Energy Distributions in Small Populations: Pascal versus Boltzmann

In the Classroom

Energy Distributions in Small Populations: Pascal versus Boltzmann Roger W. Kugel* Department of Chemistry, Saint Mary's University of Minnesota, Winona, Minnesota 55987 *[email protected] Paul A. Weiner Department of Mathematics and Statistics, Saint Mary's University of Minnesota, Winona, Minnesota 55987

The Boltzmann distribution describes how a fixed quantity of energy distributes over a collection of a large number of particles. Understanding the way energy distributes among molecules in a system is important for students of chemistry to grasp because that distribution determines many of the macroscopic attributes of a system, including its thermodynamic, kinetic, and spectroscopic properties. Derivations of the Boltzmann distribution can be found in most beginning physical chemistry textbooks (1-4). Some students find these derivations difficult to understand because they require the Lagrange method of undetermined multipliers, which is not typically covered in the basic calculus sequence. Furthermore, the Boltzmann distribution only applies to systems with very large numbers of particles, N, because the derivation requires the use of Stirling's approximation for N!, which is only valid for very large N. In addition, the approximation causes serious errors for higher energy states of large systems for which state populations, nj's, become small. A number of attempts have been made to derive the Boltzmann distribution law without using Lagrange's method or Stirling's approximation (5-17). Some of these articles take a thermodynamic approach wherein the Helmholtz energy is minimized (5-8) or the entropy is maximized (9); others (10-14) use discrete substitutions to eliminate constant N and constant E constraints, where N is the total number of particles and E is the total energy of a thermodynamic system; others eliminate Stirling's approximation by a series substitution (15, 16). A unique approach was recently offered by Friedman and Grubbs (17) who used a geometrical argument to predict Boltzmann probabilities from hyperplanes in N-dimensional state space, a line of reasoning that also led to the binomial coefficients of Pascal's triangle. All of these treatments, although interesting and mathematically sound, seem to fall short of the goal of making the energy distribution problem more transparent to beginning students. In this article, we attempt to demystify the Boltzmann distribution law by starting with model systems having small numbers of particles, N, with correspondingly small numbers of energy units, E. Such systems can be analyzed using simple combinatorics. When these systems are studied systematically, a pattern reminiscent of Pascal's triangle emerges in the predicted energy probability distributions. This pattern can be represented by a general formula that is shown to approach the Boltzmann distribution law for large values of E and N. Although the observation of binomial coefficients and Pascal's triangle in Boltzmann probabilities has been previously reported (17), the present treatment appears to be more accessible to beginning students because it arises out of the determination of expected 1200

Journal of Chemical Education

_

_

population proportions of energy states in small model systems. It should also be pointed out that an early treatment by Bent (18, 19) used a similar small-system approach to calculate the number of microstates of model systems (18) and a thermodynamic argument (T = ∂E/∂S ≈ ΔE/ΔS) for the determination of the Boltzmann distribution (19). We believe that the approach offered in this article is useful and important for a basic understanding of the way energy distributes itself among the available states of a system. It is more accessible to undergraduate chemistry students than the derivation provided by most standard physical chemistry textbooks because it is based on combinatorics of small model systems and does not require the use of Lagrange's method of undetermined multipliers or Stirling's approximation. Furthermore, it gives results that are more accurate than the Boltzmann distribution for small populations and might therefore be applicable for predicting energy distributions in nanoscale systems. Model Systems Consider a collection of a small number, N, of particles having a total energy of E that is distributed over the N particles. If the particles have quantized (nondegenerate) integral energy states with energy levels given by εj = j units, the total energy, E, may be distributed discretely over the available energy states. The lowest energy level is ε0, which has 0 units of energy, and is called the ground state of a particle. The highest possible energy level of a particle in this system is εE = E, whereupon, all the energy of the system resides in a single particle. A distribution of energy among particles may be described by Ωi = ni0, ni1, ni2, ..., niE, where nij is the population of the jth state in the ith distribution. For each distribution Ωi, in a system it must be true that X nij ¼ N ð1Þ j

and

X

nij εj ¼ E

ð2Þ

j

The statistical weight, Wi, of a distribution gives the number of distinct ways that distribution can be achieved by the exchange of particles among the energy states. The statistical weight of the ith distribution may be calculated by the formula shown in N! Wi ¼ Q nij !

