned with the description of chemical substances and their reactions-this is the concern of organic and inorganic chemistry-as it is with theoretical principles and with quantitative problems. Microscopic Properties Physicochemical Study Macroscopic Properties

  2. Physical chemistry encompasses the structure of matter at equilibrium as well as the processes of chemical change. Physical chemistry’s three principle subject areas are thermodynamics, quantum chemistry, and chemical kinetics; other topics, such as electrochemistry, have aspects that lie in all three of these categories. ~Thermodynamic, as applied to chemical problems, is primarily concerned with the position of chemical equilibrium, with the direction of chemical change, and with the associated changes in energy. ~Quantum chemistry theoretically describes bonding at a molecular level. In its exact treatments, it deals only with the simplest of atomic and molecular systems, but it can be extended in an approximate way to deal with bonding in much more complex molecular structures. ~Chemical kinetics is concerned with the rates and mechanisms with which processes occur as equilibrium is approached. ~An intermediate area, known as statistical thermodynamics, links the three main areas of thermodynamics, quantum chemistry, and kinetics and also, through computer simulations, provides a basic relationship between the microscopic and macroscopic worlds. ~Related to statistical thermodynamics is nonequilibrium statistical mechanics, which is becoming an increasingly important part of modern physical chemistry. This field includes problems in such areas as the theory of dynamics in liquids, and light scattering.

  3. 1.2 Some Concepts from Classical Mechanics Work If a force F (a vector) acts through an infinitesimal distance dl (l is the position vector), the work is This equation can be integrated to determine the work in a single direction. The force F can also be resolved into three components, Fx, Fy, Fz, one along each of the three-dimensional axes. For instance, for a constant force Fx in the X-direction, x0=initial value of x

  4. Several important cases exist where the force does not remain constant, including gravitation, electrical charges, and springs. As an example, Hooke’s law states that for an idealized spring where x is the displacement from a position (x0=0) at which F is initially zero, and kh (known as force constant) relates the displacement to the force. The work done on the spring to extend it is found as follows: Note: A particle vibrating under the influence of a restoring force that obeys Hooke’s law is called a harmonic oscillator. These relationships apply fairly well to vibrational variations in bond lengths and consequently to the stretching of a chemical bond.

  5. Kinetic and Potential Energy The energy possessed by a moving body by virtue of its motion is called its kinetic energy and can be expressed as where u (=dl/dt) is the velocity (i.e., the instantaneous rate of change of the position vector l with respect to time) and m is the mass. a is acceleration u is in the same direction as du (cos=1) This equation implies that the difference in kinetic energy between the initial and final states of a point body is the work performed in the process.

  6. Problem 1.1 Calculate the amount of work required to accelerate a 1000-kg car (typical of a Honda Civic) to 88 km hr-1 (55 miles hr-1). Compare this value to the amount of work required for a 1600-kg car (typical of a Ford Taurus) under the same conditions. Solution Note: The work required is directly proportional to the mass of the car.

  7. If we assume that the force is conservative, a new function of l can be defined as This new function Ep(l) is the potential energy, which is the energy a body possesses by virtue of its position. For the case of a system that obeys Hooke’s law, the potential energy for a mass in position x is usually defined as the work done against force in moving the mass to the position from one at which the potential energy is arbitrarily taken as zero: There is no naturally defined zero of potential energy. This means that absolute potential energy values cannot be given but only values that relate to an arbitrarily defined zero energy. The potential energy rises parabolically on either side of the equilibrium position. Conservation law

  8. Problem 1.3 Atoms can transfer kinetic energy in a collision. If an atom has a mass of 110-24 g and travels with a velocity of 500 m s-1, what is the maximum kinetic energy that can be transferred from the moving atom in a head-on elastic collision to the stationary atom of mass 110-23 g? Solution u2=0 Conservation of momentum: Conservation of energy:

  9. 1.3 Systems, States, and Equilibrium ~Physical chemist attempt to define very precisely the object of their study, which is called the system. It may be solid, liquid, gaseous, or any combination of these. ~The study may be concerned with a large number of individual components that comprise a macroscopic system. Alternatively, if the study focuses on individual atoms and molecules, a microscopic system is involved. ~We may summarize by saying that the system is a particular segment of the world (with definite boundaries) on which we focus our attention. ~Outside the system are the surroundings, and the system plus the surroundings compose a universe. No material can pass between the system and the surroundings, but there can be transfer of heat. Neither matter nor heat is permitted to exchange across the boundary. There can be transfer of heat and also material.

