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Lecture notes in Physical Chemistry

Physical Chemistry lecture notes and Kinetics and Thermodynamics | pdf free download
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Lecture Notes in Physical Chemistry Semester 2: Kinetics and thermodynamics George C. McBane Department of Chemistry Grand Valley State University ©2016 George C. McBane December 1, 20162 notes-1Chapter 1 Gases We spent last semester learning about the structure and behavior of individual molecules. This semester will be spent learning about the behavior of collec- tions of molecules. We’ll start with gases, where the molecules more or less act individually. 1.1 Kinetic-molecular theory of gases The “kinetic theory of gases” makes the following assumptions about gases: Levine §15.1 1. Gases are composed of particles in constant, random motion. 2. The particles are negligibly small compared to the distances between them and the size of the container. 3. The particles do not interact except that they have collisions with each other and the container walls. In these collisions, the average translational energy is conserved. 4. The particles move according to classical mechanics. All but the first of those assumptions are not strictly true, so we should expect the theory to disagree with observation in some circumstances. We’ll have to see whether the disagreements are frequent enough to cause trouble. 1.1.1 Pressure of an ideal gas Consider a gas of identical molecules in a cubical container with sides of length l . Levine §15.2 We want to calculate from kinetic theory the pressure the gas exerts on the walls. 34 Pressure is force per unit area, so we will calculate the force on one wall of the 2 box and divide by its area l . From elementary mechanics we have dv d(mv) dp FÆ maÆ m Æ Æ , (1.1) dt dt dt ¢p so we will evaluate the force as from the momentum change¢p at the wall ¢t during some time interval¢t. First imagine a single gas molecule in the box. The components of its velocity 1 2 2 2 2 are v , v , and v . Its speed is vÆ (v Å v Å v ) , and its translational energy is x y z x y z 1 2 " Æ mv , where m is the molecule’s mass. The x component of its momentum tr 2 is mv . When the molecule collides with a wall parallel to the y z plane, let x us assume that the x component of its velocity changes sign, and the other two components are unaVected. (This assumption corresponds to a “specular reflection”, like light oV a mirror. It is not necessarily true for a single molecule having a single collision, but it must be true on average, or else the gas could develop a net direction of travel inside the box.) The change in momentum of the particle is therefore¢pÆ 2mv . The molecule will bounce back and forth in the x box (possibly also moving in the y and z directions, but those do not aVect the x motion). The round-trip time in the x direction is¢tÆ 2l/v , so the average x 2 ¢p mv 2mv x x force must be FÆ Æ Æ . The pressure this one-molecule gas exerts ¢t 2l /v l x 2 2 3 2 on the wall is then F /AÆ F /l Æ mv /l Æ mv /V . For the one-molecule gas we x x 2 therefore have PV Æ mv . x If we have N molecules that don’t interact, their forces on the wall (since they are all in the same direction) simply add, and we have P N 2 N X v 2 iÆ1 xi 2 PV Æ mv Æ mN Æ mNhv i, (1.2) x xi N iÆ1 where the angle brackets indicate the usual average (add up the individual values 2 of v and divide by the number of molecules.) x If we neglect gravity (appropriate for gases as long as the box isn’t too big) 2 2 2 there is nothing special about the x direction, so we expect thathv iÆhv iÆhv i. x y z 2 2 2 2 The square of the speed of a molecule is v Æ v Å v Å v , so the average squared x y z notes-11.1. Kinetic-molecular theory of gases 5 speed is N