ð3Þ

j

_

Vol. 87 No. 11 November 2010 pubs.acs.org/jchemeduc r 2010 American Chemical Society and Division of Chemical Education, Inc. 10.1021/ed1003838 Published on Web 09/21/2010

In the Classroom Table 1. Distributions of Particles in Energy States for N = 3, E = 3

a

Table 2. Dot Representations for Model Systems with E = 5 and N = 2-7

Wtotal = 10.

The classic Boltzmann distribution derivation finds the maximum distribution, Wmax, the distribution of the highest statistical weight with the constraints of constant (and extremely large) values of N and E, and derives the population probability, BkT(j), of the Boltzmann distribution BkT ðjÞ ¼

nj e - ðεj =kT Þ ¼ P - ðε =kT Þ N e l

ð4Þ

l

In eq 4, BkT(j) is the probability that a given particle will have energy εj, nj is the average number of particles with energy εj, N is the total number of particles in the system, k is the Boltzmann constant, and kT is the average thermal energy of a particle in the system. In the model systems we describe, we will start with constant and small values of N and E and derive the corresponding population probability distribution, nj/N. This distribution, in general, differs from that of the Boltzmann distribution for small populations, but approaches that of the Boltzmann distribution as N and E increase. Energy Distributions in Model Systems: Dot Representation In the dot representation, energy levels are represented as boxes vertically arranged with the particles represented as dots, •, in these boxes. The number of particles (dots) in level j is The designated nj and the energy of each particle in level j is εj. P total number of particles inPan allowed distribution is N = nj and the total energy is E = njεj. Several different distributions, designated Ωi, may be possible and each one may occur in a number of different ways, designated Wi, owing to particle interchange.1 For example, consider a system of N = 3 particles with E = 3 units of energy. The particles have available four discrete nondegenerate energy levels with integral spacing, designated ε0 = 0, ε1 = 1, ε2 = 2, and ε3 = 3. There are three different distributions for this system designated Ω1, Ω2, and Ω3 (Table 1). Note that each distribution has N = 3 particles with a total energy of E = 3. For example, distribution Ω1 has one particle in the ε3 state and two in the ε0 state. In this distribution, all of the system's energy resides in one particle. Distribution Ω2 has one particle in each of the three lowest energy states, ε0, ε1, and ε2. Finally, distribution Ω3 has all three particles in the ε1 state. By interchanging particles, it is easy to see that there are three ways to achieve Ω1, six ways to achieve Ω2, and one way to achieve Ω3. That is, W1 = 3, W2 = 6, and W3 = 1. In a similar way, model systems with other values for N and E may be represented in this dot formalism. For example, Table 2 gives the dot representations for systems with E = 5 and N = 2-7.

r 2010 American Chemical Society and Division of Chemical Education, Inc.

_

Total Microstates: Method of Stars and Bars Another way of determining the total number of microstates, Wtotal, in a system with total energy E distributed over N particles may also be given. We call this the method of stars

pubs.acs.org/jchemeduc

_

Vol. 87 No. 11 November 2010

_


1201

In the Classroom Table 3. Distributions of Particles in Energy States for N = 3, E = 3 Distribution

E = 3 stars and N-1 = 2 bar

Table 4. Average Populations and Expected Population Proportions of Energy States for the System N = 3, E = 3

