Chapter 21. The Kinetic Theory of Gases

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1 Chapter The Kinetic Theory of Gases CHAPTER OUTLINE. Molecular Model of an Ideal Gas. Molar Specific Heat of an Ideal Gas.3 Adiabatic Processes for an Ideal Gas.4 The Equipartition of Energy.5 The Boltzann Distribution Law.6 Distribution of Molecular Speeds.7 Mean Free Path Dogs do not have sweat glands like huans. In hot weather, dogs pant to proote evaporation fro the tongue. In this chapter, we show that evaporation is a cooling process based on the reoval of olecules with high kinetic energy fro a liquid. (Frank Oberle/Getty Iages) 640

2 In Chapter 9 we discussed the properties of an ideal gas, using such acroscopic variables as pressure, volue, and teperature. We shall now show that such large-scale properties can be related to a description on a icroscopic scale, where atter is treated as a collection of olecules. Newton s laws of otion applied in a statistical anner to a collection of particles provide a reasonable description of therodynaic processes. To keep the atheatics relatively siple, we shall consider priarily the behavior of gases, because in gases the interactions between olecules are uch weaker than they are in liquids or solids. In our odel of gas behavior, called kinetic theory, gas olecules ove about in a rando fashion, colliding with the walls of their container and with each other. Kinetic theory provides us with a physical basis for our understanding of the concept of teperature.. Molecular Model of an Ideal Gas We begin this chapter by developing a icroscopic odel of an ideal gas. The odel shows that the pressure that a gas exerts on the walls of its container is a consequence of the collisions of the gas olecules with the walls and is consistent with the acroscopic description of Chapter 9. In developing this odel, we ake the following assuptions:. The nuber of olecules in the gas is large, and the average separation between the is large copared with their diensions. This eans that the olecules occupy a negligible volue in the container. This is consistent with the ideal gas odel, in which we iagine the olecules to be point-like.. The olecules obey Newton s laws of otion, but as a whole they ove randoly. By randoly we ean that any olecule can ove in any direction with any speed. At any given oent, a certain percentage of olecules ove at high speeds, and a certain percentage ove at low speeds. 3. The olecules interact only by short-range forces during elastic collisions. This is consistent with the ideal gas odel, in which the olecules exert no longrange forces on each other. 4. The olecules ake elastic collisions with the walls. 5. The gas under consideration is a pure substance; that is, all olecules are identical. Although we often picture an ideal gas as consisting of single atos, we can assue that the behavior of olecular gases approxiates that of ideal gases rather well at low pressures. Molecular rotations or vibrations have no effect, on the average, on the otions that we consider here. For our first application of kinetic theory, let us derive an expression for the pressure of N olecules of an ideal gas in a container of volue V in ters of icroscopic quantities. The container is a cube with edges of length d (Fig..). We shall first Assuptions of the olecular odel of an ideal gas z d d Figure. A cubical box with sides of length d containing an ideal gas. The olecule shown oves with velocity v i. y v i v xi d x 64

3 64 CHAPTER The Kinetic Theory of Gases v i v xi v yi focus our attention on one of these olecules of ass, and assue that it is oving so that its coponent of velocity in the x direction is v xi as in Figure.. (The subscript i here refers to the ith olecule, not to an initial value. We will cobine the effects of all of the olecules shortly.) As the olecule collides elastically with any wall (assuption 4), its velocity coponent perpendicular to the wall is reversed because the ass of the wall is far greater than the ass of the olecule. Because the oentu coponent p xi of the olecule is v xi before the collision and v xi after the collision, the change in the x coponent of the oentu of the olecule is p xi v xi (v xi ) v xi v yi v i Because the olecules obey Newton s laws (assuption ), we can apply the ipulseoentu theore (Eq. 9.8) to the olecule to give us F i, on olecule t collision p xi v xi v xi Active Figure. A olecule akes an elastic collision with the wall of the container. Its x coponent of oentu is reversed, while its y coponent reains unchanged. In this construction, we assue that the olecule oves in the xy plane. At the Active Figures link at you can observe olecules within a container aking collisions with the walls of the container and with each other. F i, on olecule where is the x coponent of the average force that the wall exerts on the olecule during the collision and t collision is the duration of the collision. In order for the olecule to ake another collision with the sae wall after this first collision, it ust travel a distance of d in the x direction (across the container and back). Therefore, the tie interval between two collisions with the sae wall is t d v xi The force that causes the change in oentu of the olecule in the collision with the wall occurs only during the collision. However, we can average the force over the tie interval for the olecule to ove across the cube and back. Soetie during this tie interval, the collision occurs, so that the change in oentu for this tie interval is the sae as that for the short duration of the collision. Thus, we can rewrite the ipulse-oentu theore as F i F i t v xi where is the average force coponent over the tie for the olecule to ove across the cube and back. Because exactly one collision occurs for each such tie interval, this is also the long-ter average force on the olecule, over long tie intervals containing any nuber of ultiples of t. This equation and the preceding one enable us to express the x coponent of the long-ter average force exerted by the wall on the olecule as F i v xi t v xi d v xi d Now, by Newton s third law, the average x coponent of the force exerted by the olecule on the wall is equal in agnitude and opposite in direction: F i, on wall F i vxi d v xi d The total average force F exerted by the gas on the wall is found by adding the average forces exerted by the individual olecules. We add ters such as that above for all olecules: F N i v xi d d i N vxi where we have factored out the length of the box and the ass, because assuption 5 tells us that all of the olecules are the sae. We now ipose assuption, that the nuber of olecules is large. For a sall nuber of olecules, the actual force on the