  10. Intensive and Extensive Properties ~If the value of the properties does not change with the quantity of matter present (i.e., if it does not change when the system is subdivided), we say that the property is an intensive property. Examples are pressure, temperature, and refractive index. ~If the property does change with the quantity of matter present, the property is called an extensive property. Examples are volume and mass. ~The ratio of two extensive properties is an intensive property. There is a familiar example of this; the density of a sample is an intensive quantity obtained by the division of mass by volume, two extensive properties. intensive property gravitational field Equilibrium ~A certain minimum number of properties have to be measured in order to determine the condition or state of a macroscopic system completely. For a given amount of material it is then usually possible to write an equation describing the state in terms of intensive variables. This equation is known as an equation of state and is our attempt to relate empirical data that are summarized in terms of experimentally defined variables. ~If the variables that specify the state of the system do not change with time, then we say the system is in equilibrium. Thus, a state of equilibrium exists when there is no change with time in any of the system’s macroscopic properties.

  11. 1.4 Thermal Equilibrium ~It is common experience that when two objects at different temperatures are placed in contact with each other for a long enough period of time, their temperatures will become equal; they are then in equilibrium with respect to temperature. The concept of heat as a form of energy enters here. We observe that the flow of heat from a warmer body serves to increase the temperature of a colder body. However, heat is not temperature. ~We extend the concept of equilibrium by considering two bodies A and B that are in thermal equilibrium with each other; at the same time an additional body C is in equilibrium with B. Experimentally we find that A and C also are in equilibrium with each other. This is a statement of the zeroth law of thermodynamics: Two bodies in thermal equilibrium with a third are in equilibrium with each other. This then leads to a way to measure temperature.

  12. The Concept of Temperature and its Measurement ~On the old centigrade scale (Celsius scale) the freezing point of water at 1 atm pressure was fixed at exactly 0 C, and the boiling point at exactly 100 C. ~The construction of many thermometers is based on the fact that a column of mercury changes its length when its temperature is changed. ~In the case of mercury column, we assign its length the value l100 when it is at thermal equilibrium with boiling water vapor at 1 atm pressure. The achievement of equilibrium with melting ice exposed to 1 atm pressure establishes the value of l0 for this length. ~Assuming a linear relationship between the temperature  and the thermometric property (length, in this case), and assuming 100 divisions between the fixed marks, allows us to write l is the length at temperature  and l0 and l100 are the lengths at the freezing and boiling water temperatures, respectively. ~Some thermometric properties do not depend on a length, such as in a quartz thermometer where the resonance frequency response of quartz crystal is used as the thermometric property. The above equation still applies, however. ~Thermometric properties of actual materials generally deviate from exact linearity, even over short ranges, because of the atomic or molecular interactions within the specific material, thus reducing the value of that substance to function as thermometric material over large temperature ranges.

  13. 1.5 Pressure and Boyle’s Law barometer ~Pressure is the force per unit area. ~Atmospheric pressure is often measured as a difference in height, h, of a mercury column trapped in an inverted tube suspended in a pool of mercury, as shown in Figure. ~The pressure is proportional to h, where P=gh,  is the density, and g is the acceleration of gravity. Units of Pressure 1 atm=760 mmHg=101325 Pa=760 Torr=1.0133 bar A gas contained within a closed vessel exerts a force on the walls of the vessel. This force is related to the pressure P of gas (the force F divided by the wall area A) and is a scalar quantity; that is, it is independent of direction.

