molecular driving forces 2nd edition solutions manual
TRANSCRIPT
Chapter 1Principles of Probability
1. Combining independent probabilities.
You have applied to three medical schools: University of California at San Francisco (UCSF),Duluth School of Mines (DSM), and Harvard (H). You guess that the probabilities you’llbe accepted are: p(UCSF) = 0.10, p(DSM) = 0.30, and p(H) = 0.50. Assume that theacceptance events are independent.
(a) What is the probability that you get in somewhere (at least one acceptance)?
(b) What is the probability that you will be accepted by both Harvard and Duluth?
(a) The simplest way to solve this problem is to recall that when probabilities areindependent, and you want the probability of events A and B, you can multiply them.When events are mutually exclusive and you want the probability of events A or B,you can add the probabilities. Therefore we try to structure the problem into an andand or problem. We want the probability of getting into H or DSM or UCSF. Butthis doesn’t help because these events are not mutually exclusive (mutually exclusivemeans that if one happens, the other cannot happen). So we try again. The probabilityof acceptance somewhere, P (a), is P (a) = 1− P (r), where P (r) is the probability thatyou’re rejected everywhere. (You’re either accepted somewhere or you’re not.) But thisprobability can be put in the above terms. P (r) = the probability that you’re rejectedat H and at DSM and at UCSF. These events are independent, so we have the answer.The probability of rejection at H is p(rH) = 1− 0.5 = 0.5. Rejection at DSM isp(rDSM) = 1− 0.3 = 0.7. Rejection at UCSF is p(rUCSF) = 1− 0.1 = 0.9. ThereforeP (r) = (0.5)(0.7)(0.9) = 0.315. Therefore the probability of at least one acceptance= P (a) = 1− P (r) = 0.685.
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(b) The simple answer is that this is the intersection of two independent events:
p(aH)p(aDSM) = (0.50)(0.30)
= 0.15.
A more mechanical approach to either part (a) or this part is to write out all thepossible circumstances. Rejection and acceptance at H are mutually exclusive. Theirprobabilities add to one. The same for the other two schools. Therefore all possiblecircumstances are taken into account by adding the mutually exclusive events together,and multiplying independent events:
[p(aH) + p(rH)][p(aDSM) + p(rDSM)][p(aUCSF) + p(rUCSF)] = 1,
or equivalently,
= p(aH)p(aDSM)p(aUCSF) + p(aH)p(aDSM)p(rUCSF)
+p(aH)p(rDSM)p(aUCSF) + · · ·
where the first term is the probability of acceptance at all 3, the second term representsacceptance at H and DSM but rejection at UCSF, the third term represents acceptanceat H and UCSF but rejection at DSM, etc. Each of these events is mutually exclusivewith respect to each other; therefore they are all added. Each individual termrepresents independent events of, for example, aH and aDSM and aUCSF. Therefore itis simple to read off the answer in this problem: we want aH and aDSM, but notice wedon’t care about UCSF. This probability is
p(aH)p(aDSM) = p(aH)p(aDSM)[p(aUCSF) + p(rUCSF)]
= (0.50)(0.30)
= 0.15.
Note that we could have solved part (a) the same way; it would have required addingup all the appropriate possible mutually exclusive events. You can check that it givesthe same answer as above (but notice how much more tedious it is).
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2. Probabilities of sequences.
Assume that the four bases A, C, T, and G occur with equal likelihood in a DNA sequenceof nine monomers.
(a) What is the probability of finding the sequence AAATCGAGT through randomchance?
(b) What is the probability of finding the sequence AAAAAAAAA through randomchance?
(c) What is the probability of finding any sequence that has four A’s, two T’s, two G’s,and one C, such as that in (a)?
(a) Each base occurs with probability 1/4. The probability of an A in position 1 is 1/4, ofA in position 2 is 1/4, of A in position 3 is 1/4, of T in position 4 is 1/4, and so on.There are 9 bases. The probability of this specific sequence is (1/4)9 = 3.8× 10−6.
(b) Same answer as (a) above.
(c) Each specific sequence has the probability given above, but in this case there are manypossible sequences which satisfy the requirement that we have 4 A’s, 2 T ’s, 2 G’s, and 1C. How many are there? We start as we have done before, by assuming all nine objectsare distinguishable. There are 9! arrangements of nine distinguishable objects in alinear sequence. (The first one can be in any of nine places, the second in any of theremaining eight places, and so on.) But we can’t distinguish the four A’s, so we haveovercounted by a factor of 4!, and must divide this out. We can’t distinguish the twoT ’s, so we have overcounted by 2!, and must also divide this out. And so on. So theprobability of having this composition is
[9!
4!2!2!1!
] (1
4
)9
= 0.014.
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3. The probability of a sequence (given a composition).
A scientist has constructed a secret peptide to carry a message. You know only the composi-tion of the peptide, which is six amino acids long. It contains one serine S, one threonine T,one cysteine C, one arginine R, and two glutamates E. What is the probability that thesequence SECRET will occur by chance?
The S could be in any one of the 6 positions with equal likelihood. The probability that it isin position 1 is (1/6). Given that S is in the first position, we have 2 E ′s which could occurin any of the remaining 5 positions. The probability that one of them is in position 2 is(2/5). Given those two letters in position, the probability that the 1 C is in the next of the 4remaining positions is (1/4). The probability for the R is (1/3). For the remaining E is(1/2), and for the last T is (1/1), so the probability is
(1/6)(2/5)(1/4)(1/3)(1/2) = 1/360 =
[6!
1!2!1!1!
]−1
.
4. Combining independent probabilities.
You have a fair six-sided die. You want to roll it enough times to ensure that a 2 occurs atleast once. What number of rolls k is required to ensure that the probability is at least 2/3that at least one 2 will appear?
q =5
6= probability that a 2 does not appear on that roll.
qk = probability that a 2 does not appear on k INDEPENDENT rolls.
P (k) = 1− qk = probability that at least one 2 appears on k rolls.
For
P (k) ≥ 2
3, 1− qk ≥ 2
3=⇒ qk ≤ 1
3=⇒ k ln q ≤ ln
(1
3
)
=⇒ k ≥ ln(1/3)
ln(5/6)= 6.03
Approximately six or more rolls will ensure with probability P ≥ 2/3 that a 2 will appear.
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5. Predicting compositions of independent events.
Suppose you roll a die three times.
(a) What is the probability of getting a total of two 5’s from all three rolls of the dice?
(b) What is the probability of getting a total of at least two 5’s from all three rolls of thedie?
The probability of getting x 5’s on n rolls of the dice is
(1
6
)x (5
6
)n−x n!
x!(n− x)!
Note that this is a “2-outcome” problem (getting a 5 or not getting a 5). It is not a“6-outcome” problem.
(a) So the probability of two 5’s on three dice rolls is
(1
6
)2 (5
6
)1 3!
2!1!=
(1
36
)(5
6
)3
=15
216
= 6.94× 10−2
(b) The probability of getting at least two 5’s is the probability of getting two 5’s or three5’s. Since these two situations are mutually exclusive, we seek
p(two 5’s) + p(three 5’s) =(
1
6
)2 (5
6
)3!
2!1!+(
1
6
)3 (5
6
)0 3!
3!0!
=15
216+
1
216
=16
216
= 7.41× 10−2
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6. Computing a mean and variance.
Consider the probability distribution p(x) = axn, 0 ≤ x ≤ 1, for a positive integer n.
(a) Derive an expression for the constant a, to normalize p(x).
(b) Compute the average 〈x〉 as a function of n.
(c) Compute σ2 = 〈x2〉 − 〈x〉2 as a function of n.
(a)∫ 1
0p(x) dx = 1 =⇒
∫ 1
0axn dx =
axn+1
n+ 1
∣∣∣∣∣
1
0
=a
n+ 1
= 1 =⇒ a = n + 1.
(b) 〈x〉 =∫ 1
0xp(x) dx
=∫ 1
0(n+ 1)xn+1 dx =
((n+ 1)xn+2
n + 2
)∣∣∣∣∣
1
0
=n+ 1
n+ 2.
(c) 〈x2〉 =∫ 1
0x2p(x) dx
= (n+ 1)∫ 1
0xn+2 dx
= (n+ 1)
(xn+3
n + 3
)∣∣∣∣∣
1
0
=n + 1
n + 3.
So
σ2 = 〈x2〉 − 〈x〉2
=(n+ 1
n+ 3
)−(n+ 1
n+ 2
)2
.
7. Computing the average of a probability distribution.
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Compute the average 〈i〉 for the probability distribution function shown in the figure below.
0 1 2 3 4
i
P (i )
0.0
0.2
0.4
0.3
0.1
A simple probability distribution.
〈i〉 =4∑
i=0
ip(i)
= 0(0.0) + 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4)
= 3
8. Predicting coincidence.
Your statistical mechanics class has twenty-five students. What is the probability that atleast two classmates have the same birthday?
If you first find the probability, q, that no two students have the same birthday, then thequantity you want is
p(2 students have same birthday) = 1− q
The probability that a second student does not have the same birthday as the first is(364/365). The probability that the third student has a birthday different than either of thefirst two is (363/365), and so on. It is like a sequence problem in which each possible
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birthday is one card drawn out of a barrel. The probability that no two people have the samebirthday, out of m people, is:
q =(
364
365
)(363
365
)(362
365
)· · ·
(365− (m− 1)
365
)
In factorial notation,
q =N !
(N −m)!Nm
where N = 365. (Incidentally, this expression is identical to the expression for excludedvolume in the Flory–Huggins model of polymer solutions (see Chapter 31)). Using Stirling’sapproximation x! ≈ (x/e)x, we get
q =(N/e)N
(N−me
)N−mNm
Collecting together terms in e and dividing the numerator and denominator by NN gives
q =e−m
(1− m
N
)N−m
Substituting m = 25 students and N = 365 gives
q = 0.4163, so
p = 1− q= 0.5837
There is a better than 50% chance two students will have the same birthday!
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9. The distribution of scores on dice.
Suppose that you have n dice, each a different color, all unbiased and six-sided.
(a) If you roll them all at once, how many distinguishable outcomes are there?
(b) Given two distinguishable dice, what is the most probable sum of their face valueson a given throw of the pair? (That is, which sum between two and twelve has thegreatest number of different ways of occurring?)
(c) What is the probability of the most probable sum?
(a)
6 on one die6× 6 on two dice
...6n on n dice.
(b) Number of ways a sum can occur
most probable sum
1
2
3
4
5
6
7
1 2 3 4 5 6 7 8 9 10 11 12
(1,1)
(1,2) × 2(1,6) × 2
(3,4) × 2
(2,3) × 2 (2,5) × 2(1,3) × 2
(1,4) × 2
(2,2)
sum of 2 dice
When dice show different numbers, there is a degeneracy of two. When each of the dicehas the same number, the degeneracy equals one.
(c) probability of 7 = p(7) =number of ways of getting 7
total number of ways of all outcomes
p(7) =6
1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1=
1
6
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10. The probabilities of identical sequences of amino acids.
You are comparing protein amino acid sequences for homology. You have a twenty-letteralphabet (twenty different amino acids). Each sequence is a string n letters in length. Youhave one test sequence and s different data base sequences. You may find any one of thetwenty different amino acids at any position in the sequence, independent of what you findat any other position. Let p represent the probability that there will be a ‘match’ at a givenposition in the two sequences.
(a) In terms of s, p, and n, how many of the s sequences will be perfect matches (identicalresidues at every position)?
(b) How many of the s comparisons (of the test sequence against each database sequence)will have exactly one mismatch at any position in the sequences?
(a) For comparing one sequence, each position assumed independent, the probability of aperfect match of all n residues ispn = (number of matched seqs/number of total seqs) =⇒number of matches in s sequences = spn.
(b) n− 1 positions match, so the probability is pn−1; one position doesn’t match which hasthe probability (1− p); and there are n different positions at which the mismatch couldoccur, therefore
= spn−1(1− p)n
Note in general, for k matches:
(1) P (k) = spk(1− p)n−k n!
k!(n− k)! .
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11. The combinatorics of disulfide bond formation.
A protein may contain several cysteines, which may pair together to form disulfide bondsas shown in the figure below. If there is an even number n of cysteines, n/2 disulfide bondscan form. How many different disulfide pairing arrangements are possible?
1
3
4
2
5
6
This disulfide bonding configuration with pairs 1-6, 2-5, and 3-4 is one of the manypossible pairings. Count all the possible pairing arrangements.
Number the individual sulfhydryl groups along the chain. The first sulfhydryl along thesequence can bond to any of the other n− 1. This removes two sulfhydryls fromconsideration. The third sulfhydryl can then bond to any of the remaining n− 3. Foursulfhydryls are now removed from consideration. The fifth can now bond to any of theremaining n− 5 sulfhydryls, etc., until all n/2 bonds are formed. Thus the total possiblenumber of arrangements of disulfide bonds is a product of n/2 terms:
D(n) = (n− 1)(n− 3)(n− 5) · · ·1
Another approach gives an expression that is easier to calculate. Consider placing thesulfhydryls in a sequence. The first place may be occupied by any of n sulfhydryls, thesecond place by any of n− 1 sulfhydryls, the third by any of n− 2 sulfhydryls, etc. Thus ifeach sulfhydryl were distinguishable from every other, there would be n! arrangements.However, each sulfhydryl has a mate from which it cannot be distinguished. We must divideby a factor of 2 (per bond) to correct for the indistinguishability of the two ends of eachbond. Finally, since we cannot distinguish any of the n/2 bonds from any other, we mustalso divide by (n/2)!. Hence the number of arrangements is
W (n) =n!
2n/2(n2
)!.
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Although these two equations were derived in very different ways, they are numericallyidentical for all n.
12. Predicting combinations of independent events.
If you flip an unbiased green coin and an unbiased red coin five times each, what is theprobability of getting four red heads and two green tails?
The probability of four red heads in five coin flips is
(1
2
)5(
5!
4!1!
)=
5
32.
The probability for two green tails is
(1
2
)5 5!
2!3!=
10
32.
Since the green coin flips are independent of the red coin flips, the probability we seek is(5/32)(10/32) = (50/1024) = 4.88× 10−2.
13. A pair of aces.
What is the probability of drawing two aces in two random draws without replacement froma full deck of cards?
A deck has 52 cards and four aces. The probability of getting an ace on the first draw is4/52 = 1/13. Since you draw without replacement, the probability of getting one of theremaining three aces on the second draw is 3/51, so the probability of two aces on two drawsis
(4
52
)(3
51
)= 4.5× 10−3.
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14. Average of a linear function.
What is the average value of x, given a distribution function q(x) = cx, where x rangesfrom zero to one, and q(x) is normalized?
q(x) = cx
c
0 x 1
〈x〉 =∫ 1
0xq(x) dx =
∫ 1
0cx2 dx
= c
(x3
3
)1
0
=c
3
We can also find c:
1 =∫ 1
0q(x) dx =
∫ 1
0cx dx
=
(cx2
2
)1
0
=c
2= 1
So,
c = 2
〈x〉 =c
3=
2
3
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15. The Maxwell-Boltzmann probability distribution function.
According to the kinetic theory of gases, the energies of molecules moving along the x-direction are given by εx = (1/2)mv2
x, where m = mass and vx is the velocity in thex-direction. The distribution of particles over velocities is given by the Boltzmann law,p(vx) = e−mv
2x/2kT . This is the Maxwell–Boltzmann distribution (velocities may range from
−∞ to +∞).
(a) Write the probability distribution p(vx), so that the Maxwell–Boltzmann distributionis correctly normalized.
(b) Compute the average energy 〈(1/2)mv2x〉.
(c) What is the average velocity 〈vx〉?
(d) What is the average momentum 〈mvx〉?
According to the kinetic theory of gases, the energies of molecules along the x-direction aregiven by εx = (1/2)mv2
x, where m = mass and vx is the velocity in the x-direction. Theirdistribution is given by the Boltzmann law, e−mv
2x/2kT . This is the Maxwell–Boltzmann
distribution. (Note that velocities range from −∞ to +∞.)
(a) To write the probability distribution p(vx)dvx so that the Maxwell–Boltzmanndistribution is correctly normalized, we require
c∫ ∞
−∞e−mv
2x/2kT dvx = 1
From integral tables, we see that
I =∫ ∞
−∞e−ax
2
dx =(π
a
)1/2
.
Aside To compute integrals of the form
I =∫ ∞
−∞e−ax
2
dx
we use the following trick. It is easy to see that we can write
I2 =∫ ∞
−∞e−ax
2
dx∫ ∞
−∞e−ay
2
dy =∫ ∞
−∞
∫ ∞
−∞e−a(x
2+y2) dx dy
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This is now an integral over the entire x− y plane. Converting to polar coordinates rand θ and recognizing that r2 = x2 + y2, the integral becomes
I2 =∫ ∞
0dr r
∫ 2π
0dθ e−ar
2
=∫ ∞
0dr re−ar
2∫ 2π
0dθ = 2π
∫ ∞
0dr re−ar
2
Making the substitution that u = −ar2, du = −2ar dr, we can finish the integral
I2 = −πa
∫ −∞
0du eu = −π
aeu∣∣∣∣−∞
0=π
a
hence
I =(π
a
)1/2
.
For our integral, a = m/2kT .
∫ ∞
−∞e−mv
2x/2kT dvx =
[2πkT
m
]1/2
=⇒ p(vx) dvx =[
m
2πkT
]1/2e−mv
2x/2kT dvx
(b) To compute the average energy,⟨
12mv2
x
⟩, we have
⟨1
2mv2
x
⟩=∫ ∞
−∞
1
2mv2
xp(vx) dvx =m
2
[2πkT
m
]1/2 ∫ ∞
−∞v2xe−mv2x/2kT dvx
Again consulting our table of integrals, we find
∫ ∞
−∞x2e−ax
2
dx =π1/2
2a3/2
Aside: Integrals of the form
∫ ∞
−∞x2e−ax
2
dx
can be computed by integration by parts. Recall that
∫ b
au dv = uv|ba −
∫ b
av du
Choosing the substitutions u = x and dv = xe−ax2, then du = dx and v = − 1
2ae−ax
2.
Our integral therefore becomes
∫ ∞
−∞x2e−ax
2
dx = − 1
2axe−ax
2∣∣∣∣∞
−∞+
1
2a
∫ ∞
−∞e−ax
2
=1
2a
(π
a
)1/2
= 0 +π1/2
2a3/2
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Note that we have used the result of the integral from part (a) above.
Therefore
⟨1
2mv2
x
⟩=
1
2m
[π1/2
2a3/2
] [m
2πkT
]1/2=
1
4m
(
2kT
m
)3/2[
m
π1/2kT
]1/2=
1
2kT.
(c) To find the average velocity, 〈vx〉, we recall that for functions with odd symmetry(f(x) = −f(−x)), the integral under the curve for negative x cancels with that underthe curve for positive x. Using the fact that p(x) = p(−x),
〈vx〉 =∫ ∞
−∞vxp(vx) dvx
=∫ 0
−∞vxp(vx) dvx +
∫ ∞
0vxp(vx) dvx
=∫ ∞
0(−vx)p(−vx) dvx +
∫ ∞
0vxp(vx) dvx
= −∫ ∞
0vxp(vx) dvx +
∫ ∞
0vxp(vx) dvx
= 0.
(d) What is the average momentum, 〈mvx〉?
〈mvx〉 = m〈vx〉 = 0,
from the result above.
16. Predicting the rate of mutation based on the Poisson probability distributionfunction.
The evolutionary process of amino acid substitutions in proteins is sometimes described bythe Poisson probability distribution function. The probability ps(t) that exactly s substitu-tions at a given amino acid position occur over an evolutionary time t is
ps(t) =e−λt(λt)s
s!,
where λ is the rate of amino acid substitutions per site per unit time. Fibrinopeptides evolverapidly: λF = 9.0 substitutions per site per 109 years. Lysozyme is intermediate: λL ≈ 1.0.Histones evolve slowly: λH = 0.010 substitutions per site per 109 years.
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(a) What is the probability that a fibrinopeptide has no mutations at a given site in t = 1billion years?
(b) What is the probability that lysozyme has three mutations per site in 100 millionyears?
(c) We want to determine the expected number of mutations 〈s〉 that will occur in timet. We will do this in two steps. First, using the fact that probabilities must sum toone, write α =
∑∞s=0 (λt)s/s! in a simpler form.
(d) Now write an expression for 〈s〉. Note that
∞∑
s=0
s(λt)s
s!= (λt)
∞∑
s=1
(λt)s−1
(s− 1)!= λtα.
(e) Using your answer to part (d), determine the ratio of the expected number of mutationsin a fibrinopeptide to the expected number of mutations in histone protein, 〈s〉fib/〈s〉his.
(a) What is the probability that a fibrinopeptide has no mutations at a given site in t = 1billion years?
P0(t) = e−λF t = exp (−(9.0 per 109 years)(109 years))
= e−9 = 1.23 · 10−4.
(b) What is the probability lysozyme has three mutations per site in 100 million years?
λLt = (1.0 per 109 years)(108 years) = 0.1
P3(t) =e−λLt(λLt)3
3!=
(e−0.1)(0.1)3
6
= 1.51 · 10−4.
(c) We want to know the expected number of mutations 〈s〉 that will occur in time t. Wewill do this in two steps. First, using the fact that probabilities must sum to one, write
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α =∞∑
s=0
(λt)s
s!
in a simpler form. Since the probabilities sum to 1,
∞∑
s=0
Ps(t) =∞∑
s=0
e−λt(λt)s
s!= e−λt
∞∑
s=0
(λt)s
s!= 1
Therefore
α =∞∑
s=0
(λt)s
s!= eλt.
(d) Now write an expression for 〈s〉.
Hint: Note that∞∑
s=0
s(λt)s
s!= (λt)
∞∑
s=1
(λt)s−1
(s− 1)!= λtα
〈s〉 =∞∑
s=0
sPs(t) =∞∑
s=0
se−λt(λt)s
s!
= e−λt(λt)∞∑
s=1
(λt)s−1
(s− 1)!
= e−λt(λt)∞∑
h=0
(λt)h
h!= (λtα)e−λt
so
〈s〉 = λt.
(e) Using (d), determine the ratio of the expected number of mutations in a fibrinopeptideto the expected number of mutations in histone, 〈s〉fib〈s〉his.
〈s〉fib
〈s〉his=λF t
λHt=λFλH
=9
0.01= 900.
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17. Probability in court.
In forensic science, DNA fragments found at the scene of a crime can be compared withDNA fragments from a suspected criminal to determine that the probability that a matchoccurs by chance. Suppose that DNA fragment A is found in 1% of the population, fragmentB is found in 4% of the population, and fragment C is found in 2.5% of the population.
(a) If the three fragments contain independent information, what is the probability thata suspect’s DNA will match all three of these fragment characteristics by chance?
(b) Some people believe such a fragment analysis is flawed because different DNA frag-ments do not represent independent properties. As before, suppose that fragment Aoccurs in 1% of the population. But now suppose that the conditional probability ofB, given that A is p(B|A) = 0.40 rather than 0.040, and p(C|A) = 0.25 rather than0.025. There is no additional information about any relationship between B and C.What is the probability of a match now?
(a) Since the fragments are independent
p = p(A) p(B) p(C)
= (0.01)(0.04)(0.025) = 1× 10−5.
(b)
p = p(A) p(B/A) p(C/A)
= (0.01)(0.40)(0.25) = 1× 10−3.
18. Flat distribution.
Given a flat distribution, from x = −a to x = a, with probability distribution p(x) = 1/(2a):
(a) Compute 〈x〉.
(b) Compute 〈x2〉.
(c) Compute 〈x3〉.
(d) Compute 〈x4〉.
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(a) 〈x〉 =∫ a
−axp(x) dx =
∫ a
−a
x
2adx =
∫ a
0
x
2adx−
∫ a
0
x
2adx = 0.
(b) 〈x2〉 =∫ a
−ax2p(x) dx =
(1
2a
) (x3
3
)∣∣∣∣∣
a
−a=
1
2a
[a3
3−(−a
3
3
)]=a2
3.
(c) By symmetry (as in (a)), 〈x3〉 = 0. In fact 〈xn〉 = 0 for all odd integers, n.
(d) 〈x4〉 =∫ a
−ax4p(x) dx =
(1
2a
)x5
5
∣∣∣∣∣
a
−a=a4
5.
19. Family probabilities.
Given that there are three children in your family, what is the probability that:
(a) two are boys and one is a girl?
(b) all three are girls?
The probability is about 1/2 for having either a boy or a girl. The binomial distributionshows that the probabilities are:
3 girls(
1
2
)3 3!
3!0!=
1
8
2 girls, 1 boy(
1
2
)3 3!
2!1!=
3
8
1 girl, 2 boys(
1
2
)3 3!
1!2!=
3
8
3 boys(
1
2
)3 3!
0!3!=
1
8.
20
20. Evolutionary fitness.
Suppose that the probability of having the dominant allele (D) in a gene is p and theprobability of the recessive allele (R) is q = 1 − p. You have two alleles, one from eachparent.
(a) Write the probabilites of all the possibilities: DD, DR, and RR.
(b) If the fitness of DD is fDD, the fitness of DR is fDR, and the fitness of RR is fRR, writethe average fitness in terms of p.
(a)
D R
D p2 pq
R qp q2
DD : p2
DR : 2pq
RR : q2
(b) The average fitness is
〈fitness〉 =∑
i
(fitness)iprobabilityi
= fDDp2 + 2fDRp(1− p) + fRR(1− p)2.
21. Ion-channel events.
A biological membrane contains N ion-channel proteins. The fraction of time that any oneprotein is open to allow ions to flow through is q. Express the probability P (m,N) that mof the channels will be open at any given time.
Channels are either open, with probability q or closed, with probability (1− q), so theexpression we want is the binomial distribution,
P (m,N) = qm(1− q)N−m N !
m!(N −m)!.
21
22. Joint probabilities: balls in a barrel.
For Example 1.9, two green balls and one red ball drawn from a barrel without replacement:
(a) Compute the probability p(RG) of drawing one red and one green ball in either order.
(b) Compute the probability p(GG) of drawing two green balls.
(a) We have
p(G1) = 2/3,
p(R1) = 1/3,
p(G2 | G1) = 1/2,
p(G2 | R1) = 1,
p(R2 | G1) = 1/2,
so
p(RG) = p(R1)p(G2 | R1) + p(G1)p(R2 | G1)
= (1/3)(1) + (2/3)(1/2)
= 2/3.
(b) p(GG) = p(G1)p(G2 | G1)
= (2/3)(1/2) = 1/3.
Note that since p(RR) = 0, the quantities in (a) and (b) sum to one.
22
23. Sports and weather.
The San Francisco football team plays better in fair weather. They have a 70% chance ofwinning in good weather, but only a 20% chance of winning in bad weather.
(a) If they play in the Super Bowl in Wisconsin and the weatherman predicts a 60%chance of snow that day, what is the probability that San Francisco will win?
(b) Given that San Francisco lost, what is the probability that the weather was bad?