4 SECTION. Molecular Model of an Ideal Gas 643 wall would vary with tie. It would be nonzero during the short interval of a collision of a olecule with the wall and zero when no olecule happens to be hitting the wall. For a very large nuber of olecules, however, such as Avogadro s nuber, these variations in force are soothed out, so that the average force given above is the sae over any tie interval. Thus, the constant force F on the wall due to the olecular collisions is F d To proceed further, let us consider how to express the average value of the square of the x coponent of the velocity for N olecules. The traditional average of a set of values is the su of the values over the nuber of values: v x i N vxi N vxi i N The nuerator of this expression is contained in the right-hand side of the preceding equation. Thus, cobining the two expressions, the total force on the wall can be written F d Nv x (.) Now let us focus again on one olecule with velocity coponents v xi, v yi, and v zi. The Pythagorean theore relates the square of the speed of the olecule to the squares of the velocity coponents: Hence, the average value of v for all the olecules in the container is related to the average values of v, v, and v according to the expression x y Because the otion is copletely rando (assuption ), the average values,, and v x y z are equal to each other. Using this fact and the preceding equation, we find that Thus, fro Equation., the total force exerted on the wall is Using this expression, we can find the total pressure exerted on the wall: P F A z P 3 N V v vi vxi vyi vzi v vx vy vz v 3v x F N 3 v d F d 3 N d 3 v 3 N V v (.) This result indicates that the pressure of a gas is proportional to the nuber of olecules per unit volue and to the average translational kinetic energy of the olecules,. In analyzing this siplified odel of an ideal gas, we obtain an iportant result that relates the acroscopic quantity of pressure to a icroscopic quan- v tity the average value of the square of the olecular speed. Thus, we have established a key link between the olecular world and the large-scale world. You should note that Equation. verifies soe features of pressure with which you are probably failiar. One way to increase the pressure inside a container is to increase the nuber of olecules per unit volue N/V in the container. This is what you do when you add air to a tire. The pressure in the tire can also be increased by increasing the average translational kinetic energy of the air olecules in the tire. v v Relationship between pressure and olecular kinetic energy

5 644 CHAPTER The Kinetic Theory of Gases This can be accoplished by increasing the teperature of that air, as we shall soon show atheatically. This is why the pressure inside a tire increases as the tire wars up during long trips. The continuous flexing of the tire as it oves along the road surface results in work done as parts of the tire distort, causing an increase in internal energy of the rubber. The increased teperature of the rubber results in the transfer of energy by heat into the air inside the tire. This transfer increases the air s teperature, and this increase in teperature in turn produces an increase in pressure. Molecular Interpretation of Teperature We can gain soe insight into the eaning of teperature by first writing Equation. in the for PV 3 N ( v ) Let us now copare this with the equation of state for an ideal gas (Eq. 9.0): PV Nk B T Recall that the equation of state is based on experiental facts concerning the acroscopic behavior of gases. Equating the right sides of these expressions, we find that Teperature is proportional to average kinetic energy Average kinetic energy per olecule (.3) This result tells us that teperature is a direct easure of average olecular kinetic energy. By rearranging Equation.3, we can relate the translational olecular kinetic energy to the teperature: That is, the average translational kinetic energy per olecule is, it follows that 3 v v x T 3k B ( v ) v 3 k BT v x k BT In a siilar anner, it follows that the otions in the y and z directions give us 3 (.4). Because k BT (.5) v y k BT and v z k BT Thus, each translational degree of freedo contributes an equal aount of energy,, to the gas. (In general, a degree of freedo refers to an independent eans k BT by which a olecule can possess energy.) A generalization of this result, known as the theore of equipartition of energy, states that Theore of equipartition of energy each degree of freedo contributes to the energy of a syste, where possible k BT degrees of freedo in addition to those associated with translation arise fro rotation and vibration of olecules. The total translational kinetic energy of N olecules of gas is siply N ties the average energy per olecule, which is given by Equation.4: Total translational kinetic energy of N olecules K tot trans N( v ) 3 N k BT 3 nrt (.6) where we have used k B R/N A for Boltzann s constant and n N/N A for the nuber of oles of gas. If we consider a gas in which olecules possess only translational kinetic energy, Equation.6 represents the internal energy of the gas. This result iplies that the internal energy of an ideal gas depends only on the teperature. We will follow up on this point in Section..