  14. manometer ~The pressure of a gas contained in a closed vessel may be measured using a manometer. ~Two versions are in common use (see Figure). Both consist of a U-tube filled with a liquid of low volatility (such as mercury or silicone oil). In both, the top of one leg of the U-tube is attached to the sample in its container. ~In the closed-end manometer, the sample pressure is directly proportional to the height difference of the two columns. ~In the open-end manometer, the difference in height of the two column is proportional to the difference in pressure between the sample and the atmospheric pressure.

  15. Example 1.1 Compare the length of a column of mercury to that of a column of water required to produce a pressure of 1.000 bar. The densities of mercury and water at 0.00 C are 13.596 gcm-3 and 0.99987 gcm-3, respectively. Solution The pressure exerted by both liquids is given by P=gh. Since the length of both liquids must exert the same pressure, we can set the pressure equal with the subscripts Hg and w, denoting mercury and water, respectively. 760 mmHg=1.0133 bar 1 bar = 10.199 m water

  16. Boyle’s Law The pressure of a fixed amount of gas varies inversely with the volume if the temperature is maintained constant. ~A plot of 1/P against V for some of Boyle’s original data is shown in Figure 1.7a. ~The advantage of this plot over a P against V plot is that the linear relationship makes it easier to see deviations from the law. ~Boyle’s law is surprisingly accurate for many gases at moderate pressures. ~In Figure 1.7 b, we plot P against V for a gas at several different temperature. Each curve of PV=constant is a hyperbola, and since it represents a change at constant temperature, the curve is called an isotherm. Figure 1.7

  17. Problem 1.10 The volume of a vacuum manifold used to transfer gases is calibrated using Boyle's law. A 0.251-dm3 flask at a pressure of 697 Torr is attached, and after system pumpdown, the manifold is at 10.4 mTorr. The stopcock between the manifold and flask is opened and the system reaches an equilibrium pressure of 287 Torr. Assuming isothermal conditions, what is the volume of the manifold? Solution

  18. Problem 1.14 A J-shaped tube is filled with air at 760 Torr and 22 C. The long arm is closed off at the top and is 100.0 cm long; the short arm is 40.00 cm high. Mercury is poured through a funnel into the open end. When the mercury spills over the top of the short arm, what is the pressure on the trapped air? Let h be the length of mercury in the long arm. Solution where P2 is the pressure Since h, the height of the mercury column on the trapped air side, is proportional to the volume of a uniform tube, we can write P1=760cmHg h=195.5 cmHg or 20.5 cmHg h=20.5 cmHg

  19. 1.6 Gay-Lussac’s (Charles’s) Law ~Guillaume Amontons measured the influence of temperature on the pressure of a fixed volume of a number of different gases and predicted that as the air cooled, the pressure should become zero at some low temperature, which he estimated to be -240 C. He thus anticipate the work of Jacques Alexandre Charles, who a century later independently derived the direct proportionality between the volume of a gas and the temperature. ~Since Charles never published his work, it was left to Joseph Louis Gay-Lussac, proceeding independently, to make a more careful study using mercury to confine the gas and to report that all gases showed the same dependence of V on . He developed the idea of an absolute zero of temperature and calculated its value to be -273 C. Thus, for a particular value of the temperature  and a fixed volume of gas V0 and 0 C, we have the linear relation Figure 1.8 where  is the cubic expansion coefficient. The modern value of  is 1/273.15. As shown in Figure 1.8, the curves of the experimentally determined region can be extrapolated to zero volume where  is -273.15 C.

  20. ~This fact immediately suggest that the addition of 273.15 to the Celsius temperature would serve to define a new temperature scale T that would not have negative numbers. The relationship between the two scales is best expressed as That is, the value of the absolute temperature (i.e., the temperature divided by its unit) is obtained simply by adding 273.15 to the value of the Celsius temperature. ~On the new scale, 100 C is therefore 373.15 K. Note that temperature intervals remain the same as on the Celsius scale. ~This new scale is called the absolute Kelvin temperature scale or the Kelvin temperature scale. Figure 1.9 Gay-Lussac’s (Charles’s) Law at moderate to high temperatures only as the pressure is reduced to zero.

