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Kinetic Theory of Gases
Chapter Outline
1.1 Introduction
1.2 Pressure of an Ideal Gas
1.3 Temperature and Energy
1.4 Distributions, Mean Values, and Distribution Functions
1.5 The Maxwell Distribution of Speeds
1.6 Energy Distributions
1.7 Collisions: Mean Free Path and Collision Number
1.8 Summary
Appendix 1.1   The Functional Form of the Velocity Distribution Appendix 1.2   Spherical Coordinates Appendix 1.3   The Error Function and Co-Error Function Appendix 1.4   The Center-of-Mass Frame
1.1 INTRODUCTION
The overall objective of this chapter is to understand macroscopic properties such as pressure and temperature on a microscopic level. We will find that the pressure of an ideal gas can be understood by applying Newton's law to the microscopic motion of the molecules making up the gas and that a comparison between the Newtonian prediction and the ideal gas law can provide a function that describes the distribution of molecular velocities. This distribution function can in turn be used to learn about the frequency of molecular collisions. Since molecules can react only as fast as they collide with one another, the collision frequency provides an upper limit on the reaction rate.
The outline of the discussion is as follows. By applying Newton's laws to the molecular motion we will find that the product of the pressure and the volume is proportional to the average of the square of the molecular velocity, <u2>, or equiv-alently to the average molecular translational energy e. In order for this result to be consistent with die observed ideal gas law, the temperature Tof the gas must also be proportional to <v2> or <e>. We will then consider in detail how to determine the average of the square of the velocity from a distribution of velocities, and we will use the proportionality of T with <v2> to determine the Maxwell-Boltzmann distribution of speeds. This distribution, F(v) dv, tells us the number of molecules with speeds between v and V + dv. The speed distribution is closely related to the distribution of molecular energies, G(e) de. Finally, we will use the velocity distribution
Section 1.2  Pressure of an Ideal Gas
■   Figure 1.1
All the molecules in the box that are moving toward the z-y plane will strike the wall.
Of course, not all molecules will be traveling with the same velocity vx. We will learn below how to characterize the distribution of molecular velocities, but for now let us simply assume that the pressure will be proportional to the average of the square of the velocity in the x direction, p = n*m<v2>.c The total velocity of an individual molecule most likety contains other components along y and z. Since v = ivx + jvy + kuz,A where i, j, and /rare unit vectors in the x, y, and z directions, respectively, v2 = v2 + v2 + v2 and <v2> = <i>2> + <i>2> + <v2>. In an isotropic gas the motion of the molecules is random, so there is no reason for the velocity in one particular direction to differ from that in any other direction. Consequently, <v2> = <Vy> = <v2> = <v2>/3. When we combine this result with the calculation above for the pressure, we obtain
p = ~nm<v2>. (1.1)
Of course, n in equation 1.1 is the number of molecules per unit volume and can be rewritten as nNA/V, where vVA is Avogadro's number and n is the number of moles. The result is
I
pV - -x/VAwO>. (1.2) 3
Since the average kinetic energy of the molecules is <e> = \m<v2>, another way to write equation 1.2 is
PV=yiNA<e>. (1.3)
Equations 1.2 and 1.3 bear a close resemblance to the ideal gas law,/>V = nRT. The ideal gas law tells us that the product of p and V will be constant if the temperature is constant, while equations 1.2 and 1.3 tell us that the product will be constant if <v2> or <€> is constant. The physical basis for the constancy of pV with <v2> or <e> is clear from our previous discussion. If the volume is
cIn this text, as in many others, we will use the notation <x> or x to mean "the average value ofx" dThroughout the text we will use boldface symbols to indicate vector quantities and normal weight symbols to indicate scalar quantities. Thus, v = Ivl. Note that v2 = v ■ v = l>2.
Section 1.4  Distributions, Mean Values, and Distribution Functions
(1.38 X 10"23J/K)(300K) [(28/6.02 X 1026)]
= 2.67 X 105(m/s)2 = (516 m/s)2.
To summarize the discussion so far, we have seen from equation 1.2 that pV is proportional to <v2> and that the ideal gas law is obtained if we take the definition of temperature to be that embodied in equation 1.5. Since <e> = \m<v2>, both temperature and pV are proportional to the average of the square of the velocity. The use of an average recognizes that not all the molecules will be moving with the same velocity. In the next few sections we consider the distribution of molecular speeds. But first we must consider what we mean by a distribution.