6 SECTION. Molecular Model of an Ideal Gas 645 The square root of is called the root-ean-square (rs) speed of the olecules. Fro Equation.4 we find that the rs speed is v v rs v 3k B T 3RT M (.7) where M is the olar ass in kilogras per ole and is equal to N A. This expression shows that, at a given teperature, lighter olecules ove faster, on the average, than do heavier olecules. For exaple, at a given teperature, hydrogen olecules, whose olar ass is kg/ol, have an average speed approxiately four ties that of oxygen olecules, whose olar ass is kg/ol. Table. lists the rs speeds for various olecules at 0 C. Table. Soe rs Speeds Molar ass v rs Gas (g/ol) at 0C(/s) H.0 90 He H O Ne N or CO NO O CO SO Root-ean-square speed PITFALL PREVENTION. The Square Root of the Square? Notice that taking the square root of v does not undo the square because we have taken an average between squaring and taking the square root. While the square root of (v) is v because the squaring is done after the averaging, the square root of v is not v, but rather v rs. Exaple. A Tank of Heliu A tank used for filling heliu balloons has a volue of and contains.00 ol of heliu gas at 0.0 C. Assue that the heliu behaves like an ideal gas. (A) What is the total translational kinetic energy of the gas olecules? Solution Using Equation.6 with n.00 ol and T 93 K, we find that K tot trans 3 nrt 3 (.00 ol)(8.3 J/olK)(93 K) J (B) What is the average kinetic energy per olecule? Solution Using Equation.4, we find that the average kinetic energy per olecule is v 3 k BT 3 ( J/K)(93 K) J What If? What if the teperature is raised fro 0.0 C to 40.0 C? Because 40.0 is twice as large as 0.0, is the total translational energy of the olecules of the gas twice as large at the higher teperature? Answer The expression for the total translational energy depends on the teperature, and the value for the teperature ust be expressed in kelvins, not in degrees Celsius. Thus, the ratio of 40.0 to 0.0 is not the appropriate ratio. Converting the Celsius teperatures to kelvins, 0.0 C is 93 K and 40.0 C is 33 K. Thus, the total translational energy increases by a factor of 33 K/93 K.07. Quick Quiz. Two containers hold an ideal gas at the sae teperature and pressure. Both containers hold the sae type of gas but container B has twice the volue of container A. The average translational kinetic energy per olecule in container B is (a) twice that for container A (b) the sae as that for container A (c) half that for container A (d) ipossible to deterine.

7 646 CHAPTER The Kinetic Theory of Gases Quick Quiz. Consider again the situation in Quick Quiz.. The internal energy of the gas in container B is (a) twice that for container A (b) the sae as that for container A (c) half that for container A (d) ipossible to deterine. Quick Quiz.3 Consider again the situation in Quick Quiz.. The rs speed of the gas olecules in container B is (a) twice that for container A (b) the sae as that for container A (c) half that for container A (d) ipossible to deterine. P f i Isothers f f T + T T V Figure.3 An ideal gas is taken fro one isother at teperature T to another at teperature T T along three different paths.. Molar Specific Heat of an Ideal Gas Consider an ideal gas undergoing several processes such that the change in teperature is T T f T i for all processes. The teperature change can be achieved by taking a variety of paths fro one isother to another, as shown in Figure.3. Because T is the sae for each path, the change in internal energy E int is the sae for all paths. However, we know fro the first law, Q E int W, that the heat Q is different for each path because W (the negative of the area under the curves) is different for each path. Thus, the heat associated with a given change in teperature does not have a unique value. We can address this difficulty by defining specific heats for two processes that frequently occur: changes at constant volue and changes at constant pressure. Because the nuber of oles is a convenient easure of the aount of gas, we define the olar specific heats associated with these processes with the following equations: Q nc V T Q nc P T (constant volue) (constant pressure) (.8) (.9) where C V is the olar specific heat at constant volue and C P is the olar specific heat at constant pressure. When we add energy to a gas by heat at constant pressure, not only does the internal energy of the gas increase, but work is done on the gas because of the change in volue. Therefore, the heat Q constant P ust account for both the increase in internal energy and the transfer of energy out of the syste by work. For this reason, Q constant P is greater than Q constant V for given values of n and T. Thus, C P is greater than C V. In the previous section, we found that the teperature of a gas is a easure of the average translational kinetic energy of the gas olecules. This kinetic energy is associated with the otion of the center of ass of each olecule. It does not include the energy associated with the internal otion of the olecule naely, vibrations and rotations about the center of ass. This should not be surprising because the siple kinetic theory odel assues a structureless olecule. In view of this, let us first consider the siplest case of an ideal onatoic gas, that is, a gas containing one ato per olecule, such as heliu, neon, or argon. When energy is added to a onatoic gas in a container of fixed volue, all of the added energy goes into increasing the translational kinetic energy of the atos. There is no other way to store the energy in a onatoic gas. Therefore, fro Equation.6, we see that the internal energy E int of N olecules (or n ol) of an ideal onatoic gas is Internal energy of an ideal onatoic gas E int K tot trans 3 Nk BT 3 nrt (.0) Note that for a onatoic ideal gas, E int is a function of T only, and the functional relationship is given by Equation.0. In general, the internal energy of an ideal gas is a function of T only, and the exact relationship depends on the type of gas.

8 SECTION. Molar Specific Heat of an Ideal Gas 647 If energy is transferred by heat to a syste at constant volue, then no work is done on the syste. That is, W P dv 0 for a constant-volue process. Hence, fro the first law of therodynaics, we see that Q E int (.) In other words, all of the energy transferred by heat goes into increasing the internal energy of the syste. A constant-volue process fro i to f for an ideal gas is described in Figure.4, where T is the teperature difference between the two isothers. Substituting the expression for Q given by Equation.8 into Equation., we obtain E int nc V T If the olar specific heat is constant, we can express the internal energy of a gas as E int nc V T (.) This equation applies to all ideal gases to gases having ore than one ato per olecule as well as to onatoic ideal gases. In the liit of infinitesial changes, we can use Equation. to express the olar specific heat at constant volue as C V n (.3) Let us now apply the results of this discussion to the onatoic gas that we have been studying. Substituting the internal energy fro Equation.0 into Equation.3, we find that C V 3 R de int dt (.4) This expression predicts a value of C V 3 R.5 J/olK for all onatoic gases. This prediction is in excellent agreeent with easured values of olar specific heats for such gases as heliu, neon, argon, and xenon over a wide range of teperatures (Table.). Sall variations in Table. fro the predicted values are due to the P i f Isothers f T + T T V Active Figure.4 Energy is transferred by heat to an ideal gas in two ways. For the constant-volue path i : f, all the energy goes into increasing the internal energy of the gas because no work is done. Along the constant-pressure path i : f, part of the energy transferred in by heat is transferred out by work. At the Active Figures link at you can choose initial and final teperatures for one ole of an ideal gas undergoing constant-volue and constant pressure processes and easure Q, W, E int, C V, and C P. Table. Molar Specific Heats of Various Gases Molar Specific Heat ( J/ol K) a Gas C P C V C P C V C P /C V Monatoic Gases He Ar Ne Kr Diatoic Gases H N O CO Cl Polyatoic Gases CO SO H O CH a All values except that for water were obtained at 300 K.