X 1 2 2 2 2 hv iÆ (v Å v Å v ) (1.3) xi yi zi N iÆ1 " N N N X X X 1 2 2 2 Æ v Å v Å v (1.4) xi yi zi N iÆ1 iÆ1 iÆ1 2 2 2 Æhv iÅhv iÅhv i (1.5) x y z 2 2 hv iÆ 3hv i (1.6) x 2 1 2 sohv iÆ hv i. Substituting into Eq. (1.2) gives x 3 1 1 2 2 PV Æ mNhv iÆ nMhv i, (1.7) 3 3 where n is the number of moles and M is the molar mass. This formula gives the pressure of the gas in terms of microscopic properties of the molecules (their masses and average squared speed). We see that the pressure is directly related to the average of the squared speed of the gas molecules: the faster they go, the higher the pressure. 1.1.2 RMS speed and average translational energy ¡23 The ideal gas law is PV Æ nRTÆ N k T , where k Æ 1.38£ 10 J/K is the Boltz- Levine §15.3 B B mann constant. (Note that the ordinary gas constant RÆ N k , where N is A B A Avogadro’s number.) From that we immediately obtain a formula for the average squared speed: 3RT 3k T B 2 hv iÆ Æ . (1.8) M m The “root-mean-square” or RMS speed c (called v in Levine) is just the square rms root of the average squared speed, and gives a measure of the typical speed of molecules in a sample of gas: 1 1 µ ¶ µ ¶ p 2 2 3k T 3RT B 2 cÆ hv iÆ Æ . (1.9) m M In problems the second form is more convenient, but you must remember to put in the molar mass M in kg/mol. Example What is the RMS speed of an N molecule at 300 K? We have 2 1 µ ¶ 1 µ ¶ ¡1 ¡1 2 2 3RT 3(8.314Jmol K )(300K) cÆ Æ Æ 517m/s. ¡1 M 0.028kgmol GCM December 1, 20166 1.1.3 Average speed, translational energy, and temperature By arguments similar to Eq. (1.3), you can show that the average translational 1 2 energy per molecule ish" iÆ mhv i, so from Eq. (1.8) we find immediately tr 2 3 h" iÆ k T. (1.10) tr B 2 So the average translational energy in a gas is indeed proportional to the absolute temperature, and now we know the proportionality constant. This proportionality holds not only for ideal gases but for any fluid whose translational motion can be described classically, that is, any ordinary liquid or vapor. (It does not hold accurately for solids at normal temperatures, or very low temperature fluids such as liquid helium.) 3 3 For the whole sample of N particles we have E Æ N k T Æ nRT . It is tr B 2 2 important to notice that even though the molecules have a wide distribution of speeds (more on that later), the translational energy of the whole sample is very well defined if N is large. Substituting PV from the ideal gas law gives 2 2 PV Æ Nh" iÆ E . (1.11) tr tr 3 3 1.1.4 The distribution of speeds Levine §15.4 The molecules do not all move with the same speed. To describe the distribution of speeds, we need to use a probability density function, just as in quantum me- 2 chanics (where the probability density function,à , described the distribution of position). Remarkably, we can find the distribution of speeds assuming only that all directions in space are equivalent, and that the diVerent velocity components for a molecule are independent: what a molecule’s speed is in the x direction says nothing about its speed in the y or z directions. One-dimensional velocity distribution First, let’s seek the one-dimensional distribution of speeds, f (v ), such that the x fraction of molecules with x-components of speed between a and b is Z b P(a· v · b)Æ f (v )dv . (1.12) x x x a (Levine, starting on p. 463, calls this one-dimensional distribution not f (v ) but x g (v ).) Because all directions in space are equivalent, the function of f (v ) must x x be the same one that describes the probability distributions in v and v as well. y z notes-11.1. Kinetic-molecular theory of gases 7 You can think of this one-dimensional function of v as giving the probability x that a molecule will have its x component of velocity between v and v Å dv . x x x The