1.4   DISTRIBUTIONS, MEAN VALUES, AND DISTRIBUTION FUNCTIONS
Suppose that five students take a chemistry examination for which the possible grades are integers in the range from 0 to 100. Let their scores be ,S, = 68, S2 = 76, S3 = 83, S4 = 91, and S5 = 97. The average score for the examination is then
S, + S2 + S} + 54 + S5     1 *k <s> = _J-2__^-4-5 = J_ (1.7)
where iVT = 5 is the number of students. In this case, the average is easily calculated to be 83.
Now suppose that the class had 500 students rather than 5. Of course, the average grade could be calculated in a manner similar to that in equation 1.7 with an index i running from 1 to Nr = 500. However, another method will be instructive. Clearly, if the examination is still graded to one-point accuracy, it is certain that more than one student will receive the same score. Suppose that, instead of summing over the students, represented by the index in equation 1.7, we form the average by summing over the scores themselves, which range in integer possibilities from j = 0 to 100. In this case, to obtain the average, we must weight each score Sj by the number of students who obtained that score, Nf.
1 100
<S> = ]7 2W (1.8)
Note that the definition of JV; requires that XNj = NT. The factor l/NT in equation 1.8 is included for normalization, since, for example, if all the students happened to get the same score S, = S then
Now let us define the probability of obtaining score Sf as the fraction of students receiving that score:
NJ
P, = ~. (1.10)
1 NT
Section 1.4   Distributions, Mean Values, and Distribution Functions
or, more generally for any observable quantity,
'(.» = \P(q)q<iq.
11.16)
Equation 1.16 will form the basis for much of our further work. The probability function p(q) is sometimes called a distribution function, and the range of the integral is over all values of q where the probability is nonzero. Note that normalization of the probability requires
\p{Q)*Q = 1.
(1.17)
The quantity \ip(x)\2dx is simply a specific example of a distribution function. Although knowledge of quantum mechanics is not necessary to solve it, you may recognize a connection to the particle in the box in Problem 1.7, which like Example 1.3 is an exercise with distribution functions.
example
Determining Distribution Functions
Objective Bees like honey. A sphere of radius r0 is coated with honey and hanging in a tree. Bees are attracted to the honey such that the average number of bees per unit volume is given by Kr~5, where K is a constant and /• is the distance from the center of the sphere. Derive the normalized distribution function for the bees. They can be at any distance from the honey, but they cannot be inside the sphere. Using this distribution, calculate the average distance of a bee from the center of the sphere.
Method First we need to find the normalization constant K by applying
equation 1.17, recalling that we have a three-dimensional problem and that in spherical coordinates the volume element for a problem that does not depend on the angles is 4-rrr2 dr. Then, to evaluate the average, we apply equation 1.16.
Solution Recall that, by hypothesis, there is no probability for the bees being at r < r0, so that the range of integration is from r0 to infinity. To determine K we require
(Kr~5) Airr dr = 1,
(1.18)
or
AttK\   r~3dr = 1 = AirK
°° _ AttK
Or2 ' r0 £l 0
(1.19)
so that
K =
A
2tt'
Section 1.5   The Maxwell Distribution of Speeds
1.5.2  The Velocity Distributions Are Independent and Uncorrelated
We now consider the relationship between the distribution of x-axis velocities and y- or z-axis velocities. In short, there should be no relationship. The three components of the velocity are independent of one another since the velocities are uncorrelated. An analogy might be helpful. Consider the probability of tossing three honest coins and getting "heads" on each. Because the tosses t; are independent, uncorrelated events, the joint probability for a throw of three heads, P{tl = heads, r2 = heads, f3 = heads), is simply equal to the product of the probabilities for the three individual events, P{}x = heads) X P(t2 = heads) X P(t3 = heads) = | X 5 X I. In a similar way, because the x-, y-, and z-axis velocities are independent and uncorrelated, we can write that
F{vx,vy,vz) = F{vx)F(vy)F{vz). (1.21)
We can now use the conclusion of the previous section. We can write, for example, that F(ux) = f(u2) and similarly for the other directions. Consequently,
F(vx,vy,vz) = F(vx)F(uy)F(vz) = f(v2x )f(v2y)f(v2). (1.22)
What functional form has the property that/(a + b + c) = j{a)f(b)j{c)l A little thought leads to the exponential form, since exp(a + b + c) = eaef'ec. It can be shown, in fact, that the exponential is the only form having this property (see Appendix 1.1), so that we can write
F(vz) =f(v2x) = Kexp(±kv% (1.23)
where K and k are constants to be determined. Note that although k can appear mathematically with either a plus or a minus sign, we must require the minus sign on physical grounds because we know from common experience that the probability of very high velocities should be small.