Microstates

Ω1

{***| |}

{|***|}

{| |***}

3

Ω2

{**|*|}

{*|**|}

{|**|*}

6

{**| |*}

{*| |**}

{|*|**}

Ω3

{*|*|*}

1

Wtotal

10 arrangements

10 states

j = εj

and bars. It is based on simple combinatorics and is described below. The total number of microstates for a system with N (distinguishable) particles2 and E (indistinguishable) units of energy is given by EþN -1 ð5Þ N -1 which represents the number of combinations of E þ N - 1 items taken N - 1 at a time. To see this, imagine putting the N particles in a row and separating them with N - 1 vertical bars: fParticle1jParticle2jParticle3j:::jParticle N - 1jParticle N g Thus, the bars ( | ) represent a dividing line between particles. Now each particle has some number of units of energy. Each unit of energy is represented by a star (*). So a microstate may be represented as a sequence of N - 1 bars and E stars. The number of stars in each region determined by the bars gives the energy level of the corresponding particle. For instance, for the case of N = 3 and E = 3, the sequence fjjg represents the microstate with the first particle having 0 units of energy; the second particle, 2 units; and the third particle, 1 unit. In other words, the first particle is in the ε = 0 state, the second particle is in the ε = 2 state, and the third particle is in the ε = 1 state. Or, as another example, consider the sequence fj jg in which the first particle has all three units of energy, and the second and third particles have no units of energy, that is, the first particle is in the ε = 3 state and the other two are in the ε = 0 state. In summary, Table 3 shows the stars and bars representations of all 10 microstates for the N = 3, E = 3 system. Each microstate corresponds uniquely to a sequence of N - 1 bars and E stars. Thus, to count the number of microstates, we need only count the number of these sequences. Any such sequence has E þ N - 1 positions. We may choose any N - 1 of these positions for the bars, and then the remaining positions must be filled with the E stars. The number of ways of choosing N - 1 positions for bars from the E þ N - 1 total positions is given by the binomial coefficient: ðE þ N - 1Þ! E þ N 1 ð6Þ ¼ Wtotal ¼ N -1 ðN - 1Þ!E! In the example in Table 3 5! Wtotal ¼ 3 þ 3 - 1 ¼ 5 ¼ ¼ 10 3-1 2 ð2!Þð3!Þ

1202


_


ð7Þ

_

nj

P3,3(j)

3

3/10

1/10

2

6/10

2/10

1

9/10

3/10

0

12/10

4/10

Total

30/10 = 3 = N

10/10 = 1

Table 5. Expected Population Proportions for E = 5, N = 2-7 P5,N(j) Nf

2

3

4

5

6

7

5

1/6

1/21

1/56

1/126

1/252

1/462

4

1/6

2/21

3/56

4/126

5/252

6/462

3

1/6

3/21

6/56

10/126

15/252

21/462

2

1/6

4/21

10/56

20/126

35/252

56/462

1

1/6

5/21

15/56

35/126

70/252

126/462

j

0

1/6

6/21

21/56

56/126

126/252

252/462

Sum

1

1

1

1

1

1

This method may be used to verify that all of the distributions and statistical weights have been correctly determined. The reader can verify that Wtotal for the model systems given in Table 2 may be calculated using eq 6. Average Populations and Expected Population Proportions of Energy States Assuming that the particles can rapidly and freely exchange energy with each other so that all microstates are equally likely, the average populations of the available energy states may be determined from the distributions and statistical weights: P nij Wi nj ¼ iP ð8Þ Wi i

Thus, the average population of the jth state in a system is the sum over distributions of the products of the populations of the jth state in the ith distribution times the statistical weight of each distribution divided by the sum of all statistical weights. These values are simply the distribution-weighted average populations of the energy states. The probability that a particular particle is in the jth energy state is simply the average population of the jth energy state divided by the total number of particles: nj ð9Þ PE , N ðjÞ ¼ N The average populations and proportions calculated for the example system with N = 3 and E = 3 are shown in Table 4. Using eqs 8 and 9, the average state population proportions were calculated for all the model systems given in Table 2. These values in fractional form are shown in Table 5. When the array of numerators for these population probabilities is displayed alone, a recognizable pattern emerges.