9 648 CHAPTER The Kinetic Theory of Gases fact that real gases are not ideal gases. In real gases, weak interolecular interactions occur, which are not addressed in our ideal gas odel. Now suppose that the gas is taken along the constant-pressure path i : f shown in Figure.4. Along this path, the teperature again increases by T. The energy that ust be transferred by heat to the gas in this process is Q nc P T. Because the volue changes in this process, the work done on the gas is W P V where P is the constant pressure at which the process occurs. Applying the first law of therodynaics to this process, we have E int Q W nc P T (P V ) (.5) In this case, the energy added to the gas by heat is channeled as follows: Part of it leaves the syste by work (that is, the gas oves a piston through a displaceent), and the reainder appears as an increase in the internal energy of the gas. But the change in internal energy for the process i : f is equal to that for the process i : f because E int depends only on teperature for an ideal gas and because T is the sae for both processes. In addition, because PV nrt, we note that for a constant-pressure process, P V nr T. Substituting this value for P V into Equation.5 with E int nc V T (Eq..) gives nc V T nc P T nr T C P C V R (.6) Ratio of olar specific heats for a onatoic ideal gas This expression applies to any ideal gas. It predicts that the olar specific heat of an ideal gas at constant pressure is greater than the olar specific heat at constant volue by an aount R, the universal gas constant (which has the value 8.3 J/ol K). This expression is applicable to real gases, as the data in Table. show. 3 Because C V R for a onatoic ideal gas, Equation.6 predicts a value C P 5 R 0.8 J/ol K for the olar specific heat of a onatoic gas at constant pressure. The ratio of these olar specific heats is a diensionless quantity (Greek gaa): C P 5R/ C V 3R/ (.7) Theoretical values of C V, C P and are in excellent agreeent with experiental values obtained for onatoic gases, but they are in serious disagreeent with the values for the ore coplex gases (see Table.). This is not surprising because the 3 value C V R was derived for a onatoic ideal gas and we expect soe additional contribution to the olar specific heat fro the internal structure of the ore coplex olecules. In Section.4, we describe the effect of olecular structure on the olar specific heat of a gas. The internal energy and, hence, the olar specific heat of a coplex gas ust include contributions fro the rotational and the vibrational otions of the olecule. In the case of solids and liquids heated at constant pressure, very little work is done because the theral expansion is sall. Consequently, C P and C V are approxiately equal for solids and liquids. Quick Quiz.4 How does the internal energy of an ideal gas change as it follows path i : f in Figure.4? (a) E int increases. (b) E int decreases. (c) E int stays the sae. (d) There is not enough inforation to deterine how E int changes. Quick Quiz.5 How does the internal energy of an ideal gas change as it follows path f : f along the isother labeled T T in Figure.4? (a) E int increases. (b) E int decreases. (c) E int stays the sae. (d) There is not enough inforation to deterine how E int changes.

10 SECTION.3 Adiabatic Processes for an Ideal Gas 649 Exaple. Heating a Cylinder of Heliu A cylinder contains 3.00 ol of heliu gas at a teperature of 300 K. (A) If the gas is heated at constant volue, how uch energy ust be transferred by heat to the gas for its teperature to increase to 500 K? Solution For the constant-volue process, we have Q nc V T Because C V.5 J/ol K for heliu and T 00 K, we obtain Q (3.00 ol)(.5 J/olK)(00 K) (B) How uch energy ust be transferred by heat to the gas at constant pressure to raise the teperature to 500 K? Solution Making use of Table., we obtain Q nc P T (3.00 ol)(0.8 J/olK)(00 K) J Note that this is larger than Q, due to the transfer of energy out of the gas by work in the constant pressure process J.3 Adiabatic Processes for an Ideal Gas As we noted in Section 0.6, an adiabatic process is one in which no energy is transferred by heat between a syste and its surroundings. For exaple, if a gas is copressed (or expanded) very rapidly, very little energy is transferred out of (or into) the syste by heat, and so the process is nearly adiabatic. Such processes occur in the cycle of a gasoline engine, which we discuss in detail in the next chapter. Another exaple of an adiabatic process is the very slow expansion of a gas that is therally insulated fro its surroundings. Suppose that an ideal gas undergoes an adiabatic expansion. At any tie during the process, we assue that the gas is in an equilibriu state, so that the equation of state PV nrt is valid. As we show below, the pressure and volue of an ideal gas at any tie during an adiabatic process are related by the expression PV constant (.8) where C P /C V is assued to be constant during the process. Thus, we see that all three variables in the ideal gas law P, V, and T change during an adiabatic process. Relationship between P and V for an adiabatic process involving an ideal gas Proof That PV Constant for an Adiabatic Process When a gas is copressed adiabatically in a therally insulated cylinder, no energy is transferred by heat between the gas and its surroundings; thus, Q 0. Let us iagine an infinitesial change in volue dv and an accopanying infinitesial change in teperature dt. The work done on the gas is PdV. Because the internal energy of an ideal gas depends only on teperature, the change in the internal energy in an adiabatic process is the sae as that for an isovoluetric process between the sae teperatures, de int nc V dt (Eq..). Hence, the first law of therodynaics, E int Q W, with Q 0 becoes de int nc V dt P dv Taking the total differential of the equation of state of an ideal gas, PV nrt, we see that PdV VdP nr dt Eliinating dt fro these two equations, we find that P dv V dp R C V P dv