The constant K can be detennined from normalization since, using equation 1.17, the total probability that vx lies somewhere in the range from — °° to +°° should be unity:
F(vx)dvx = l. (1.24)
o
Substitution of equation 1.23 into equation 1.24 leads to the equation 1 = K
exp(-KWj)dwx = Ky—J , (1.25) where the integral was evaluated using Table 1.1. The solution is then K = (k/tt)112. 1.5.3   <v2> Should Agree with the Ideal Gas Law
The constant k is determined by requiring <v2> to be equal to 3kT/m, as in equation 1.6. From equation 1.16 we find
<v2>= f  vlF(vx)dvx= (jV f  vUxv(-kv2x)dvx. (1.26) ■Loo ^   ' J-03
The integral is a standard one listed in Table 1.1, and using its value we find that
Section 1.5   The Maxwell Distribution of Speeds
i—i—i—i—i—i—i—i—i—i—i—i—i—i—i—i—i—i—r
vx (m/s)
■   Figure 1.2
One-dimensional velocity distribution for a mass of 28 amu and two temperatures.
1.5.4  The Distribution Depends Only on the Speed
Note that the right-hand side of equation 1.30 depends on v2 and not on the directional property of v. When we have a function that depends only on the length of the velocity vector, i> = Ivl, and not on its direction, we can be more precise by saying that the function depends on the speed and not on the velocity. Since F(vx,vy,vJ = fiv2) depends on the speed, it is often more convenient to know the probability that molecules have a speed in a particular range than to know the probability that their velocity vectors will terminate in a particular volume. As shown in Figure 1.3, the probability that the speed will be between v and v + dv is simply the probability that velocity vectors will terminate within the volume of a spherical shell between the radius v and the radius v + dv. The volume of this shell is dur dvy dvz = 4irv2 dv, so that the probability that speed will be in the desired range isf
rAn alternate method for obtaining equation 1.31 is (0 note that dvxdvydvz can be written as v2s'md d0 d<f> du in spherical coordinates (see Appendix 1.2) and then to integrate over (he angular coordinates. Since the distribution does not depend on Ihe angular coordinates, the integrals over d9 and dtf> simply give 4-tt and we are left with the factor v2 dv.
F(v)dv =
sin0dudfld$
=w(2^?rexp(^^)dy-
A more complete description of spherical coordinates is found in Appendix 1.2.
Section 1.5   The Maxwell Distribution of Speeds
1.6: = <v2>m = (3kT/m)m. Another speed is the mean speed defined by using equation 1.16 to calculate <v>:
<v> =
vF(v)dv
(1.32)
where the integral was evaluated using Table 1.1 as described in detail in Example 1.4. Finally, the distribution might also be characterized by the most probable speed, c, the speed at which the distribution function has a maximum (Problem 1.8):
2kT in
(1.33)
example
Using the Speed Distribution
Objective Method
The speed distribution can be used to determine averages. For example, find the average speed, <u>.
Once one has the normalized distribution function, equation 1.16 gives the method for finding the average of any quantity. Identifying Q as the velocity and p(Q) dQ as the velocity distribution function given in equation 1.31, we see that we need to integrate vF(u) dv from limits v = 0 to v = ~>.
Solution
" CO f CO /       % \
vF(v)dv = J  4irvyy~^Jtxp(-a2v2)dv aVexp(—a2v2)dv,
(1.34)
Vír"
where a = (m/2kT)U2. We now transform variables by letting x = av. The limits will remain unchanged, and du = Axla. Thus the integral in equation 1.34 becomes
4    i™ 4 1
—-t= x3&T^{~x2)dx = —y=-aVTT Jq ay it z
(1.35)
__2_fmyi1 _ (mi V^v m ) \™n
where we have used Table 1.1 to evaluate the integral.