(a) These chances of winning given in the problem are conditional probabilities (i.e., theprobability of winning) given that the weather is good or bad. The approach here is toelucidate the four mutually exclusive and collectively exhaustive outcomes, winning andgood weather, P (W,G), winning and bad weather, P (W,B), losing and good weather,P (L,G), and losing and bad weather, P (L,B). These joint probabilities can be relatedto the conditional probabilities, P (W |G), etc., and the weather probabilities, P (G) andP (B) by the following equations:
P (W,G) = P (W |G)P (G) = (0.7)(0.4) = 0.28
P (W,B) = P (W |B)P (B) = (0.2)(0.6) = 0.12
P (L,G) = P (L|G)P (G) = (0.3)(0.4) = 0.12
P (L,B) = P (L|B)P (B) = (0.8)(0.6) = 0.48
(Note that P (L|G) was computed from using the fact that P (W |G) + P (L|G) = 1.)
(b) P (B|L) =P (L,B)
P (L)
P (L) = P (L|G)P (G) + P (L|B)P (B)
= (0.3)(0.4) + (0.8)(0.6)
= 0.6
Therefore, P (B|L) = 0.48/0.6 = 0.8—there is an 80% chance there was bad weather, giventhat they lost.
23
24. The Monty Hall Dilemma.
You are a contestant on a game show. There are three closed doors: one hides a car, andtwo hide goats. You point to one door, call it C. The gameshow host, knowing what’sbehind each door, now opens either door A or B, to show you a goat; say it’s door A. Towin a car, you now get to make your final choice: should you stick with your original choiceC, or should you now switch and choose door B? (New York Times, July 21, 1991; SciAmer, August 1998.)
A good way to illustrate how people sometimes try to tackle this problem is to consider asimilar one: Suppose three cards are lying face down on a table, only one of which is an ace.The first card is turned over, and is not an ace, so
p(B = ace) =13
1− 13
=1
2p(C = ace) =
13
1− 13
=1
2
So both remaining face-down cards are equally likely to be an ace. Using this type ofreasoning, many people will say that switching isn’t any more likely to win than staying withthe door you initially chose. But these problems are not equivalent – in the Monty Hall case,the host has knowledge of both which door you initially picked and which door contains thecar.
A simple way to come to arrive at the correct solution is to break it down into two separatequestions:1. What is the probability of winning if you don’t switch doors?If you don’t switch doors, it means that you make no use of the information given by the hostrevealing a goat. The only way you can win is if the door you initially chose has the carbehind it, and hence the probability of winning is 1/3.2. What is the probability of winning if you do switch?If you do switch doors, the only way you can lose is if the door you initially picked had thecar behind it, so you have a 1/3 probability of losing and your probability of winning istherefore 2/3. So you should switch doors.
24
25. Probabilities of picking cards and rolling dice.
(a) What is the probability of drawing either a Queen or a Heart in a normal deck of 52cards?
(b) What is the probability P of getting three 2’s and two 4’s on 5 independent rolls of adie?
(a) P (Q of Hearts) = 152
. P (Q not of Hearts) = 352
. P (Heart and not a Q) = 1252
.P = 1
52+ 3
52+ 12
52= 16
52.
(b) P (2) = P (4) = 16. P (2)3P (4)2 = (1
6)5 = 1
7776= 1.29 · 10−4.
P =(
52
)1.29 · 10−4 = 1.29 · 10−3.
26. Probability and translational start codons.
In prokaryotes, translation of mRNA messages into proteins is most often initiated at startcodons on the bacterial mRNA having sequence ”AUG.” Assume that the mRNA is single-stranded and consists of a sequence of bases, each described by a single letter. The alphabetof letters for mRNA consists of ”A”, ”C”, ”U”, ”G”.Consider the set of all random pieces of bacterial mRNA of length six bases.
(a) What is the probability of having either no A’s OR no U’s in the mRNA sequence ofsix base pairs long?
(b) What is the probability of a random piece of mRNA having exactly 1 A, 1 U, and 1G?
(c) What is the probability of a random piece of mRNA of length six base pairs havingan ”A” directly followed by a ”U” directly followed by a ”G”; in other words, havingan ”AUG” in the sequence?
(d) What is the total number of random pieces of mRNA of length six base pairs thathave exactly one ”A”, exactly one ”U”, and exactly one ”G”, with ”A” appearingfirst, then the ”U”, then the ”G”? (ex: AXXUXG)
(a) P(no A’s OR no U’s) = P(no A) + P(no U) - P(no A AND no U)
25
= (3
4)6 + (
3
4)6 − (
1
2)6
= 0.34
(b) Use the Multinomial Probability Distribution (eq. 1.31)
W =6!
1!1!1!3!, P = (
1
4)6 · 6!
1!1!1!3!= 0.0293
(c) Let us find W(AUG):6 positions total:3 positions are fixed: Wf = 13
3 positions are variable: Wv = 43
4 positions ”AUG” can take in the sequence: Wpos = 3
A U GA U G
A U G
Note: We do not count this next ”AUG” position because it has already been includedin the first A U G when we count the multiplicity of the variable positions:
A U G
Because Wtotal is the product of independent sources of multiplicity,W (AUG) = Wf ·Wv ·Wpos = 13 · 43 · 3P (AUG) =
Wf
Wtotal= 3·43
46 = 343 = 3
64≈ 0.0469
(d) We will calculate W for 1A, 1U, and 1G for any ordering of the A, U, G with respect toeach other. Then we will divide this result by the number of ways of ordering A, U, Gwith respect to each other, because we wish only to count the cases where A appearsbefore U appears before G.W (1A, 1U, 1G) = 6!
1!1!1!3!
The number of ways of ordering A, U, G with respect to each other : 3!
W (1A, 1U, 1G) =6!
1!1!1!3!
3!= 6·5·4
3·2·1 = 20
27. DNA synthesis.
26
Suppose you synthesize a strand of DNA that is 1000 bases long. A process you use intro-duces a wrong base on average every 1000 bases synthesized.
(a) Calculate and draw a bar graph indicating the yield (probability) of each productDNA, containing 0, 1, 2 and 3 mutations.
(b) Calculate how many combinations of DNA sequences of 1000 bases long contain ex-actly 2 mutant bases.
(c) What is the probability of having the 500th base and the 888th base mutated in thepool of DNA which has only two mutations?
(d) What is the probability of having two mutations side-by-side in the pool of DNAwhich has only two mutations?
(a) Let letter R stand for a right base, and M stand for a mutant one.p(1000R, 1000) = (0.999)1000(0.001)0 1000!
1000!≈ 0.368
p(999R, 1M, 1000) = (0.999)999(0.001)1 1000!999!1!
≈ 0.368p(998R, 2M, 1000) = (0.999)998(0.001)2 1000!
998!2!≈ 0.18
p(997R, 2M, 1000) = (0.999)997(0.001)3 1000!997!3!
≈ 0.06
(b) W = 1000!998!2!
(c) The probability of having a DNA strand with only 2 mutated bases is 0.18. Theprobability of having mutations at places 500 and 888 of the strand isp(...M(500)......M(888)...) = 0.18 · 998!2!
1000!= 0.18 · 2 · 10−6 = 3.6 · 10−7
(d) p(...MM...) = 0.18 · 998!2!1000!
· 999 = 0.18 · 1500
= 3.6 · 10−4
28. Presidential election.
27
Two candidates are running for President. Candidate A has already received 80 electoralvotes and only needs 35 more to win. Candidate B already has 50 votes, and needs 65 moreto win.
Five states remain to be counted. Winning a state gives a candidate 20 votes; losing givesthe candidate zero votes. Assume both candidates otherwise have equal chances to win inthose 5 states:
(a) Write an expression for WA, total, the number of ways A can succeed at winning 40more electoral votes.
(b) Write the corresponding expression for WB, total.
(c) What is the probability candidate A beats candidate B?
(a) For candidate A, this is like flipping a coin 5 times. Each head is like winning 20 points;each tail is zero. Candidate A can win by getting 2 ore more states (’heads’), so:
WA,total =5!
2!3!+
5!
3!2!+
5!
4!1!+
5!
5!0!= 10 + 10 + 5 + 1 = 26.
(b) Candidate B can win by getting 3 or more states, so:
WB, total =5!
4!1!+
5!
5!0!= 5 + 1 = 6.
(c)
P (A wins) =WA,total
WA,total +WB,total=
26
26 + 6= 0.81.
28
Chapter 2Extremum Principles PredictEquilibria
1. A lattice gas.
How many arrangements are there of fifteen indistinguishable lattice gas particles distributedon:
(a) V = 20 sites?
(b) V = 16 sites?
(c) V = 15 sites?
(a) W (N = 15, V = 20) =20!
15!5!=
20 · 19 · 18 · 17 · 16
5 · 4 · 3 · 2 = 15, 504
(b) W (N = 15, V = 16) =16!
15!1!= 16
(c) W (N = 15, V = 15) =15!
15!0!= 1
29
2. Maximum of binomial distribution.
Find the value n = n∗ that causes the function
W =N !
n!(N − n)!pn(1− p)N−n
to be at a maximum, for constants p andN . Use Stirling’s approximation, x! ' (x/e)x. Notethat it is easier to find the value of n that maximizes lnW than the value that maximizesW . The value of n∗ will be the same.
W is maximal where lnW is maximal, and
lnW = n ln p+ (N − n) ln(1− p) + lnN !− lnn!− ln(N − n)!
Now using Stirling’s approximation, lnN ! ≈ N lnN −N , we obtain
lnW ≈ n ln p+ (N − n) ln(1− p) + (N lnN −N)− (n lnn− n)
−[(N − n) ln(N − n)− (N − n)]
= n ln p+ (N − n) ln(1− p) +N lnN − n lnn− (N − n) ln(N − n)
This function is maximal where
d lnW
dn= 0
d lnW
dn= ln p− ln(1− p)−
(n · 1
n+ lnn
)
−(
(N − n) · 1
(N − n)+ ln(N − n) · (−1)
)
= ln
(p
1− p
)− ln (n) + ln (N − n)− 1 + 1
We add lnN − lnN to the right-hand side and rearrange terms to allow us to write ln(n/N)
30
and ln(N − n)/N :
d lnW
dn= ln
(p
1− p
)− ln
(n
N
)+ ln
(N − nN
)= 0
ln(
n∗
N − n∗)
= ln
(p
1− p
)
N∗
N − n∗ =p
1− p
n∗(
1 +p
1− p
)=
Np
1− pn∗
N=
p/(1− p)1/1− p = p.
3. Finding extrema.
V (x) =x3
3+
5x2
2− 24x :
(a) Where is the maximum?
(b) Where is the minimum?
To find the extrema, determine the values x = x∗ that cause the derivative to equal zero
dV
dx
∣∣∣∣∣x∗
= (x2 + 5x− 24)∣∣∣x∗
= (x∗ − 3)(x∗ + 8) = 0
=⇒ x∗ = 3,−8.
To determine whether the extrema are maxima or minima, evaluate the second derivative atthe x∗ points
d2V
dx2= 2x+ 5
31
d2V
dx2
∣∣∣∣∣x∗=3
= 2(3) + 5 = 11.
Since this value is positive, x∗ = 3 is a minimum.
d2V
dx2
∣∣∣∣∣x∗=−8
= 2(−8) + 5 = −9.
This value is negative, so x∗ = −8 is a maximum.
4. The binomial distribution narrows as N increases.
Flip a coin 4N times. The most probable number of heads is 2N , and its probability isp(2N). If the probability of observing N heads is p(N), show that the ratio p(N)/p(2N)diminishes as N increases.
p(N)
p(2N)=
((4N)!
(N !)(3N)!
)
((4N)!
(2N)!(2N)!
)
=[(2N)!]2
N !(3N)!≈
[(2Ne
)2N]2
(Ne
)N (3Ne
)3N
=24NN4N
33NN4N
=
(24
33
)N=(
16
27
)N
Note that as N →∞
limN→∞
p(N)
p(2N)= lim
N→∞
(16
27
)N= 0
32
5. De-mixing is improbable.
Using the diffusion model of Example 2.3, with 2V lattice sites on each side of a permeablewall and a total of 2V white particles and 2V black particles, show that perfect de-mixing (allwhite on one side, all black on the other) becomes increasingly improbable as V increases.
The ratio r of the perfectly demixed to perfectly mixed configurations is
r =1(
2V !V !V !
)(2V !V !V !
) =
(V !V !
(2V )!
)2
≈
(Ve
)2V
(2Ve
)2V
2
=(
1
24
)V=(
1
16
)V
Note that as V →∞
limr→∞ r = lim
N→∞
(1
16
)N= 0
6. Stable states.
For the energy function V (θ) = cos θ for 0 ≤ θ ≤ 2π, find the values θ = θs that identifystable equilibria, and the values θ = θu that identify unstable equilibria.
dV (θ)
dθ= − sin θ.
This derivative is zero when sin θ is zero. This occurs at θ = nπ, where n is an integer. Todetermine which of these points are maxima and which are minima, we need to compute thesecond derivative.
d2V
dθ2= − cos θ
This is negative for all even-numbered multiples of π and positive for all odd-numberedmultiples. Therefore, the unstable equilibria are given by θu = 2kπ and the stable equilibriaby θs = π + 2kπ, where k is an integer.
7. One dimensional lattice.
33
You have a 1-dimensional lattice that contains NA particles of type A and NB particlesof type B. They completely fill the lattice, so the number of sites is NA + NB. Write anexpression for the multiplicity W (NA, NB), the number of distinguishable arrangements ofthe particles on the lattice.
W (NA, NB) =
(NA +NB
NA
)=
(NA +NB
NB
)=
(NA +NB)!
NA!NB!
34
Chapter 3Heat, Work and Energy
1. The time dependence of a mass on a spring.
(a) For the harmonic motion of a mass on a spring, the kinetic energy is K = (1/2)mv2,and the potential energy is V = (1/2)ksx
2, where ks is the spring constant. Using theconservation of energy find the time-dependent spring displacement, x(t).
(b) Compute the force f(t) on the mass.
Harmonic motion is the model for a bond stretching in molecules and for the motions ofatoms in crystals around their equilibrium positions. Harmonic motion can be pictured as aspring motion. The potential energy of a spring is given by V (x) = (1/2)ksx
2, where k is aconstant that depends on the characteristics of the spring. Writing the equation forconservation of energy, we have
mx(t)2
2+ksx(t)
2
2= E.
Here, x(t) = dx(t)dt
. Rearranging terms, we get
x(t)2 =2E
m− ksmx(t)2
35
To simplify things, we introduce a constant ω2 = ks/m
x(t)2 = ω2(
2E
mω2− x(t)2
)= ω2
(2E
ks− x(t)2
)
Introducing another constant A = 2Eks
, we arrive at
x(t)2 = ω2(A2 − x(t)2
)
We can integrate this with another trick that is sometimes possible to use on simpledifferential equations:
(dx
dt
)2
= ω2(A2 − x2
)
dx
dt= ω
(A2 − x2
)1/2
dx
(A2 − x2)1/2= ωdt
∫ x
0
dx′
(A2 − x′2)1/2=
∫ t
0ω dt′
The integral on the left is an inverse trigonometric integral, and the integral on the right istrivial.
ωt+ φ = sin−1(x
A
)
Therefore, the equation of motion is
x(t) = A sin(ωt+ φ)
The force on the particle as a function of time can be determined from the spatial derivativeof the potential
f(t) = −dVdx
∣∣∣∣∣x(t)
= −ksx(t) = −ksA sin(ωt+ φ)
36
or from the time derivative of the position
f(t) = mx(t)
= mdx(t)
dt
= −mAω2 sin(ωt+ φ)
= −ksA sin(ωt+ φ)
2. Equalizing energies.
For the two 10-particle two-state systems of Example 3.4, suppose the total energy to beshared between the two objects is U = UA + UB = 4. What is the distribution of energiesthat gives the highest multiplicity?
We can write
W (UA) =
(10!
UA!(10− UA)!
)(10!
(4− UA)!(10− 4 + UA)!
).
The following table lists all the possibilties
U W (U)
0 210
1 1200
2 2025
3 1200
4 210
This shows that the highest multiplicity occurs, in this case, when the energy is dividedequally between the two objects.
37
3. Energy conversion.
When you drink a beer, you get about 100 cal (1 food cal = 1 kcal). You can work this offon an exercise bicycle in about 10 minutes. If you hook your exercise bicycle to a generator,what wattage of light bulb could you light up, assuming 100% efficiency?
(100 cal)(4.18
J
cal
)(1
10 min
)(1 min
60 sec
)= 697
J
sec= 697.6 watts.
An interesting comparison is that 1 horsepower = 746 watts.
4. Kinetic energy of a car.
How much kinetic energy does a 1700 kg car have, if it travels 100 km h−1?
Kinetic energy =1
2mv2
=(
1
2
)(1700 kg)
[(105 meters
hr
)(1 hr
3600 sec
)]2
= 655.8 kJ
5. Rms velocity of a gas.
Using 12kT = 1
2m〈vx2〉, for T = 300 K, compute the rms velocity, 〈vx2〉
12 , of O2 gas.
〈vx2〉 = kTM
M = 32 gmol· 1mol
6.02·1023molecules· kg
1000g= 5.316 · 10−26 kg
molecule
〈vx2〉 = 1.381 · 10−23 JK· 300K · 1
5.316·10−26molecules
kg= 77, 934.5m
2
s2
(1J = 1kgm2
s2)
〈vx2〉12 = 279.2m
s
6. Earth’s energy balance
38
The figure below shows energy inputs and outputs to the earth, in Watts/m2, which is ameasure of energy per unit area per unit time (1 W = 1 V · 1 A).
(a) Suppose your backyard has an area of 1000 m2. Your yard only receives sunlight aboutone third of each day, and because of overcast skies, let’s say the total flux is reducedby another factor of one half. What average energy flux, in Watts/m2, does your yardreceive?
(b) Your cell phone uses about 2 W. Assuming 20% efficiency, how big a solar cell wouldyou need to power it?
(c) If we had no natural greenhouse gases (CO2 and water vapor), how would that shiftearth’s energy balance?
(d) How much energy, in Watt-hours, is in a gallon of gasoline? Assume that 100% of thegasoline is octane (C8H18). Gasoline is about 1 g/cc, and each carbon-carbon covalentbond is worth about 60 kcal/mol. Assume 50% efficiency in burning.
(a) Energy flux absorbed by every m2 of earth: 168 + 324 ≈ 500 Wm2
Energy flux absorbed by the backyard: (500 Wm2 · 1000m2) · (1
3) · (1
2) ≈ 83333W
(b) The amount of power a solar cell can convert per m2 is ≈ 16667W1000m2 = 16.67 W
m2
For 2 watts, one would need a 216.67≈ 0.12m2 solar cell
39
(c) The earth would lose 324324+168
≈ 66% of the radiation absorbed at the surface.Essentially, the earth would cool down significantly.
(d) 100% octane, gasoline ∼ 1 g/ml , a C-C bond ∼ 60 kcalmol
= 2.51 · 105 Jmol
, 50% efficiency
C8H18 = 114 gmol
, 7 C-C bonds in one octane molecule
1gallon · 3.785L1gallon
· 1kg1L
= 3.785kg of octane.
3785g octane ·moloctane114g
· 7molC−Cbonds1moloctane
· 251000JmolC−Cbonds
= 5.83 · 107J
At 50% efficiency, we have 2.92 · 107J = 2.92 · 107W · s = 8102W · hr
7. Taking the earth’s temperatureA physical law, called the Stefan-Boltzmann (SB) law, relates the transfer rate of electro-magnetic radiation of a body (called black-body radiation) to its temperature. The SB law isa relative of the Wein law, the principle that explains why an object that is hot enough willemit visible light, as when a burning ember in a fire glows red, or glows blue if is even hotter.An object having a non-zero temperature will emit electromagnetic radiation. Conversely,an object that absorbs energy via radiation, increases its temperature. The SB law saysthat the electromagnetic power P is relat ed to the temperature T by
P = (5.67× 10−8 W
m2K4)T 4 (1)
(a) If the the incoming solar power hitting the earth is P = 342 Wm−2, what would youpredict should be the earth’s temperature?
(b) Now, turn around this argument around. The earth’s actual average temperature isT = 288K or +15◦ C. Use the SB law to compute the power o f radiated heat fromthe earth’s core.
(a) Using the SB law, the earth’s estimated temperature would be T = 255K, or -18◦ C.
(b) Using the SB law, the power radiated from the earth’s core would be P = 390 Wm−2.
8. Why do elephants live longer than mice?Let L represent the lifespan of an organism. It is known that all mammalian species live, onaverage, to have about 1.5 × 109 heartbeats (HB), independent of their size. Assume thatan organism will have a total of HB = L ×HR = constant heartbeats during its lifetime,where HR is its heart-rate. What assumptions would you need to make to explain theobservation that the lifespans of organisms grow with body mass M as L ∝ HB
HR∝M1/4?
40
Assume the rate of energy expenditure per unit mass (BMR/M) of biological tissue isproportional to energy flow into that tissue. This energy flow is proportional to the rate ofblood flow. And, assume that blood flow correlates with the heart rate (HR) of an organism.Then, you have HR ∝ BMR/M ∝M−1/4, indicating that heart rates decrease in largerorganisms, consistent with observations. Combining these observations gives
L ∝ HB
HR∝M1/4, (2)
rationalizing why elephants live longer than mice. This, too, is one of the 1/4-power scalinglaws observed in nature.
41
Chapter 4Math Tools: Multivariate Calculus
1. Applying the Euler test.
Which of the following are exact differentials?
(a) 6x5dx + dy
(b) x2y2dx+ 3x2y3dy
(c) (1/y)dx− (x/y2)dy
(d) ydx+ 2xdy
(e) cos xdx− sin ydy
(f) (x2 + y)dx+ (x+ y2)dy
(g) xdx+ sin ydy
If you have an exact differential, then dg(x, y) = u(x, y)dx+ v(x, y)dy.
42
∂u
∂y
∂v
∂xg(x,y)
(a) 0 0 exact x6 + y
(b) 2x2y 6xy3 inexact
(c) −y−2 −y−2 exact xy−1
(d) 1 2 inexact
(e) 0 0 exact sin x+ cos y
(f) 1 1 exact x3
3+ y3
3+ xy
(g) 0 0 exact x2
2− cos y
2. Differentiating multivariate functions.
Compute the partial derivatives, (∂f/∂x)y and (∂f/∂y)x, for the following functions.
(a) f(x, y) = 3x2 + y5
(b) f(x, y) = x10y1/2
(c) f(x, y) = x+ y2 + 3
(d) f(x, y) = 5x
(a)
(∂f
∂x
)
y
= 6x
(∂f
∂y
)
x
= 5y4
(b)
(∂f
∂x
)
y
= 10x9y1/2
(∂f
∂y
)
x
=1
2x10y−1/2
(c)
(∂f
∂x
)
y
= 1
(∂f
∂y
)
x
= 2y
(d)
(∂f
∂x
)
y
= 5
(∂f
∂y
)
x
= 0
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3. Minimizing a single variable function subject to a constraint.
Given the function y = (x − 2)2, find x∗, the value of x that minimizes y subject to theconstraint y = x.
You do not need anything sophisticated like Lagrange multipliers for this problem becauseyou have two equations in two unknowns. It can be solved directly. Lagrange multipliers areneeded only when you have more unknowns than equations.
y = x
y
x
1
1 2
y = x24
4
Substitute y = x into y = (x− 2)2
y = (y − 2)2 = y2 − 4y + 4
or y2 − 5y + 4 = 0
(y − 1)(y − 4) = 0 =⇒ y = 1 or 4
The two solutions are
(x, y) = (1, 1) or (4, 4)
The lowest of these is (x, y) = (1, 1).
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4. Maximizing a multivariate function without constraints.
Find the maximum (x∗, y∗, z∗) of the function f(x, y, z) = d− (x− a)2− (y− b)2− (z− c)2.
(∂f
∂x
)
y,z
= 0 − 2(x∗ − a) = 0 =⇒ x∗ = a
(∂f
∂y
)
x,z
= 0 − 2(y∗ − b) = 0 =⇒ y∗ = b
(∂f
∂z
)
x,y
= 0 − 2(z∗ − c) = 0 =⇒ z∗ = c
The extremum is (x∗, y∗, z∗) = (a, b, c) and the maximum value is f(a, b, c) = d.
5. Extrema of multivariate functions with constraints.
(a) Find the maximum of the function f(x, y) = −(x − a)2 − (y − b)2 subject to theconstraint y = kx.
(b) Find the minimum of the paraboloid f(x, y) = (x − x0)2 + (y − y0)
2 subject to theconstraint y = 2x.
(a) Use the Lagrange multiplier method. Let g(x, y) = y − kx. We want
(∂f
∂x
)− λ
(∂g
∂x
)= 0 (3)
and (∂f
∂y
)− λ
(∂g
∂y
)= 0 (4)
−2(x∗ − a)− λ(−k) = 0 (5)
−2(y∗ − b)− λ = 0 (6)
45
Substituting y = kx into Equation (6) gives:
−2(kx∗ − b) = λ
Substituting this expression for λ into Equation (5) gives
x∗ =a + kb
1 + k2y∗ =
k(a+ kb)
1 + k2
(b) g(x, y) = y − 2x
2(x∗ − x0) + 2λ = 0
2(y∗ − y0)− λ = 0
x∗ − x0 = −λ = −2(y∗ − y0).
Combined with the constraint y∗ = 2x∗, this gives
x∗ =x0 + 2y0
5y∗ =
2
5(x0 + 2y0)
6. Composite functions.
(a) Given the functions f(x, y(u)) = x2 + 3y2 and y(u) = 5u + 3, express df in terms ofchanges dx and du.
(b) What is
(∂f
∂u
)
x,u=1
?
(a) df =
(∂f
∂x
)dx+
(∂f
∂y
)(dy
du
)du
df = 2xdx + (6y)(5)du
= 2xdx + 30ydu = 2xdx+ 30(5u+ 3)du
(b)
(∂f
∂u
)=
(∂f
∂y
)(∂y
du
)= 30(5u+ 3)|u=1 = 240.
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7. Converting to an exact differential.
Given the expression dx+ (x/y)dy, show that dividing by x results in an exact differential.What is the function f(x, y) such that df is dx+ (x/y)dy divided by x?
Dividing dx+ (x/y)dy by x gives (dx/x) + (dy/y). Euler’s test
(∂(1/x)
∂y
)= 0 =
(∂(1/y)
∂x
)= 0.
Thus (dx/x) + (dy/y) is path independent, and can be integrated to give∫(dx/x) + (dy/y) = ln(xy) + constant = f(x, y).
8. Propagation of error.
Suppose that you can measure independent variables x and y and that you have a dependentvariable f(x, y). The average values are x, y, and f . We define the error in x as thedeviations εx = x− x, in y as εy = y − y, and in f as εf = f − f .
(a) Use a Taylor series expansion to express the error, εf , in f , as a function of the errorsεx and εy in x and y.
(b) Compute the mean-squared error 〈ε2f〉 as a function of 〈ε2
x〉 and 〈ε2y〉.
(c) Consider the ideal gas law with f = p, x = T , and y = V . If T and V have 10%random errors, how big will be the error in p?
(d) How much increase in ε(f) comes from adding n measurements that all have the sameuncorrelated error ε(x)?