11 650 CHAPTER The Kinetic Theory of Gases P f P i P V f f Isothers Adiabatic process i V i T f T i Figure.5 The PV diagra for an adiabatic copression. Note that T f T i in this process, so the teperature of the gas increases. Relationship between T and V for an adiabatic process involving an ideal gas V Substituting R C P C V and dividing by PV, we obtain Integrating this expression, we have dv V dp P C P C V C V dv V dp P which is equivalent to Equation.8: dv V 0 ln P PV constant The PV diagra for an adiabatic copression is shown in Figure.5. Because, the PV curve is steeper than it would be for an isotheral copression. By the definition of an adiabatic process, no energy is transferred by heat into or out of the syste. Hence, fro the first law, we see that E int is positive (work is done on the gas, so its internal energy increases) and so T also is positive. Thus, the teperature of the gas increases (T f T i ) during an adiabatic copression. Conversely, the teperature decreases if the gas expands adiabatically. Applying Equation.8 to the initial and final states, we see that P i V i lnv constant P f V f Using the ideal gas law, we can express Equation.9 as T i V i T f V f ( ) dv V (.9) (.0) Exaple.3 A Diesel Engine Cylinder Air at 0.0 C in the cylinder of a diesel engine is copressed fro an initial pressure of.00 at and volue of c 3 to a volue of 60.0 c 3. Assue that air behaves as an ideal gas with.40 and that the copression is adiabatic. Find the final pressure and teperature of the air. Solution Conceptualize by iagining what happens if we copress a gas into a saller volue. Our discussion above and Figure.5 tell us that the pressure and teperature both increase. We categorize this as a proble involving an adiabatic copression. To analyze the proble, we use Equation.9 to find the final pressure: P f P i V i V f 37.6 at (.00 at) c c 3.40 Because PV nrt is valid throughout an ideal gas process and because no gas escapes fro the cylinder, P i V i T i Pf Vf T f T f P f V f P i V i T i (37.6 at)(60.0 c3 ) (.00 at)(800.0 c 3 ) 86 K 553C (93 K) To finalize the proble, note that the teperature of the gas has increased by a factor of.8. The high copression in a diesel engine raises the teperature of the fuel enough to cause its cobustion without the use of spark plugs..4 The Equipartition of Energy We have found that predictions based on our odel for olar specific heat agree quite well with the behavior of onatoic gases but not with the behavior of coplex gases (see Table.). The value predicted by the odel for the quantity C P C V R, however, is the sae for all gases. This is not surprising because this difference is the result of the work done on the gas, which is independent of its olecular structure. In the adiabatic free expansion discussed in Section 0.6, the teperature reains constant. This is a special process in which no work is done because the gas expands into a vacuu. In general, the teperature decreases in an adiabatic expansion in which work is done.

12 SECTION.4 The Equipartition of Energy 65 To clarify the variations in C V and C P in gases ore coplex than onatoic gases, let us explore further the origin of olar specific heat. So far, we have assued that the sole contribution to the internal energy of a gas is the translational kinetic energy of the olecules. However, the internal energy of a gas includes contributions fro the translational, vibrational, and rotational otion of the olecules. The rotational and vibrational otions of olecules can be activated by collisions and therefore are coupled to the translational otion of the olecules. The branch of physics known as statistical echanics has shown that, for a large nuber of particles obeying the laws of Newtonian echanics, the available energy is, on the average, shared equally by each independent degree of freedo. Recall fro Section. that the equipartition theore states that, at equilibriu, each degree of freedo contributes k B T of energy per olecule. Let us consider a diatoic gas whose olecules have the shape of a dubbell (Fig..6). In this odel, the center of ass of the olecule can translate in the x, y, and z directions (Fig..6a). In addition, the olecule can rotate about three utually perpendicular axes (Fig..6b). We can neglect the rotation about the y axis because the olecule s oent of inertia I y and its rotational energy I y about this axis are negligible copared with those associated with the x and z axes. (If the two atos are taken to be point asses, then I y is identically zero.) Thus, there are five degrees of freedo for translation and rotation: three associated with the translational otion and two associated with the rotational otion. Because each degree of freedo contributes, on the average, k B T of energy per olecule, the internal energy for a syste of N olecules, ignoring vibration for now, is x x z (a) z (b) z y y E int 3N ( k BT ) N ( k BT ) 5 Nk BT 5 nrt We can use this result and Equation.3 to find the olar specific heat at constant volue: C V n de int dt n Fro Equations.6 and.7, we find that C P C V R 7 R d dt (5 nrt ) 5 R (.) x (c) Figure.6 Possible otions of a diatoic olecule: (a) translational otion of the center of ass, (b) rotational otion about the various axes, and (c) vibrational otion along the olecular axis. y C P C V 7 R 5 R 7 5 These results agree quite well with ost of the data for diatoic olecules given in Table.. This is rather surprising because we have not yet accounted for the possible vibrations of the olecule. In the odel for vibration, the two atos are joined by an iaginary spring (see Fig..6c). The vibrational otion adds two ore degrees of freedo, which correspond to the kinetic energy and the potential energy associated with vibrations along the length of the olecule. Hence, classical physics and the equipartition theore in a odel that includes all three types of otion predict a total internal energy of E int 3N ( k BT ) N ( k BT ) N ( k BT ) 7 Nk BT 7 nrt and a olar specific heat at constant volue of.40 C V n de int dt n d dt (7 nrt ) 7 R (.) This value is inconsistent with experiental data for olecules such as H and N (see Table.) and suggests a breakdown of our odel based on classical physics. It ight see that our odel is a failure for predicting olar specific heats for diatoic gases. We can clai soe success for our odel, however, if easureents of olar specific heat are ade over a wide teperature range, rather than at the