_


In the Classroom Table 6. Expected Population Proportion Numerator Array for E = 5, N = 2-7 P5,N(j) numerators Nf

2

3

4

5

6

7

j 5

1

1

1

1

1

1

4

1

2

3

4

5

6

3

1

3

6

10

15

21

2

1

4

10

20

35

56

1

1

5

15

35

70

126

0

1

6

21

56

126

252

Sum

6

21

56

126

252

462

This array is shown in Table 6. Notice that the sum of each column of numerators is the last entry of the next column, the array is symmetric, that is, each row is the same as the corresponding column, and each numerator entry is the sum of the entry above it and the one to its left. The array is thus recognized as the array of binomial coefficients commonly known as Pascal's triangle. When larger E values are used, the Pascal's triangle array continues downward and the j column offsets so that j = E is always the first row of the array and j = 0 is the E þ 1 row. When larger N values are used, the Pascal's triangle array continues to the right and the kth column in the array corresponds to N = k þ 1particles in the system. The entries in Pascal's triangle (nth row, kth column) may be computed from the binomial coefficients: n! n ¼ ð10Þ k k!ðn - kÞ! The probability for the energy level j in a system of N particles with total energy E may thus be determined from the Pascal triangle entries: E-jþN -2 N -2 PE, N ðjÞ ¼ EþN -1 N -1 ¼

ðE - j þ N - 2Þ!E!ðN - 1Þ! ðE þ N - 1Þ!ðE - jÞ!ðN - 2Þ!

ð11Þ

where j = 0, 1, 2, ..., E. The equations may also be justified using the “stars and bars” argument used for the total number of microstates. The probability that a specific particle has j units of energy is the same as the probability that the other N - 1 particles have E - j units of energy. That probability is given by ðE - jÞ þ ðN - 1Þ - 1 ðN - 1Þ - 1 PE, N ðjÞ ¼ EþN -1 N -1 EþN -j-2 N -2 ð12Þ ¼ EþN -1 N -1


_

Figure 1. Plot of the function y = 1/t. The shaded area is equal to the Rxþ1 1 xþ1 ¼ - ln x þx 1 , which is approximated by integral: x t dt ¼ ln x the rectangle whose area is: ½ðx þ 1Þ - x 1=ðx þ 0:5Þ ¼ ½1=ðx þ 0:5Þ.

The denominator of this fraction gives the total number of microstates, as calculated in eq 5. The numerator is calculated by first imagining giving the specified particle its j units of energy and then counting the number of ways of distributing the remaining E - j units of energy to the remaining N - 1 particles, using the same type of “stars and bars” argument as we used for counting the total number of microstates. Thus, eq 11, derived from the binomial coefficients of Pascal's triangle, and eq 12, derived from the “stars and bars” method of determining average energy distribution, are identical. Relationship of the Pascal Distribution to the Boltzmann Distribution The Pascal distribution of particles in energy states is clearly different from the Boltzmann distribution for small populations, for example, a plot of PE,N(j) versus j for N = 3 decreases linearly for the Pascal distribution but exponentially for the Boltzmann distribution. This disagreement is not surprising because the Boltzmann derivation uses Stirling's approximation for N!, which is only valid for very large values of N. It is shown below that the Pascal distribution shown in eq 12 approaches the exponential behavior typical of a Boltzmann distribution under suitable conditions, that is, in the limit of large E and N. We will use the approximation [x/(x þ 1)] ≈ e-[1/(xþ0.5)]. This may be developed as follows: ln(x þ 1) - ln(x) is the area under the curve y = 1/t from t = x to t = x þ 1. This region has a width of 1 unit, and the height may be approximated by the height of the curve above the midpoint t = x þ 0.5. Hence, ln(x þ 1) - ln x ≈ 1/(x þ 0.5). Equivalently, ln[x/(x þ 1)] ≈ -(1/(x þ 0.5)). The desired approximation follows by exponentiating both sides. A graphical illustration of this point is shown in Figure 1. The shaded area in Figure 1 is exactly equal to Z xþ1 1 xþ1 x dt ¼ ln ¼ - ln t x xþ1 x However, the shaded area also may be approximated by a rectangle whose width is 1 and whose height is the height above the midpoint of the segment on the x axis. This approximating rectangular area is 1/(x þ 0.5). This gives ln[x/(x þ1)] ≈ -[1/ (x þ 0.5)], with the approximation improving as x gets larger in the positive direction.