  21. 1.7 The Ideal Gas Thermometer If the temperature and volume of a fixed amount of gas held at some low pressure are T1 and V1, respectively, before the addition of heat, the ratio of the temperature of T2 to T1 after the addition of heat is given by the ratio of the initial volume and the final volume, V2, of the gas. low-pressure limit of a gas volume is infinite, If a gas can obey these relationship exactly for all values of P, then we have defined an ideal gas; the thermometer using such a gas is known as the ideal gas thermometer. Using the defined value of the triple point for T1, a working definition for the new Celsius scale becomes where limP0(PV)T=0 when T2=0 triple point of water

  22. Problem 1.26 An ideal gas thermometer and a mercury thermometer are calibrated at 0 C and at 100 C. The thermal expansion coefficient for mercury is where  is the value of the Celsius temperature and V = V0 at  = 0. What temperature would appear on the mercury scale when the ideal gas scale reads 50 C? Solution

  23. 1.8 The Equation of State for an Ideal Gas ~Experimentally, the macroscopic properties (pressure, volume, and temperature) of a gas cannot be arbitrarily chosen to describe the state of a fixed amount of gas. (The effects of gravitational, electric, and magnetic fields are neglected in this treatment.) ~For a fixed amount of gas these basic properties are related by an equation of state. However, one finds experimentally that the linear relationship of Boyle’s law for P-V data is attained only at very low pressure. Hence, in the limit of zero pressure, all gases should obey Boyle’s law to the same degree of accuracy. Thus, Boyle’s law could be better written as Avogadro’s hypothesis R is the universal gas constant ~A given volume of any gas (at a fixed temperature and pressure) must contain the same number of independent particles. ~The particles of gas could be atoms or combinations of atoms (molecule). One mole is the amount of any substance containing the same number of elementary entities (atoms, molecules, ions, and so forth) as there are in exactly 0.012 kg of carbon-12. L (Avogadro constant):6.0221371023 mol-1

  24. Equation of State for an Ideal Gas What it states is that in any sample of gas behaving ideally, if one of the four variables (amount, pressure, volume, or temperature) is allowed to change, the values of the other three variable will always be such that a constant value of R is maintained. where n, the amount of substance, is the mass m divided by the molar mass M, and m/V is the density  of the gas.

  25. Example 1.2 Calculate the average molar mass of air at sea level and 0 C if the density of air is 1.29 kgm-3. Solution At sea level the pressure may be taken equal to 1 atm or 10132 Pa.

  26. Problem 1.7 Vacuum technology has become increasingly more important in many scientific and industrial applications. The unit torr, defined as 1/760 atm, is commonly used in the measurement of low pressures. a.Find the relation between the older unit mmHg and the torr. The density of mercury is 13.5951 g cm-3 at 0.0 C. The standard acceleration of gravity is defined as 9.80665 m s-2. b.Calculate at 298.15 K the number of molecules present in 1.00 m3 at 1.0010-6 Torr and at 1.00 10-15 Torr (approximately the best vacuum obtainable). Solution a. A column of mercury 1 m2 in cross-sectional area and 0.001 m in height has a volume of 0.001 m3 and a mass of 0.001 m313595.1 kg m-3. Then 1 mmHg= 0.001 m313595.1 kg m-39.80665 m s-2/1 m2=133.3223874 Pa By definition, 1 atmosphere=101325 Pa 1 Torr=101325(Pa)/760=133.3223684 Pa Thus 1 mmHg=133.3223874/133.3223684=1.00000014 Torr

  27. b. For P=10-6 Torr: For P=10-15 Torr:

  28. Problem 1.22 A gas mixture containing 5 mol % butane and 95 mol % argon (such as is used in Geiger-Muller counter tubes) is to be prepared by allowing gaseous butane to fill an evacuated cylinder at 1 atm pressure. The 40.0-dm3 cylinder is then weighed. Calculate the mass of argon that gives the desired composition if the temperature is maintained at 25.0 C. Calculate the total pressure of the final mixture. The molar mass of argon is 39.9 g mol-1. Solution The amount of butane present is given by n=PV/RT 5 mol % butane and 95 mol % argon Mass of argon = 31.2 mol39.9 g mol-1 = 1240 g The final pressure, Pf, is proportional to the total amount of gas; thus Pf=1 atm100/5=20 atm