13 65 CHAPTER The Kinetic Theory of Gases C V ( J/ol K) Translation Rotation Vibration 7 R 5 R 3 R Teperature (K) Figure.7 The olar specific heat of hydrogen as a function of teperature. The horizontal scale is logarithic. Note that hydrogen liquefies at 0 K. single teperature that gives us the values in Table.. Figure.7 shows the olar specific heat of hydrogen as a function of teperature. There are three plateaus in the curve. The rearkable feature of these plateaus is that they are at the values of the olar specific heat predicted by Equations.4,., and.! For low teperatures, the diatoic hydrogen gas behaves like a onatoic gas. As the teperature rises to roo teperature, its olar specific heat rises to a value for a diatoic gas, consistent with the inclusion of rotation but not vibration. For high teperatures, the olar specific heat is consistent with a odel including all types of otion. Before addressing the reason for this ysterious behavior, let us ake a brief reark about polyatoic gases. For olecules with ore than two atos, the vibrations are ore coplex than for diatoic olecules and the nuber of degrees of freedo is even larger. This results in an even higher predicted olar specific heat, which is in qualitative agreeent with experient. For the polyatoic gases shown in Table. we see that the olar specific heats are higher than those for diatoic gases. The ore degrees of freedo available to a olecule, the ore ways there are to store energy, resulting in a higher olar specific heat. A Hint of Energy Quantization Our odel for olar specific heats has been based so far on purely classical notions. It predicts a value of the specific heat for a diatoic gas that, according to Figure.7, only agrees with experiental easureents ade at high teperatures. In order to explain why this value is only true at high teperatures and why the plateaus exist in Figure.7, we ust go beyond classical physics and introduce soe quantu physics into the odel. In Chapter 8, we discussed quantization of frequency for vibrating strings and air coluns. This is a natural result whenever waves are subject to boundary conditions. Quantu physics (Chapters 40 to 43) shows that atos and olecules can be described by the physics of waves under boundary conditions. Consequently, these waves have quantized frequencies. Furtherore, in quantu physics, the energy of a syste is proportional to the frequency of the wave representing the syste. Hence, the energies of atos and olecules are quantized. For a olecule, quantu physics tells us that the rotational and vibrational energies are quantized. Figure.8 shows an energy-level diagra for the rotational and vibrational quantu states of a diatoic olecule. The lowest allowed state is called the ground state. Notice that vibrational states are separated by larger energy gaps than are rotational states.

14 SECTION.4 The Equipartition of Energy 653 Vibrational states Rotational states ENERGY Rotational states Figure.8 An energy-level diagra for vibrational and rotational states of a diatoic olecule. Note that the rotational states lie closer together in energy than the vibrational states. At low teperatures, the energy that a olecule gains in collisions with its neighbors is generally not large enough to raise it to the first excited state of either rotation or vibration. Thus, even though rotation and vibration are classically allowed, they do not occur at low teperatures. All olecules are in the ground state for rotation and vibration. Thus, the only contribution to the olecules average energy is fro translation, and the specific heat is that predicted by Equation.4. As the teperature is raised, the average energy of the olecules increases. In soe collisions, a olecule ay have enough energy transferred to it fro another olecule to excite the first rotational state. As the teperature is raised further, ore olecules can be excited to this state. The result is that rotation begins to contribute to the internal energy and the olar specific heat rises. At about roo teperature in Figure.7, the second plateau has been reached and rotation contributes fully to the olar specific heat. The olar specific heat is now equal to the value predicted by Equation.. There is no contribution at roo teperature fro vibration, because the olecules are still in the ground vibrational state. The teperature ust be raised even further to excite the first vibrational state. This happens in Figure.7 between 000 K and K. At K on the right side of the figure, vibration is contributing fully to the internal energy and the olar specific heat has the value predicted by Equation.. The predictions of this odel are supportive of the theore of equipartition of energy. In addition, the inclusion in the odel of energy quantization fro quantu physics allows a full understanding of Figure.7. Quick Quiz.6 The olar specific heat of a diatoic gas is easured at constant volue and found to be 9. J/ol K. The types of energy that are contributing to the olar specific heat are (a) translation only (b) translation and rotation only (c) translation and vibration only (d) translation, rotation, and vibration. Quick Quiz.7 The olar specific heat of a gas is easured at constant volue and found to be R/. The gas is ost likely to be (a) onatoic (b) diatoic (c) polyatoic. The Molar Specific Heat of Solids The olar specific heats of solids also deonstrate a arked teperature dependence. Solids have olar specific heats that generally decrease in a nonlinear anner with decreasing teperature and approach zero as the teperature approaches