probability density must be normalized, so that Z 1 f (v )dv Æ 1 (1.13) x x ¡1 Now, what is the probability that a particular molecule will have its x compo- nent of velocity between v and v Ådv , its y component of velocity between v x x x y and v Ådv , and its z component of velocity between v and v Ådv ? Because y y Z Z z the speeds in the various directions are assumed to be independent, that must be the product of the three one-dimensional probabilities: dN v v v x y z Æ f (v )f (v )f (v )dv dv dv (1.14) x y z x y z N This is a three-dimensional function of the three velocity components. But, by the assumption of equivalence of directions, it cannot depend on the direction of the velocity; it can only depend on speed. Therefore, f (v )f (v )f (v )ÆÁ(v) (1.15) x y z ³ ´ 1 2 2 2 2 a function of speed v only, where vÆ v Å v Å v . Now, what kind of function x y z satisfies this requirement, that a product of functions of diVerent arguments is equal to a single function involving a sum of functions of the arguments? There’s a b c aÅbÅc only one function that does that: the exponential, because e e e Æ e . In 2 this case a must be v and so on, so the candidate function is x 1 2 ¡ bv x 2 f (v )Æ Ae , (1.16) x a Gaussian (Levine gives a more thorough argument for this result in equations 1 15.30–15.34.) I have inserted the¡ for later convenience (this only changes the 2 definition of b.) A and b are yet to be determined, but if we choose this distribu- tion function, we can be assured that the requirements of our assumptions will be satisfied. To find A, we normalize: the particle must have some x component of velocity, between¡1 and1. So Z 1 1 2 ¡ bv x 2 Ae dv Æ 1 (1.17) x ¡1 We can do this using the standard integral r Z 1 2 ¼ ¡ax e dxÆ , (1.18) a ¡1 GCM December 1, 20168 where aÈ 0, so that 1 µ ¶ 2 2¼ A Æ 1 (1.19) b µ ¶ 1 2 b AÆ (1.20) 2¼ Notice that b must be positive for this normalization to work; otherwise the integral is infinite and our function is not an acceptable probability density. 2 2 2 Now we need to find b. We havehv iÆ 3hv iÆ 3kT /m, so thathv iÆ kT /m. x x We can also calculate that average from the probability density function, using the usual formula for the average of a function (compare to a quantum mechanical ¤ expectation value, whereà à gives f (v )): x Z 1 2 2 hv iÆ v f (v )dv (1.21) x x x x ¡1 µ ¶ 1 Z 1 2 1 2 b 2 ¡ bv x 2 Æ v e dv (1.22) x x 2¼ ¡1 We need the standard integral 1 Z 1 2 2 (2n)¼ 2n ¡ax x e dxÆ , (1.23) 2n nÅ1/2 2 na ¡1 which with nÆ 1 and aÆ b/2 gives us µ ¶ 1 1 2 2 b 2¼ 2 hv iÆ (1.24) x 2 3/2 2¼ 2 (b/2) 1 Æ (1.25) b so we now have 2 ¡1 bÆ (hv i) (1.26) x µ ¶ ¡1 k T B Æ (1.27) m m Æ (1.28) k T B so that finally µ ¶ 1 µ ¶ 2 2 mv m x f (v )Æ exp ¡ . (1.29) x 2¼k T 2k T B B notes-11.1. Kinetic-molecular theory of gases 9 This expression gives us the one-dimensional distribution of velocity. Examples are plotted in Figure 1.1. Any particular molecule could have a velocity component (or projection) along the x axis anywhere between¡1 and1; this distribution function shows us that the most likely velocity component is zero, and that the probability density falls oV with increasingjv j in a Gaussian way. Note that the Gaussian will be x wider for larger T and for smaller m. Also note that the average speed along the x-axis is zero: there is no net tendency for the molecules to be moving either left 2 or right. (That is why we had to evaluate b usinghv i rather thanhv i; the latter x x quantity is zero no matter what value b has.) 0.0014 0.0014 300K N 0.0012 0.0012 2 N 300 K 0.0010 2 0.0010 0.0008 0.0008 0.0006 0.0006 0.0004 0.0004 He 700 K 0.0002 0.0002 0.0000 0.0000 -2000 -1000 0 1000 2000 -2000 -1000 0 1000 2000 v /(m/s) v /(m/s) x x Figure 1.1: The one-dimensional velocity distribution, showing variations with