  29. 1.9 The Kinetic-Molecular Theory of Ideal Gases ~An experimentally study of the behavior of a gas, such as that carried out by Boyle, cannot determine the nature of the gas or why the gas obeys particular laws. ~In order to understand gases, one could first propose some hypothesis about the nature of gases. Such hypothesis are often referred to as constituting a “model” for a gas. ~The properties of the gas that are deduced from this model are then compared to the experimental properties of the gas. The validity of the model is reflected in its ability to predict the behavior of gases. ~We will simple state three postulates of the kinetic-molecular model and show how this model leads to the ideal gas laws. This model of an idealized gas will be seen to fit the behavior of many real gases. 1.The gas is assumed to be composed of individual particles (atoms or molecules) whose actual dimensions are small in comparison to the distances between them. 2.These particles are in constant motion and therefore have kinetic energy. 3.Neither attractive nor repulsive forces exist between the particles.

  30. ~In order to see how this model predicts the observed behavior quantitatively, we must arrive at an equation relating the pressure and volume of a gas to the important characteristic of the gas, namely the number of particles present, their mass, and the velocity with which they move. ~We will first focus our attention on a single molecule of mass m confined in an otherwise empty container of volume V. ~The particles traverses the container with velocity u (a vector quantity indicating both speed and direction). Figure 1.10 ~Because the particle’s X-component of velocity (i.e., the component ux as shown in Figure 1.10) is normal to the YZ wall, the molecule will traverse the container of length x in X-direction, collide with the wall YZ, and then rebound. ~In the impact, the molecule exerts a force Fw on the wall. This force is exactly counterbalanced by the force F exerted on the molecule by the wall.

  31. The force F is equal to the change of momentum p of the molecule in the given direction per unit time, in agreement with Newton’s second law of motion, The molecule’s momentum in the X-direction when it strikes the wall is mux. Since we assume that the collision is perfectly elastic, the molecule bounces off the wall with a velocity –ux in the opposite direction. The change in velocity of the molecule on each collision is The corresponding change of momentum is -2mux. Since each collision with this wall occurs only after the molecule travels a distance of 2x (i.e., one round trip), the number of collisions a molecule makes in unit time may be calculated by dividing the distance ux the molecule travels in unit time by 2x. The result is

  32. The change of momentum per unit time is thus The force Fw exerted on the wall by the particle is exactly equal in magnitude to this but with opposite sign: Since the pressure is force per unit area and the area A is yz, we may write the pressure in the X-direction, Px, as xyz=V is the volume of the container.

  33. The Pressure of a Gas Derived from Kinetic Theory ~For an assembly of N molecules there will be a distribution of molecular velocities, since even if the molecules all again with the same velocity, collisions would occur altering the original velocity. ~If we define ui2 as the square of the velocity component in the X-direction of molecule i and take average of this over molecules rather than the sum of over ui2, we have is the mean of the squares of the normal component of velocity in the X-direction This is the equation for the pressure of a one-dimensional gas. ~It is more convenient to write these expressions in terms of the magnitude of the velocity u rather than in terms of the squares of the velocity components. ~The word speed is used for the magnitude of the velocity; speed is defined as the positive square root of u2 and is given by If we average over all molecules, we obtain

  34. Since there is no reason for one direction to be favored over the others, the mean of the ux2 values will be the same as the mean of the uy2 and the mean of the uz2 values. Hence, ~This is the fundamental equation as derived from the simple kinetic theory of gases. ~This equation is in the form of Boyle’s law and is consistent with Charles’s law if is directly proportional to the absolute temperature. N/n is equal to the Avogadro constant, L. mL=M, molar mass

  35. Kinetic Energy and Temperature Boltzmann constant kB is the gas constant per molecule. is independent of the kind of substance, the average molecular kinetic energy of all substances is the same at a fixed temperature. kB=1.38062210-23 J K-1 Considering a number of different gases all at the same temperature and pressure, Thus, N1=N2=…=Ni when the volumes are equal. In other words, equal volumes of gases at the same pressure and temperature contain equal numbers of molecules. This is just a statement of Avogadro’s hypothesis.