(a) To find the error in f if you know the error in x, use a Taylor series expansion about x:
f(x)− f =df
dx
∣∣∣∣∣x
(x− x) +1
2
d2f
dx2
∣∣∣∣∣x
(x− x)2 + · · ·
=⇒ εf =df
dx
∣∣∣∣∣x
εx +1
2
d2f
dx2
∣∣∣∣∣x
ε2x + · · ·
If the error in x is small, εx � 1 and we can neglect terms that are second-order or
47
higher in εx:
εf ≈df
dx
∣∣∣∣∣x
εx
(b) Generalizing this to errors in x and y gives
εf =
(∂f
∂x
)
y
εx +
(∂f
∂y
)
x
εy
We are usually less interested in a single error and more interested in the average of thesquare of all the errors in a set of measurements, 〈ε2
f〉. (The average error itself, 〈εf〉, isless informative than 〈ε2
f〉 because positive and negative errors cancel resulting in〈εf〉 ≈ 0. 〈ε2
f〉 is more useful because it tells us about the magnitudes of the errors.)
To get 〈ε2f〉, we first square εf :
ε2(f) =
[(∂f
∂x
)εx +
(∂f
∂y
)εy
]2
=
(∂f
∂x
)2
ε2x +
(∂f
∂y
)2
ε2y +
(∂f
∂x
)(∂f
∂y
)εxεy
Now, we take the expectation value. We use the fact that x and y are independent, sotheir errors will be uncorrelated and 〈εxεy〉 = 0:
〈ε2f〉 =
(∂f
∂x
)2
〈ε2x〉+
(∂f
∂y
)2
〈ε2y〉.
(c) For one mole of an ideal gas
(∂f
∂x
)=
(∂p
∂T
)=R
Vand
(∂f
∂y
)=
(∂p
∂V
)= −RT
V 2.
We assume errors of 10%, so εT/T = 0.1 and εV /V = 0.1. Hence
ε2p =
(R
V
)2
T 2(0.10)2 +(RT
V 2
)2
V 2(0.10)2 = 2(RT
V
)2
0.01 = 0.02p2
(d) n measurements gives ε2f = n
(∂f∂x
)2ε2x so εf grows as (
√n)εx.
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9. Small differences of large numbers can lead to nonsense.
Using the results from Problem 8, show that the propagated error is larger than the differenceitself, for f(x, y) = x− y, with x = 20± 2 and y = 19± 2.
Since (∂f/∂x)2 = 1 and (∂f/∂y)2 = 1, we have ε2f = ε2
x + ε2y = 8. Since εx = εy = z so
εf =√
8 = 2.83. Therefore (20± 2)− (19± 2) = 1± 2.83.
10. Finding extrema.
Find the point (x∗, y∗, z∗) that is at the minimum of the function
f(x, y, z) = 2x2 + 8y2 + z2
subject to the constraint equation
g(x, y, z) = 6x + 4y + 4z − 72 = 0.
Use the Lagrange multiplier method:
(∂f
∂x
)− λ
(∂g
∂x
)= 0
(∂f
∂y
)− λ
(∂g
∂y
)= 0
(∂f
∂z
)− λ
(∂g
∂z
)= 0
(4x)− λ(6) = 0
(16y)− λ(4) = 0
(2z)− λ(4) = 0
λ =2x
3
λ = 4y
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λ =z
2
g(x, y, z) = 6x + 4y + 4z − 72 = 0
Find x:
0 = 6x+ 4y + 4z − 72
= 6x+ 4(
2x
12
)+ 4
(4x
3
)− 72
=(
36x
3
)− 72
x∗ = 6
Find y:
0 = 6x+ 4y + 4z − 72
= 6(6) + 4y + 4(8y)− 72
= −36 + 36y
y∗ = 1
Find z:
0 = 6x + 4y + 4z − 72
= 6(6) + 4(1) + 4z − 72
= −32 + 4z
z∗ = 8
11. Derivative of a multivariable composite function.
50
For the function f(x, y(v)) = x2y+ y3, where y = mv2, compute dfdv
around the point wherem=1, v=2, x=3.
dFdv
= ∂F∂y· ∂y∂v
∂F∂y
= x2 + 3y2, ∂y∂v
= 2mvdFdv
= 2mv(x2 + 3y2)plugging in the values:2 · 1 · 2(32 + 3(mv2)2) = 4(9 + 3(4)2) = 4(9 + 3 · 16) = 4(9 + 48) = 228
OR
F (x,m, v) = x2mv2 + (mv2)3 = x2mv2 +m3v6
dFdv
= 2x2mv + 6m3v5 = 2mv(x2 + 3m2v4) = 2mv(x2 + 3(mv2)2) = same as above.
12. Volume of a cylinder.
For a cylinder of radius r and height h, the volume is V = πr2h and the surface area isA = 2πr2 + 2πrh.
(a) Derive the height h(r0) that maximizes the volume of a cylinder with a given areaA = a0 and given radius r0.
(b) Compute the change in volume, ∆V , from (r1, h1) = (1, 1) to (r2, h2) = (2, 2).
(c) Compute the component volume changes, ∆Va and ∆Vb, which sum to ∆V , where∆Va is the change from (r1, h1) = (1, 1) to (r2, h1) = (2, 1) and (r2, h1) = (2, 1) to(r2, h2) = (2, 2).
(d) Should (b) equal (c)? Why or why not?
(a) Use the expression A = a0 = 2πr20 + 2πr0h to solve for h: h(r0) =
a0−2πr202πr0
= a02πr0− 1.
Since V = πr20h, V (a0, r0) = πr2
0
(a0
2πr0− 1
)= a0r0
2− πr2
0.
(b) V = πr2h∂V∂r
= 2πrh, ∂V∂h
= πr2
∂2V∂r∂h
= 2πr, ∂2V∂h∂r
= 2πr =⇒ This is a state function.dV = ∂V
∂rdr + ∂V
∂hdh = 2πrhdr + πr2dh∫ 2
1
∫ 21 2πrhdr + πr2dh = πr2h|2,21,1
π(22 · 2− 12 · 1) = π(8− 1)
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∆V = 7π
(c) ∆Va =∫ 21 2πrhdr = 2πr2h
2|21 = πr2 · 1|21 = π(22 − 12) = 3π
∆Vb =∫ 21 pir
2dr = πr2h2|21 = π22 · h|21 = 4π(2− 1) = 4π
∆V = ∆Va + ∆Vb = 3π + 4π = 7π
(d) Yes, (b) and (c) should be equal. V is a state function, as shown in part (b). Therefore,the path of integration should not affect the final result.
13. Equations of State.
Which of the following could be the total derivative of an equation of state?
(a) 2nRT(V−nb)2dV + R(V −nb)
nb2dT
(b) − nRT(V−nb)2dV + nR
V−nbdT
Cross derivatives for equations of state must be equal.
(a) Is ∂∂T
( 2nRT(V−nb)2 ) = ∂
∂V(R(V −nb)
nb2) ?
2nRT(V−nb)2 6= R
nb2
No, this could not be the total derivative of an equation of state.
(b) Is ∂∂T
(− nRT(V −nb)2 ) = ∂
∂V( nRV−nb)?
− nR(V −nb)2 = − nR
(V−nb)2Yes, this could be the total derivative of an equation of state.
14. Distance from the circumference of a circle.
52
The circle shown below satisfies the equation x2 + y2 = 4.
(3,2)2
0
20–2
–2
x
y
Find the point (x∗, y∗) on the circle that is closest to the point (3, 2). (That is, minimizesthe distance f(x, y) = ∆r2 = (x− 3)2 + (y − 2)2.)
Minimize the distance f(x, y) = ∆r2 = (x− 3)2 + (y − 2)2 under the constraint x2 + y2 = 4.Use Lagrange multipliers.∂g∂x
= 2x, ∂g∂y
= 2y∂f∂x
= 2(x− 3), ∂f∂y
= 2(y − 2)The Lagrange method then states that the solution will be given by:∂f∂x
= λ( ∂g∂x
) =⇒ 2(x ∗ −3) = 2λx∗ =⇒ λ = x∗−3x∗ = 1− 3
x∗∂f∂y
= λ( ∂g∂y
) =⇒ 2(y ∗ −2) = 2λy∗ =⇒ λ = y∗−2y∗ = 1− 2
y∗Combining these results with the constraint x2 + y2 = 4 gives us:λ = 1− 3
x∗ = 1− 2y∗ =⇒ 3
x∗ = 2y∗ =⇒ x∗ = 3
2y∗
(32y∗)2 + y∗2 = 4 =⇒ 13
4y∗2 = 4 =⇒ y∗2 = 16
13=⇒ y∗ =
√1613
= 4√13
x∗ = 32
4√13
= 6√13
=⇒ (x∗, y∗) = ( 6√13, 4√
13)
15. Find df and ∆f .
(a) If f(x, y) = x2 + 3y, express df in terms of dx and dy.
(b) For f(x, y) = x2 + 3y, integrate from (x, y) = (1, 1) to (x, y) = (3, 3) to obtain ∆f .
(a) f(x, y) = x2 + 3y
∂f
∂x= 2x;
∂f
∂y= 3
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df =∂f
∂xdx+
∂f
∂ydy = 2x dx+ 3 dy
(b) f(x, y) = x2 + 3y
∫ x=3
x=1df(x, y = 1) +
∫ y=3
y=1df(x = 3, y) =
∫ 3
12x dx+
∫ 3
13 dy = 32 − 12 + 3 ∗ 3− 3 ∗ 1 = 14
To check this, ∆f = f(3, 3)− f(1, 1) = (32 +3 ∗ 3)− (12− 3 ∗ 1) = (9+9)− (1− 3) = 14.
16. Derivative of a composite function.
For f(x, y) = x2y + y3 where y = mr2, find dfdr
.
df
dr=∂f
∂x
df
dr+∂f
∂y
dy
dr=⇒ df
dr= 2mr(x2 + 3m2r4)
17. Maximum volume of a solid with constraints.
You have a rectangular solid of length x, width y, and height z. Derive the values ofx, y, and z that give the maximum volume, subject to a fixed surface area: g(x, y, z) =2xz + 2yz + 2xy − constant = 0.
Use the method of Lagrange multipliers:
f(x, y, z) = xyz; g(x, y, z) = 2xz + 2yz + 2xy − constant = 0
∂f
∂x= λ
∂g
∂x=⇒ yz = λ(2y + 2z)
∂f
∂y= λ
∂g
∂y=⇒ xz = λ(2x+ 2z)
∂f
∂z= λ
∂g
∂z=⇒ xy = λ(2x+ 2y)
1
2λ=x + y
xy=x + z
xz=y + z
yz=⇒ xy + xz
xyz=xy + zy
xyz=xz + yz
xyz
54
x∗ = y∗ = z∗
18. Short answer questions.
(a) Compute the partial derivatives, ( ∂f∂x
)y, (∂f∂y
)x, for the following functions.
(a) f(x, y) = ln(2x) + 5y3
(b) f(x, y) = (x + a)8y12
(c) f(x, y) = exp7y2 +9
(d) f(x, y) = 13x+ 6xy3
(b) Which of the following are exact differentials?
(a) 5x2dx+ 6ydy
(b) 5 ln(y)dx+ 5x0dy
(c) (sin(x)− y cos(x))dx + sin(−x)dy(d) (exp2x +y)dx+ (x + exp2y)dy
(e) y2dy + x2dx
(a) (a) (∂f∂x
)y = 1x
, (∂f∂y
)x = 15y2
(b) (∂f∂x
)y = 8y12 (x+ a)7 , (∂f
∂y)x = (x+a)8
2√y
(c) (∂f∂x
)y = 0 , (∂f∂y
)x = 14ye7y2
(d) (∂f∂x
)y = 13 + 6y3 , (∂f∂y
)x = 18xy2
(b) Exact Differentials
Let f(x, y)= u(x, y)dx+v(x, y)dy, where f(x, y) = dg(x, y) if f(x, y) is an exact differential.∂u∂y
∂v∂y
(a) 0 0 exact(b) 5
y0 inexact
(c) −cos(x) −cos(x) exact(d) 1 1 exact(e) 0 0 exact
55
Chapter 5Entropy and the Boltzmann Law
1. Calculating the entropy of dipoles in a field.
You have a solution of dipolar molecules with a positive charge at the head and a negativecharge at the tail. When there is no electric field applied to the solution, the dipolespoint north (n), east (e), west (w), or south (s) with equal probabilities. The probabilitydistribution is shown in part (a) of the figure below. However when you apply a field tothe solution, you now observe a different distribution, with more heads pointing north, asshown in part (b) of the figure below.
(a) What is the polarity of the applied field? (In which direction does the field have itsmost positive pole?)
(b) Calculate the entropy of the system in the absence of the field.
(c) Calculate the entropy of the system in the applied field.
(d) Does the system become more ordered or disordered when the field is applied?
n e w s n e w s
(a) Electric Field Absent (b) Electric Field Present
PP
14
14
14
14
14
14
116
716
56
(a) Since the field causes more “+”-charged heads to point north and fewer south, then itmeans that the field is negative at the north pole and positive at the south pole.
(b) The entropy is
S/k = −4∑
i=1
p(i) ln p(i)
= −pN ln pN − pS − ln pS − pE − pW ln pW
= −[4(
1
4ln
1
4
)= ln 4 = 1.386
S = 1.386k
(c) In the presence of the applied field,
S
k= −
[(7
16
)ln(
7
16
)+
1
4ln(
1
4
)+
1
4ln(
1
4
)+
1
16ln(
1
16
)]
= 1.228 so S = 1.228k
(d) The entropy is reduced by the applied field, so the system is more ordered.
2. Comparing the entropy of peaked and flat distributions.
Compute the entropies for the spatial concentration shown in figures (a) and (b) below.
P
(a) Peaked Distribution (b) Flat Distribution
0 0 0 0
1
P
x x
15
15
15
15
15
(a) S/k = −∑ p(i) ln p(i) = −1 ln 1 = 0
(b) S/k = −∑ p(i) ln p(i) = −5(0.2 ln 0.2) = ln 5 = 1.609
Thus the entropy is greater for distribution (b). Concentration gradients tend to become
57
smoother and more uniform over time as a result of trying to maximize entropy.
3. Comparing the entropy of two peaked distributions.
Which of the two distributions shown in the figure below has the greater entropy?
P
� � � ��������� ��
� � � ���������� ��
����� ��������� ������! #" �%$ ��&'��( � )*� �����+�� ,�-�.�! #" �%$ �/&0��(
P
12
13
12
12
13
12
12
12
Both have the same entropy. For either one S/k = −∑4i=1 p(i) ln p(i)
= −[1
6ln
1
6+
1
6ln
1
6+
1
6ln
1
6+
1
2ln
1
2
]
=1
2ln 6 +
1
2ln 2 =
1
2ln 12 = 1.242
4. Calculating the entropy of mixing.
Consider a lattice with N sites and n green particles. Consider another lattice, adjacent tothe first, with M sites and m red particles. Assume that the green and red particles cannotswitch lattices. This is state A.
(a) What is the total number of configurations WA of the system in state A?
(b) Now assume that all N +M sites are available to all the green and red particles. Theparticles remain distinguishable by their color. This is state B. Now what is the totalnumber of configurations WB of the system?
58
Now take N = M and n = m for the following two problems.
(c) Using Stirling’s approximation, what is the ratio WA/WB?
(d) Which state, A or B, has the greatest entropy? Calculate the entropy difference givenby
∆S = SA − SB = k ln(WA
WB
).
(a) Since the two lattices are independent: Wa = WNWM = N !n!(N−n)!
M !m!(M−m)!
(b) Wb =(N +M)!
n!m!(N +M − n−m)!
(c)Wa
Wb=
(N !)2(2N − 2n)!
(2N)![(N − n!)]2≈ N2N(2N − 2n)2N − 2n
(2N)2N(N − n)2N − 2n=
22N−2n
22N= 2−2n
(d) ∆S = k ln(Wa
Wb
)= k ln 2−2n = −nk ln 4
There is greater entropy in B, when all the particles are unconstrained and have accessto the full volume.
5. Proof of maximum entropy for two outcomes.
Example 6.4 is simple enough that you can prove the answer is correct even without usingthe maximum entropy principle. Show this.
Let pH be the probability of the outcome tails, and pT the probability of heads. Because theyare probabilities, we require pH + pT = 1. Let εT = 1, εH = 2. Then 〈ε〉 = pT + 2pH . Wetherefore have a system of two equations
pT + 2pH = 〈ε〉 pT + pH = 1
with two unknowns, pT and pH . Subtracting the two equations gives us pH = 〈ε〉 − 1 andpT = 1− pH = 1− 〈ε〉+ 1 = 2− 〈ε〉.
For 〈ε〉 = 1.5, pT = pH = 0.5, but for 〈ε〉 = 1.2, pT = 0.8, pH = 0.2.
59
6. Other definitions of entropy don’t satisfy the product rule.
In contrast to problem 6, show that if the entropy were defined by a least-squares criterion,as S/k =
∑ti=1 p
2i , the multiplication rule would not be satisfied when S is maximal.
The table used for Problem 6 applies, but now the entropy would be
S
k= q2 + (u1 − q)2 + (v1 − q)2 + (v2 − u1 + q)2
and
(∂(S/k)
∂q
)= 2 [q − (u1 − q)− (v1 − q) + (v2 − u1 + q)] = 0
=⇒ 4q = 2u1 + v1 − v2
which does not give q = u1v1, as we expected.
7. The maximum entropy distribution is Gaussian when the second moment isgiven.
Prove that the probability distribution pi that maximizes the entropy for die rolls subjectto a constant value of the second moment 〈i2〉 is a Gaussian function. Use εi = i.
Maximize the entropy subject to two constraints,
t∑
i=1
pi = 1
t∑
i=1
i2pi = 〈i2〉 = α.
Now Equation (6.11) becomes
(a) − 1− ln p∗i − λ− βi2 = 0 =⇒ p∗i = e−1−λβi2 .
60
Dividing both sides of Equation (a) by
t∑
i=1
e−1−λ−βi2
gives
p∗i =e−βi
2
∑ti=1 e
−βi2
which is a gaussian function of i.
8. Maximum entropy for three-sided die.
You have a three-sided die, with numbers 1, 2 and 3 on the sides. For a series of N dicerolls, you observe an average score α per roll using the maximum entropy principle.
(a) Write expressions that show how to compute the relative probabilities of occurrenceof the three sides, n∗1/N , n∗2/N , and n∗3/N , if α is given.
(b) Compute n∗1/N , n∗2/N , and n∗3/N if α = 2.
(c) Compute n∗1/N , n∗2/N , and n∗3/N , if α = 1.5.
(d) Compute n∗1/N , n∗2/N , and n∗3/N if α = 2.5.
(a)n∗iN
=xi
qq = x + x2 + x3 α =
x+ 2x2 + 3x3
x + x2 + x3=
1 + 2x+ 3x2
1 + x + x2.
=⇒ (3− α)x2 + (2− α)x + (1− α) = 0. Use quadratic equation
x =(α− 2)± [(2− α)2 − 4(1− α)(3− α)]
1/2
2(3− α).
(b) α = 2 =⇒ n∗1N
=n∗2N
=n∗3N
=1
3.
(c) α = 1.5 =⇒ n∗1N
= 0.62n∗2N
= 0.27n∗3N
= 0.11.
(d) α = 2.5 =⇒ n∗1N
= 0.11n∗2N
= 0.27n∗3N
= 0.62.
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9. Maximum entropy in Las Vegas.
You play a slot machine in Las Vegas. For every $1 coin you insert, there are three outcomes:(1) you lose $1, (2) you win $1, so your profit is $0, (3) you win $5, so your profit is $4.Suppose you find that your average expected profit over many trials is $0 (what an optimist!).Find the maximum entropy distribution for the probabilities p1, p2, and p3 of observing eachof these three outcomes.
We have three possible payoffs, f1 = −1 (we don’t win anything and so have lost $1), f2 = 0(we win $1 for a net profit of zero), and f3 = 4 (we win $5 for a net profit of $4). If each ofthese possible outcomes has associated with it a probability pi, then the expected net profitwill be given by:
〈f〉 = f1p1 + f2P2 + f3p3.
Let x = e−β. Then pi = xfi/q where q = xf1 + xf2 + xf3 = x−1 + x0 + x4 and
〈f〉 =−1x−1 + 0x0 + 4x4
x−1 + x0 + x4= 0
=⇒ −1
x+ 4x4 = 0 =⇒ 4x5 = 1 =⇒ x = 0.758.
So you have
pi =e−βfi
q=xfi
q
q =(
1
0.758
)+ 1 + (0.758)4
q = 2.65.
The probabilities of the outcomes are therefore:
p1 =x−1
q= 0.498
p2 =x0
q= 0.377
62
p3 =x4
q= 0.125
10. Flat distribution, high entropy.
For four coin flips, for each distribution of probabilities, (pH , pT ) =(0, 1), (1/4, 3/4), (1/2, 1/2), (3/4, 1/4), (1, 0), compute W , and show that the flattestdistribution has the highest multiplicity.
(a) For (pH , pT ) = (0, 1) or (1, 0), W = 4!0!4!
= 1.
(b) For (pH , pT ) = (1/4, 3/4) or (3/4, 1/4), W = 4!1!3!
= 4.
(c) For (pH , pT ) = (1/2, 1/2), W = 4!2!2!
= 6.
Therefore the flattest distribution of probabilities, (c), corresponds to the highest multiplicity.
63
Chapter 6Thermodynamic Driving Forces
1. One dimensional lattice.
You have a 1-dimensional lattice that contains NA particles of type A and NB particles oftype B. They completely fill the lattice, so the number of sites is NA +NB.
(a) Express the entropy S(NA, NB) as a function of NA and NB.
(b) Give the relationship between the chemical potential µA and the quantity(∂S∂NA
)NB
.
(c) Express µA(NA, NB).
(a) W (NA, NB) =(NA +NB)!
NA!NB!,
S = k lnW = k [(NA +NB) ln(NA +NB)−NA lnNA −NB lnNB] .
(b) dS =1
TdU +
P
TdV − µA
TdNA −
µBTdNB,
µAT
= −(∂S
∂NA
)
NB ,U,V
.
(c)
(∂S
∂NA
)
NB
= k [ln(NA +NB) + 1− lnNA − 1] = −k lnNA
NA +NB,
µA = kT lnNA
NA +NB= kT lnC,
where C is the mole fraction concentration of A’s in B’s.
64
2. The entropy of an ideal gas.
Show that the entropy for an ideal gas is
S(V,N) = Nk lnV.
Using Equation (6.6) we have
(∂S
∂V
)
U,N
=p
T=Nk
V.
Rearranging and integrating gives dS = Nk∫dV
V
=⇒ ∆S = S2 − S1 = Nk ln(V2
V1
).
3. Entropy changes don’t depend on a process pathway.
For an ideal gas, the entropy is S = Nk lnV (see above).
(a) Express ∆SV = S2(V2)− S1(V1), the entropy change upon changing the volume fromV1 to V2, at fixed particle number N .
(b) Express ∆SN = S2(N2) − S1(N1), the entropy change upon changing the particlenumber from N1 to N2, at fixed volume V .
(c) Write an expression for the entropy change, ∆S, for a two-step process (V1, N1) →(V2, N1) → (V2, N2) in which the volume changes first at fixed particle number, thenthe particle number changes at fixed volume.
(d) Show that the entropy change ∆S above is exactly the same as for the two-step processin reverse order: changing the particle number first, then the volume.
(a) ∆S = Nk ln(V2
V1
).
(b) ∆S = (N2 −N1)k lnV .
65
(c)
∆S = N1k ln(V2
V1
)+ (N2 −N1)k lnV2
= k[N1 lnV2 −N1 lnV1 +N2 lnV2 −N1 lnV2]
= k[N2 lnV2 −N1 lnV1].
(d)
∆S = (N2 −N1)k lnV1 +N2k ln(V2
V1
)
= k[N2 lnV2 −N1 lnV1].
4. Compute ∆S(V ) for an ideal gas.
What is the entropy change if you double the volume from V to 2V in a quasi-staticisothermal process at temperature T ?
∆S = S2 − S1 = Nk ln(
2V
V
)= Nk ln 2.
66
Chapter 7The Logic of Thermodynamics
1. The work of compression.
One mole of a van der Waals gas is compressed quasi-statically and isothermally from volumeV1 to V2. For a van der Waals gas, the pressure is
p =RT
V − b −a
V 2,
where a and b are material constants, V is the volume and RT is the gas constant ×temperature.
(a) Write the expression for the work done.
(b) Is more or less work required than in the low-density limit than for an ideal gas? Whatabout the high-density limit? Why?
(a) w = − ∫ V2V1pextdV In a quasi-static process
pext = pgas =RT
V − b −a
V 2
w =∫ V2
V1
[(−RTV − b
)+(a
V 2
)]dV = −RT ln
(V2 − bV1 − b
)− a
(1
V2− 1
V1
)
(b) For large volumes, the logarithm term is small and for
V2 < V1, −a(
1
V2
− 1
V1
)< 0.
Therefore less work is required to compress a real gas. The van der Waals gas hasinternal attractive energy which aids compression. On the other hand, when the
67
volumes are small (high densities), then the logarithm term is dominant and the vander Waals gas has stronger repulsion; in that case more work is required to compress itthan an ideal gas.
2. Deriving the ideal gas law in two dimensions.
Molecules at low density on a surface, such as surfactants at an interface of air and water, of-ten obey a two-dimensional equivalent of the ideal gas law. The two-dimensional equivalentof p is π, where π is a lateral two-dimensional pressure. A is area. Using
π = T
(∂S
∂A
)
N
(7)
and assuming no energetic interactions, derive the two-dimensional equivalent of the idealgas law by using a lattice model in the low-density limit.
As in the three-dimensional problem, we consider the N molecules randomly distributed on alattice of M sites. In this case, we use a two-dimensional lattice. There are M possibleplacements of the first particle, M − 1 possible placements for the second particle, M − 2 forthe third, and so on. Therefore the number of arrangements, W , is
W =(M !)
(M −N)!N !
We use the Boltzmann equation
S = k lnW = k ln
[M !
(M −N)!N !
]
and Stirling’s approximation (x! ≈ (xe)x) to get:
S = k ln
[MM
(M −N)M−NNN
]= −k
[N ln
(N
M
)+ (M −N) ln
(M −NM
)]
Using the thermodynamic expression given in the problem,
Π = T
[∂S
∂A
]
U,N
; also
[∂S
∂A
]
U,N
=
(∂S
∂M
)(∂M
∂A
)
68
so we calculate each term
(∂S
∂M
)= k
[N
M+M −NM
− 1 + lnM − ln(M −N)]
= k ln(
M
M −N)
= −k ln(1− N
M
)
(∂M
∂A
)=
M
A;
At low densities,
N �M, soN
M� 1,
and therefore
Π = −kT(M
A
)ln(1− N
M
)≈ NkT
A,
since
ln(1− N
M
)≈ −N
M+ · · ·
This is the two-dimensional ideal gas law.
3. The work of expansion in freezing an ice cube.
At 1 atm, freeze an amount of liquid water that is 2 cm×2 cm×2 cm in volume. The density(mass per unit volume) of liquid water at 0◦ C is 1.000 g cm−3 and the density of ice at 0◦Cis 0.915 g cm−3.
(a) What is the work of expansion upon freezing?