molecular mass and with temperature. The area under each curve is 1. Notice that the one-dimensional distribution can be written 1 µ ¶ µ ¶ 2 m " tr,x f (v )Æ exp ¡ . (1.30) x 2¼k T k T B B The argument of the exponential is the ratio of two terms, each with dimensions of energy: the “one-dimensional translational energy” of the molecule, " , tr,x and the “characteristic energy” kT . It is relatively easy for molecules to have translational energies less than or similar to kT , while it is quite improbable that they will have energies much greater than kT . This is our first example of the extremely important Boltzmann distribution. GCM December 1, 2016 f(v /(s/m) x10 Three-dimensional speed distribution Levine §15.4 Now we want to go on to find the distribution of molecular speeds in three dimensions. Note that while the velocity component v in a single dimension can x have any value between¡1 and1, the speed of a molecule must be nonnegative, 2 2 2 2 because v Æ v Å v Å v . We will therefore expect to find a probability density x y z function F (v) that is nonzero only for positive v. (Levine calls this function G(v).) Eq. (1.14) gave the probability that a molecule has its x-component of velocity between v and v Å dv , y-component of velocity between v and v Å dv , x x x y y y and its z-component of velocity between v and v Å dv , as the product of z z z the three independent probabilities. (Think of the probability of three people 1 1 1 simultaneously flipping coins all getting heads: it’s £ £ .) That is, 2 2 2 à µ ¶ 3 2 2 2 m(v Å v Å v ) d N 2 v v v m x y z x y z Æ exp ¡ dv dv dv (1.31) x y z N 2¼k T 2k T B B (Notice that the exponent on the normalization factor is now 3/2.) If you think of the function d N /N as living in a three-dimensional “velocity space” whose v v v x y z axes are v , v , and v , then the dv dv dv part of Eq. (1.31) describes the x y z x y z volume of a small rectangular box, which is located a distance v from the origin. We are looking for a distribution in speed only, and we don’t care what direction the molecule is moving. The most straightforward way to find that distribution is to convert Eq. (1.31) to the spherical polar coordinates v,µ,Á. The two angles specify the direction of motion and the v variable (which corresponds to r in ordinary spatial coordinates) is exactly the speed variable we are interested in. We can then integrate over the angular coordinatesµ andÁ. This problem is exactly analagous to the problem of finding the probability that an electron in an H atom is in some range of distances from the nucleus, independent of direction.) To change the distribution to spherical polar coordinates, we use the sub- 2 2 2 2 stitution v Æ v Å v Å v , and we must also remember to convert the “volume x y z 2 element” dx dy dz to the spherical polar element v sinµ dv dµ dÁ. We then inte- grate over the angles: 3 µ ¶ Z Z µ ¶ ¼ 2¼ 2 2 m mv 2 F (v)dvÆ sinµdµ dÁ exp ¡ v dv (1.32) 2¼k T 2k T B 0 0 B The only angular dependence is the simple sinµ, so the integration over bothµ andÁ is easy and just gives 4¼. So our final distribution of molecular speeds is 3 µ ¶ µ ¶ 2 2 m mv 2 F (v)Æ 4¼v exp ¡ (1.33) 2¼k T 2k T B B notes-11.1. Kinetic-molecular theory of gases 11 Eq. (1.33) is called the Maxwell distribution of speeds. 2 I think of this distribution in three parts: there’s a normalization part, a 4¼v “degeneracy” part that counts all the possible velocities that correspond to the same speed, and there is an exponential “Boltzmann factor” that compares the kinetic energy of the molecule to kT , the average energy available at temperature T . What do these curves look like? The normalization part does not depend on 2 v; the v part is a parabola; the Boltzmann part is a Gaussian centered at zero. 2 So at low speeds the curve looks like a rising parabola, then as v increases the curve turns over and dives back into the baseline as the Gaussian becomes small. Figure 1.2 shows examples corresponding to the 1D distributions we saw before. 