15 654 CHAPTER The Kinetic Theory of Gases C V ( J/ol K) Lead Aluinu Silicon Diaond T(K) Figure.9 Molar specific heat of four solids. As T approaches zero, the olar specific heat also approaches zero. absolute zero. At high teperatures (usually above 300 K), the olar specific heats approach the value of 3R 5 J/ol K, a result known as the DuLong Petit law. The typical data shown in Figure.9 deonstrate the teperature dependence of the olar specific heats for several solids. We can explain the olar specific heat of a solid at high teperatures using the equipartition theore. For sall displaceents of an ato fro its equilibriu position, each ato executes siple haronic otion in the x, y, and z directions. The energy associated with vibrational otion in the x direction is Total internal energy of a solid Molar specific heat of a solid at constant volue E v x kx The expressions for vibrational otions in the y and z directions are analogous. Therefore, each ato of the solid has six degrees of freedo. According to the equipartition theore, this corresponds to an average vibrational energy of 6( per k BT ) 3k B T ato. Therefore, the internal energy of a solid consisting of N atos is E int 3Nk B T 3nRT (.3) Fro this result, we find that the olar specific heat of a solid at constant volue is C V n de int dt 3R (.4) This result is in agreeent with the epirical DuLong Petit law. The discrepancies between this odel and the experiental data at low teperatures are again due to the inadequacy of classical physics in describing the world at the atoic level. PITFALL PREVENTION. The Distribution Function Notice that the distribution function n V (E ) is defined in ters of the nuber of olecules with energy in the range E to E de rather than in ters of the nuber of olecules with energy E. Because the nuber of olecules is finite and the nuber of possible values of the energy is infinite, the nuber of olecules with an exact energy E ay be zero..5 The Boltzann Distribution Law Thus far we have considered only average values of the energies of olecules in a gas and have not addressed the distribution of energies aong olecules. In reality, the otion of the olecules is extreely chaotic. Any individual olecule is colliding with others at an enorous rate typically, a billion ties per second. Each collision results in a change in the speed and direction of otion of each of the participant olecules. Equation.7 shows that rs olecular speeds increase with increasing teperature. What is the relative nuber of olecules that possess soe characteristic, such as energy within a certain range? We shall address this question by considering the nuber density n V (E ). This quantity, called a distribution function, is defined so that n V (E ) de is the nuber of olecules per unit volue with energy between E and E de. (Note that the ratio of the nuber of olecules that have the desired characteristic to the total nuber of olecules is the probability that a particular olecule has that characteristic.) In general,

16 SECTION.6 Distribution of Molecular Speeds 655 the nuber density is found fro statistical echanics to be n V (E ) n 0 e E/k BT (.5) Boltzann distribution law where n 0 is defined such that n 0 de is the nuber of olecules per unit volue having energy between E 0 and E de. This equation, known as the Boltzann distribution law, is iportant in describing the statistical echanics of a large nuber of olecules. It states that the probability of finding the olecules in a particular energy state varies exponentially as the negative of the energy divided by k B T. All the olecules would fall into the lowest energy level if the theral agitation at a teperature T did not excite the olecules to higher energy levels. Exaple.4 Theral Excitation of Atoic Energy Levels Interactive As we discussed in Section.4, atos can occupy only certain discrete energy levels. Consider a gas at a teperature of 500 K whose atos can occupy only two energy levels separated by.50 ev, where ev (electron volt) is an energy unit equal to J (Fig..0). Deterine the ratio of the nuber of atos in the higher energy level to the nuber in the lower energy level. Solution Equation.5 gives the relative nuber of atos in a given energy level. In this case, the ato has two possible energies, E and E, where E is the lower energy level. Hence, the ratio of the nuber of atos in the higher energy level to the nuber in the lower energy level is n () V (E ) n V (E ) n 0e E /kbt n 0 e E /k B T In this proble, E E.50 ev, and the denoinator of the exponent is k B T ( J/K)( 500 K) ev.60 0 J ev E e(e E )/k B T Therefore, the required ratio is n V (E ) n V (E ) e.50 ev/0.6 ev e This result indicates that at T 500 K, only a sall fraction of the atos are in the higher energy level. In fact, for every ato in the higher energy level, there are about 000 atos in the lower level. The nuber of atos in the higher level increases at even higher teperatures, but the distribution law specifies that at equilibriu there are always ore atos in the lower level than in the higher level. What If? What if the energy levels in Figure.0 were closer together in energy? Would this increase or decrease the fraction of the atos in the upper energy level? Answer If the excited level is lower in energy than that in Figure.0, it would be easier for theral agitation to excite atos to this level, and the fraction of atos in this energy level would be larger. Let us see this atheatically by expressing Equation () as r e (E E )/k B T E.50 ev Figure.0 (Exaple.4) Energy-level diagra for a gas whose atos can occupy two energy states. where r is the ratio of atos having energy E to those with energy E. Differentiating with respect to E, we find dr de d e (E E )/k B T de k B T e(e E )/k B T 0 Because the derivative has a negative value, we see that as E decreases, r increases. At the Interactive Worked Exaple link at you can investigate the effects of changing the teperature and the energy difference between the states..6 Distribution of Molecular Speeds In 860 Jaes Clerk Maxwell (83 879) derived an expression that describes the distribution of olecular speeds in a very definite anner. His work and subsequent developents by other scientists were highly controversial because direct detection of olecules could not be achieved experientally at that tie. However, about 60 years later, experients were devised that confired Maxwell s predictions.