(b) Is work done on the system or by the system?
69
At constant external pressure, pext = 1 atm, the work is given by Equation (7.37).
w = −pext(Vice − Vliquid water)
= −pext
(mice
ρice− mliquid
ρliquid
)
= − (1 atm) (8 g)
(1
0.915 g cm−3− 1
1 g cm−3
)(2.422× 10−2 cal
cm3atm
)
= −0.018 cal
Since the sign is negative, this work is done by the system expanding against the atmosphere.
4. The work of expanding a gas.
Compute the total work performed when expanding an ideal gas, at constant temperature,from volume V to 2V .
The work is given by Equation (7.42)
w = −NkT ln(
2V
V
)= −NkT ln 2.
5. Pulling out the partition between two chambers of a gas.
A partition separates equal volumes containing equal numbers N of ideal gas molecules ofthe same species at the same temperature. Using a simple lattice model for ideal gases,evaluate the relevant multiplicity terms to show that the entropy of the composite systemdoes not change when the partition is removed (hint: use Stirling’s approximation).
Let 1 refer to the left side, and 2 refer to the right side. Before removing the partition,composite W = W1W2. Since each side is identical, W (before) = W 2
1 .W1 = M !/(N !(M −N)!) where M is the number of lattice sites and N is the number ofparticles. Using Stirling’s approximation, we can write
W1 =
(Me
)M
(Ne
)N (M−Ne
)M−N =MM
NN (M −N)M−Nso
70
W (before) = W 21 =
M2M
N2N (M −N)2M−2N.
After removing the partition, there are 2M lattice sites and 2N particles. This leads to
W (after) =(2M)!
(2N)!(2M − 2N)!=
(2Me
)2M
(2Ne
)2N (2M−2N
e
)2M−2N
=M2M
N2N (M −N)2M−2N= W 2
1 .
Note that the factors (2/e) all cancel out in the above expression, in the same manner thatthe factor (1/e) cancels out in the equation for W1. Thus W (after) = W (before) and sinceS = k lnW , we have S(after removing partition) = S(before removing partition) and thus
∆S = k lnW (after)
W (before)= k ln 1 = 0.
6. The work in a thermodynamic cycle.
A thermodynamic cycle is a series of steps that ultimately returns to its beginning point.Compute the total work performed around the thermodynamic cycle of quasi-static processesin the figure below.
p (atm)
V (cm3)
100
300
1 2
A
B C
D
Step 1
Step 2
Step 3
Step 4
71
Using Equation (7.36), we have
W = −∫ B
Ap dV −
∫ C
Bp dV −
∫ D
Cp dV −
∫ A
Dp dV
= 0− (300 atm)(2 cm3 − 1 cm3)− 0− (100 atm)(1 cm3 − 2 cm3)
= −(200 atm cm3)× (2.422× 10−2cal/(cm3 atm)) = −4.84 cal
Since steps 1 and 3 involve no volume change, there is no work performed during those steps.
72
7. Engine efficiencies.
Consider a Carnot engine that runs at Th = 380 K.
(a) Compute the efficiency if Tc = 0◦C = 273 K.
(b) Compute the efficiency if Tc = 50◦C = 323 K.
Equation (7.45) gives
η = 1− TcTh.
(a) η = 1− 273
380= 28% efficient
(b) η = 1− 323
380= 15% efficient
Big difference!
8. Hadley Cycles - what powers the trade winds?
The Earth’s trade winds arise from the differences in buoyancy between hot air and coldair. Air is heated at the Earth’s surface near the equator (see (1) in figure below), loweringit’s density; it rises (2), pushing upper air away toward the Northern latitudes where theair cools (3), then drops back down to Earth (4), pushing the cold surface air back towardthe equator.
Consider an imaginary balloon containing 1 m3 of an ideal gas.
73
(a) At p = 1 atm and T = 300 K, what is the number of moles n of air contained in theballoon?
(b) If that balloon of n moles of air remains at p = 1 atm but is now heated to T = 330K, its volume increases. What is the new density, ρ = n/V ?
(c) Assuming no heat transfer, how high will the balloon in part (b) rise? Use Figure 10.2in the book to make a rough guess. (Useful conversion: 1 atm ≈ 1 bar)
(a) Use the ideal gas law: p = nRTV
so n = pVRT
=(1atm)(1m3)( 1liter
10−3m3 )
(8.21·10−2 liter·atmK·mol )(300K)
n = 40.6moles
(b) Using the ideal gas law in the form: p = ρRT , we haveρ1T1 = p
R= ρ2T2 =⇒ ρ2
ρ1= T2
T2
=⇒ ρ2 = (300330
)ρ1 = (300330
)(40.6molesm3 ) = 36.9moles
m3
(c) p ≈ 300330≈ 0.9atm is at an altitude of 1 mile.
9. Stirling Engine.
A Stirling engine has the following pV cycle and uses an ideal gas working fluid. The stepsare quasi-static.
74
(a) How much work is done in the constant volume segments, W12 and W34?
(b) What is the energy change around the cycle, ∆Utot = ∆U12 + ∆U23 + ∆U34 + ∆U41?
(c) What is the total work performed around the cycle, Wtot = W12 +W23 +W34 +W41?
(d) If T2 = 2000K, T1 = 300K, V2 = 1L, V1 = 0.01L and n = 0.1 mole of ideal gas,compute Wtot.
(a) Constant volume =⇒ w = 0
(b) U is a state function, so ∆Utot = 0
(c) w12 = 0 = w34
w23 = − ∫ V2V1pdV = −NkT2
∫ V2V1
1VdV = −NkT2ln(V2
V1) for an ideal gas
w41 = − ∫ V1V2pdV = −NkT1
∫ V1V2
1VdV = −NkT1ln(V1
V2)
wtot = −NkT2ln(V2
V1)−NkT1ln(V1
V2)
(d) T2 = 2000K, T1 = 300K, V2 = 1L, V1 = 0.01L and n = 0.1 mole of ideal gas
wtot = −(0.1mol)( 8.314JK·mol )(2000K)ln( 1L
0.01L)− (0.1mol)( 8.314J
K·mol )(300K)ln(0.01L1L
)≈ −7657.5 + 1148.6 = −6508.9J = −1555.7cal
10. Ideal efficiency of a car engine.
75
Suppose the compression ratio in your car engine is V2/V1 = 8. For a diatomic gas CV =(5/2)Nk and for ethane CV ≈ 5Nk.
(a) What is the efficiency of your engine if you burn a diatomic gas?
(b) Which is more efficient: a diatomic gas or ethane?
(c) Would your engine be more or less efficient with a higher compression ratio?
(a) Use Equation 8.53,
η = 1−(V1
V2
)(Nk)/CV
= 1−(
1
8
)0.4∼= 0.56.
(b) For ethane,
η = 1−(
1
8
)0.2
= 0.34,
so it is less efficient than a diatomic gas.
(c) Higher compression ratios lead to higher efficiency.
76
Chapter 8Laboratory Conditions and FreeEnergies
1. Finding a fundamental equation.
While the Gibbs free energy G is the fundamental function of the natural variables (T, p,N),growing biological cells often regulate not the numbers of molecules N, but the chemicalpotentials µi. That is, they control concentrations. What is the fundamental function Z ofnatural variables (T, p, µ)?
Start with G = G(T, p,N), and its differential form:
dG = −SdT + V dp+ µdN − d(µN) = −Ndµ− µdN
so
d(G− µN) = −SdT + V dp−Ndµ
and the fundamental function is Z = G− µN .
2. Why does increasing temperature increase disorder?
Systems become disordered as the temperature is increased. For example, liquids and solidsbecome gases, solutions mix, adsorbates desorb. Why?
Increasing temperature increases disorder, because the entropy dominates the free energy at
77
high temperatures, whereas enthalpy dominates at low temperatures. That is, in theexpression
∆G = ∆H − T∆S,
the entropy term (T∆S) becomes more important (either for ∆S positive or ∆S negative) asT increases, hence at high temperatures the system will become more disordered to increaseits entropy, even at the cost of increasing enthalpy.
3. The difference between the energy and enthalpy changes in expanding anideal gas.
How much heat is required to cause the quasi-static isothermal expansion of one mole of anideal gas at T = 500 K from PA = 0.42 atm, VA = 100 liters to PB = 0.15 atm?
(a) What is VB?
(b) What is ∆U for this process?
(c) What is ∆H for this process?
(a) For an ideal gas, in any isothermal process ∆U = 0. Therefore
∆U = q + w = 0 =⇒ q = −w.
For such a process, when it is also quasi-static:
w = −∫p dV = −
∫ V2
V1
nRT
VdV = −nRT ln
V2
V1
so
q = nRT ln(V2
V1
)
= (1 mole)(2cal
mol− deg)(500◦ ln
(V2
100l
)
Using the ideal gas law:
p1V1 = nRT = p2V2 =⇒ V2
V1=p1
p2
78
Therefore, we can write
q = nRT lnp1
p2
= (1000 cal) ln(
0.42 atm
0.15 atm
)
= 1030 cal ≈ 1000 cal.
(b) From the ideal gas law:
V2 =p1
p2V1 =
(0.42 atm
0.15 atm
)(100 L)
= 280 liters
(c) For any isothermal process of an ideal gas
∆U = 0 [∆U = Cv(T2 − T1)]
(d) For any isothermal process of an ideal gas
∆H = 0
This can be shown as follows. ∆H = ∆U + ∆(pV ). But ∆U = 0; ∆(pV ) = p2V2− p1V1.Using the ideal gas law
P2V2 = nRT
P1V1 = nRT
Therefore
∆H = 0 + (nRT − nRT )
= 0.
79
4. The work and the heat of boiling water.
For the reversible boiling of five moles of liquid water to steam at 100◦C and 1 atm pressure,calculate q. Is w positive or negative?
From tables that give the heat of vaporization of water,
qboiling = −qcondensation = −∆Hcondensation
=
(540
cal
g
)(18
g
mole
)(5moles)
= 48, 600cal
qboil = 48.6kcal
At constant pressure,
w = −∫ V2
V1
p dV
= −p(V2 − V1) = −p(Vgas − Vliquid)
so
Vgas � Vliquid =⇒ Vgas − Vliquid
is positive, so
w < 0.
80
5. The entropy and free energy of gas expansion.
Two moles of an ideal gas undergo an irreversible isothermal expansion from VA = 100 litersto VB = 300 liters at T = 300 K.
(a) What is the entropy change for this process?
(b) What is the Gibbs free energy change?
(a) ∆S =∫ dqrev
T.
Since S and U are state functions, we needn’t be concerned that the actual process isirreversible (read: complicated). We can choose any path to calculate state functions sowe invent an isothermal, quasi-static process to get qrev.
∆U = 0 =⇒ qrev = −wrev = nRT lnV2
V1
.
For an isothermal process,
∆S =qrevT
= nR lnV2
V1
= (2moles)
(2
cal
mol deg
)ln
300
100= 4.39
cal
deg
(b) ∆G = ∆H − T∆S. For an isothermal process on an ideal gas, ∆H = 0.
(Note that ∆H = ∆U + ∆(PV ) : For an isothermal process, ∆U = 0
and ∆(PV ) = ∆(nRT ) = 0, therefore ∆H = 0).
∆G = −T∆S = −(300K)
(4.39
cal
deg
)= −1318.33 cal
81
6. The free energy and entropy of membrane melting.
Pure membranes of dipalmitoyl lecithin phospholipids are models of biological mem-branes. They melt at Tm = 41◦ C. Reversible melting experiments indicate that ∆Hm =9 kcal mol−1. Calculate
(a) the entropy of melting ∆Sm, and
(b) the free energy of melting ∆Gm.
(c) Does the membrane become more or less ordered upon melting?
(d) There are 32 rotatable CH2–CH2 bonds in each molecule. What is the increase inmultiplicity on melting one mole of bonds?
(a) For a phase transition,
∆Sm =∆Hm
Tm
=9 kcal/mol
(273 + 41)K
= 28.66cal
mol K
(b) For reversible phase transitions, the two phases are in equilibrium, therefore
∆Gm = 0
(c) ∆S > 0, therefore the entropy increases, thus the disorder also increases.
(d)∆Sm cal
mol bonds ·K =∆Smcal
mol lipids ·K
(1 mol lipids
nB mol bonds
)
=
(NAbonds
mol bonds
)(k J
K
)(cal
4.184 J
)ln(Wfluid
Wsolid
)
82
so
Wfluid
Wsolid= exp
[∆SmNknB
]
= exp
[(28.66 cal
mol lipids ·K
)
(1 mol lipids
nB mol bonds
)(1mol bonds)
(K
1.381 · 10−23 J
)(4.184 J
cal
)]
=⇒ log(Wfluid
Wsolid
)≈ 2.713 · 1023
7. State and path dependent functions.
Which quantities sum to zero around a thermodynamic cycle?
(a) q, heat
(b) w, work
(c) −pdV
(d) U
(e) G
State functions sum to zero around a thermodynamic cycle. That includes (d) U and (e) G,but not heat and work (a), (b) or (c). (Remember, −pdV = δw.)
83
8. Computing enthalpy and entropy with a temperature dependent heatcapacity.
The heat capacity for liquid n-butane depends on temperature:
Cp(T ) = a + bT,
where a = 100 J K−1 mol−1 and b = 0.1067 J K−2 mol−1, from its freezing temperatureTf ≈ 140 K to Tb ≈ 270 K, its boiling temperature.
(a) Compute ∆H for heating liquid butane from TA = 170 K to TB = 270 K.
(b) Compute ∆S for the same process.
(a) Compute ∆H for heating liquid butane from T1 = 170K to T2 = 270K.
∆H =∫ T2
T1
Cp dT =∫ 270
170a dT +
∫ 270
170bT dT
= a∆T +bT 2
2
∣∣∣∣∣
T2
T1
= a(270K − 170K) +b
2
[(270K)2 − (170K)2
]
= 10.0kJ
mol+ 2.35
kJ
mol
= 12.35kJ
mol
(b) Compute ∆S for the same process.
∆S =∫ T2
T1
CpTdT =
∫ T2
T1
(a
T+ b
)dT
= a ln(T2
T1
)+ b∆T
=(100
J
K mol
)ln(
270K
170K
)+(0.1067
J
K2 mol
)(100K)
= 56.93J
K mol
84
9. Cycle for determining the enthalpy of vaporization of water at 473K.
Suppose that you want to know how much heat it would take to boil water at 473 K, ratherthan 373 K. At T = 373 K, ∆Hboiling(100◦C) = 40.7 kJmol−1 is the enthalpy of vaporization.Assuming that the heat capacities of the liquid (Cp,liquid = 75 J K−1mol−1) and the vapor(Cp,vapor = 3.5 J K−1mol−1) are constant over this temperature range, calculate the enthalpyof vaporization at 473 K using the thermodynamic cycle shown in the figure below.
∆H 465'7 8 7 9�:�;=<�>%<�?�@
∆H 4A507 8 7 9�:�; B�>'<�?C@
∆H 8 7 D'E�7GF ∆H HJILKNM�O
P�QLR�STQLUVXWCV.Y
Z6[�\�]+^_ WCV.Y
P�QLR�STQLU_ WCV.Y
Z6[�\�]+^VXWCV.Y
A thermodynamic cycle for calculating the enthalpy of boiling water at a temperature higherthan the boiling point.
∆Hliquid + ∆H373K = ∆H473K + ∆Hsteam
∆H373K = 40.7 kJ/mol
∆Hliquid =∫ 473
373CpliquiddT = Cp,liquid(T2 − T1)
= (75 J/K mol)(100 K) = 7.5 kJ/mol
∆Hsteam =∫ 473
373Cp,steamdT = (3.5 J/K mol)(100 K) = .35 kJ/mol
∆H473K = (40.7 kJ/mol) + (.35 kJ/mol)− (7.5 kJ/mol) = 33.55 kJ/mol.
85
10. Heating a house.
If your living room, having a volume of 6 m× 6 m× 3 m ≈ 100 m3, were perfectly insulated,how much energy would be needed to raise the temperature inside the room from Tinitial = 0◦
C to Tfinal = 25◦ C? Note that CV = Cp − nR for an ideal gas.
Assuming constant heat capacity over the temperature range, the energy required to heat theroom is given by
(8.a) ∆U = CV (Tfinal − Tinitial).
To get the heat capacity of the air in the room, assume a mixture of O2 and N2. For an idealgas, CV = Cp −Nk and is constant over the given temperature range for both gasses. Table8-2 gives Cp = 7 cal/mol deg for either gas, so CV = 7 cal/mol deg - 2 cal/mol deg =5 cal/mol deg. If we assume the room was sealed at T = 300 K at 1 atm, the number ofmoles of gas in the room can be estimated from the ideal gas law,
n =PV
RT=
(1 atm)(100 m3)(8.2× 10−2 l atm
K mol
) (1000 cm3
1 l
) (1 m
100 cm
)3(300 K)
= 4067 moles.
So Equation (8.a) gives
∆U =
(5 cal
mol K
)(4065 mol)(25 K) = 508 kcal
11. Objects in thermal contact.
Suppose two objects A and B, with heat capacities CA and CB and initial temperatures TAand TB, are brought into thermal contact. If CA � CB, is the equilibrium temperature Tcloser to TA or to TB?
Use Equation 8.43:
T =CATA + CBTBCA + CB
.
86
Divide the numerator and denominator by CB to get
T =
(CACb
)TA + TB
CACB
+ 1−→ TA if
CACB� 1.
12. ∆S for an adiabatic expansion of a gas.
In an adiabatic quasi-static expansion of an ideal gas, how do you reconcile the followingtwo facts: (1) the increase in volume should lead to an increase in entropy, but (2) in anadiabatic process, δq = 0 so there should be no change in entropy (since dS = δq/T = 0)?
Adiabatic expansion of a gas involves both an increase in volume, which increases theentropy, and a decrease in temperature, which decreases the entropy. To show this morequantitatively, consider the total entropy change as a volume increase at constant T and thena temperature change at constant V :
S3 − S1 =∫dS =
∫δq
T= 0 =
∫ [(∂S
∂V
)
T
dV +
(∂S
∂T
)
V
dT
]
S
V
1
2
3
change Tchange V
The two terms in the integral are
∫ 2
1
(∂S
∂V
)
T
dV =∫ (
Nk
V
)dV = Nk ln
(V2
V1
)= Nk ln
(V3
V1
)since V3 = V2.
∫ 3
2
(∂S
∂T
)
V
dT =∫ (
CvdT
T
)= Cv ln
(T3
T2
)= Cv ln
(T3
T1
)since T1 = T2.
But according to Equation (8.48), these terms must exactly cancel in a quasi-static adiabaticexpansion.
87
13. A thermodynamic cycle for mutations in protein folding.
Suppose that you can measure the stability of a wild-type protein, ∆G1 = Gfolded−Gunfolded,the free energy difference between folded and unfolded states. A mutant of that protein hasa single amino acid replacement. Design a thermodynamic cycle that will help you find thefree energy difference ∆G2 = Gunfolded, mutant − Gunfolded, wildtype, the effect of the mutationon the unfolded state.
D′
∆G1
∆G2∆G2
D N
N′
′
′ ∆G1
From this thermodynamic cycle, with the arrows in the directions indicated,
∆G1 + ∆G′2 = ∆G′1 + ∆G2
so
∆G2 = ∆G1 −∆G′1 + ∆G′2.
You can measure ∆G1 and ∆G′1. If you can determine or estimate ∆G′2, then you can use theabove equation to get ∆G2. Such estimates are sometimes made by computer simulations.
88
14. Free energy of an ideal gas.
(a) For an ideal gas, calculate F (V ), the free energy versus volume, at constant temper-ature.
(b) Compute G(V ).
(a) dF = −SdT − pdV +t∑
i=1
µidNi = −pdV
for constant T and Ni’s. So
F (V ) = −∫pdV = −
∫NkT
VdV = −NkT ln(V )
(b) G = H − TS = (U + pV )− TS = (U − TS) + pV = F + pV
So
G(V ) = F (V ) + pV = −NkT ln(V ) +NkT = −NkT (ln(V )− 1)
15. Heat capacity of an ideal gas.
The energy of an ideal gas does not depend on volume,(∂U
∂V
)= 0.
Use this fact to prove that the heat capacities Cp = (∂H/∂T )p and Cv = (∂U/∂T )V for anideal gas are both independent of volume.
(a) Use
∂2U
∂V ∂T=
∂2U
∂T∂V
89
and the definition CV = (∂U/∂T )V to get
∂CV∂V
=∂
∂T
(∂U
∂V
)= 0,
and (∂U/∂V ) = 0, which proves that Cv does not depend on volume V .
(b) Cp =∂H
∂T=∂(U + pV )
∂T=∂U
∂T+∂(pV )
∂T=∂U
∂T+
∂
∂T(nRT ) =
∂U
∂T+ nR.
We proved in (a) that ∂U/∂T = CV is independent of V , and so is nR, so the lastequatino shows that Cp is also independent of V .
16. Computing Entropies.
The heat capacity of an ideal gas is Cv = 32Nk for N molecules. Assume T = 300K where
needed.
(a) One mole of O2 gas fills a room of 500m3. What is the entropy change, ∆S, forsqueezing the gas into 1cm3 in the corner of the room?
(b) An adult takes in about 3000 kcal/day from food (1 food cal = 1 kcal). What is ∆Sfor this process?
(c) One mole of O2 gas is in a room of 500m3. What is the entropy change, ∆S, forheating the room from T = 270K to 330K?
(d) The free energy of a conformational motion of a loop in a protein is ∆G = 2 kcal/mol.The enthalpy change is ∆H = 0.5 kcal/mol. Compute ∆S.
(a) ∆S = nRln(V2
V1)
n = 1mole, R = 2cal/(mol ·K), V1
V2= 500m3
1cm3 (100cmm
)3
V2
V1= 2.0 · 10−9
=⇒ ∆S = (2 calmol·K )ln(2 · 10−9) = −40 cal
mol·K(At T = 300K, this is equivalent to a free energy of ∆F = −T∆S ≈ 12 kcal
mol)
(b) ∆S = ∂qT
= 3000kcal300K
= 10 kcalmol·K
(c) ∆S =∫ T2T1
CvTdT = Cvln(T2
T1) = 3
2kln(330
270) = 0.6 cal
mol·K
(d) ∆G = ∆H − T∆S=⇒ ∆S = ∆H−∆G
T= 500−2000
300= −5 cal
mol·K
90
17. What powers a hurricane?
A hurricane is powered by a Hadley cycle (A) of rising hot air and falling cold air, coupledto the earth’s Coriolis forces (B) that cause the wind circulation.
The reason a hurricane is so strong is that its Hadley cycle is augmented by an additionalingredient, the vaporization of warm water. You can saturate 1 kg of air with 20 grams ofwater vapor. (You can consider air as consisting of N2 gas.)
(a) If air’s density is 1.25 kgm3 , how many grams of water vapor will occupy 1 m3 of air?
(b) If the enthalpy of vaporization of water is 2.3 · 106 Joules/(kg of water), how muchheat, q, is given off by the water when its vapor condenses into rain, per m3 of air?
(c) If that heat q, that is given off by water’s condensation, all goes into raising thetemperature of the surrounding air (at p=1atm), what would be the temperatureincrease, ∆T ?
(a) (20gwaterkg air
)(1.25kg airm3 air
) = 25gwaterm3 air
20gwaterkg air
(b) ( 2.3·106Jkgwater
)( 1kg1000g
)(25gwaterm3 air
) = 5.75·104Jm3 air
(c) q = Cp∆T
Cp = (7 calmol
)(40.6moles) for N2 or O2
∆T = (5.75·104J7 calmol
)( 1cal4.2J
)( 140.6moles
)
=⇒ ∆T = 48 degreesThis extra heating drives the density lower, which contributes additional force to theHadley cycle.
18. Lattice model of ligand binding to a protein.
91
You have a ligand L and a protein P . When the ligand binds, the energy change is favorableby 5ε0 (ε0 < 0) because it makes 5 contacts with the protein. As (crudely) drawn below,the protein is modeled as a simple two-dimensional lattice and consists of a rigid portion(squares containing a ’P’) with a flexible loop (circles containing a ’P’) at one end. Theligand (rectangle containing an ’L’) is rigid and occupies two lattice sites.
(a) Counting only lateral contacts (no diagonal connections), find the number of conform-ers, W , of the flexible loop.
(b) What are the energetic and entropic contributions to the free energy change due toligand binding? Write an expression for ∆Fbinding.
(c) For T = 300 K, what is ∆Fbinding = Fbound − Funbound if ε0 = −1 kcal/mol?
(d) What is the dissociation temperature, Tdissoc?
(e) If the flexible loop were made rigid instead (and in the proper configuration for ligandbinding), what would be the new ∆Fbinding?
(f) Does the ligand bind more tightly or more weakly with the rigid loop?
(a) Since we’re only counting lateral contacts, there are 3 possible locations of the firstflexible link, and each of these has 3 possible locations for the second flexible link. So, thenumber of conformers is W = 9.(b) The Helmholtz free energy can be calculated by summing together the energetic term(favorable) and the entropic term (unfavorable, due to the loss of conformer multiplicity):
F = U − TS,
∆U = 5ε0; −T∆S = −T · k ln 9,
∆Fbinding = 5ε0 + kT ln 9.
92
(c) Plugging values into the expression found in part (b):
∆Fbinding = 5 · (−1 kcal/mol) + (0.002 kcal/(mol ·K))(300 K) · ln 9
= −3.7 kcal/mol.
(d) At the dissociation temperature, the free energy change is zero, so:
Tdissoc =|5ε0|k ln 9
= 1138 K.
(e)
∆Fbinding,rigid = −5 kcal/mol.
(f) If the loop is rigid, then the binding will be tighter because it does not incur the entropicpenalty from the loss of the loop’s conformational freedom. So the free energy is morenegative in the rigid case.
93
Chapter 9Maxwell’s Relations and Mixtures
1. How do thermodynamic properties depend on surface area?
The surface tension of water is observed to decrease linearly with temperature (in experi-ments at constant p and a): γ(T ) = b− cT , where T = temperature◦ C, b = 75.6 erg cm−2
(the surface tension at 0◦ C) and c = 0.1670 erg cm−2deg−1.
(a) If γ is defined by dU = TdS − pdV + γda, where da is the area change of a purematerial, give γ in terms of a derivative of the Gibbs free energy at constant T and p.
(b) Using a Maxwell relation, determine the quantitative value of (∂S/∂a)p,T from therelationships above.
(c) Estimate the entropy change ∆S from the results above if the area of the water/airinterface increases by 4 A2 (about the size of a water molecule).
(a) Since G = H − TS = U + pV − TS, we have
dG = dU + pdV + V dp− TdS − SdT= −SdT + V dp+ γda
so
γ =
(∂G
∂a
)
T,p
(b)
(∂S
∂a
)
p,T
= −(∂γ
∂T
)
p,a
= c = +0.167erg
cm2 deg
94
(c) Estimate the entropy change ∆S from the results above if the area of the water/airinterface increases by 4 A2, about the size of a water molecule.