0.0020 0.0020 N 300K 300 K N 0.0015 2 0.0015 2 0.0010 0.0010 He 700 K 0.0005 0.0005 0.0000 0.0000 0 500 1000 1500 2000 0 500 1000 1500 2000 v/(m/s) v/(m/s) Figure 1.2: The Maxwell distribution of speeds. The area under each of the curves is 1. 1.1.5 Testing the Maxwell distribution I know of two good methods for experimentally checking the Maxwell distribution of speeds: time-of-flight methods, including the use of slotted-disk “velocity selectors”, probably described in your textbook, and Doppler spectroscopy. In a velocity selector experiment, molecules leave a source through a small hole, and then pass through a series of disks with slots in them. The disks are arranged on a rotating shaft and the slots are oVset, so that for a particular speed of rotation only molecules of a particular speed can make it through all the slots. Which speed makes it through is controlled by the rotation rate of the shaft. These experiments were first done by Eldridge in 1927 (J. A. Eldridge, Phys. Rev. 30, GCM December 1, 2016 F(v)/(s/m)12 931 (1927).) A thorough analysis of slotted-disk velocity selectors by C. J. B. van den Meijdenberg appears in Atomic and Molecular Beam Methods, G. Scoles, ed., (Oxford, 1988). In Doppler spectroscopy, the absorption spectrum of gas molecules is mea- sured with very high resolution. Nowadays such spectroscopy is often done with lasers since they can provide the required resolution easily. A molecule moving toward a laser source will “see” a frequency that is higher than the frequency of the laser because of the Doppler eVect. The shift is proportional to v /c, where x v is the component of the molecule’s velocity along the laser beam direction x and c is the speed of light. The absorption spectrum that appears therefore has lines that are broadened by the motion of the molecules, and if the line shape is measured carefully, the distribution f (v ) can be determined directly. x 1.1.6 Applications of the Maxwell distribution Average speedhvi Levine §15.5 We use the usual approach to averaging things: Z 1 hviÆ vF (v)dv (1.34) 0 Z µ ¶ 3 ³ ´ 1 2 m mv 2 3 Æ 4¼ v exp ¡ dv (1.35) 2¼kT 2kT 0 Use the standard integral Z 1 2 n 2nÅ1 ¡ax x e dxÆ (1.36) nÅ1 2a 0 m with nÆ 1 and aÆ to give 2kT 3 ³ ´ m 1 2 hviÆ 4¼ (1.37) ¡ ¢ 2 m 2¼kT 2 2kT µ ¶ 1 2 8kT Æ (1.38) ¼m p 2 The average speedhvi diVers from the root-mean-square speed hv iÆ c p p p because it contains the numerical factor 8/¼Æ 2.546 rather than 3. Most probable speed v mp The most probable speed is the speed at which F (v) reaches a maximum. We find it by diVerentiating F (v), setting the derivative equal to 0, and solving for notes-11.1. Kinetic-molecular theory of gases 13 v ; I ask you to work it out in a homework problem. The result is mp 1 µ ¶ 2 2k T B v Æ (1.39) mp m 1 ³ ´ p 2 kT The most probable speed has 2 as the numerical factor multiplying ; it is m the smallest of the three measures of speed we have considered. Fractions of molecules within finite speed or velocity ranges The natural interpretation F (v) is “the fraction of molecules with speeds between R b a and b is P(a· v Ç b)Æ F (v)dv”, and similarly for f (v ). However, that x a integral cannot be evaluated analytically for values of a or b diVerent from zero or infinity. Such finite integrals can be expressed in terms of the error function, erf(x), defined by Z x 2 2 ¡t erf(x)Æ e dt. (1.40) p ¼ 0 The error function erf(x) can be evaluated easily in Excel with the notationERF(). In terms of it the two most useful indefinite integrals for evaluating probabilities of speeds or velocities in finite