17 656 CHAPTER The Kinetic Theory of Gases N v v p v v rs N v v dv Active Figure. The speed distribution of gas olecules at soe teperature. The nuber of olecules having speeds in the range v to v dv is equal to the area of the shaded rectangle, N v dv. The function N v approaches zero as v approaches infinity. At the Active Figures link at you can ove the blue triangle and easure the nuber of olecules with speeds within a sall range. Let us consider a container of gas whose olecules have soe distribution of speeds. Suppose we want to deterine how any gas olecules have a speed in the range fro, for exaple, 400 to 40 /s. Intuitively, we expect that the speed distribution depends on teperature. Furtherore, we expect that the distribution peaks in the vicinity of v rs. That is, few olecules are expected to have speeds uch less than or uch greater than v rs because these extree speeds result only fro an unlikely chain of collisions. The observed speed distribution of gas olecules in theral equilibriu is shown in Figure.. The quantity N v, called the Maxwell Boltzann speed distribution function, is defined as follows. If N is the total nuber of olecules, then the nuber of olecules with speeds between v and v dv is dn N v dv. This nuber is also equal to the area of the shaded rectangle in Figure.. Furtherore, the fraction of olecules with speeds between v and v dv is (N v dv)/n. This fraction is also equal to the probability that a olecule has a speed in the range v to v dv. The fundaental expression that describes the distribution of speeds of N gas olecules is N v 4N k B T 3/ v e v /k B T (.6) where is the ass of a gas olecule, k B is Boltzann s constant, and T is the absolute teperature. Observe the appearance of the Boltzann factor e E/k BT with E. v As indicated in Figure., the average speed is soewhat lower than the rs speed. The ost probable speed v p is the speed at which the distribution curve reaches a peak. Using Equation.6, one finds that v rs v 3k B T 8k v B T.60 k BT v p k B T.73 k BT.4 k BT (.7) (.8) (.9) Equation.7 has previously appeared as Equation.7. The details of the derivations of these equations fro Equation.6 are left for the student (see Probles 39 and 65). Fro these equations, we see that v rs v v p Ludwig Boltzann Austrian physicist ( ) Boltzann ade any iportant contributions to the developent of the kinetic theory of gases, electroagnetis, and therodynaics. His pioneering work in the field of kinetic theory led to the branch of physics known as statistical echanics. (Courtesy of AIP Niels Bohr Library, Lande Collection) Figure. represents speed distribution curves for nitrogen, N. The curves were obtained by using Equation.6 to evaluate the distribution function at various speeds and at two teperatures. Note that the peak in the curve shifts to the right as T increases, indicating that the average speed increases with increasing teperature, as expected. The asyetric shape of the curves is due to the fact that the lowest speed possible is zero while the upper classical liit of the speed is infinity. (In Chapter 39, we will show that the actual upper liit is the speed of light.) Equation.6 shows that the distribution of olecular speeds in a gas depends both on ass and on teperature. At a given teperature, the fraction of olecules with speeds exceeding a fixed value increases as the ass decreases. This explains why lighter olecules, such as H and He, escape ore readily fro the Earth s atosphere than do heavier olecules, such as N and O. (See the discussion of escape speed in Chapter 3. Gas olecules escape even ore readily fro the Moon s surface than fro the Earth s because the escape speed on the Moon is lower than that on the Earth.) The speed distribution curves for olecules in a liquid are siilar to those shown in Figure.. We can understand the phenoenon of evaporation of a liquid fro this distribution in speeds, using the fact that soe olecules in the liquid are ore For the derivation of this expression, see an advanced textbook on therodynaics, such as that by R. P. Bauan, Modern Therodynaics with Statistical Mechanics, New York, Macillan Publishing Co., 99.

18 SECTION.6 Distribution of Molecular Speeds 657 N v, nuber of olecules per unit speed interval (olecules//s) T = 300 K v p Curves calculated for N = 0 5 nitrogen olecules v v rs T = 900 K At the Active Figures link at you can set the desired teperature and see the effect on the distribution curve v (/s) Active Figure. The speed distribution function for 0 5 nitrogen olecules at 300 K and 900 K. The total area under either curve is equal to the total nuber of olecules, which in this case equals 0 5. Note that v rs v v p. energetic than others. Soe of the faster-oving olecules in the liquid penetrate the surface and leave the liquid even at teperatures well below the boiling point. The olecules that escape the liquid by evaporation are those that have sufficient energy to overcoe the attractive forces of the olecules in the liquid phase. Consequently, the olecules left behind in the liquid phase have a lower average kinetic energy; as a result, the teperature of the liquid decreases. Hence, evaporation is a cooling process. For exaple, an alcohol-soaked cloth often is placed on a feverish head to cool and cofort a patient. Quick Quiz.8 Consider the qualitative shapes of the two curves in Figure., without regard for the nuerical values or labels in the graph. Suppose you have two containers of gas at the sae teperature. Container A has 0 5 nitrogen olecules and container B has 0 5 hydrogen olecules. The correct qualitative atching between the containers and the two curves in Figure. is (a) container A corresponds to the blue curve and container B to the brown curve (b) container B corresponds to the blue curve and container A to the brown curve (c) both containers correspond to the sae curve. Exaple.5 A Syste of Nine Particles Nine particles have speeds of 5.00, 8.00,.0,.0,.0, 4.0, 4.0, 7.0, and 0.0 /s. (A) Find the particles average speed. Solution The average speed of the particles is the su of the speeds divided by the total nuber of particles: v ( ) /s 9.7 /s (B) What is the rs speed of the particles? Solution The average value of the square of the speed is v Hence, the rs speed of the particles is (C) What is the ost probable speed of the particles? Solution Three of the particles have a speed of.0 /s, two have a speed of 4.0 /s, and the reaining have different speeds. Hence, we see that the ost probable speed v p is.0 /s. ( ) /s 78 /s 9 v rs v 78 /s 3.3 /s

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