1/2
The molecular speed is related to the speed of sound, since sound vibrations cannot travel faster than the molecules causing the pressure waves. For example, in Example 1.5 we find that the most probable speed for 02 is 322 m/s, while the
Section 1.5   The Maxwell Distribution of Speeds
3000
2400
1800
1200
600
Mass (amu). 20 40 60
k i\   1 i li—r~nr
\\\
80
"1   i   i r
r=300K
100 ,500
I
- 400
\\\
s \
300 „ 200 "
100
OL-j_i_i_i_I_i_i_i_i_I_i_1_i_i_I_1_i_i_i_I_i_i_i_1_J o
0 5 10 15 20 25
Mass (amu)
■   Figure 1.6
Various average speeds as a function of mass for T = 300 K.
atoms having speeds in excess of ue, while minuscule (about 10~31), is still 10'175 times larger than the fraction of oxygen molecules having speeds in excess of vj As a consequence, the composition of the atmosphere, is changing; much of the helium released during the lifetime of the planet has already escaped into space. A plot of various speeds as a function of mass for T = 300 K is shown in Figure 1.6.
1.5.5   Experimental Measurement of the Maxwell Distribution of Speeds
Experimental verification of the Maxwell-Boltzmann speed distribution can be made by direct measurement using the apparatus of Figure 1.7. Two versions of the measurement are shown. In Figure 1.7a, slits (S) define a beam of molecules moving in a particular direction after effusing from an oven (O). Those that reach the detector (D) must successfully have traversed a slotted, multiwheel chopper by traveling a distance d while the chopper rotated through an angle <j>. In effect, the chopper selects a small slice from the velocity distribution and passes it to the detector. The speed distribution is then measured by recording the integrated detector signal for each cycle of the chopper as a function of the angular speed of the chopper.
A somewhat more modem technique, illustrated in Figure 1.7b, clocks the time it takes for molecules to travel a fixed distance. A very short pulse of molecules leaves the chopper at time t = 0. Because these molecules have a distribution of speeds, they spread out in space as they travel toward the detector, which records as a function of time the signal due to molecules arriving a distance L from the chopper.
Section 1.6   Energy Distributions
I_:_I_I_[_
0 200 400 600
Flight time (/is)
■   Figure 1.8
Time-of-flight measurements: intensity as a function of flight time.
From I. F. C. Wang and H. Y. Wachman, as illustrated in F. O. Goodman and H. Y. Wachrnan, Dynamics of Gas-Surface Scattering (Academic Press, New York, 1976). Figure from "Molecular Beams" in DYNAMICS OF GAS-SURFACE SCATTERING by F. O. Goodman and H. Y. Wachmann, copyright © 1976 by Academic Press, reproduced by permission of the publisher. All rights or reproduction in any form reserved.
using this "time-of-flight" technique. The open circles are the detector signal, while the smooth line is a fit to the data of a function of the form expected for S(t). The best fit parameter gives a temperature of 300 K.
1.6   ENERGY DISTRIBUTIONS
It is sometimes interesting to know the distribution of molecular energies rather than velocities. Of course, these two distributions must be related since the molecular translational energy e is equal to \mv2. Noting that this factor occurs in the exponent of equation 1.31 and that de = mv dv = (2me)1/2 dv, we can convert velocities to energies in equation 1.31 to obtain
277*77    '   \   AT / V2
(1.37)
The function G(e) de tells us the fraction of molecules which have energies in the range between e and € 4- de. Plots of G(e) are shown in Figure 1.9.
The distribution function G(e) can be used to calculate the average of any function of e using the relationship of equation 1.16. In particular, it can be shown as expected that <e> = 3£772 (see Problem 1.9).
Let us pause here to make a connection with thermodynamics. In the case of an ideal monatomic gas, there are no contributions to the energy of the gas from internal degrees of freedom such as rotation or vibration, and there is normally very
Section 1.7   Collisions: Mean Free Path and Collision Number
0 1 2 3 4 5 6
a2 = e*/ir
■   Figure 1.10
The fraction of molecules having energy in excess of e' as a function of z'/kT.