∆S ≈(∂S
∂a
)∆a
=
(+0.167
erg
cm2 deg
) 4A
2
molec
(1 cm
108A
)2
×(
1 cal
4.18× 107 ergs
)(6.02× 1023 molec
mole
)
∆S = +0.96cal
mol deg
2. Water differs from simple liquids?
Figures 9.5 and 9.6 show that the thermal expansion coefficient α = (1/V )(∂V/∂T )p andisothermal compressibility κ = −(1/V )(∂V/∂p)T are both much smaller for water, whichis hydrogen-bonded, than for simpler liquids like benzene, which are not. Give a physicalexplanation for what this implies about molecular packing and entropies in water versussimple liquids.
The observation most directly implies that the volume of water doesn’t change as much asother liquids with temperature and pressure. Using the Maxwell relations, it implies theentropy doesn’t change as much with pressure as for other liquids. Other liquids are morereadily “squeezed” into smaller volume, losing translational entropy, but water isn’t soreadily squeezed because of its hydrogen bond network.
3. The heat capacity of an ideal gas.
For an ideal gas (∂U/∂V )T = 0. Show that this implies the heat capacity CV of an idealgas is independent of volume.
Since U(T, V ) is a state function,
(∂2U
∂T∂V
)=
(∂2U
∂V ∂T
),
95
so we have
∂
∂T
(∂U
∂V
)= 0 =⇒ ∂
∂V
(∂U
∂T
)=∂CV∂V
= 0
which shows that CV is independent of V .
4. Using Maxwell relations.
Show that (∂H/∂p)T = V − T (∂V/∂T )p.
Start with dH = TdS + V dp.
Now divide by dp, holding T constant:
dH
dp[at constant T ] =
(∂H
∂p
)
T
= T
(∂S
∂p
)
T
+ V.
Use the Maxwell relation (Table 9.1),
(∂S
∂p
)
T
= − (∂V/∂T )p
to get the result
(∂H
∂p
)
T
= −T(∂V
∂T
)
p
+ V.
96
5. Pressure dependence of the heat capacity.
(a) Show that, in general, for quasi-static processes
(∂Cp∂p
)
T
= −T(∂2V
∂T 2
)
p
.
(b) Show that (∂Cp/∂p)T = 0 for an ideal gas.
(a) Begin with the definition of the heat capacity,
Cp =δq
dT= T
dS
dT
for a quasi-static process. Take the derivative
(a)
(∂Cp∂p
)
T
= T
(∂2S
∂p∂T
)= T
(∂2S
∂T∂p
)
since S is a state function. Substitute the Maxwell relation
(∂S
∂p
)
T
= −(∂V
∂T
)
p
into Equation (a) to get
(∂Cp∂p
)
T
= −T(∂2V
∂T 2
)
p
.
(b) For an ideal gas V (T ) = (NkT )/p, so
(∂V
∂T
)
p
=Nk
pand
(∂2V
∂T 2
)
p
= 0
97
6. Relating G to µ.
Prove that G =∑Ni=1 µiNi.
Substitute Equation (9.29)
U = TS − pV +M∑
i=1
µiNi
into G = H − TS = U + pV − TS to get
G =M∑
i=1
µiNi.
7. Piezoelectricity.
Apply a mechanical force f along the x-axis of a piezoelectric crystal, such as quartz, whichhas dimension ` in that direction. The crystal will develop a polarization p0, a separationof charge along the x-axis, positive on one face and negative on the other. Applying anelectric field E along the x-axis causes a mechanical deformation. Such devices are used inmicrophones, speakers, and pressure transducers. For such systems the energy equation isdU = TdS + fd`+ Edp0. Find a Maxwell relation to determine (∂p0/∂f)T,E.
You can’t use the energy equation to get cross-terms to find (∂p0/∂f)T,E becauseE = (∂U/∂p0)S,` and f = (∂U/∂`)S,p0 are derivatives for constant entropy, not constanttemperature. So first you need to start with a fundamental function of T , not S. UsingLegendre transforms, construct the function G(T,E, f), which is called the piezoelectricGibbs function,
G = U − (TS + f`+ Ep0).
Substituting dU gives
dG = TdS + fd`+ Edp0 − TdS − SdT − fd`− `df − Edp0 − p0dE
= −SdT − `df − p0dE.
98
Taking cross-derivatives gives
(∂`
∂E
)
T,f
=
(∂p0
∂f
)
T,E
.
(∂`/∂E) requires a measurement of how the length of the crystal changes with the appliedfield E.
8. Relating CV and Cp.
Show that Cp = CV +Nk for an ideal gas.
There are various ways to do this. One way is to recognize that
Cp = T
(∂S
∂T
)
p
and CV = T
(∂S
∂T
)
V
so you can start with the equation S(T, p),
(a) dS =
(∂S
∂T
)
p
dT +
(∂S
∂p
)
T
dp.
For constant volume, divide Equation (a) by dT and multiply by T
T
(∂S
∂T
)
V
= T
(∂S
∂T
)
p
+ T
(∂S
∂p
)
T
(∂p
∂T
)
V
(b) =⇒ CV = Cp + T
(∂S
∂p
)
T
(∂p
∂T
)
V
.
Substitute the Maxwell relation
(∂S
∂p
)
T
= −(∂V
∂T
)
p
99
into Equation (b) to get
(c) CV = Cp − T(∂V
∂T
)
p
(∂p
∂T
)
V
.
For an ideal gas, pV = NkT , so
(∂V
∂T
)
p
=Nk
pand
(∂p
∂T
)
V
=Nk
V.
Substituting these expressions into Equation (e) gives
CV = Cp −(NkT
p
)(Nk
V
)= Cp −Nk.
9. Rubber bands are entropic springs.
Experiments show that the retractive force f of polymeric elastomers as a function of tem-perature T and expansion L is approximately given by f(T, L) = aT (L− L0) where a andL0 are constants.
(a) Use Maxwell relations to determine the entropy and enthalpy, S(L) and H(L), atconstant T and p.
(b) If you adiabatically stretch a rubber band by a small amount its temperature increases,but its volume does not change. Derive an expression for its temperature T as afunction of L, L0, a, and its heat capacity C = (∂U/∂T ).
(a) To use the Maxwell relation, Equation (9.12), we take the derivative of the equation ofstate to get
(∂f
∂T
)
p,L
= a(L− L0).
Using Equation (9.18) gives
(∂S
∂L
)
T,p
= −a(L− L0).
100
Integrating gives
S(L) =−a(L − L0)
2
2
which shows that the entropy follows a square-law, like a Hooke’s law spring. ForH(L), use Equation (9.20):
(∂H
∂L
)
T,p
= f − T(∂f
∂T
)
L,p
= aT (L− L0)− aT (L− L0) = 0.
(b) For an adiabatic process, the derivation parallels that of Example 8.8:
dU = δq + δw = fdL
since δq = 0 and the pV work component is zero. Also dU = CV dT , so
CV dT = aT (L− L0)dL
=⇒∫ T2
T1
CVT
dT =∫ L
L0
a(L− L0) dL
=⇒ CV ln(T2
T1
)=
a(L− L0)2
2
=⇒ T2
T1
= exp
[a(L− L0)
2
2CV
]
10. Metal elasticity is due to energy, not entropy.
Experiments show that the retractive force f of metal rods as a function of temperature Tand extension L relative to undeformed length L0 is given by f(T, L) = Ea∆L/L0, where∆L = L[1 − α(T − T0)] − L0 = L − Lα(T − T0) − L0. a is the cross-sectional area of therod, E (which has the role of a spring constant) is called Young’s modulus, and α ≈ 10−5
is the linear expansion coefficient. Compute H(L) and S(L). Is the main dependence on Ldue to enthalpy H or entropy S?
To get the enthalpy, use Table 8.1:
dH = TdS + V dp+ fdL = CpdT +Ea(L− L0)
L0dL.
101
Integrating gives
H(L) = Cp(T − T0) +Ea
2L0
(L− L0)2.
To get the entropy, use
dS =
(∂S
∂T
)
L
dT +
(∂S
∂L
)
T
dL
=CLTdT −
(∂f
∂T
)
L
dL
where CL is the heat capacity of a rod held at constant length, and we have used Maxwell’srelation Equation (9.19). Since
(∂f
∂T
)
L
= −EaLαL0
,
equation (a) becomes
dS =CLTdT +
EaαL
L0dL
and integrating gives
S(T, L) =∫ T
T0
CL(T )
TdT +
EaαL2
2L0+ constant.
Since α = 10−5 appears in S(L) but not in H(L), the dominant L dependence is in theenthalpy for metals. For polymers, stretching mostly affects the entropy.
11. Pressures in Surfactant Monolayers
102
A Langmuir trough is shown below. N surfactant molecules float on wa-ter on the left side of a bar. When you push the bar to the left withpressure π, the N surfactant molecules occupy an equilibrium area A, where
π = −(∂G∂A
)T,P,N
Consider a 2-dimensional van der Waals model in which the entropy is given by
S = kln(A− A0)
where A0 is the minimum area the molecules can occupy when they crowd together, andthe enthalpy by
H = − ba
(a) You can express the pressure in terms of components: enthalpic (πH) and entropic(πS):
π = πH + πS
Write a Maxwell’s relation for
πS = T ( ∂S∂A
)T,P,N
(b) Express πH in terms of measurable quantities, T and A.
(c) Write the ”equation of state” of the model, which is the quantity π(N,A, T ).
(d) Write an expression for the free energy ∆G(N, T,A) at constant p.
(a) dG = −SdT + V dP + µdN − πdA
103
∂G∂T
= −S, ∂G∂P
= V, ∂G∂N
= µ, ∂G∂A
= −π∂2G∂T∂A
= ∂2G∂A∂T
=⇒ − ∂π∂T
= − ∂S∂A
∂S∂A
= ∂π∂T
=⇒ πS = T ∂π∂T
(b) G = H − TS =⇒ ∂G∂A
= ∂H∂A− T ∂S
∂A− S ∂T
∂A
At constant T and for π = −∂G∂A
, −∂G∂A
= T ∂S∂A− ∂H
∂A
π = πS + πH =⇒ πH = −∂H∂A
πH = π − πSπH = π − T ∂π
∂T
ORH = −B
A=⇒ ∂H
∂A= B
A2 =⇒ πH = −∂H∂A
= − BA2
(c) π = T ∂S∂A− ∂H
∂A
S = kln(A− A0) =⇒ ∂S∂A
= kA−A0
H = −BA
=⇒ ∂H∂A
= BA2
π = kTA−A0
− BA2
(d) π = −∂G∂A
= kTA−A0
− BA2 =⇒
∆G = − ∫ kTA−A0
− BA2dA
∆G = [−kT ln(A− A0)− BA
]A2A1
∆G = [−kT ln(A2 − A0)− BA2
] + [kT ln(A1 − A0) + BA1
]
∆G = kT ln(A1−A0
A2−A0) + B
A1− B
A2
∆G = kT ln(A1−A0
A2−A0) + B(A2−A1)
A1A2
12. Torque on DNA.
104
One end of a DNA molecule is fixed to a surface. Using single-molecule methods, a torque(twisting force) is applied to the other end, to twist the molecule.
When the molecule is twisted through an angle θ, the DNA gives a restoring torque τ thatdepends on temperature T,
τ = (k0 + k1T )θ
where k0 and k1 are measured constants.
(a) Write an expression for the free energy G(θ) at fixed T and p.
(b) Use a Maxwell relation to express ( ∂S∂θ
)T,p in terms of experimental observables, whereS is entropy and p is pressure.
(c) Derive an expression for S(θ), at constant T and p.
(d) Derive an expression for the enthalpy, H(θ), at constant T and p.
(e) Under what conditions, i.e. for what values of k0, k1, and T, will this torsionalretractive force be entropy-driven?
(a) τ = (k0 + k1T )θdG = −SdT + V dp+ µdN + τdθ=⇒ τ = (∂G
∂θ)T,p,N =⇒ G(T, θ) =
∫τdθ
=⇒ G(T, θ) = 12(k0 + k1T )θ2
(b) From the equation for dG above, the Maxwell relation is:(∂S∂θ
)T,p,N = −( ∂τ∂T
)θ,p,N
(c) Since ( ∂τ∂T
)θ,p,N = k1θ = −∂S∂θ
, you haveS(θ) = − ∫ k1θdθ = −k1
2θ2
105
(d) Since τ = ∂G∂θ
= ∂H∂θ− T ∂S
∂θ, rearranging gives
∂H∂θ
= τ + T ∂S∂θ
= (k0 + k1T )θ − k1Tθ = k0θ
=⇒ H(θ) =∫k0θdθ = k0θ2
2
(e) If k0 is more negative than k1T , the torque is entropy-driven.
13. The Massieu Function: J(β, V, N).
Whereas the entropy is an extremum function S(U, V, N), it is sometimes useful to usea related extremum function, J(β, V, N) where β = 1/T and T is temperature. For thisfunction, β and U are conjugate variables.
(a) Derive an expression for the differential quantity dJ in terms of variations dβ, dV,and dN. (Hint: as an initial guess for J, try either J = S + βU or J = S - βU).
(b) Using additional reduced variables π = p/T and m = µ/T, write Maxwell relationsfor:
( ∂π∂N
)β,V =(∂U∂V
)β,N =( ∂U∂N
)β,V =
(c) What is the relation between J and the Helmholtz free energy F(T, V, N)?
(a) Let J = S − βU , thendJ = dS − βdU − Udβ = 1
TdU + p
TdV − µ
TdN − βdU − Udβ
=⇒ dJ = −Udβ + pβdV + µβdN
(b) In reduced quantities,dJ = −µdβ + πdV −mdNSo the Maxwell’s relations, which are just the cross-derivatives are( ∂π∂N
)β,V = −∂m∂V
(∂U∂V
)β,N = −∂π∂β
( ∂U∂N
)β,V = ∂m∂β
(c) Multiply the equation, J = S − βU , by (-T) to get−JT = −TS + TβU = U − TS = F=⇒ J = −βFThe utility of the Massieu function comes from the relationship that F
kT= −lnK, so
J ∼ lnK is proportional to the experimentally observable equilibrium constant.
106
14. Thermodynamics of a single molecule.
One end of a molecule is fixed to a surface. Another end is pressed down by the tip of anatomic force microscope.
Using the tip of an atomic force microscope as a probe, we measure the force of moleculeretraction. The force depends on the temperature at which the experiment is performed asfollows
f = (aT + b)x, where a and b are constants
(a) Produce a fundamental equation describing the energy of the molecule. Write anexpression for dG at fixed T, P. Derive an expression showing how G depends on x.
(b) Use a Maxwell relation to express ( ∂S∂x
)T,P .
(c) Derive an expression for S(x), at fixed T, P.
(d) Derive an expression for enthalpy H(x) at fixed T, P.
(e) What values of a and b will produce a condition that the retraction force is driven bya change in entropy?
(a) dU = TdS + PdV + µdN + FdxdG = −SdT + V dP + FdxF = (∂G
∂x)T,P
G(x) =∫(aT + b)xdx = 1
2(aT + b)x2
107
(b) (∂S∂x
)T,P = −(∂F∂T
)x,P
(c) (∂S∂x
)T,P = −(∂F∂T
)x,P = −axS(x) =
∫ −axdx = −12ax2
(d) F = (∂G∂x
)T,P = (∂H∂x
)T,P − T (∂S∂x
)T,P = (∂H∂x
)T,P + T (∂F∂T
)x,P(∂H∂x
)T,P = F + T (∂S∂x
)T,P = (aT + b)x + T (−ax) = aTx+ bx− Tax = bx=⇒ H(x) =
∫bxdx = 1
2bx2
(e) Entropy of the molecule has to decrease when x decreases (we press on the molecule).Thus, the coefficient a has to be negative for a < 0, and(∂H∂x
)T,P << T (∂F∂T
)x,Pbx << aTxb << aT < 0b is more negative than aT.
15. Gibbs-Helmholtz Equation.
From the definition of G=H-TS, derive the Gibbs-Helmholtz equation:
∂∂T
(GT) = − H
T 2
which relates the temperature variation of the Gibbs free energy G to the enthalpy H.
Starting with G = H − T (∂G∂T
)p,N , and multiplying both sides by (− 1T 2 )
− GT 2 − 1
T(∂G∂T
)p,N = − HT 2
∂∂T
(GT) = − H
T 2
16. Surface tension components.
108
The figure shows the surface tension of water as a function of temperature.
(a) From the figure, determine the numerical value of ∂γ∂T p,N,a
, where T = temperature, p
= pressure, and a = surface area.
(b) Find the Maxwell relation for the partial derivative equal to ∂γ∂T p,N,a
.
(c) Write an expression for the enthalpic and entropic components of the surface tensionγ.
(d) Combining the results from above, compute the numerical values of the enthalpic andentropic parts of γ at T = 300 K, and comment on which component dominates thesurface tension.
(a) From the slope:
∂γ
∂T= −0.15
dyn
cm · deg.
(b) Use Eq. (9.6) to get the appropriate cross term:
(∂γ
∂T
)
p,N,a
= −(∂S
∂a
)
109
(c)
γ =
(∂G
∂a
)
T,p,N
=
(∂H
∂a
)
T,p,N
− T(∂S
∂a
)
T,p,N
.
(See Eq. (9.14) for an equivalent problem.)
(d)
Entropic part = −T ∂S
∂a= T
∂γ
∂T= −0.15
dyn
cm · deg· 300 K
= 45 dyn/cm.
(e) At T = 300 K, γ ≈ 70 dyn/cm, so the enthalpic part is:
∂H
∂a= γ − T ∂γ
∂T= 70− 45 = 25 dyn/cm.
Therefore water’s surface tension is dominated by entropy. This is because water is astructured liquid.
17. Exploring the quantity F/T.
110
Sometimes a quantity of interest is the Helmholtz Free Energy F (T, V,N) divided by T .(For example, this quantity is proportional to the logarithms of equilibrium constants orsolubilities.)
(a) Derive a relationship showing that
∂(F/T )∂T
∝ U .
Find the constant of proportionality.
(b) Suppose F(T) depends on temperature in the following way: F (T ) = 2aT 2 + bT (soF/T = 2aT + b). Find S(T ) and U(T ).
(c) For this and the next parts of this problem, consider a transition between a helix andcoil state of a polymer. F/T is related to the ratio of populations as F/T = −k ln(Keq),where Keq = Ncoil/Nhelix. Using F/T = 2aT + b from above, at what temperature arethe helix and coil in equilibrium (e.g., half of the molecules are helical and half areunfolded).
(d) Assuming a > 0, does increasing the temperature favor the helix or the coil?
(a) There are several ways to solve this problem. Consider a couple of them:FT
= UT− S
∂∂T
(UT− S) = ∂U
∂T· 1T− U
T 2 − ∂S∂T
= ∂U∂S· ∂S∂T· 1T− U
T 2 − ∂S∂T
= T · ∂S∂T· 1T− U
T 2 − ∂S∂T
= ∂S∂T− U
T 2 − ∂S∂T
=⇒ ∂(F/T )∂T
= − UT 2
OR∂(F/T )∂T
= 1T· ∂F∂T− F
T 2
= 1T(−S)− U−TS
T 2
= −ST− U
T 2 + ST
=⇒ ∂(F/T )∂T
= − UT 2
(b) ∂F∂T
= −S(T ) = −(4aT + b)=⇒ S(T ) = 4aT + bF = U + TS2aT 2 + bT = U(T ) + 4aT 2 + bT=⇒ U(T ) = −2aT 2
(c) FT
= −kln(Keq)2aT+b=−kln(1/2
1/2)
111
2aT + b = 0=⇒ T = − b
2a
(d) ∂(F/T )∂T
= 2a, for a > 0 is always positive. Thus,∂(kln(Keq))
∂T< 0
and this favors a coil.
112
Chapter 10The Boltzmann Distribution Law
1. The statistical thermodynamics of a cooperative system.
Perhaps the simplest statistical mechanical system having ‘cooperativity’ is the three-levelsystem in the following table.
Energies 2ε0 ε0 0
Degeneracies γ 1 1
(a) Write an expression for the partition function q as a function of energy ε, degeneracyγ, and temperature T .
(b) Write an expression for the average energy 〈ε〉 versus T .
(c) For ε0/kT = 1 and γ = 1, compute the populations, or probabilities, p∗1, p∗2, p
∗3 of the
three levels.
(d) Now if ε0 = 2 kcal mol−1 and γ = 1000, find the temperature T0 at which p1 = p3.
(e) Under condition (d), compute p∗1, p∗2, and p∗3 at temperature T0.
(a) q = 1 + e−ε0/kT + γe−2ε0/kT
(b) 〈ε0〉 = 0p0 + ε0pi + 2ε0p2
=1
q
[ε0e−ε0/kT + 2ε0γe
−2ε0/kT]
= −∂ ln q
∂β
113
(c) q = 1 + e−1 + e−2 = 1.5032
p1 =1
q= 0.6652
p2 =e−1
q= 0.2447
p3 =e−2
q= 0.0900
(d) Since p1 = 1/q and p3 = γe−2ε0/kT/q, we want
γe−2ε0/kT0 = 1 =⇒ − 2ε0
kT0= ln(1/γ) = −6.9078
=⇒ T0 =2(2000 cal/mol)
6.9078(2cal/mol)= 289.5 K
(Use R in place of k for molar calculations)
(e)ε0
kT=
2000 cal/mol
(2 cal/(mol deg)) (289.5 deg)= 3.4539
Now q = 1 + e−3.45 + 1000e−6.9078 = 2.0317
p1 =1
q= 0.4922 = p3
p2 = e−3.45/q = 0.0156
114
2. Speed of sound.
The speed of sound in air is approximately the average velocity 〈v2x〉1/2 of the gas molecules.
Compute this speed for T = 0◦ C, assuming that air is mostly nitrogen gas.
An equation on p. 183 gives
〈v2x〉1/2 =
(kT
m
)1/2
=(RT
M
)1/2
where R is the gas constant and M = 28 gm/mol is the molecular weight. We have
〈v2x〉1/2 =
(8.314 J
Kmol
)(273K)
0.028 kg/mol
1/2
= 285m
sec.
The actual speed of sound in air is 346 ms−1 at 300 K.
3. Properties of a two-state system.
Given a two-state system in which the low energy level is 600 cal mol−1, the high energylevel is 1800 cal mol−1, and the temperature of the system is 300 K,
(a) What is the partition function q?
(b) What is the average energy 〈ε〉?
(a) q = 1 + e−(ε2−ε1)/kT = 1 + e−1200/(2)(300) = 1 + e−2 = 1.135
(b) 〈ε〉 = ε1p1 + ε2p2
= 600cal
mol
(1
q
)+ 1800
cal
mol
(e−2
q
)
= 600cal
mol
(1 + 3e−2
1 + e−2
)= 743
cal
mol
115
4. Binding to a surface.
Consider a particle that has two states: bonded to a surface, or non-bonded (released). Thenon-bonded state is higher in energy by an amount ε0.
(a) Explain how the ability of the particle to bond to the surface contributes to the heatcapacity, and why the heat capacity depends on temperature.
(b) Compute the heat capacity CV in units of Nk if T = 300 K and ε0 = 1.2 kcal mol−1
(which is about the strength of a weak hydrogen bond in water).
(a) This binding process adds an energy ladder (in this case it has only two states) towhatever energy levels exist. This provides an energy storage mechanism, thus a heatcapacity (since CV = ∂U/∂T ). This heat capacity depends on temperature becauseequipartition does not apply to a two-state system. At low temperatures, only thelowest level is populated (all particles are bonded) and small increments of thermalenergy cannot be stored as excitations. At high temperatures, only the highest level ispopulated (all particles are unbound and free) and no further energy storage ispossible. Only at intermediate temperatures can the system absorb energy by releasingbound molecules.
(b) Using the Schottky two-state model,
CVNk
=(ε
kT
)2 e−ε/kT
(1 + e−ε/kT )2 .
ε
kT=
1.2× 103 calmol
1.987 calmolK 300K
= 2.013 =⇒ −eε/kT = 0.134, so
CVNk
= (2.013)2 0.134
(1 + 0.134)2= 0.421.
116
5. Entropy depends on distinguishability
Given a system of molecules at T = 300 K, q = 1× 1030, and ∆U = 3740 J mol−1,
(a) What is the molar entropy if the molecules are distinguishable?
(b) What is the molar entropy if the molecules are indistinguishable?
(a) For distinguishable particles, where Q = qN , the molar entropy is
S
n= R ln q +
∆U
nT
∆U
n= 3740 J/mol =⇒ S
n=(8.314
J
mol K
)ln(1030) +
(3740 J)
300 mol K
=⇒ S
n= 586.8
J
K mol
(b) For indistinguishable particles, Q = qN
N !≈ qN
(Ne )N =⇒ lnQ ≈ N ln q −N(lnN − 1), so
S
n= R[ln q − lnN + 1] +
∆U
nT
N = nNA where N = no. of molecules,n = no. of moles and NA = Avogadro’s number.
=⇒ S
n= R ln
(q
nNA+ 1
)+
∆U
nT
=(8.31
J
molK
)[ln
(1030
6× 1023
)+ 1
]+
3740
300= 139.9
J
molK
Note that S is larger if particles are distinguishable.
117
6. The Boltzmann distribution of uniformly-spaced energy levels.
A system has energy levels uniformly spaced at 3.2× 10−20 J apart. The populations of theenergy levels are given by the Boltzmann distribution. What fraction of particles is in theground state at T = 300 K?
The energies of this system are 0, ε, 2ε, . . . , where ε = 3.2× 10−20J. Then
q =∞∑
i=0
e−iε/kT,
which is a geometric series of the form
∞∑
k=0
axk
with a=1, and x = e−ε/kT < 1. Using Equation (4.4), we get
q =1
1− e−ε/kT=
1
1− e−7.72= 1.0004.
So p∗0 = 1q
= 1− e−7.72 = 0.9996—only 0.04% of the molecules are not in the ground state.
7. The population of spins in a magnetic field.
The nucleus of a hydrogen atom, a proton, has a magnetic moment. In a magnetic field, theproton has two states of different energy; spin up and spin down. This is the basis of protonNMR. The relative populations can be assumed to be given by the Boltzmann distribution,where the difference in energy between the two states is ∆ε = gµB, g = 2.79 for protons,and µ = 5.05× 10−24 J Tesla−1. For a 300 MHz NMR instrument, B = 7 Tesla.
(a) Compute the relative population difference, |N+ −N−|/(N+ +N−), at room temper-ature for a 300 MHz machine.
(b) Describe how the population difference changes with temperature.
(c) What is the partition function?
118
(a)N↑N↓
= eεkT =⇒ N↑ = N↓e
εkT
=⇒ N↑ +N↓ = N↓(1 + eεkT )
N↑ −N↓ = N↓(eεkT − 1)
|N↑ −N↓|N↑ +N↓
=1− e ε
kT
1 + eεkT.
For small εkT
, 1− e εkT ≈ ε
kT, so
|N↑ −N↓|N↑ +N↓
≈εkT
2− εkT
≈ ε
2kT
=(2.79)(5.05× 10−24 J/Tesla)(7 Tesla)
2(300K)(1.38× 10−23 J/K)
= 1.19× 10−2
This is the fractional population difference.
(b) As T increases, the population difference decreases (populations become equal). AsT 0, all of the population goes to the lower energy state.
(c) q = 1 + eεkT (if the lower energy state is taken as the ground state, with respect to which
the other energy is measured).