  36. Problem 1.30 Nitrogen gas is malntained at 152 kPa in a 2.00-dm3 vessel at 298.15 K. If its molar mass is 28.0134 g mol-1 calculate: a. The amount of N2 present. b. The number of molecules present. c. The root-mean-square speed of the molecules. d. The average translational kinetic energy of each molecule. e. The total translational kinetic energy in the system. Solution a. b. The number of molecules = amount of substance  Avogadro’s constant = 0.123(mol)6.0221023(mol-1)=7.411022 c.

  37. d. e. Ek(total)=0.123(mol)6.0221023(mol-1)6.17410-21(J)=457 J

  38. Dalton’s Law of Partial Pressures The total pressure observed for a mixture of gases is equal to the sum of the pressures that each individual component gas would exert had it alone occupied the container at the same temperature. Note: In order for the law to be obeyed, no chemical reactions between component gases may occur. The term partial pressure, Pi, is used to express the pressure exerted by one component of the gas mixture and is defined as xiis the mole fraction of the gas(i). Pt is the total pressure.

  39. Example 1.3 The composition in volume percent of standard “1976” dry air at sea level is approximately 78.084% N2, 20.948% O2, 0.934% Ar, and 0.031% CO2. All other gases, including (in decreasing order) Ne, He, Kr, Xe, CH4, and H2, constitute only 2.710-3% of the atmosphere. What is the partial pressure of each of the first four gases listed at a total pressure of 1 atmosphere? What the partial pressures (in units of kPa) if the total pressure is 1 bar? Solution 78.084% N2, 20.948% O2, 0.934% Ar, 0.031% CO2, 2.710-3% other gases weight percent 75.520% N2, 23.142% O2, 1.288% Ar, 0.031% CO2, 4.7610-2% other gases mole fraction 0.77898 N2, 0.20872 O2, 0.009316 Ar, 0.000312 CO2 partial pressure 0.77898atm N2, 0.20872atm O2, 0.009316atm Ar, 0.000312atm CO2 Pt=1 atm partial pressure 77.898kPa N2, 20.872kPa O2, 0.9316kPa Ar, 0.0312kPa CO2 Pt=1 bar=100kPa

  40. Graham’s Law of Effusion ~Graham measured that the movement of gases through plaster, fine tubes, and small orifices in plates where the passages for the gas are small as compared with the average distance that the gas molecules travel between collisions. Such movement is known as effusion. ~Effusion is sometimes mistakenly referred to as diffusion, but in that process there are no small passage ways to inhibit the moving particles. ~Graham showed that the rate of effusion of a gas was inversely proportional to the square root of its density . Later, he showed that the rate of effusion was inversely proportional to the square root of the molar mass M. This is know as Graham’s law of effusion. ~In the case of the gases oxygen and hydrogen at equal pressures, oxygen molecule are 32/2=16 times more dense than those of hydrogen. Therefore, hydrogen effuses 4 times as fast as oxygen: As a comparison, the speed of sound in dry air of density 1.29 gL-1 is 346.2 m s-1 at 298.15 K. Graham’s law can be explained by the simple kinetic molecule theory.

  41. Molecular Collisions ~We now apply the kinetic theory to the investigation of the collisions of molecules in order to have a clearer understanding of the interactions in a gas. ~We will be interested in three aspects of molecular interaction: the number of collisions experienced by a molecule per unit time, the total number of collisions in a unit volume per unit time, and how far the molecules travel between collisions. ~We consider that the molecules behave as rigid spheres in elastic collisions and that there are two kinds of molecules, A and B, with diameters dA and dB. ~Suppose that a molecule of A travels with an average speed of in a container that contains both A and B molecules. ~We will first assume that the molecules of B are stationary and later remove this restriction. ~As shown in Figure 1.11, a collision will occur each time the distance between the center of a molecule A and that of a molecule B becomes equal to dAB=(dA+dB)/2, where dAB is called the collision diameter.