_


_


1203

In the Classroom

Upon algebraic simplification and cancellation of factorials in eq 11, we obtain PE, N ð jÞ ¼

N -1 E E-1 E-2 EþN -13EþN -23EþN -33EþN -4 E-jþ1 ð13Þ 3 3 3 3 3EþN -j-1

In each fraction, we may divide top and bottom by N to get PE , N ðjÞ ¼ j-1 N 23 33 13 jþ1 43 333 3 ε þ1ε þ1ε þ1ε þ1ε þ1N N N N N 1-

1 N

ε-

ε

1 N

ε-

2 N

ε-

ð14Þ

where ε ¼ E=N is the average energy per particle. We note that for j , N, all expressions with a denominator of N are quite small. These small values of j relative to N correspond to low energy levels, which should be more common in the distribution. For larger values of j, both the Boltzmann and Pascal distributions rapidly decay to 0. Also, in the expression for the Boltzmann distribution,

l ¼0

we note that the sum in the denominator forms a geometric series. So for large E we have that E X

e - ðl=kT Þ

l ¼0

¥ X

e - l=kT ¼

l ¼0

1 1 - e - ð1=kT Þ

which gives BkT ð jÞ ð1 - e - ð1=kT Þ Þe - ð j=kT Þ . Therefore for large N and j , N, PE;N ðjÞ

1 ε ε ε ε 3 3 3 3 3 3 3 3 εþ1 εþ1 εþ1 εþ1 ε þ1 !j 1 ε ¼ ε þ13 ε þ1 ¼

ð15Þ

One of us (R.W.K.) has presented the Pascal distribution of energy in the undergraduate physical chemistry classroom two times in the last 2 years with great success. The presentation of statistical distributions of energy was accomplished as follows:

! !j ε ε 1εþ1 3 εþ1

This suggests letting kT = εh þ 0.5 (kT should represent average energy, and the 0.5 adjustment works out nicely mathematically). This yields ð16Þ PE;N ðjÞ ð1 - e - ð1=kT Þ Þ 3 e - ðj=kT Þ e - ð j=kT Þ ¼ P ¥ e - ðl=kT Þ l ¼0

BkT ð jÞ


_


This treatment is useful because it provides a more accessible route to understanding the Boltzmann distribution for beginning chemistry students. Specifically, our derivation does not involve calculus of variations or Lagrange's method of undetermined multipliers, which are not typically covered in the beginning calculus sequence. Furthermore, by avoiding the use of Stirling's approximation for N!, the Pascal distribution provides a method of determining the energy distributions in small systems that is more accurate than the Boltzmann distribution. One can easily imagine potential applications of the Pascal distribution in nanoscale systems where N is relatively small. As one example, consider the distribution of vibrational energy among a small number of thermalized I2 molecules whose lower energy levels are equally spaced. Application Notes

ð1 - e - 1=ðεþ0:5Þ Þ 3 e - j=ðεþ0:5Þ

1204

The fact that the Pascal distribution approaches the Boltzmann distribution as N increases is shown in Figure 2, wherein selected Pascal distributions for E/N = 5 and N = 2 - 100 are compared to the Boltzmann distribution for kT = E/N þ 1/2 = 5.5. Conclusion

e - ð j=kT Þ E P e - ðl=kT Þ

BkT ð jÞ ¼

Figure 2. Probability curves for the Pascal distribution with E/N = 5, N = 2 - 100, and the Boltzmann distribution for E/N = 5, kT = E/N þ 0.5 = 5.5.