8. Energy and entropy for indistinguishable particles.
Equations (10.34) for 〈ε〉 and (10.37) for S apply to distinguishable particles. Compute thecorresponding quantities for systems of indistinguishable particles.
For indistinguishable particles, Q = qN
N !≈ qN
(Ne )N , so
∆U = −(∂ lnQ
∂β
)≈ − ∂
∂β(N ln q −N lnN +N) = −N
(∂ ln q
∂β
),
119
so 〈ε〉 =∆U
N=
∆U
N= −
(∂ ln q
∂β
)=⇒ 〈ε〉 doesn’t depend on distinguishability.
Now S = k lnQ+∆U
T= Nk(ln q − lnN + 1)− N
T
(∂ ln q
∂β
)
= Sdistinguishable − lnN + 1 < Sdistinguishable for N > 2 particles.
As seen in problem 5, the entropy is larger for distinguishable systems.
9. Computing the Boltzmann distribution.
You have a thermodynamic system with three states. You observe the probabilities p1 = 0.9,p2 = 0.09, p3 = 0.01 at T = 300 K. What are the energies ε2 and ε3 of states 2 and 3 relativeto the ground state?
q = 1 + e−ε2kT + e−
ε3kT
p1 =1
q= 0.9 =⇒ q =
10
9
p2 =e−
ε2kT
q= 0.9e−
ε2kT = 0.09 =⇒ e−
ε2kT =
1
10
=⇒ ε2 = kT ln 10 =
(2
cal
K mol
)(300K) ln 10 = 1.38
kcal
mol
p3 =e−
ε3kT
q= 0.9e−
ε3kT = 0.01 =⇒ e−
ε3kT =
1
90
=⇒ ε3 = kT ln 90 =
(2
cal
K mol
)(300K) ln 90 = 2.70
kcal
mol
120
10. The pressure reflects how energy levels change with volume.
If energy levels εi(V ) depend on the volume of a system, show that the pressure is theaverage
p = −N⟨∂ε
∂V
⟩.
From the fundamental energy equation, p = −(∂∆U∂V
)S,N
. But ∆U = N〈ε〉 so
p = −N(∂〈ε〉∂V
)
S,N
= −N(∂
∂V
N∑
1
εi(V )pi
)= −N
(N∑
1
∂
∂Vεi(V )pi
)
= −N(
N∑
1
(∂εi(V )
∂V
)pi
)= −N〈 ∂ε
∂V〉.
11. End-to-end distance in polymer collapse.
Use the two-dimensional four-bead polymer of Example 10.3. The distance between thechain ends is 1 lattice unit in the compact conformation, 3 lattice units in the extendedconformation, and
√5 lattice units in each of the other three chain conformations. Plot the
average end-to-end distance as a function of temperature, if the energy is
(a) ε = 1 kcal mol−1.
(b) ε = 3 kcal mol−1.
To get the average end-to-end length, 〈d〉,use the Boltzmann probabilities pj for each of the j = 1, 2, . . . , 5 conformations:
〈d〉 =5∑
j=1
pjdj = (1)
(1
q
)+√
5
3e
−εkT
q
+ 3
e
−εkT
q
where q = 1 + 4e−εkT ,
so 〈d〉 =1 + 9.71e
−εkT
1 + 4e−εkT
.
121
The maximum end-to-end distance (T ∞) is
(1 + 3√
5 + 3)
5= 2.14.
The midpoint is at 〈d〉 = 1.57. You can find the midpoint temperature T0 from
(1.57)(1 + 4e
−εkT0
)= 1 + 9.71e
−εkT0
=⇒ 0.57
3.43= e
−εkT0 .
(a) T0 =
(1000 cal
mol
)
(2 cal
molK
)ln(
3.430.57
) = 278K
(b) T0 = 835K
12. The lattice model of dimerization.
Use the lattice model for monomers bonding to form dimers, and assume large volumesV � 1.
(a) Derive the partition function.
(b) Compute p1(T ) and p2(T ), the probabilities of monomers and dimers as a function oftemperature, and sketch the dependence on temperature.
(c) Compute the bond breakage temperature T0 at which p1 = p2.
(a) If you choose the convention that the ground state (dimer) has zero energy
and the dissociated state has energy ε > 0, then the partition function is
Q =∑
ε
g(ε)e−ε
kT
= (V − 1) +(V
2− 1
)(V − 1)e−
εkT
122
(b) The Boltzmann populations are
Pmonomer =
(V2− 1
)e−
εkT
1 +(V2− 1
)e−
εkT
Pdimer =1
1 +(V2− 1
)e−
εkT
(c) The populations are equal when
(V2− 1
)e− ε
kT0 = 1
=⇒ − εkT0
= − ln(V
2− 1
)
=⇒ T0 =ε
k ln(V2− 1
)
which is the same result we found in Equation (8.8)
13. Equivalent premises for Boltzmann distribution.
Use Equation (10.6) to show that the distribution of probabilities p∗j that minimizes the freeenergy F at constant (T, V,N) is the same distribution that maximizes the entropy S atconstant (U, V,N) = (〈E〉, V, N).
It is clear that requiring dS = 0, subject to the constraint that dU = 0, is the same asrequiring dF = dU − TdS = 0, if T is constant.
14. Protein conformations.
123
Assume a protein has six different discrete conformations, with energies given in the figurebelow.
(a) Write an expression for the probability, p(i) of finding the protein in conformation i.
(b) Write an expression for the probability, p(E) of finding the protein having energy E.
(c) Use the expressions you wrote in (a) and (b) to calculate the following probabilities:
(a) p(StateB).
(b) p(StateA).
(c) p(StateD).
(d) p(1 kcal/mol).
(e) p(−5 kcal/mol).
(d) What is the average energy of the ensemble of conformations?
(a) p(Statei) = e
−EStateiRT
e3000RT +e
5000RT +e
3000RT +e
−1000RT +e
−2000RT +e
−1000RT
RT = 1.987 calK·mol300K = 596.1 cal
mol
=⇒ Q = e5.03 + e8.39 + e5.03 + e−1.68 + e−3.36 + e−1.68
Q = 152.9 + 4399 + 152.9 + 0.19 + 0.03 + 0.19 = 4705.21
(b) P (E) = W (E)e−ERT
Q
where Q = 4705.21W (−3kcal
mol) = 2,W (−5kcal
mol) = 1,W (1kcal
mol) = 2,W (2kcal
mol) = 1
(c) (a) p(StateB) = e5000596.1
4705.21= 4399
4705.21= 0.935
(b) p(StateA) = e3000596.1
4705.21= 152.9
4705.21= 0.032
(c) p(StateD) = e−1000596.1
4705.21= 0.19
4705.21= 4.04 · 10−5
(d) p(1kcal/mol) = 2·e−1000596.1
4705.21= 2·0.19
4705.21= 8.08 · 10−5
(e) p(−5kcal/mol) = 1·e5000596.1
4705.21= 4399
4705.21= 0.935
124
(d) 〈E〉 =∑iEipi
=−3000·0.032−5000·0.935−3000·0.032+1000·4.04·10−5+2000·8.08·10−5+1000·4.04·10−5
= −96− 4675− 96 + 0.0404 + 0.1616 + 0.0404 = −4866.6 calmol
15. Modeling ligand binding.
A ligand is bound to a protein with a spring-like square-law energy ε(x), where x is thedistance between the ligand and protein as shown in the figure.
ε(x) = 12cx2
(a) For constant (T,V,N), write an expression for the probability distribution p(x) of theligand separation from the protein.
(b) Sketch a plot of p(x) vs x.
(c) Write an expression for the average location of the ligand, 〈x〉.
(d) Write an expression for the second moment of the location of the ligand, 〈x2〉.
(e) Calculate the average energy, 〈ε〉 of the system.
(a) p(x) = e−βε(x)Q
, x ≥ 0
p(x) = e− cx2
2kT∫∞0
e− cx2
2kT dx=√
2cπkT
e−cx2
2kT
(b) Sketch
125
(c) 〈x〉 =∫∞0 xp(x)dx =
∫∞0
xe− cx2
2kT dx∫∞0
e− cx2
2kT dx
Look up these integrals in the appendix D:∫∞0 e−ax
2dx = 1
2
√πa
Here, a = c2kT
=⇒ 〈x〉 = ( 12a
)(2√
aπ) = 1√
πa=√
2kTπc
(d) 〈x2〉 =
∫∞0
x2e− cx2
2kT dx∫∞0
e− cx2
2kT dx
=14a
√πa
12
√πa
= 12a
= kTc
(e) 〈ε〉 = 〈12cx2〉 = 1
2c〈x2〉 = kT
2
This is also expected from the equipartition theorem.
16. Distribution of torsional angles.
In a computer simulation that samples a molecular torsion angle θ, you observe a gaussiandistribution, p(θ), shown in the figure below.
p(θ) = p0e−ks(θ−θ0)2
What is the underlying energy function E(θ) that gives rise to it?
Since the Boltzmann law is: p(θ) = p0e−E(θ)kT , you have:
126
E(θ)kT
= −ln(p(θ)p0
) = ks(θ − θ0)2
This is a spring-like square-law energy.
17. 3-bead polymer chain model.
Consider a 3-bead polymer that can undergo conformational change from a non-linear to alinear form, as shown in Figure A. Both conformations have the same energy. Now supposethe X and Y atoms of the polymer can bind a ligand L (Figure B). Breaking one bondincreases the energy by ε to breaking two bonds increases the energy by 2ε. Assume thatthe ligand-bound conformation has the lowest energy.Figure A:
Figure B:
(a) Draw a picture showing the possible binding states.
(b) Calculate the equilibrium probabilities of these conformations.
(c) Plot the population distribution of the conformations at temperature kT = 0 andkT = 1.4ε. What is the difference in entropy between these two temperatures.
(d) Plot the distribution at high temperatures (T→ inf) and explain its shape.
(e) What is the average energy of the system at each temperature in part (c)?
(a) Possible conformations:
127
(a)
(b)
(c)
(b) Probabilities:p1 = 1
1+2e−ε/kT+e−2ε/kT
p2 = 2e−ε/kT
1+2e−ε/kT+e−2ε/kT
p3 = e−2ε/kT
1+2e−ε/kT+e−2ε/kT
(c) Distributions:p1 = 1
1+2e−ε/kT+e−2ε/kT = 11+2e−1/1.4+e−2/1.4 ≈ 0.45
p2 = 2e−ε/kT
1+2e−ε/kT+e−2ε/kT ≈ 0.44
p3 = e−2ε/kT
1+2e−ε/kT+e−2ε/kT ≈ 0.11
S = −k∑3i=1 piln
pigi
= k(0.45ln0.45 + 044ln0.22 + 0.11ln0.11) ≈ 1.27k
(d) Partition function at T → inf is Q = 1 + 2 + 1
128
(e) Average energy:For T = 0: 〈E〉 = 0For T = 1.4ε/k: 〈E〉 = 0 ∗ 0.45 + 1 ∗ 0.44 + 2 ∗ 0.11 = 0.66
18. Protein hinge motions.
A protein has two domains, connected by a flexible hinge. The hinge fluctuates around anangle θ0. The distribution of the angles is gaussian, around θ0, as shown in the second figurebelow.
µµ0
p(θ) = 1σ√
2πexp(− (θ−θ0)2
2σ2 ).
where σ is the standard deviation.
(a) Assume a Boltzmann distribution of probabilities: p(θ) = b · exp(− c(θ−θ0)2
kT), with
energy ε = c(θ − θ0)2. Derive the value of b.
(b) If you measure a standard deviation σ = 30 deg, what is the value of c?
(c) Derive an expression relating 〈θ2〉 to c. Assume T = 300 K.
(d) Derive an expression for the entropy in terms of kT and c: Sk
= − ∫ +∞−∞ p(θ)ln(p(θ))dθ.
(e) If a mutant of the protein has fluctuations that are smaller than the wild type, 〈θ2〉M =12〈θ2〉WT , what is the change in entropy, ∆S = SM − SWT , where SM is the entropy
of the mutant and SWT is the entropy of the original wild-type protein?
129
(a) We must have∫∞−∞ p(θ)dθ = 1,
so 1b
=∫∞−∞ exp
−c(θ−θ0)2
kT dθ =√
πkTc
=⇒ b =√
cπkT
You can also see this by noticing that
2σ2 = kTc, so σ =
√kT2c
b = 1σ√
2π=√
cπkT
(b) You have c = RT2σ2 =
600 calmol
2(900deg2)= 1
3cal
mol·deg2 = 2.4 · 10−24 Jdeg2
(c) 〈(θ − θ0)2〉 =
∫∞−∞(θ − θ0)
2√
cπkT
e−c(θ−θ0)2
kT dθ
Use∫∞−∞ x
2e−ax2dx = 1
2a
√πa
to get
〈(θ − θ0)2〉 = kT
2c= σ2
(d) Sk
= − ∫∞−∞[√
cπkT
e−cθ2
kT ][ln(√
cπkT
)− cθ2
kT]dθ
= −√
cπkT
ln(√
cπkT
)∫∞−∞ e
− cθ2kT dθ +
√c
πkT
∫∞−∞( cθ
2
kT)e−
cθ2
kT dθ
= −√
cπkT
ln(√
cπkT
)√
πkTc
+ 12
Sk
= −ln(√
cπkT
) + 12
(e) ∆S = Sm − Swt = −kln(√
cmcwt
)
= −kln(
√〈θ2wt〉〈θ2m〉 )
= −kln√
2
130
Chapter 11Statistical Mechanics of Simple Gasesand Solids
1. The heat capacity of an ideal gas.
What is the heat capacity CV of an ideal gas of argon atoms?
∆U = NkT 2
(∂ ln q
∂T
)
=NkT 2
a
(∂q
∂T
)
=3
2NkT
CV =∂∆U
∂T=
3
2Nk.
2. The statistical mechanics of oxygen gas.
Consider a system of one mole of O2 molecules in the gas phase at T = 273.15 K in a volumeV = 22.4× 10−3 m3. The molecular weight of oxygen is 32.
(a) Calculate the translational partition function qtranslation.
(b) What is the translational component of the internal energy per mole?
(c) Calculate the constant-volume heat capacity.
131
(a) qtranslation =
(2πmkT
h2
)3/2
V
m ≈ 32gm
mol= 5.31× 10−26 kg
molecule
k = 1.38× 10−23 JK−1
T = 273.15 K
h = 6.626× 10−34 J s
qtranslation =
[(6.28)(5.32× 10−26 kg)(3.77× 10−21 J)
(6.63× 10−34 J s)2
]3/2
(22.4× 10−3 m3)
= 3.43× 1030 accessible states
molecule
(note :
kg m2
s2= J
)
(b) ∆Utranslation =3
2RT (per mole) = 3.4
kJ
mole
(c) CV (per mole) =
(∂∆U
∂T
)
V
=3
2R = 12.47
J
Kmol
3. The statistical mechanics of a basketball.
Consider a basketball of mass m = 1 kg in a basketball hoop. To simplify, suppose the hoopis a cubic box of volume V = 1 m3.
(a) Calculate the lowest two energy states using the particle-in-a-box approach.
(b) Calculate the partition function at T = 300 K. Show whether quantum effects areimportant or not. (Assume that they are important only if q is smaller than about10.)
(a) Three-dimensional particle-in-a-box
ε =h2
8mL2(n2
x + n2y + n2
z)
132
two lowest energy states:
1. nx = ny = nz = 1 (n2x + n2
y + n2z) = 3
2. 2 n’s = 1 and 1 n = 2, (12 + 12 + 22) = 6, so
ε1 =3h2
8mL2ε2 =
6h2
8mL2
h2
mL2=
(6.63× 10−34 J s)2
(1kg)(1 m)2= 4.39× 10−67 J
ε1 = 1.65× 10−67
ε2 = 3.30× 10−67 J
(b) qtranslation =
(2πmkT
h2
)3/2
V
=
[(6.28)(1 kg)(1.38× 10−23J/K)(300 K)
(6.63× 10−34 J s)2
]3/2
(1/m)3
= 1.44× 1070 accessible states
molecule
Quantum effects are clearly not important for objects as large as basketballs.
4. The statistical mechanics of an electron.
Calculate the two lowest energy levels for an electron in a box of volume V = 1 A3 (This is anapproximate model for the hydrogen atom). Calculate the partition function at T = 300 K.Are quantum effects important?
(a) As above
ε1 =
(3h2
8mL2
), ε2 =
3h2
4mL2
133
where m = mass of the electron = 9.11× 10−31 kg
h2
mL2=
(6.63× 10−34 J S)2
(9.11× 10−31 kg)(10−10m)2= 4.82× 10−17 J
ε1 = 1.81× 10−17 J
ε2 = 3.62× 10−17 J
(b) qtranslation =
[(6.28)(9.11× 10−31 kg)(4.14× 10−21 J)
(6.63× 10−34 J s)2
]3/2
(10−10 m)3
= 1.25× 10−5;
The classical approximation is not valid (since q < 10), and quantum effects are important.Since q cannot be smaller than 1, the value obtained above is an artifact of the assumptionthat the sum in the partition function can be approximated as an integral for this problem.
5. The translational partition function in two dimensions.
When molecules adsorb on a two-dimensional surface, they have one less degree of freedomthan in three dimensions. Write the two-dimensional translational partition function for anotherwise structureless particle.
For each dimension
qtranslation,1D =
[2πmkT
h2
]1/2
L
and qtranslation,2D = q2translation,1D so if A = area = L2, then
qtranslation,2D =
(2πmkT
h2
)A
134
6. The accessibility of rotational degrees of freedom.
Diatomic ideal gases at T = 300 K have rotational partition functions of approximatelyq = 200. At what temperature would q become small (say q < 10) so that quantum effectsbecome important?
qrot =8π2IkT
σh2
If qrot = 200 at T = 300 K, then
8π2Ik
σh2=
2
3. qrot =
2
3T = 10,
about when quantum effects become important, when T = 15 K.
7. The statistical thermodynamics of harmonic oscillations.
Write the internal energy, entropy, enthalpy, free energy, and pressure for a system of Nindependent distinguishable harmonic oscillators.
For harmonic oscillators, q = e−hνβ1−e−hνβ . For distinguishable particles, Q = qN .
(a)∆U
N= −1
q
∂q
∂β=
hνe−hνβ
1− e−hνβ
(b)S
kN= ln q +
∆U
NkTS
kN= − ln(1− e−hνβ) +
(hν
kT
)(e−hνβ
1− e−hνβ)
(c) F = −kT lnQ =⇒ F
NkT= − ln q = + ln(1− e−hνβ).
135
(Alternatively F = ∆U − TS, and the result in (c) is obtained directly from (a) and (b).)
(d) p = −(∂F
∂V
)
T
= 0
since q doesn’t depend on V .
(e) H = ∆U + pV = ∆U =⇒ H
NkT=
(hν
kT
)(e−hνβ
1− e−hνβ)
8. Orbital steering in proteins.
To prove that proteins do not require ‘orbital steering,’ a process once proposed to orienta substrate with high precision before binding, T. Bruice has calculated the dependenceof the total energy on the rotational conformation of the hydroxymethylene group of 4-hydroxybutyric acid at T = 300 K. Assume that the curve in the figure below is approxi-mately parabolic, ε = (1/2)k(α − α0)
2, where α is the dihedral angle of rotation. Use theequipartition theorem.
(a) Determine the spring constant k.
(b) What is the average energy 〈ε〉?
(c) What is the rms dihedral angle 〈α2〉1/2?
28162
28154
20 40 60 800
−ε (kcal mol−1)
Dihedral Angle α
α
Source: TC Bruice, Cold Spring Harbor Symposia on Quantitative Biology 36, 21–27 (1972).
136
Define the point α = 0 to be at about 54◦, the minimum in the curve. Take two points:
ε0 = 0 α0 = 0
ε1 = 5.3 α1 = 30◦ to determine the parabola
5.3kcal
mol=
1
2c(30◦)2
(a) =⇒ c = 1.2× 10−2 kcal
mole deg2
(b) 〈ε〉 = kT2
by equipartition since
〈ε〉 =
∫+30−30
cα2
2e−
cα2
2kT dα∫+30−30 e
− cα2
2kT dα=
c2
√π(
2kTc
)3/2
2√
2πkTc
so 〈ε〉 = 0.3kcal
mol
(c) Note that α2 is the same integral as above, except without the c/2 in the numerator,i.e.,
〈α2〉 =
∫α2e−
cα2
2kT
∫e−
cα2
2kT dαso
〈α2〉 =〈ε〉c2
=kT
2(c2
) =kT
c
〈α2〉1/2 =
(kT
c
)1/2
=
(0.6kcal/mol
1.2× 10−2 kcal/mol deg2
)1/2
= 7.07 degrees
137
9. The entropy of crystalline carbon monoxide at T = 0K.
Carbon monoxide doesn’t obey the ‘third law’ of thermodynamics: that is, its entropy isnot zero when the temperature is zero. This is because molecules can pack in either the C—- O or O —- C direction in the crystalline state. For example, one packing arrangementof twelve CO molecules could be:
C —- O C —- O C —- O C —- OC —- O C —- O O —- C O —- CO —- C O —- C C —- O C —- O
Calculate the partition function and the entropy of a carbon monoxide crystal at T = 0 K.
Since each CO molecule is distinguishable and has 2 possible orientations, then q = 2 andSCO,crystal
R= lnQ = ln 2n = n ln 2, where n is the number of moles of CO.
10. Temperature dependent quantities in statistical thermodynamics.
Which quantities depend on temperature?
(a) Planck’s constant h
(b) partition function q
(c) energy levels εj
(d) average energy 〈ε〉
(e) heat capacity CV for an ideal gas.
The temperature dependent quantities are (b) and (d).
138
11. Heat capacities of liquids.
(a) CV for liquid argon (at T = 100 K) is 18.7 J(K mol)−1. How much of this heat capacitycan you rationalize on the basis of your knowledge of gases?
(b) CV for liquid water at T = 10◦ C is about 75 J(K mol)−1. Assuming water has threevibrations, how much of this heat capacity can you rationalize on the basis of gases?What is responsible for the rest?
(a) Liquid argon is spherical (no rotations and no vibrations), and has only translationalfreedom, so
CV =3
2nR
= n(
3
2
)(8.314
J
K mol
)
= 12.47J
K mol
can be rationalized. The rest must be due to intermolecular interactions.
(b) 32nR due to translations, plus 0 or 3nR for vibrations depending whether they are
strong or weak, plus 32nR for rotations equals 3nR = 24.94 J
K mol. The rest must be
rationalized based on intermolecular interactions and hydrogen bonding.
12. The entropies of CO.
(a) Calculate the translational entropy for carbon monoxide CO (C has mass m = 12 amu,O has mass m = 16 amu) at T = 300 K, p = 1 atm.
(b) Calculate the rotational entropy for CO at T = 300 K. The CO bond has lengthR = 1.128× 10−10 m.
Here the gas molecules are indistinguishable. This means that we need to include thecorrection 1/N ! in the translation partition function of the N -particle system.
139
(a) S = Nk ln(qe
N
)+
∆U
T
= nR(1 + ln(q
N
)) +
3nR
2
S = nR
5
2+ ln
(2πmkT
h2
)3/2 (V
N
)
S
n= R
5
2+ ln
(2πmkT
h2
)3/2 (kT
p
)
m = (28)(1.78× 10−27kg)
kT = (1.38× 1023J/K)(300K)
p = 1.013× 105N/m2
h2 =[6.626× 10−34Js
]2
R = 8.314 J/K mol
=⇒ S
n= 150.3
J
K mol
(b)S
n= R[ln q + 1]
q =8π2µR2kT
σh2
σ = 1
µ =(12)(16)
12 + 16
= (6.86)× (1.67× 10−27 kg)
Srot
n= 47.28 J/K mol.
140
13. Conjugated polymers: why the absorption wavelength increases with chainlength.
Polyenes are linear double-bonded polymer molecules (C —- C—C)N , where N is the num-ber of C —- C—C monomers. Model a polyene chain as a box in which π-electrons areparticles that can move freely. If there are 2N carbon atoms each separated by bond lengthd = 1.4 A, and if the ends of the box are a distance d past the end C atoms, then the lengthof the box is ` = (2N + 1)d. An energy level, representing the two electrons in each bond,is occupied by two paired electrons. Suppose the N lowest levels are occupied by electrons,so the wavelength absorption of interest involves the excitation from level N to level N +1.Compute the absorption energy ∆ε = εN+1 − εN = hc/λ where c is the speed of light andλ is the wavelength of absorbed radiation, using the particle-in-a-box model.
The particle-in-a-box Equation (11.12) gives
εN =N2h2
8m`2
where m is the mass of an electron, so
∆ε = εN+1 − εN =h2
8md2
[(N + 1)2 −N2
(2N + 1)2
]
=h2
8md2(2N + 1)
=(6.626× 10−34Js)2(6.02× 1023molec/mole)
8(9.11× 10−31kg)(1.4× 10−10m)2(2N + 1)
=1.8× 103 kJ/mol
(2N + 1)=hc
λ
so the wavelength λ increases in proportion to N . The quantitative accuracy of this model isnot high, but the qualitative trend is correct.
141
14. Why are conjugated bonds so stiff?
As in problem 13, model polyene chain boxes of length ` ≈ 2Nd, where d is the averagelength of each carbon–carbon separation, and 2N is the number of carbons. There are 2Nelectrons in N energy levels, particles distributed throughout ‘boxes,’ according to the Pauliprinciple, with at most two electrons per level.
(a) Compute the total energy.
(b) Compute the total energy if the chain is ‘bent,’ that is if there are two boxes, each oflength `/2 containing N electrons each.
(a) From equation (11.8), we have
εn =(nh)2
8m`2n = 1, 2, 3, . . .
so the total energy E of 2N electrons is
E =N∑
n=1
εn =2h2
8m(nd)2
N∑
n=1
n2 =h2
4m`2N(N + 1)(2N + 1)
6
=h2
4md2
(N + 1)(2N + 1)
6N.
(b) For the bent chain,
Eb =4h2
8m(N2d2
4
)N/2∑
n=1
n2 =h2
2md2
(N + 1)(N + 2)
6N.
The difference is
Eb − E =h2
8md2
[4(N + 1)(N + 2)− 2(N + 1)(2N + 1)
6N
]
=h2
8md2
(N + 1)
N.
Since h2/(8md2) ' 1.8× 103 kJ/mol (see Problem 13), these bonds are predicted to bevery stiff, according to this simple model.
142
15. Electrons flowing in wires carry electrical current.
Consider a wire 1 m long and 10−4 m2 in cross-sectional area. Consider the wire to be abox, and use the particle-in-a-box model to compute the translational partition function ofan electron at T = 300 K.
The particle-in-a-box partition function is (Eq 11.18):
q =
[2πmkT
h2
]3/2
V
=
[2π(9.11× 10−31 kg)(1.38× 10−23 J
K)(300K)
(6.63× 10−34 J s)2
]3/2
(10−4 m3)
= 1.25× 1021 states
16. Fluctuations.
A stable state of a thermodynamic system can be described by the free energy G(x) as afunction of the degree of freedom x. Suppose G obeys a square law, with spring constantks, G(x)/kT = ksx
2 as shown in figure below.