  42. Figure 1.11 ~A convenient way to visualize this is to construct around the center of A an imaginary sphere of radius dAB, which is the sum of two radii. ~In unit time this imaginary sphere, represented by the dashed circle in Figure 1.11, will sweep out a volume of . ~If the center of a B molecule is in this volume, there will be a collision. ~If NB is the total number of B molecules in the system, the number per unit volume is NB/V, and the number of centers of B molecules in the volume swept out is .

  43. The number ZA of collision experienced by the one molecule of A in unit time is collision frequency ~If there is a total of NA/V molecules of A per unit volume in addition to the B molecules, the total number ZAB of A-B collision per unit volume per unit time is ZA multiplied by NA/V: collision density or collision number ~If only A molecules are present, we are concerned with the collision density ZAA for the collisions of A molecules with one another. 1/2counting each collision twice

  44. ~An error in this treatment of collision is that we have only considered the average speed of the A molecules. More correctly, we should consider the relative speeds of the molecules. ~When we have a mixture containing two kinds of molecules A and B of different masses, the values are different. The average relative speed is equal to . only A molecules present

  45. Example 1.4 Nitrogen and oxygen are held in a 1.00 m3 container maintained at 300 K at partial pressure of PN2=80 kPa and PO2=21 kPa. If the collision diameters are dN2=3.7410-10 m and dO2=3.5710-10 m, calculate ZA, the average number of collisions experienced in unit time by one molecule of nitrogen and by one molecule of oxygen. Also calculate ZAB, the average number of collisions per unit volume per unit time. Do this last calculation both at 300 K and 3000 K on the assumption that values for d and N do not change. At 300 K, (uN22+uO22)1/2 is 625 m s-1; at 3000 K, it is 2062 m s-1. Solution

  46. At 300 K, At 3000 K, From this example it can be seen that the effect of T on Z is not large since it enters as Tand the effect of d is much more pronounced since it enters as d2.

  47. Mean Free Path ~ average distance that a molecule travels between two successive collisions Example 1.5 Molecular oxygen has a collision diameter of 3.5710-10 m. Calculate  for oxygen at 300 K and 101.325 kPa. Solution

  48. 1.10 The Barometric Distribution Law ~In this section we will consider the effect of a gravitational field. ~In a laboratory experiment the effect of the gravitational field can usually be ignored, except in cases involving surface and interfacial tensions. However, for a large-scale system such as our atmosphere or an ocean, gravity can cause appreciable variation in properties. ~The effect of gravity on the pressure can be determined by considering a column of fluid (either liquid or gas) at constant temperature as shown in Figure 1.12. Figure 1.12 The change in pressure is thus proportional to the length of the column, and since dz is positive, the pressure decreases with an increase in height.

  49. For liquid, P0 is the reference pressure at the base of the column and P is the pressure at height z. This quantity P-P0 is the familiar hydrostatic pressure in liquids. For an ideal gas, Barometric Distribution Law This expression describes the distribution of gas molecules in atmosphere as function of their molar mass, height, temperature, and the acceleration due to gravity. For a given gas, a smaller relative pressure change is expected at high temperature than at low temperatures. At a given temperature, a gas having a higher molar mass is expected to have a larger relative decrease in pressure than a gas with a lower molar mass.

  50. In the upper reaches of the earth’s atmosphere the partial pressure will be relatively higher for a very light gas such as helium. This fact explains why helium must be extracted from a few helium-producing natural-gas wells in the United States and in Russia where helium occurs underground. Mgz is the gravitational potential energy, Ep density  is directly proportional to pressure 0 represents the density at the reference state height of z=0 These equations, in which the property varies exponentially with Ep/RT, are special cases of the Boltzmann distribution law.

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