_

1. First, a standard and traditional lecture presentation of the rationale for and derivation of the Boltzmann distribution was made. This presentation included the Lagrange method of undetermined multipliers and Stirling's approximation for N!, as do most physical chemistry textbooks. The limitation of accuracy that requires very large N was emphasized. 2. Second, hypothetical model systems showing how, for example, E = 3 units of energy distributes statistically over a system containing N = 3 molecules. Students easily grasp this conceptual picture of a small system. 3. The probability of populating the energy states was then computed for the class using eqs 8 and 9. The distribution (probability versus energy state) was plotted on the board and


_


In the Classroom

4.

5.

6.

7.

shown to be linear. When students were asked why this distribution did not resemble the exponential Boltzmann distribution, they eventually came up with the large N approximation of the latter. Students were then asked to come to the board (individually or in pairs) and create dot diagrams like those presented in Table 2 for a larger system (all with the same E but different values of N). Each individual or group drew the dot diagrams, computed the value of W for each distribution, found Wtotal, and computed the probability fractions for the energy states of their system. The student developed population probabilities were then displayed as fractions on the board in a single table like that shown in Table 5. Students were then asked to discover the pattern shown by these fractions. With some hints, they eventually saw Pascal's triangle. They showed surprise and satisfaction at this “aha moment”. Alternatively (or additionally), a lab simulation exercise was developed for students to compute the population probabilities of a model system on their own. The handout used for this exercise is appended in the supporting information.

Students appreciated both the in-class, hands-on exercise and the lab simulation exercise. These activities helped them to understand and internalize the statistical nature of energy distributions in molecular systems. The exercises also helped the students to appreciate the brilliance and utility of the Boltzmann distribution while realizing its limitations. Notes

Literature Cited 1. Atkins, P.; de Paula, J. Physical Chemistry, 8th ed.; W. H. Freeman and Company: New York, 2006; pp 582-584. 2. Noggle, J. Physical Chemistry, 3rd ed.; Harper Collins College Publishers: New York, 1996; pp 223-230. 3. Laidler, K. J.; Meiser, J. H.; Sanctuary, B. C. Physical Chemistry, 4th ed.; Houghton Mifflin Company: Boston, 2003; pp 793-795. 4. Moore, W. J. Physical Chemistry, 2nd ed.; Prentice-Hall: Englewood Cliffs, NJ, 1955; pp 349-352. 5. David, C. W. J. Chem. Educ. 2006, 83, 1695–1697. 6. Lie, G. C. J. Chem. Educ. 1981, 58, 603–604. 7. Russell, D. K. J. Chem. Educ. 1996, 73, 299–300. 8. Wall, F. T. Proc. Natl. Acad. Sci. U.S.A. 1971, 68, 1720–1724. 9. McDowell, S. A. C. J. Chem. Educ. 1999, 76, 1393–1394. 10. Waite, B. A. J. Chem. Educ. 1986, 63, 117–120. 11. Kleppner, D. Am. J. Phys. 1968, 36, 843. 12. Lorimer, J. W. J. Chem. Educ. 1966, 43, 39–40. 13. Nash, L. K. J. Chem. Educ. 1982, 59, 824–826. 14. Gibbs, J. H. J. Chem. Educ. 1971, 48, 542. 15. Hakala, R. W. J. Chem. Educ. 1961, 38, 33–35. 16. Hakala, R. W. J. Chem. Educ. 1962, 39, 526–527. 17. Friedman, E.; Grubbs, W. T. Chem. Educator 2003, 8, 116–121. 18. Bent, H. A. The Second Law; Oxford University Press: New York, 1965; pp 144-154. 19. Bent, H. A. The Second Law; Oxford University Press: New York, 1965; pp 163-166.

Supporting Information Available

1. Note that the model systems are believed to be at sufficiently high temperatures for Boltzmann statistics to apply.


2. We treat the particles as distinguishable so as to correctly count the number of microstates.

_

Student handout. This material is available via the Internet at http://pubs.acs.org.


_


_


1205

Energy Distributions in Small Populations: Pascal versus Boltzmann

Recommend Documents