Problem 1.10 shows that this integral is given by
f(e) = —j= ae-"1 + erfc(a), (1.41)
where a = (&*/kT)[l2 and erfc(a) is the co-error function defined in Appendix 1.3. A plot of/(e*) as a function of e*/kT is shown in Figure 1.10. Note that for e* > 2>kT. the function _/{€*) is nearly equal to the first term in equation 1.41, 2\Z(e*/irkT) exp(—£*/kT), shown by the dashed line in the figure. Thus, the fraction of molecules with energy greater than e* falls off as Ve* exp(—e/kT), provided that e* > 3kT.
1.7   COLLISIONS: MEAN FREE PATH AND COLLISION NUMBER
One of the goals of this chapter is to derive an expression for the number of collisions that molecules of type 1 make with molecules of type 2 in a given time. We will argue later that this collision rate provides an upper limit to the reaction rate, since the two species must have a close encounter to react.
The principal properties of the collision rate can be easily appreciated by anyone who has ice skated at a local rink. Imagine two groups of skaters, some rather sedate adults and some rambunctious 13-year-old kids. If there is only one kid and one adult in the rink, then the hkelihood that they will collide is small, but as the number of either adults or kids in the rink increases, so does the rate at which collisions
Section 1.7   Collisions: Mean Free Path and Collision Number
■   Figure 1.12
Molecule 1 sweeps out a cylinder of area Trb^„. Any molecule of type 2 whose center is within the cylinder will be struck.
Consider a molecule of type 1 moving through a gas with a speed equal to the average magnitude of the relative velocity <ur>. Figure 1.12 shows that any molecule of type 2 located in a cylinder of volume ■nb7mM<vr> Ar will then be struck in the time At.' If the density of molecules of type 2 is n^> then the number of collisions one molecule of type 1 will experience with molecules of type 2 per unit time is
Z2 = nbl„<vT>n2. (1.42)
Of course, for a molecule of type 1 moving through other molecules of the same type,
= 7rfo;~mN<i'r> n\ = 7rd~<vT> n], (1.43)
where has been replaced by d2 since r, + r2 = 2rt = d. The quantity trb2^^ is known as the hard-sphere collision cross section. Cross sections are generally given the symbol cr.
Equation 1.42 gives the number of collisions per unit time of one molecule of type 1 with a density n\ of molecules of type 2. The total number of collisions of molecules of type 1 with those of type 2 per unit time and per unit volume is found simply by multiplying by the density of type 1 molecules:
Z12 = Zjji'i = Trb2mm <yr> n\n2. (1-44)
Note that the product n\n\ is simply proportional to the total number of pairs of collision partners.
By a similar argument, if there were only one type of molecule, the number of collisions per unit time per unit volume is given by
Z„ = X-Z,n\ = i TTfeL, <»P> (n,)2. d-45)
The factor of \ is introduced for the following reason. The collision rate should be proportional to the number of pairs of collision partners. If there are n molecules, then the number of pairs is n{n - l)/2, since each molecule can pair with n — 1 others and the factor of 2 in the denominator corrects for having counted each pair twice. If n is a large number, then we can approximate n(n — 1) as n2, and since the number of molecules is proportional to the number density, we see that the number of pairs goes as {n\)2l2.
It remains for us to determine the value of the relative speed, averaged over the possible angles of collision and averaged over the speed distribution for each molecule. One way to arrive quickly at the answer for a very specific case is shown in
'Because of the collisions, the molecule under consideration will actually travel along a zigzag path, but the volume swept out per unit time will be the same.
Section 1.7   Collisions: Mean Free Path and Collision Number
Method Use equation 1.44, remembering to convert the abundances to
number densities at 300 K and calculating the average relative velocity by use of equation 1.46.
Solution First find the total number density n at 1 arm: n = {nlV)NK = (plRT)NA = (1 atm)(6.02 X 1023 molec/mole)/[(0.082 L armmor1 K-')(300 K)] = 2.45 X 1022 molec/L. Next determine the number densities of NO and 03, each being the total density times 0.2 X 10~6: «*(NO) = n(03) = (0.2 X 10"6)(2.45 X 1022) = 4.9 X 1015 molec/L. The average relative velocity is <ur> = {?>kTlirpC)m = [8(1.38 X 10"23 J K-^POO K)(6.02 X 1023 amu/g)(1000 g/kg)/ (tt(48 X 30/78) amu)f/2= 586 m/s. The average diameter is (300 + 375 pm)/2 = 337.5 pm. Then Z12 = tt(337.5 X 10"12 m)2 (586 m/s)(4.9 X 1015 molec/L)2(l L/10~3 m3)2 - 5.0 X 1021 colhsions s"1 m~3. If every collision resulted in a reaction, this would be the number of reactions per unit second per cubic meter.