ranges are ¡ ¢ p p Z ¼erf x a 2 ¡ax e dxÆ ÅC (1.41) p 2 a ¡ ¢ p p Z 2 ¡ax ¼erf x a 2 xe 2 ¡ax x e dxÆ ¡ ÅC (1.42) 3/2 4a 2a So, for example, we might ask “What fraction of oxygen molecules have speeds between 100 and 200 m/s?” Comparing Eq. (1.33) with those standard integrals, we see that we need the second one with aÆ m/2k T Æ M/2RT Æ B p ¡6 2 2 6.4149£ 10 s /m so aÆ 0.00253276s/m. We have for our probability " ¡ ¢ p p 2 200 Z ³ ´ ¡ax 200 3/2 a ¼erf x a xe P(100· vÇ 200)Æ F (v)dvÆ 4¼ ¡ (1.43) 3/2 ¼ 2a 4a 100 100 p p Excel givesERF(100 a)Æ 0.279796 andERF(200 a)Æ 0.526239. Plugging in the rest of the numbers (which I did in a simple Excel spreadsheet, shown in Figure 1.3) gives PÆ 0.07225, or 7.225%. GCM December 1, 201614 0.0025 M 0.032 kg/mol R 8.314 J/mol K 0.0020 T 300 K a 6.41488E-06 s2/m2 sqrt(a) 0.002532762 s/m 0.0015 lowlim 100 m/s uplim 200 m/s 0.0010 lowlimsqrt(a) 0.253276184 erf(lowlimsqrt(a)) 0.279796 uplimsqrt(a) 0.506552368 erf(uplimsqrt(a)) 0.526239 0.0005 prefactor 3.66664E-08 integral at lower limit 320792.7 integral at upper limit 2291361 0.0000 0 200 400 600 800 1000 1200 prefactor(intupper-intlower) 0.072254 v/(m/s) Figure 1.3: Spreadsheet and plot showing calculation of fraction of O molecules 2 with speeds between 100 and 200 m/s at 300 K. The spreadsheet is available on the Blackboard site as o2frac.xls . Wall collision rates and eVusion Collision rate with a wall Let’s work out the number of molecules that hit a Levine §15.6 container wall in a time¢t. We’ll do this by considering the contribution of each velocity separately, then adding them up, using the same cubical box of side l we used before. Let’s say the wall is perpendicular to the x axis. The number of molecules in the gas with x component of velocity between v and v Å dv is just N f (v )dv , x x x x x where f (v ) is the one-dimensional velocity distribution and N is the total num- x ber of molecules. Those molecules will hit the wall in time¢t if v is toward x the wall and they start out within a distance v ¢t of the wall. Since the length x of the box is l, the fraction of those molecules close enough to hit the wall is v ¢t/l, so the number of collisions they contribute is (v ¢t/l)N f (v )dv . The x x x x total number of wall collisions during time¢t is the sum of the contributions from all velocities, or Z 1 N¢t N Æ v f (v )dv . (1.44) wall x x x l 0 Notice that we only integrate from 0 up to1, not from¡1; molecules that are moving away from the wall don’t hit it. Putting in Eq. (1.29) for f (v ) we have x µ ¶ 1 µ ¶ Z 2 1 2 mv N¢t m x N Æ v exp ¡ dv . (1.45) wall x x l 2¼k T 2k T B 0 B notes-1 F(v)1.1. Kinetic-molecular theory of gases 15 The integral can be done with straightforward substitution and equals k T /m, B so we have µ ¶ 1 µ ¶ 1 2 2 N¢t m k T N¢t k T B B N Æ Æ . (1.46) wall l 2¼k T m l 2¼m B A more generally useful quantity is the collision rate per unit area, which is N wall 3 simply ; since l Æ V , that rate is 2 l ¢t 1 dN 1 N 1 P N wall A Æ hviÆ hvi (1.47) A dt 4 V 4 RT where N is Avogadro’s number, and we have used the ideal gas law andhviÆ A p 8k T /¼m. So the number of collisions per second with a wall is proportional B to P and tohvi, as we expected. EVusion rates Now, say we punch a small hole in the wall of area a, and we put the box in a vacuum system so there is no gas to leak into the container. This hole must be small enough that leaks through it don’t disturb the distribution of speeds or densities in the main gas; eVectively, the molecules must escape one at a time. At what rate does gas leak out? Each molecule that hits the hole will escape, so the escape rate will be just the area of the hole times the collision frequency. The rate of change of the number of molecules N in the container will be just dN aP N ¡aP N A A Æ¡ hviÆ . (1.48) 1/2 dt 4RT (2¼MRT ) EVusion from a mixture of gases can be used to separate the gas molecules by mass. The eVusion rate is proportional to the inverse square root of the molecular mass; light molecules escape through the hole more quickly than heavier ones. The eVusing gas is therefore enriched in the light component. This eVect is the basis of the gaseous-diVusion technique used at Oak Ridge to separate uranium isotopes for the first atomic bombs. It is convenient to rewrite Eq. (1.48) in terms of mass. Converting the number of molecules N to the mass by dividing by Avogadro’s number and multiplying by the molecular weight M, we find for the rate of mass loss 1 µ ¶ 2 dm M gas Æ¡aP (1.49) dt 2¼RT This expression provides a standard way of determining the vapor pressures of not-very-volatile substances. If you put a sample of solid inside a cell with a small hole of known area a (determined with a microscope), apply a vacuum GCM December 1, 201616 pump to the other side of the hole, let the material eVuse for a while (typically minutes to hours), and then measure the loss of mass of your sample, you can determine its vapor pressure P. Irving Langmuir at General Electric measured vapor pressure curves for tungsten this way more than fifty years ago, and his results still appear in standard reference books; they have not been improved upon. 1.1.7 Translational energy distribution Levine §15.5 Let’s consider the Maxwell distribution in terms of translational energy rather than speed. We need to make a change of variable. Any time you convert a distribution from one set of variables to another, you must be careful: you must make sure that probabilities calculated from the distributions written in terms of the two diVerent variables match up. In other words, if the energy distribution is G(" ), we must have G(" )d" Æ F (v)dv. Therefore, we must be careful to tr tr tr change variables in the accompanying diVerential dv as well as in F (v) itself. When we converted the Maxwell distribution from Cartesian to spherical polar coordinates above, we did this matching “automatically” by knowing beforehand 2 that the new volume element in spherical polar coordinates was r sinµdrdµdÁ. Here we are making a diVerent change so I will show the conversion in more detail. We want to change from v to" , starting from tr G(" )d" Æ F (v)dv (1.50) tr tr 3 µ ¶ µ ¶ 2 2 m mv 2 G(" )d" Æ 4¼v exp ¡ dv (1.51) tr tr 2¼k T 2k T B B ³ ´ 1 ³ ´ 1 ³ ´ 1 ¡ 2 2 2 1 2" 1 2" 2 1 2 tr tr We have" Æ mv , so vÆ and dvÆ d" Æ d" . In tr tr tr 2 m 2 m m 2m" tr ³ ´ 1 ³ ´ 1 2" 2 2 1 tr the Maxwell distribution F (v)dv, we replace v with and dv with d" tr m 2m" tr to get µ ¶ 3 µ ¶ µ ¶µ ¶ 1 1 2 2 m 2" " 1 ¡ tr tr 2 G(" )d" Æ 4¼ exp ¡ " d" (1.52) tr tr tr tr 2¼k T m k T 2m B B 3 µ ¶ µ ¶ 1 2 1 " tr 2 Æ 2¼ " exp ¡ d" (1.53) tr tr ¼k T k T B B so we say that µ ¶ 3 µ ¶ 1 2 1 " tr 2 G(" )Æ 2¼ " exp ¡ . (1.54) tr tr ¼k T k T B B notes-11.1. Kinetic-molecular theory of gases 17 All the dependence on mass has canceled; the translational energy distribution is the same for all molecules at the same temperature. Figure 1.4 shows this distribution for temperatures of 300 and 700 K. -4 2.0x10 300K 1.5 1.0 700 K 0.5 0.0 0 5 10 15 20 25 ε /(kJ/mol) tr Figure 1.4: Translational energy distributions for gases. The distribution is inde- pendent of mass. The translational energy distribution rises very steeply from the origin; it has infinite slope at the origin, while the speed distribution has zero slope there. If you draw a vertical line at any energy, the area under the distribution to the right of that line gives the fraction of molecules with translational energy equal to or greater than that amount. In a simple theory of chemical kinetics, it is only those molecules than can surmount an “activation barrier” and react; this distribution therefore plays an important role in kinetics. GCM December 1, 2016 G(ε )/(mol/kJ) tr18 1.1.8 Hard-sphere collision rates Levine §15.7 Let’s begin thinking about molecules colliding with each other. Clearly that can be a complicated field; most of the richness of chemical reactions occurs in some sequence of bimolecular collisions, and if a single simple theory could describe everything about those collisions chemistry wouldn’t be nearly so interesting. But, for starters, let’s use a simple theory: think of molecules as little tiny marbles. The “hard-sphere” model can teach us a remarkable amount about molecular collisions. I’ll start out by thinking about one molecule as moving with speed v through rel a forest of other, stationary, molecules. All the molecules are hard spheres with diameter d. As our one molecule moves along, if its trajectory takes its center within a distance d of the center of any other molecule, the two will hit. (See Figure 1.5.) In a time t, our molecule carves out a “collision cylinder” of volume 2 ¼d v t; any other molecules whose centers are in that cylinder will collide with rel it. The number of such molecules is just the volume of the cylinder times the number density of the gas,N Æ N /V . So the number of collisions one molecule makes per second, z, is 2 zƼd v N . (1.55) rel miss d hit hit Figure 1.5: The collision cylinder. Molecules whose centers lie within the cylinder will be hit by the moving molecule. If we rewriteN with the ideal gas law we find N nN P N P A A N Æ Æ Æ Æ (1.56) V V RT kT so that in terms of the pressure the collision rate of a single molecule is P 2 zƼd v (1.57) rel kT 2 The eVective “target area” of the molecule,¼d , is often called the collision cross section and given the symbol¾. This idea of an eVective size can be usefully notes-11.1. Kinetic-molecular theory of gases 19 extended to many kinds of events other than hard-sphere collisions. Events that are less likely than simply bouncing—for example, chemical reaction—will have smaller cross sections. Of course, all the molecules are moving, and not all with the same speed. When you include all the molecules’ motions, the appropriate value for v is just rel the average speed, but calculated with the reduced mass of the colliding pair: µ ¶ 1 2 8kT v Æ (1.58) rel ¼¹ where, as usual, ¹Æ m m /(m Å m ) and m and m are the masses of the 1 2 1 2 1 2 colliding molecules. (Once again, you can express¹ in kg/mol and use R in the numerator rather than k.) If two diVerent kinds of molecules are colliding, they might have diVerent sizes as well as diVerent masses; in that case, you use the average diameter dÆ (d Å d )/2 in Eq. (1.57). 1 2 The formulas I have given so far describe the number of collisions a single molecule makes with other molecules (either the same kind or diVerent) in a gas. In a gas that contains molecule types A and B, the number of A–B collisions per second per unit volume is Z Æ z N (1.59) AB AB A 1 µ ¶ µ ¶µ ¶ 2 8kT P P A B Z ƾ . (1.60) AB AB ¼¹ kT kT AB The number of B-B collisions per second per unit volume is calculated simi- larly, but we must divide by 2 to avoid counting the same collision twice: 1 Z Æ z N (1.61) BB BB B 2 µ ¶ 1µ ¶ 2 2 1 8kT P B Z Æ ¾ (1.62) BB BB 2 ¼¹ kT BB 2 where¹ Æ m /2,¾ Ƽd , and P is the partial pressure of B. BB B BB B B Mean free path We have seen how to calculate the number of collisions a particular molecule makes with other molecules per second, and also how to calculate the average speed of the molecule. With those two results it is easy to find the average distance a molecule travels between collisions, the mean free path¸: kT ¸Æhvi/zÆ . (1.63) p 2¾P At one atmosphere and 300 K, for nitrogen and oxygen¸» 160 nm. GCM December 1, 201620 notes-2