G(x)
x
(a) Compute the mean square thermal fluctuations 〈x2〉 in terms of ks.
143
(b) Some systems have two single minima, with large spring constants k1, and others havea single broad minimum with small spring constant k2 as shown in figures (a) and (b)below. For example, two-state equilibria may have two single minima, and the freeenergies near critical points have a single broad minimum. If k2 = 1/4k1, what is theratio of fluctuations 〈x2
2〉/〈x21〉 for individual energy wells?
G(x)
x x
G(x)
k1 k1 k2
(a) Two Single Minima (b) One Broad Minimum
(a) 〈x2〉 =2∫∞0 x2e−ksx
2dx
2∫∞0 e−ksx2dx
=
14ks
√πks
12
√πks
=1
2ks
(b)〈x2
2〉〈x2
1〉=
1k21k1
=
1( 14)k11k1
= 4
144
17. Heat capacity for Cl2
What is CV at 800 K for Cl2 treated as an ideal diatomic gas in the high-temperature limit?
For a diatomic molecule in the classical limit, there is a contribution to U of (3/2)NkT fromtranslation, NkT from rotation, and NkT from vibration in the high-temperature limit. Thentotal U = (7/2)NkT. Thus
CV =
(∂U
∂T
)
V
=7
2Nk.
For one mole, CV is
7
2R =
7
2
(1.987
cal
K mol
)
= 6.95cal
K mol.
18. Protein-in-a-box.
Consider a protein of diameter 40 A trapped in the pore of a chromatography column. Thepore is a cubic box, 100 A on a side. The protein mass is 104 g mol−1. Assume the box isotherwise empty and T = 300 K.
(a) Compute the translational partition function. Are quantum effects important?
(b) If you deuterate all the hydrogens in the protein and increase the protein mass by10%, does the free energy increase or decrease?
(c) By how much?
(a) Assuming that the protein is ideal and takes up no volume, so the availabletranslational space is 100 Angstroms on each side:
145
qt =
(2πmkT
h2
)3/2
V
=
2π(104 g
mol)(10−3 kg
g)(1.38× 10−23 J
K)(300K)
(6.02× 1023mol
)(6.63× 10−34 J s)2
3/2
(10−10 m)3
= 9.77× 1011 quantum effects are not important
If we consider, though, that the protein is 40 Angstroms in diameter, that leaves only60 Angstroms free for motion in each direction. This can be corrected for bymultiplying by (0.6)3, giving qt = 2.11× 1011, so quantum effects are still not importanteven with excluded volume taken into account.
(b) F = −kT lnQ, so
∆F = −kT (lnQ2 − lnQ1)
= −kT ln
(Q2
Q1
)
= −3NkT
2ln(1.1)
=⇒ F2 − F1
NkT= −0.143
=⇒ free energy decreases (becomes more negative).
(c)∆F
n= −0.143RT = −86
cal
mol
19. Vibrational partition function of Iodine.
Compute the value of the vibrational partition function for iodine, I2, at T = 308K. (HINT:see table 11.2)
qvib = e− hν
2kT
1−e−hνkT
= e− θ
2T
1−e−θT, θI2 = 308
=⇒ qvib = e−308
2·308
1−e−308308
= e−0.5
1−e−1 ≈ 0.60651−0.3689
= 0.60650.6321
= 0.9595
146
20. Electron in a quantum-dot box.
An electron moving through the lattice of a semiconductor has reduced inertia. Assumethat the effective mass of the electron is only 10% of its actual mass at rest. Calculatethe translational partition function of the electron at room temperature (273◦K) in a smallsemiconductor particle of a cubic shape with a side
(a) 1 mm (10−3m),
(b) 100 A(100 · 10−10m);
(c) To which particle would the term quantum dot, i.e. system with quantum mechanicalbehavior, be applied, and why?
(a) qtransl = (2πm′ekT
h2 )3/2Vm′e = 0.1me = 0.1 · 9.11 · 10−31kgk = 1.38 · 10−23 J
K
h = 6.63 · 10−34Jsqtransl = (2·3.14·0.1·9.11·10−31·1.38·10−23·273
(6.63·10−34)2)3/2V = 3.43 · 1023 · V states
qtransl = 3.43 · 1023 · (10−3)3 = 3.43 · 1014 states
(b) qtransl = 3.43 · 1023 · (10−8)3 = 0.343 states.
(c) The latter system has just a few states and will behave in a quantum mechanical way.Thus, the term quantum dot can be applied to the smaller system.
21. A protein, quantum mechanics and the cell.
Let us assume for a moment that a protein of mass 50,000 g/mol can freely move in thecell. Approximate the cell as a cubic box 10 µm on a side.
147
(a) Compute the translational partition function for the protein in the whole cell. Arequantum effects important?
(b) The living cell, however, is very crowded with other molecules. Now assume that theprotein can freely move only 5 A along each, x-, y-and z-direction before it bumpsinto some other molecule. Compute the translation partition function and concludewhether quantum mechanical effects are important in this case.
(c) Now assume that we deuterate all the hydrogens in the protein (replace hydrogenswith deuterium atoms). If the protein mass is increased by 10%, what happens to thefree energy of the modified protein? By how much does it change?
(a) qt = (2πmkTh2 )3/2V = (2·3.1415·(50kg/mol)(1.38·10−23J/K)(300K)
(6.02·1023)2)3/2(10−15m3)
= (4.9 · 10−23−23)3/2 · 10−15 ≈ 1034 · 10−15 = 1019
Quantum effects are not important.
(b) qt = 1034 · (5 · 10−10m)3 ≈ 1034 · 125 · 10−30 = 1.25 · 106
Quantum effects can still be ignored.
(c) F = −kT lnQ∆F = −kT (lnQ2 − lnQ1) = −kT ln(Q2
Q1) = −NkT (m2
m1)3/2 = F2−F1
NkT= −0.143
Decreases.Change per one molecule:F2−F1
NkT= −0.143 · (1.38 · 10−23J/K) · 300K = 5.92 · 10−22J
22. Electron in benzene.
Consider an electron that can move freely throughout the aromatic orbitals of benzene.Model the electron as a particle in a 2-dimensional box 4 A ×4 A.
(a) Compute ∆ε, the energy change from the ground state to the first excited state,nx = ny = 2.
(b) Compute the wavelength λ of light that would be absorbed in this transition, if ∆ε =hcλ
, where h = Planck’s constant and c = speed of light.
(c) Will this transition be in the visible part of the electromagnetic spectrum (i.e., isliquid benzene colored or transparent), according to this simple model?
148
(a) ε(nx, ny) = h2
8ml2(n2
x + n2y).
∆ε = ε(2, 2)− ε(1, 1) = 3h2
4ml2= 3(6.626·10−34Js)2
4(9·10−31kg)(4·10−10m)2
∆ε = (2.27 · 10−18 Jelectron
)(6 · 1023 electronsmol
) = 1.37 · 106 Jmol
(b) λ = hc∆ε
= (6.626·10−34Js)(3·108m/s)2.27·10−18J
= 8.8 · 10−8m.
(c) Using Figure 10.2, this is in the UV-visible range.
23. Vibrational in insulin.
What is the average energy stored in the vibration degrees of freedom of one molecule ofinsulin, a protein with C256H381N65O76S6, at room temperature?
There are N = 256 + 381 + 65 + 76 + 6 = 784 atoms in the molecule.〈ε〉vib = (3N − 6) · 1
2kT = (3 · 784− 6) · 150 · 1.38 · 10−23 = 4.85 · 10−18J
〈θ〉vib = 4.85·10−18
1.38·1023 = 351, 449K
24. Escape velocity of gases from the moon.
The escape velocity of an object to leave the moon is 3.4 km/s. The temperature on thesunny surface of the moon is 400 K. What is the weight of the gas which will escape themoon with an average velocity 〈V 2
x 〉1/2. Draw a short conclusion from the result.
12m〈V 2
x 〉1/2 = 12kT
m = kT〈V 2x 〉2 = 400·1.38·10−23
34002 = 4.78 · 10−28kg = 4.78 · 10−25g
m (in g/mol) = 4.78 · 10−25 · 6.022 · 1023 = 0.29g/molNo gas will escape the Moon.
149
Chapter 12Temperature, Heat Capacity
1. Heat capacity peaks from phase transitions.
The peak heat capacity in the figure below shows that helium gas adsorbed on graphiteundergoes a transition from an ordered state at low temperature to a disordered state athigh temperature. Use the figure below to estimate the energy associated with the transition.
C/Nk
02
4
6
1
2
3 4T (K)
Source: HJ Kreuzer and ZW Gortel, Physisorption Kinetics, Springer Verlag, Heidelberg, 1986.Data are from RL Elgin and DL Goodstein, Phys Rev A 9, 2657–2675 (1974); Proceedings ofthe 13th International Conference on Low Temperature Physics, edited by KD Timmerhaus,WJ O’Sullivan and EF Hammel, Plenum, New York.
150
The peak in the heat capacity suggests that some kind of bond or interaction breaks at atemperature of about T = 3K. A characteristic energy can be computed from
ε = kT =
(1.987
cal
mol deg
)(3K) ∼= 6
cal
mol.
Since this heat capacity peak is much narrower than in the two-state model, it shows thatthe two-state model would not be an adequate way to model this. We also need to accountfor cooperativity.
2. Two-state model of a hydrogen bond.
Suppose a hydrogen bond in water has an energy of about 2 kcal mol−1. Suppose a ‘made’bond is the ground state in a two-state model and a ‘broken’ bond is the excited state. AtT = 300 K, what fraction of hydrogen bonds are broken?
Use Equation (12.4),
ln
(fgnd
fexc
)=
ε
kT
=2000 cal/mol
(2 cal/mol)(300 K)
= 3.33
=⇒(fgnd
fexc
)∼= 29 so
fexc
fexc + fgnd
=1
1 + fgnd/fexc
=1
29∼= 3.4%
of the bonds are broken.
151
3. A random energy model of glasses.
Glasses are materials that are disordered—and not crystalline—at low temperatures. Here’sa simple model. Consider a system that has a Gaussian distribution of energies E, accordingto Equation (12.16):
p(E) = p0 exp[−(E − 〈E〉)2/2∆E2
],
where 〈E〉 is the average energy and ∆E characterizes the magnitude of the fluctuations,that is, the width of the distribution.
(a) Derive an expression showing that the entropy S(E) is an inverted parabola,
S(E) = S0 −k(E − 〈E〉)2
2∆E2.
(b) An entropy catastrophe happens (left side of the curve in the figure below whereS = 0. For any physical system, the minimum number of accessible states is W = 1,so S = k lnW implies that the minimum entropy is S = 0. At the point of the entropycatastrophe, the system has no states that are accessible below an energy E = E0.Compute E0.
(c) The glass transition temperature Tg is the temperature of the entropy catastrophe.Compute Tg from this model.
E0
S
S0
E〈E〉0
(a) Substitute p(E) into S(E) = k ln p(E) to get:
S
k= ln p0 −
(E − 〈E〉)22∆E2
.
152
Defining S0 = k ln p0, this becomes
S(E) = S0 −k (E − 〈E〉)2
2∆E2.
(b) At S = 0, the entropy equation gives
0 = S0 −k (E0 − 〈E〉)2
2∆E2
12(a) =⇒ (E0 − 〈E〉)2 =2∆E2S0
k
=⇒ E0 = 〈E〉 −∆E
√2S0
k.
(c) Substitute the entropy equation S(E) into the definition of temperature, evaluated atthe point E = E0:
1
Tg=
∂S
∂E
∣∣∣∣∣E0
=−k (E − 〈E〉)
∆E2
∣∣∣∣∣E0
.
Now substituting the square root of Equation 12(a) into this expression gives
12(b)1
Tg=
k
∆E2
√2∆E2S0
k.
(The minus sign has disappeared because Equation 12(a) has two roots, one positiveand one negative. Since our interest is in the left side of the S(E) curve, Tg > 0.)Rearranging Equation 12(b) gives
Tg =
√∆E2
2kS0.
For more details, see Onuchic et al (1997).
153
4. Fluctuations in enthalpy.
The mean square fluctuations in enthalpy for processes at constant pressure are given byan expression similar to Equation (12.17) for processes at constant pressure:
〈δH2〉 =⟨(H − 〈H〉)2
⟩= kT 2Cp.
What is the root-mean-square enthalpy fluctuation 〈δH2〉1/2 for water at T = 300 K?
〈∂H2〉1/2 = (kT 2Cp)1/2
=
[(2
cal
mol K
)(300 K)2
(18
cal
mol K
)]1/2
= 1.8kcal
mol
5. Oxygen vibrations.
Oxygen molecules have a vibrational frequency of 1580 cm−1 (see Example 11.3). If therelative populations are fground/fexcited = 100 in the Schottky model, what is the temperatureof the system?
1
T=
(k
ε0
)ln
(fgnd
fexc
)
=
(k
hν
)ln
(fgnd
fexc
)
=(
1
θvib
)ln
(fgnd
fexc
)
=1
2274ln(1000)
=⇒ T =2274
ln(1000)= 329 K
154
6. Modeling a population inversion.
Population inversion, when more particles of a system are in an excited state than a groundstate, is used to produce laser action. Consider a system of atoms at 300 K with 3 energylevels: 0 kcal/mol, 0.5 kcal/mol, and 1.0 kcal/mol. For this problem, please use 5 decimalplaces.
(a) Compute the probabilities p1*, p2*, and p3* that an atom is in each energy level.
(b) What is the average energy 〈ε〉 of the system? What is the entropy (in cal/mol·K) ofthe system?
Now suppose you excite the system to cause a population inversion, resulting in a newpopulation distribution p** as follows: p1** = p3*, p2** = p2*, p3**= p1*.
(c) What is the average energy 〈ε〉 of the system after the population inversion? What isthe entropy (in cal/mol·K) of the system after the population inversion?
(d) What is the temperature of the system after the population inversion?.
(a) Q =∑tj=1 e
− εjkT = 1 + 0.434233 + 0.18691 = 1.61924
p1* = 11.61924
= 0.61757, p2*= 0.4342331.61924
= 0.26700, p3*= 0.186911.61924
= 0.11543
(b) 〈ε〉 =∑tj=1 pj ·Ej = 248.93 cal
mol
S = −k∑tj=1 pjln(pj) = 1.78 cal
mol·K
(c) Using new p**, 〈ε〉 = 751.07 calmol
S = 1.78 calmol·K (Same as in part 2 since the entropy should not change.)
(d) p3**= p1* = e−εjkT
q= e
− 503.14779T
1+e−251.57390
T +e−503.14779
T= 0.61757
=⇒ T = −300KWould also accept:1t
= kεln(
fgndfexc
) since p2** = p2*
7. Schottky model for hydrogen.
Hydrogen molecules have a vibrational frequency of 4300cm−1. If 0.5% of the molecules arein the excited state, what is the temperature of the system?
155
1T
= 1θvib
ln[99.50.5
] where θvib = hνk
θvib = (6.626·10−34)(4300)(2.994·1010 )1.38·10−23 = 6187.67
So: T = 1168K
8. Finding the Cv of a gas.
A gas is placed in an airtight container. A piston decreases the volume by 10% in anadiabatic process, and the temperature is observed to increase by 10%. What is the constant-volume heat capacity of the gas?
T2
T1= (V1
V2)NkCv
110100
= (10090
)NkCv
Take the Log of both sides:Log(1.10) = Log(1.11) · R
Cv
Cv = 0.04530.041
(2 calmol·K ) = 2.21 cal
mol·K
156
Chapter 13Chemical Equilibria
1. Iodine dissociation.
Compute the dissociation constant Kp for iodine, I2 → 2I, at T = 300 K.
Follow Example 13.3. But now
kT =(
300
1000
)1.363× 10−25 m3atm = 4.09× 10−26m3atm
e−∆ε0RT = e−
35,600(1.987)(300) = 1.16× 10−26
qrI2 =(
300
1000
)9310 = 2793
qvI2 =1
1− e− 308300
= 1.56
q2tI
qtI2=
(300
1000
)3/2
3.01× 1033m−3 = 4.95× 1032m−3
so Kp = (4.09× 10−26m3atm)(1.15× 10−26)(
1
2793
)(1
1.56
)(16)(4.95× 1032m−3)
= 8.5× 10−22 atm
157
2. Temperature dependence of a simple equilibrium.
In a reaction
AK−→ B,
the equilibrium constant is K = 10 at T = 300 K.
(a) What is ∆µ◦?
(b) If ∆h◦ = 10 kcal mol−1, what is K at T = 310 K?
(c) What is ∆s◦ at T = 300 K?
(a) ∆µ◦ = −RTlnK
= −(1.987cal
K mol)(300K) ln(10)
= −1.37kcal
mol
(b) The van’t Hoff equation gives
lnK2 = lnK1 −∆h◦
R
(1
T2− 1
T1
)
= ln(10)− 10, 000 calmol
2 calK mol
(1
310− 1
300
)
= 2.303 + 0.5376 = 2.84
=⇒ K2 = 17
(c) ∆s◦ =∆h◦ −∆µ◦
T
=(10, 000 + 1, 370) cal
mol
300K
= 37.9cal
K mol
158
3. Dissociation of oxygen, O2 2O.
Compute Kp, the pressure-based equilibrium constant for this dissociation reaction at T =3000 K. The electronic ground-state degeneracy for O is g0(0) = 9.
Similar to Example 13.3
Kp = kT
[(2πmOkT
h2
)3/2]2
(2πmO2
kT
h2
)3/2
2Θr(O2)
T
(1− e−(Θv(O2)/T )
) g2(O)
g(O2)e−[D0/(kT )]
Since mO2 = 2mO, this simplifies to
Kp =
(πmOkT
h2
)3/2 (2ΘrR
NA
)(1− e−(Θv/T)
) g2(O)
g(O2)e−(D0/kT )
From Table 11.2
mO =0.016 kg
6.02× 1023= 2.66× 10−26 kg
Θr = 2.08 K
Θv = 2274 K
D0 = 492kJ
mol
g0(O2) = 3
g0(O) = 9 =⇒ g2(O) = 81
So we compute that
(πm0kT
h2
)3/2
=
[π(2.66× 10−26 kg)(1.38J/K)(3000 K)
[6.626× 10−34 Js]2
]3/2
= (7.88× 1021 m−2)3/2 ≈ 7.00× 1032 m−3
(2ΘrR
NA
)=
2(2.08 K)(8.21× 10−5 m3 atm mol−1 K−1)
6.02× 1023 molecules mol−1
159
= 5.67× 10−28 m3 atm
(1− e−Θv/T
)=(1− e−2274 K/3000 K
)= 0.531
g2(O)
g(O2)=
81
3= 27
e−D0/kT = exp
(−492 kJ/mol
(8.314 J)(3000 K)
)= e−19.726 = 2.71× 10−9
Substituting these values gives
Kp = (7.00× 1032 m−3)(5.67× 10−28 m3 atm)(0.531)(27)(2.71× 10−9) = 0.0154 atm
4. Temperature dependence of Kp.
For the dissociation of O2, derive an expression for d lnKp/dT near T = 300 K from theexpression for Kp that you used in problem 3.
Start with
Kp =
(πmOkT
h2
)3/2 (2ΘrR
NA
) (1− e−
ΘvT
) g2(O)
g(O2)e−
D0kT
Since Θv
T>> 1, (1− e−
ΘvT ) ≈ 1, so the temperature dependence can be expressed as
Kp(T) = C0T3/2e−
D0kT .
Therefore ln Kp = ln C0 +3
2ln T− D0
kT
andd ln K
dT=
3
2T+
D0
kT 2
160
Near T = 300 K,
d ln K
dT=
3
2(300K)+
492, 000 Jmol
(8.314 Jmol K
)(300K)2
= 5× 10−3 K−1 + 0.65 K−1 ≈ 0.66 K−1
It is clear that the term D0
kTdominates the temperature dependence.
5. Polymerization.
Consider a polymerization reaction in the gas phase in which n moles of identical monomersare in equilibrium with one mole of chains of n-mers:
n( )K−→ − − − −︸ ︷︷ ︸
n monomers
(a) Do you expect typical polymerization processes to be driven, or opposed, by enthalpy?By entropy? What are the physical origins of these enthalpies and entropies?
(b) Do you expect polymerizations to be exothermic (giving off heat) or endothermic(taking up heat)? Explain the explosions in some polymerization processes.
(c) Are polymerization equilibrium constants for long chains more or less sensitive totemperature than for shorter chains?
(a) The polymerization equilibrium nAK−→ An is described by Eq 13.19
K =qnqn1e−
(n−1)∆ε0kT
where q1 is the partition function for the monomer, qn is for a polymer chain, ∆ε0 is theenergy of forming one monomer-monomer bond in the chain, and n is the number ofmonomers in a chain.(We’re simplifying here - most polymers have distributions of molecular weight, ratherthan a single value n.)Bond formation typically leads to ∆ε0 = εbonded − εnotbonded < 0, so it is driven byenthalpy and opposed by the translational entropy involved in bringing the monomerstogether in space and orienting them appropriately to react to form a bond. (See Ex.13.3 for iodine dissociation.)
161
(b) Since ∆ε0 < 0, the energy of the system is reduced by bond formation, so according tothe first law, that energy must have left the system either as heat or work. Thesereactions are typically exothermic. The amount of heat given off is often large becauseof the factor (n− 1) in the exponent of the above equation, where n can be greaterthan 103; this resulted in explosions in the early days of polymer chemistry.
(c) The van’t Hoff relation gives (approximately)
∂ lnK
∂T=
∆h◦
kT2=
(n− 1)∆ε0
kT2
so the slope of lnK versus T gets steeper as n gets larger. Hence, formation of longerchains is more sensitive to temperature than for shorter chains.
6. Hydrogen dissociation.
A hydrogen atom can dissociate in the gas phase:
HK−→ H+ + e−.
Calculate the equilibrium constant K for temperature T = 5000 K. There is no rotationalor vibrational partition function for H, H+, or e−, but there are spin partition functions:qs = 1 for H+, and qs = 2 for e−. ∆D = −311 kcal mol−1.
Kp =qH+qe−
qHe−
∆ε0kT
= kT
[2πmH+kT
h2
]3/2 [2πmekT
h2
]3/2e−
∆ε0kT
[2πmHkT
h2
]3/2
= kT
[2πmekT
h2
]3/2
e−∆ε0kT
=(8.21× 10−5 m3atm
K mol)(5000K)
6.02× 1023 moleculesmol
{(2π)(9.11× 10−31kg)(1.38× 10−23 J
K)(5000K)
(6.626× 10−34Js)2
}3/2
e−311,000/9935
= (6.82× 10−25 m3atm)(8.53× 1026 m−3)(2.54× 10−14)
= 1.48× 10−11 atm
162
7. Free energy along the reaction coordinate.
Consider the ideal gas-phase equilibrium 2A B, at constant temperature and a constantpressure of 1 atm. Assume that there is initially 1 mole of A and no B present, and thatµ◦A = 5 kcal mol−1 and µ◦B = 10 kcal mol−1 at this temperature.
(a) Show that G, at any time during the reaction, can be written as
G =
5 + RT ln
(
1− 2ξ
1− ξ
)1−2ξ (ξ
1− ξ
)ξ kcal
mol,
where ξ is the extent of reaction.
(b) What is the value of the extent of reaction ξ at equilibrium?
(a) G =2∑
i=1
niµi = nAµA + nBµB.
Let ξ represent the progress toward B:
dξ = dnB = −dnA2.
Integrating gives
nB = nB(0) + ξ
nA = nA(0)− 2ξ
where nB(0) means the value of nB when ξ = 0. Since there is initially 1 mole of A,nA(0) = 1, nB(0) = 0, so
nB = ξ
nA = 1− 2ξ
ntotal = nA + nB = 1− ξ.
Now let’s convert these quantities into pressures. The total pressure of the gas is thesum of the partial pressures
p = (nA + nB)(RT
V
)= 1 atm.
163
Therefore you can express the partial pressures as
pAp
=nA
nA + nB=
1− 2ξ
1− ξ
andpBp
=nB
nA + nB=
ξ
1− ξ
The chemical potentials are
µA = µ◦A +RT ln
(pAp
), µB = µ◦B +RT ln
(pBp
)
where µ◦A|B is the chemical potential when pA|B = p = 1 atm. Therefore Eq 13.a gives
G = nAµ◦A + nART ln
(pAp
)+ nBµ
◦B + nBRT ln
(pBp
)
= (5kcal
mol(nA + 2nB) +RT ln
[(pAp
)nA (pBp
)nB]
= 5kcal
mol+RT ln
(
1− 2ξ
1− ξ
)1−2ξ (ξ
1− ξ
)ξ
(b) At equilibrium, ∂G/∂ξ = 0. Taking the derivative of G/RT gives
2 ln
(1− ξ1− 2ξ
)+ ln
(ξ
1− ξ
)− 2 +
(1− 2ξ
1− ξ
)+ 1 +
(ξ
1− ξ
)= 0
=⇒ ξ(1− ξ)(1− 2ξ)2
= 1
=⇒ 1− 5ξ + 5ξ2 = 0
=⇒ ξ = 0.276
The other solution, ξ = 0.724, gives a maximum in G, not a minimum, as you can seeby checking the second derivative.
164
8. Pressure denaturation of proteins.
For a typical protein, folding can be regarded as involving two states, native (N) anddenatured (D),
NK−→ D.
At T = 300 K, ∆µ◦ = 10 kcal mol−1. Applying about 10,000 atm of pressure can denaturea protein at T = 300 K. What is the volume change ∆v for the unfolding process?
We assume here that with the added 10,000 atm pressure, K2 = 1 =⇒ ∆µ◦2 = 0. Therefore,following Example 13.6
∆v = vD − vN = −RT(lnK2 − lnK1)
∆p= −RT(−∆µ◦2
RT−−∆µ◦1
RT)
∆p
= −(0 + 10, 000) calmol
10, 000 atm
8.21× 10−5 m3 atm
K mol
2 calK mol
(
1010 A
m
)3 (1 mole
6.02× 1023 molecules
)
= 68A3
molecule
Since a protein may have about 100 amino acids, and each amino acid has a volume of about100 A3, the volume of a protein is about 104 A3. Hence ∆v is a small change.
9. Clusters.
Sometimes isolated molecules of type A can be in a two-state equilibrium with a cluster of
m monomers, mAK−→ Am, where Am represents an m-mer cluster.
(a) At equilibrium, what is the relationship between µ1, the chemical potential of themonomer, and µm, the chemical potential of A in the cluster?
(b) Express the equilibrium constant K in terms of the partition functions.
(a) Use Equation (13.16) with a = m, b = 0, c = 1 to get µm = mµ1
(b) Equation (13.19) gives K =
(qmqm1
)e−∆ε0/kT
10. Two-state protein folding model.
165
In Example 8.2, we considered a two-state model of proteins: the folded state has energy−ε0 (where ε0 > 0) and the unfolded state has γ = 4 open conformations. The partitionfunction is
q = 4 + eε0kT .
(a) Write an expression for the equilibrium constant for folding, Kf =(populationfolded)/(populationunfolded) in terms of ε0 and temperature T .
(b) Using the van ’t Hoff expression, (∂lnK)/(∂(1/T )), derive the enthalpy of folding,∆Hf , for this model.