A quantity related to Zj is the mean free path, A. This is the average distance a molecule travels before colliding with another molecule. If we divide the average speed <v> in meters per second by the collision number Z, in collisions per second, we obtain the mean free path in meters per collision:
A =
1
<v>
ird V2 <u> ii.
(1.47)
V2 ndn,
Note that the mean free path is inversely proportional to pressure. The mean free path will be important in Chapter 4, where we will see that the transport of heat, momentum, and matter are all proportional to the distance traveled between collisions.
example
The Mean Free Path of Nitrogen
Objective      Find Zl and the mean free path of N2 at 300 K and 1 atm given that the molecular diameter is 218 pm.
Method Use equation 1.46 to calculate <vr>, equation 1.43 to calculate
Z(, and equation 1.47 to calculate A.
Solution        We start by calculating <vt> = (8/t777r/u.)l/2, where    = 28 X 28/(28 + 28) = 14 amu.
<v> -
I 8(1.38 X 1Q-23JK-')(300K)(6.02 X 1Q23 amu/g)(1000 g/kg) (3.1415 X 14 amu)
= 673 m/s.
11/2
'(1.48)
Section 1.8 Summary
25
the relative velocity,
<v,> = V2<v> = f — J   , (1.46)
and the mean free path,
<v> 1 A =----= —p-. (1.47)
These concepts form the basis for further investigation into transport properties and chemical reaction kinetics.
appendix LI
The Functioual Form of the Velocity Distribution
We demonstrate in this appendix that the exponential form used in equation 1.23 is the only function that satisfies the equation f{a + b + c) = f{a)fib)f{c). Consider first the simpler equation
f(z)~f(a)/(b), (1.50)
where z = a + b. Taking the derivative of both sides of equation 1.50 with respect to a we obtain
^|=/W). (LSD
On the other hand, taking the derivative of both sides of equation 1.50 with respect to b, we obtain
Since z — a + b, dzJda = dzldli = 1. Consequently,
~~=Aa)f(b)=f(a)f'(b). (1.53) dz
Division of both sides of the right-hand equality by f(a)f{b) yields
/'(«) f(b)
(1.54)
Now the left-hand side of equation 1.54 depends only on a, while the right-hand side depends only on b. Since a and b are independent variables, the only way that equation 1.54 can be true is if each side of the equation is equal to a constant, ±k, where k is defined as nonnegative:
"7TT = ±K = ±k- (1-55)
f(a) f(b)
Appendix 1.3
sphere times the thickness dr (for clarity, the thickness dr is not shown in the diagram). The surface area is given by the arc length on the longitude, r d9, times the arc length on the latitude, r sin 6 d<£. Thus, the volume element is dV = rsin 6 d6 d<f>dr.
appendix
The Error Function and Co-Error Function
It often occurs that we need to evaluate integrals of the form of those listed in Table 1.1 but for limits less than the range of 0 to infinity. For such evaluations it is useful to define the error function:
2  r ,
erf(x) = —=   e-'rdu. (1.58) Vtt J0
From Table 1.1 we see that for x = °°, the value of the integral is vtt/2, so that erf(°=) = 1. Note that if we "complement" the error function by 2/vn times the integral from x to °°, we should get unity:
du + —j=    e~'f du = erffx) + —f= e^du
V 7T L VTT J.v
TTJ0
(1.59)
e""!d« = 1.
Appendix 1.4
■   Figure 1.17
Vector diagram for center-of-mass conversion.
The virtue of this transformation is that the total momentum of the system p = mivi ~*~ rrH^2 is also equal to the momentum of the center of mass, defined as Mvcom. Because we assume that no external forces are acting on the system, F = A/acom = (dPcon/dO = 0» so ma'tne momentum of the center of mass does not change during the interaction between the two particles.
Note that since (mt/M) + (m2/M) = 1 we can write
However,
--1--I v, — v„
M M
77 V2 + 77 V2 ~ Vcom-
M M
»?]V, 4- m2v2 = M\c,
(1.62)
(1.63)
so that
Consequently,
M
m2v2
M
V2
M
In a similar way, we find that
m2
M
(1.64)
(1.65)
We now note an important point, that the velocities of the particles with respect to the center of mass are just given by the two pieces of the vector vr: u, = —(w2/7tf)vr, and u, = (m,/M)\v as shown in Figure 1.18. Note also that in the moving frame of the center of mass, there is no net momentum for the particles; that is, /W|U, + ^Uj = 0. This important property enables us to calculate the velocity of one particle in the center-of-mass frame given just the mass and the velocity of the other particle.
problems
Problems
1.1 Molecules all of mass m and speed v exert a pressure p on the walls of a vessel. If half the molecules are replaced by ones of another type all with mass \m and speed 2v, will the pressure (a) increase, (b) decrease, (c) remain constant?
1.2 Suppose the probability of obtaining a score between 0 and 100 on an exam increases monotonically between 0 and 1.00. Is the average score on the exam (a) greater than 50, (b) equal to 50, (c) less than 50?
1.3 Suppose some property q of a gas is proportional to (0.326 s3 irT3)^ + (tt s9 m~9)u9. What is the average value of ql
1.4 Without referring to any formula, decide whether at constant density the mean free path (a) increases, (b) decreases, or (c) stays constant with increasing temperature and explain your answer.
1.5 Consider a deck of cards. With aces valued at one and jacks, queens, and kings valued at 11, 12, and 13, respectively, calculate the average value of a card drawn at random from a full deck.
1.6 The distribution of the grades s (where 0^X< 100) for a class containing a large number of students is given by the continuous function p(s) = K(50 — \S — 501), where bcl is the absolute value of x and K is a normalization constant. Determine the normalization constant and find out what fraction of the students received grades greater than or equal to 90.
1.7 A pair of dancers is waltzing on a one-dimensional dance floor of length L. Since they tend to avoid the walls, the probability of finding them at a position x between walls at x = 0 and x = L is proportional to sin2(7rA/L). What is the normalized distribution function for the position of the waltzers? Using this distribution function, calculate the most probable position for the waltzers. Calculate the average position of the waltzers. {Hint: The integral of y sinry dy is [y2!4] — [(y sin 2y)/4] — [(cos 2y)/8]; this is also the probability for finding a particle in a box at a particular position.)
1.8 By setting the derivative of the formula for the Maxwell-Boltzmann speed distribution equal to zero, show that the speed at which the distribution has its maximum is given by equation 1.33.
1.9 Show using equations 1.16 and 137 that the average molecular energy is 3fcT/2.
1.10 Prove equation 1.41 from equation 1.40. Integration can be accomplished by making the following change of variable. Let e = kTx2, so that de = kTdix2) and em = {kT)mx. Substitute these into equation 1.40 and integrate by parts, recalling that since d(uv) — udv + v du, then /d(i/i>) = fudv + Jvdu, so that fudv = |limits — fvdu, where the notation |liraits indicates that the product {uv) should be evaluated at the limits used for the integrals.
1.11 The Maxwell-Boltzmann distribution may not be quite valid! Calculate the fraction of N2 molecules having speeds in excess of the speed of light.
1.12 The object of this problem is to show more rigorously that <ur> = (&kT/irpS)m, where p,, the reduced mass, is defined as p, = minvj{ml + m^).
Problems 33
1.18 In a group of molecules all traveling in the positive z direction, what is the probability that a molecule will be found with a z-component speed between 400 and 401 m/s if m/(2kT) = 5.62 X 10"6 s2/m2? (Hint: You need to find and normalize a one-dimensional distribution function first!)
1.19 We will see in Chapter 3, equation 3.4, that the rate constant for a reaction as a function of temperature is given by the average of a(er)vr over the thermal energy distribution G(er), where er = \ mv2 and o-(e^) is the energy-dependent cross section for the reaction. The thermal relative kinetic energy disni-bution G(er) has the same functional form as the kinetic energy distribution G(e) given in equation 1.37, except that all energies e = ynv1 are replaced by relative kinetic energies er = \ya)2.
a. Suppose that for a particular reaction cr(er) = ce^, where c is a constant. Calculate k(T).
b. Suppose that for another reaction cr(er) = c/€r; calculate k(T).