(c) What is the free energy of folding vs. temperature T , ∆Ff (T ) = Ffolded − Funfolded?
(d) This model predicts that ∆Ffold(T ) is a linearly increasing function of T . This is oftena good approximation near the denaturation midpoint, Tm.
For myoglobin, which has about 150 amino acids, the slope is 5 kcal/(mol - 10◦ C) atTm = 70◦C. If instead of choosing γ = 4, as in our simple model, you chose γ to fitthe experimental slope in the figure, what value of γ is needed?
(a) Since the q’s cancel, Kf = eε0kT
4
(b) ∂lnK∂( 1T
)= −∆H
K
∂( 1T
)
∂T= − 1
T 2 =⇒ ∂lnK∂T
= ∆HKT 2
K = 14eε0kT =⇒ lnK = ln(1
4) + ln(e
ε0kT ) = ln(1
4) + ε0
kT
166
=⇒ ∂[ln( 14)+
ε0kT
]
∂T= ∆H
KT 2 =⇒ − ε0kT 2 = ∆H
KT 2
=⇒ ∆Hf = −ε0(c) ∆F = −kT ln(Kf )
∆F = −kT ln(14eε0kT ) = −kT [ln(1
4) + ln(e
ε0kT )]
∆F = kT ln(4)− kTε0kT
=⇒ ∆F = kT ln(4)− ε0OR∆F = ∆Ff −∆FunFf = Uf − TSf = −ε0 − Tkln(1)Fun = Uun − TSun = 0− Tkln(4)=⇒ ∆F = −ε0 + kT ln(4)
(d) ∂∆F∂T
= kln(γ) = 5000cal10oC·mol = 500cal
K·mol = Rln(γ)500calK·mol
mol·K1.987cal
= ln(γ) = 251.6=⇒ γ = e251.6 = 1.81 · 10109
167
Chapter 14Phase Equilibria
1. Applying the Clausius–Clapeyron equation.
(a) The vapor pressure of water is 23 mm Hg at T = 300 K, and 760 mm Hg at T = 373 K.Calculate the enthalpy of vaporization, ∆hvap.
(b) Assuming that each water has z = 4 nearest neighbors, calculate the interactionenergy, wAA.
(a) ln
(p2
p1
)= −∆hvap
R
(1
T2− 1
T1
)
=⇒ln(
76023
)
(1
373− 1
300
)(
1.987cal
mol
)= −∆hvap
=⇒ ∆hvap = 10.653kcal
mol
(b) Since
−zw2
= ∆hvap,
w = −∆hvap
2≈ −5.3
kcal
mol.
168
2. How does surface tension depend on temperature?
If the surface tension of a pure liquid is due entirely to energy (and no entropy), then willthe surface tension increase, decrease, or not change with increasing temperature?
The surface tension will not change with increasing temperature. γ = − w2a
is independent ofT if w is independent of T.
3. Why do spray cans get cold?
Explain why an aerosol spray can gets cold when you spray the contents.
The contents of a spray can are under high pressure. Pushing the button opens the systemsto the atmosphere, reducing the pressure below the vapor pressure of the fluid. This causesthe fluid to vaporize. Vaporization absorbs heat from the can, causing it to become cold.
4. The surface tension of water.
The surface tension of water γ(T ) decreases with temperature as shown in the figure below.
(a) As a first-order approximation, estimate the water–water pair attraction energy wAA,assuming that the interaction is temperature independent.
(b) Explain the basis for this attraction.
(c) Explain why γ decreases with T .
0 50 100
70
80
50
60
T ( C)
γ (dyn cm−1)
Surface tension of water versus temperature. Source: drawn from data in CRC Handbook ofChemistry and Physics, 47th edition, RC Weast, editor, The Chemical Rubber Co, Cleveland,1962.
169
(a) From the lattice model of surface tension
γ = −wAA2a
so wAA = −2aγ
Assume the area of a water molecule is about a = 5 A2, and take a value for γ, sayγ = 70 dynes/cm, then
wAA ≈ −2(5A2)(70
g
s2
)(1 J s2
107g cm2
)(1 cm
108 A
)2
×(
6.022× 1023 molecules
mole
)
≈ −4.214kJ
mol
(b) Hydrogen bonds and van der Waals attractions.
(c) One simple explanation is that hydrogen bonds in water are broken as temperatureincreases, which weakens the water–water attractions and thus the surface tension.
5. Applying the Clausius–Clapeyron equation again.
Suppose you want to add a perfuming agent to a liquid aerosol spray. You want its vaporpressure to double if the temperature increases from 25◦ C to 35◦ C. Calculate the pairinteraction energy that this agent should have in the pure liquid state. Assume that thenumber of nearest–neighbor molecules in the liquid is z = 6.
The vapor pressure is given by
p = p0 exp(zw
2RT
)
We want
p2
p1= exp
zw
2R(
1T2− 1
T1
)
= 2
170
Using z = 6, R = 2 cal/mol K, we have
zw
2R
(1
T2− 1
T1
)= ln 2
w =2R ln 2
z(
1T2− 1
T1
)
=2(2 cal
mol K) ln 2
6(
1308K− 1
298K
)
w = −4.24kcal
mol
6. How does the boiling temperature depend on the enthalpy of vaporization?
A fluid has a boiling temperature T = 300 K at p = 1 atm; ∆hvap = 10 kcal mol−1. Supposeyou make a molecular modification of the fluid that adds a hydrogen bond so that the newenthalpy is ∆hvap = 15 kcal mol−1 at p = 1 atm. What is the new boiling temperature?
p
p0= exp
(−∆hvap2
RT2
)
andp
p0= exp
(−∆hvap1
RT1
)
=⇒ exp
(−∆hvap2
RT2
)= exp
(−∆hvap1
RT1
)
=⇒ ∆hvap2
RT2
=∆hvap1
RT1
=⇒ T2 = T1
(∆hvap2
∆hvap1
)
= (300 K)
(15 kcal/mol
10 kcal/mol
)
= 450 K
171
7. Trouton’s Rule.
(a) Using Table 14.1, show that the entropy of vaporization, ∆svap = ∆hvap/Tb, is rel-atively constant for a broad range of different substances. ∆hvap is the enthalpy ofvaporization and Tb is the boiling temperature. This is called Trouton’s rule.
(b) Why are the values of ∆svap for water and CO2 in Table 14.1 larger than for othersubstances?
(a) Here are some examples from Table 14.1:
C6H6 ∆S =30, 800 J/mol
353 K= 87.5
J
K mol
C2H6 ∆S =14, 700 J/mol
184.55 K= 79.6
J
K mol
H2S ∆S =18, 673 J/mol
212.8 K= 88.1
J
K mol
Ar ∆S =6, 506 J/mol
87.29 K= 74.5
J
K mol
Na ∆S =98, 010 J/mol
1156 K= 84.7
J
K mol
Trouton’s rule says that the typical ∆Svap is about 85 J/mol.
(b) H2O ∆S =40, 656 J/mol
373.15 K= 108.95
J
K mol
CO2 ∆S =25, 230 J/mol
194.64 K= 129.62
J
K mol
The rationale is that water is more ordered in its liquid state than simpler materials. Thevalue for CO2 is for sublimation from the crystal state, which is more ordered than a typicalliquid.
172
8. Sublimation of graphite.
The heat of sublimation of graphite is ∆hsub = 716.7 kJ mol−1. Use this number to estimatethe strength of a carbon–carbon bond.
Equation (15.24) applies to sublimation as well as to vaporization, so
∆hsub = −zwAA2
In graphite, carbon is in the sp2 hybridization state, with three bonds and a delocalized πbond. For simplicity, we take z = 4 neighbors, so
wAA = −∆hsub
2= −358.35 kJ/mol = −85.6 kcal/mol
The experimental value of a C–C single bond is −342 kJ/mol, but it can typically vary fromabout −300 to −400 kJ/mol depending on neighboring bond substituent effects.
9. Surface tension of mercury.
The surface tension of water is 72 ergs cm−2 and of liquid mercury is 487 ergs cm−2. If awater-water attraction is about 5 kcal-mol−1, what is the mercury-mercury attraction?
wAA2
wAA1= γ2
γ1
=⇒ wAAHg = (48772
)(5kcalmol
) = 33kcalmol
10. Squeezing ice.
Use the Clapeyron relation to compute the pressure that is required to lower the meltingtemperature of ice by 10 K. For water, ∆h = 6.008 kJ-mol−1 and ∆v = −1.64 cm3-mol−1.
∫ P2P1dP = ∆h
∆v
∫ T2T1
dTT
P2 = P1 + ∆h∆vln(T2
T1)
The melting temperature of water at 1atm is 273.15K.P2 = 1.013 · 105 Pa + 6.008·103J
−1.64·10−6m3 ln(263.15273.15
)
P2 = 1.366 · 108 Pa ( 1atm1.013·105Pa
) = 1348atm = 0.137 kJcm3
173
Chapter 15Solutions and Mixtures
1. Ternary lattice mixtures.
Consider a lattice model liquid mixture of three species of spherical particles, A, B, and C.As with binary mixtures, assume that all N = nA + nB + nC sites are filled.
(a) Write an expression for the entropy of mixing.
(b) Using the Bragg–Williams approximation, write an expression for the energy of mixingU in terms of the binary interaction parameters χ.
(c) Write an expression for the chemical potential µA of A.
(a) W = N !nA!nB !nC !
. Use Stirling’s approximation ln x! ≈(xe
)xto get
S
Nk=
1
NlnW
= −xA ln xA − xB ln xB − xC ln xC ,
where xA = nA/N , xB = nB/N , xC = nC/N .
(b) U = mAAwAA +mBBwBB +mCCwCC +mABwAB +mACwAC +mBCwBC
Conservation relations:
znA = 2mAA +mAB +mAC
znB = 2mBB +mAB +mBC
znC = 2mCC +mAC +mBC
174
so we have:
mAA =znA −mAB −mAC
2
mBB =znB −mAB −mBC
2
mCC =znC −mAC −mBC
2
Substituting into the expression for U and rearranging terms, we get:
U =zwAA
2nA +
zwBB2
nB +zwCC
2nC +mAB
(wAB −
wAA + wBB2
)
+mAC
(wAC −
wAA + wCC2
)+mBC
(wBC −
wBB + wCC2
)
Defining χ parameters and using the Bragg–Williams approximation gives
U
NkT=
zwAA2kT
nAN
+zwBB2kT
nBN
+zwCC2kT
nCN
+nAnBN2
χAB +nBnCN2
χBC
+nAnCN2
χAC
=zwAA2kT
xA +zwBB2kT
xB +zwCC2kT
xC + χABxAxB
+χBCxBxC + χACxAxC
so
∆Umix
NkT= χABxAxB + χBC xBxC + χACxAxC
(c) From the definition of chemical potential,
µAkT
=∂(FkT
)
∂nA
∣∣∣∣∣∣nB ,nC
We want to hold nB and nC constant, but not N , so write in terms of nA, nB, and nConly:
F
kT=U − TSkT
=zwAA2kT
nA +zwBB2kT
nB +zwCC2kT
nC +nAnB
(nA + nB + nC)χAB
175
+nAnC
(nA + nB + nC)χAC +
nBnC(nA + nB + nC)
χBC
−nA ln(nA + nB + nC
nA
)− nB ln
(nA + nB + nC
nB
)
−nC ln(nA + nB + nC
nC
)
so
µAkT
=zwAA2kT
+ χAB
(NnB − nAnB · 1
N2
)+ χAC
(NnC − nAnC · 1
N2
)+ ln xA
− nA(xA)
(nA · 1−N · 1
n2A
)− nB(xB)
(1
nB
)− nC(xC)
(1
nC
)
=zwAA2kT
+ χABxB(1− xA) + χACxC(1− xA) + ln xA − xA(1− 1
xA)− xB − xC
=zwAA2kT
+ χABxB(xB + xC) + χACxC(xB + xC) + ln xA − xA + 1− xB − xC
=zwAA2kT
+ χABxB(xB + xC) + χACxC(xB + xC) + ln xA
2. Enthalpy of mixing.
For a mixture of benzene and n-heptane having equal mole fractions, x = 1/2, and temper-ature T = 300 K, the enthalpy of mixing is ∆Hmix = 220 cal mol−1. Compute χAB .
∆Hmix ≈ ∆Umix = x(1− x)RTχAB
so χAB ≈ 4∆Hmix
RT=
(4)(220 cal
mol
)
(1.987 cal
K mol
)(300 K)
= 1.48
176
3. Plot µ(x).
Plot the chemical potential versus x for:
(a) χAB = 0,
(b) χAB = 2,
(c) χAB = 4.
Equation (15.15) gives
µ
kT=zwAA2kT
+ ln x+ χAB(1− x)2
1
χ = 4
χ = 0
χ = 2
zwAA
2kT
x
.69
µ
kT
1
177
4. Limit of x lnx terms in mixing entropies and free energies.
What is the value of x ln x as x 0?
As x 0, x ln x 0. More specifically,
limx 0
ln x1x
= limx 0
1x
− 1x2
= limx 0
(−x) = 0,
using l’Hospital’s rule. This rule says that the limit (x− > 0) of the ratio of functionsf(x)/g(x) equals the limit of their derivatives, f ′(x)/g′(x).
5. Hydrophobic entropy.
The entropy of dissolving benzene in water at high dilution is approximately 14 cal mol−1
K−1 at T = 15◦C.
(a) How does this compare with the mixing entropy?
(b) Speculate on the origin of this entropy.
(a) The maximum possible mixing entropy occurs for x = 0.5, where
S
Rn= 2(−0.5 ln 0.5)
= − ln 0.5 = −0.69,
so
S
n= (−0.69)(1.987 cal/mol K)
= −1.37 cal/mol K.
The mixing entropy is much smaller than the hydrophobic entropy.
(b) The hydrophobic entropy involves rotational entropy in the waters that surround thesolute, which is not taken into account in the mixing entropy.
178
6. Solubility parameters.
The quantity χAB describes AB interactions relative to AA and BB interactions. If insteadof a 2-component mixture, you had a mixture of N different species A,B, . . ., then, to obtainthe pairwise quantities χij for all of them, you would need ∼ N 2 experiments involvingmixtures of all the components i = 1, 2, . . . , N with all the components j = 1, 2, . . . , N .However, sometimes this can be much simpler. If all the species are nonpolar, then you canmake the approximation wAB ≈
√wAAwBB (see Chapter 24). Show how this reduces the
necessary number of experiments to only ∼ N .
χAB =z
kT
(wAB −
wAA + wBB2
).
If
wAB =√wAAwBB ,
then
χAB =z
kT
(√wAAwBB −
wAA + wBB2
).
Let
δA =
√|wAA|
2
and
δB =
√|wBB|
2,
soχAB = − z
kT(δA− δB)2.
An experiment on pure A gives wAA and δA, and an experiment on pure B gives wBB andδB, so the derived equation gives a way to estimate the interaction parameter χAB withoutperforming experiments on mixtures.
179
7. Self-assembly of functional electrical circuits.
G Whitesides and colleagues have pioneered the formation of nano- and meso-scale struc-tures based on the self-assembly of patterned components [Gracias:2000]. Consider a circuitmade from the self-assembly of N different building block components. When shaken uptogether, each component must find its proper place in a two-dimensional square latticefor the circuit to function correctly. Using the figure below, suppose that if the letters arecorrectly ordered as on the left, each unit interacts pairwise with its neighbor through aninteraction energy wmatch. However, if a letter is surrounded by an incorrect neighbor, itinteracts more weakly, with an energy wmismatch. Find the contact energy difference
∆w = wmatch − wmismatch,
necessary to ensure that the circuit can be reliably self-assembled. That is, the fraction ofworking circuits should be higher than δ, where δ is some small value. Ignore shape andsurface effects (assume each of the N components makes z = 4 contacts) and only considerthe equilibrium between a perfectly-formed circuit and all possible misformed circuits inwhich two components have swapped places. Assume that N is large enough that thedefective components are not in contact with one another:
Comparing a correctly self-assembled two-dimensional circuit of parts vs. anincorrectly assembled circuit [Gracias:2000]. (left) A perfect Whitesides
circuit. (right) One of the possible malformed Whitesides circuits.
Reference: ”Forming electrical networks in three dimensions by self-assembly”. Gracias DH,Tien J, Breen TL, Hsu C, Whitesides GM. Science. 289(5482):1170-1172. 2000.
Energy of a perfect circuit: 4Nwmatch, W = 1Energy of a perfect circuit: 4(N − 2)wmatch + 8wmismatch, W = (N − 1)(N − 2)
Q = e−4Nwmatch
kT + (N − 1)(N − 2)e−4(N−2)wmatch+8wmismatch
kT
We want the probability of a perfectly formed circuit to be greater than δ:
δ < pmatch = e− 4Nwmatch
kT
e− 4Nwmatch
kT +(N−1)(N−2)e− 4(N−2)wmatch+8wmismatch
kT
1δ> 1
pmatch= 1 + (N − 1)(N − 2)e
8kT
(wmatch−wmismatch)
180
1δ> 1 + (N − 1)(N − 2)e
8kT
∆w
=⇒ ∆w > kT8ln( 1−δ
δ(N−1)(N−2))
181
Chapter 16Solvation and Transfer of Molecules
1. The mechanism of anesthetic drugs.
Anesthetic drug action is thought to involve the solubility of the anesthetic in the hy-drocarbon region of the lipid bilayer of biological membranes. According to the classical‘Meyer–Overton hypothesis,’ anesthesia occurs whenever the concentration of drug is greaterthan 0.03 mol kg−1 membrane, no matter what the anesthetic.
(a) For gaseous anesthetics like nitrous oxide or ether, how would you determine what gaspressure of anesthetic to use in the inhalation mix for a patient in order to achievethis membrane concentration?
(b) Lipid bilayers ‘melt’ from a solid-like state to a liquid-like state. Do you expect intro-duction of the anesthetic to increase, decrease, or not change the melting temperature?If the melting temperature changes, how would you predict the change?
(a) As a first approximation, assume that the biomembranes in the body act like a beakerof hydrocarbon, and that the anesthetic gas is in equilibrium with the solution. Henry’sLaw tells us the relationship between the vapor pressure and the solute concentrationin solution at equilibrium. Therefore we must look up, or measure, the Henry’s Lawconstant for the specific drug over whatever specific hydrocarbon we consider to bemost representative of membranes, and then the required applied pressure will be:
pappl = (kdrug/oil)
(.03
moles
kg membrane
)(z kg membrane)
(b) If anesthetics in membranes obey the colligative laws (there is good evidence that theydo, at least approximately), then we expect solute to partition predominantly into theliquid phase and thus to reduce the melting temperature. This is observed to be the
182
case for most anesthetics. The degree to which melting temperature is reduced can becalculated from the usual colligative law equations where the relevant enthalpy offreezing is that of the pure membranes.
2. Divers get the bends.
Divers, returning from deep dives, can get the bends from nitrogen gas bubbles in theirblood. Assume that blood is largely water. The Henry’s law constant for N2 in water at25◦C is 86,000 atm. The hydrostatic pressure is 1 atm at the surface of a body of water andincreases by approximately 1 atm for every 33 feet in depth. Calculate the N2 solubility inthe blood as a function of depth in the water, and explain why the bends occur.
From Henry’s Law we can determine the relationship between the vapor pressure over asolution, in this case in the form of the nitrogen gas bubbles over its solution in thebloodstream, p = kx, where k is the Henry’s Law constant for nitrogen gas in water. Thepressure of the gas being breathed D feet deep is
(1 + D
33 ft
)atm. Nitrogen comprises about
80% of the air, so its partial pressure is 0.80(1 + D
33 ft
)atm. Therefore, the mole fraction of
the solution which is nitrogen gas is
x =p
k=
(0.80
(1 + D
33 ft
)atm
)
(86, 000 atm).
Much more gas would dissolve in the diver’s blood at the high pressures deep below thesurface than at p = 1 atm at the water’s surface. Therefore, as the diver rises to the surface,the gas that dissolved deep underwater can no longer remain dissolved and bubbles out of hisblood. This causes the physiological problems.
3. Hydrophobic interactions.
Two terms describe the hydrophobic effect: (i) hydrophobic hydration, the process of trans-ferring a hydrophobic solute from the vapor phase into a very dilute solution in which thesolvent is water, and (ii) hydrophobic interaction, the process of dimer formation from twoidentical hydrophobic molecules in a water solvent.
(a) Using the lattice model chemical potentials, and the solute convention, write thestandard state chemical potential differences for each of these processes, assumingthat these binary mixtures obey the regular solution theory.
(b) Describe physically, or in terms of diagrams, the driving forces and how these twoprocesses are similar or different.
183
(a) Leaving out factors of the internal partition functions, which we assume do not changein the process, we have for hydration:
∆µ0hydration
kT=
µ0liq − µ0
vap
kT=
(zwφφ2kT
+ χφw − ln
(p
p◦φ,int
))
=(z
kT
)(wφφ2
+ wφw −wφφ + www
2
)− ln
(p
p◦φ,int
)
=(z
kT
)(wφw −
www2
)− ln
(p
p◦φ,int
)
where φ = solute and w = water.
Interaction: From dimerization theory,
∆µ0interaction
kT=
µ0ww + µ0
φφ − 2µ0φw
kT= − 1
kT(2wφw − www − wφφkT )
= − 2
kT
(wφw −
www + wφφ2
)= −2
zχφw
(b) Hydration is favored by stronger φw interactions, weaker ww interactions, or higherpressures. φφ interactions do not contribute.
φ
w
w
w
w
w
soln
vapor
Hydrophobic interaction is favored by weaker φw interactions, or stronger ww or φφinteractions. The other difference is that the driving force for hydrophobic interactionis weaker by about 2
zrelative to hydration.
w
w
w
ww
ww
w
w w
w
φ
φ
4. Solutes partition into lipid bilayers.
184
Robinson Crusoe and his friend Friday are stranded on a desert island with no whiskey,only pure alcohol. Crusoe, an anesthesiologist, realizes that the effect of alcohol, as withother anesthetics, is felt when the alcohol reaches a particular critical concentration, c0, inthe membranes of neurons. Crusoe and Friday have only a supply of propanol, butanol, andpentanol, and a table of their free energies of transfer for T = 300 K.
(a) A concentration of c1 of ethanol in water will result in a concentration c0 in themembrane, which is a hydrocarbon-like environment. Use the table below to predictwhat concentrations in water of the other alcohols would produce the same degree ofanesthesia.
Partitioning quantities in cal mol−1.
µ◦w − µ◦hc h◦w − h◦hc s◦w − s◦hc
ethanol 760 −2430 −10.7propanol 1580 −2420 −13.4butanol 2400 −2250 −15.6pentanol 3222 −1870 −17.1
(b) Mythical cannibals find Crusoe and Friday and throw them into a pot of boiling water.Will the alcohol stay in their membranes and keep them drunk at 100◦ C?
(c) Which alcohol has the greatest tendency to partition into the membrane per degreeof temperature rise?
(a) For all 3 alcohols, we need to achieve a concentration of c0 in the membrane. From thechemical potentials of each, we can get the partition coefficients of each (at T = 300 K):
∆µ0 = −RT lnK K = e−∆µ0/RT
µ0hc − µ0
w K(hc← w)
eth −760 3.6prop −1580 14.2but −2400 56.0pent −3222 223
Therefore the relative concentrations needed to achieve this membrane concentrationare
propanol(
3.6
14.2
)c1 = 0.25c1
185
butanol(
3.6
56.0
)c1 = 0.064c1
pentanol(
3.6
223
)c1 = 0.016c1
(b) Using the van’t Hoff equation, the temperature dependence of the equilibrium constantis
∂ lnK(hc←w)
∂T=
∆h(hc←w)
RT 2> 0
so the alcohol concentration in the membrane increases with temperature.
(c) It follows that ethanol has the greatest increase in partition coefficient withtemperature (2430 > 2420 > 2250).
5. Global warming.
CO2 in the earth’s atmosphere prevents heat from escaping, and is responsible for roughlyhalf of the greenhouse effect, that causes global warming. Would global warming cause afurther increase in atmospheric CO2 through vaporization from the oceans? Assume thatthe ocean is a two-component solution of water plus CO2, and that CO2 is much morevolatile than water. Give an algebraic expression for the full temperature dependence ofHenry’s law constant kH for the CO2 in water, that is, derive an equation for ∂kH/∂T .
Let B represent volatile CO2, highly dilute in solvent water (A). The vapor pressure of B is
pB = p0B,int xB exp
[z
kT
(wAB −
wAA2
)]
= p0B,int xB ec/T
where
c =z
k
(wAB −
wAA2
).
Since
∂
∂T(ec/T ) = − c
T 2ec/T ,
186
∂pB∂T
= −p0B,intcxB
T 2exp
[c
T
]
= −kHxBcT 2
Since kH = p0B,inte
c/T , then we also have
∂kH∂T
= −kHcT 2
=⇒ ∂ ln kH∂T
= − c
T 2=
∆H
RT 2
where ∆H is the enthalpy for transferring CO2 from a water phase to the vapor phase. Sincethe enthalpy of formation of CO2 (aq) is −98.69 kcal/mol and of CO2 (gas) is −94.05kcal/mol, ∆H = 4.64 kcal/mol and kH increases with temperature. What complicates theunderstanding of CO2 release with temperature is that salts dissolved in the ocean andchemical reactions that form bicarbonates are also involved in the equilibrium.
187
6. Modelling cavities in liquids.
Assume that you have a spherical cavity of radius r in a liquid. The surface tension of theliquid is γ, in units of energy area−1.
(a) Write an expression for the energy ε(r) required to create a cavity of radius r.
(b) Write an expression for the probability p(r) of finding a cavity of radius r.
(c) What is the average energy 〈ε〉 for cavities in the liquid?
(d) Write an expression for the average cavity radius.
(e) If RT = 600 cal mol−1, and γ = 25 cal mol−1A2, then compute 〈r〉.
(a) e(r) = 4πr2γ
(b) p(r) =e−e(r)/kT∫∞
0 e−e(r)/kTdr=
exp[−4πγr2
kT
]
12
√kTπ4πγ
= 4
√γ
kTexp
[−4πγr2
kT
]
(c) By equipartition 〈e〉 = 12kT .
(d) 〈r〉 =
∫∞0 re−4πγr2/kT
∫∞0 e−4πγr2/kT
=
12
(kT4πγ
)
12
(πkT4πγ
)1/2=
1
2π
(kT
γ
)1/2
(e) 〈r〉 =1
2π
(600
25
)1/2
= 0.78A
188
7. Sparkling drinks.
CO2 in water has a Henry’s law constant kH = 1.25× 106 mm Hg. What mole fraction ofCO2 in water will lead to ‘bubbling up’ and a vapor pressure equal to 1 atm?
Since the vapor pressure is
p = kHx
then x =p
kH=
760 mm Hg
1.25× 106 mm Hg
=⇒ x = 6.08× 10−4
8. Modelling binding sites.
You have a two-dimensional molecular lock and key in solvent s, as shown in the figurebelow. Different parts of each molecule have different chemical characters, A, B, and C.
(a) In terms of the different pair interactions, (wAB, wAC , wAS, . . . etc.) write an expressionfor the binding constant K (i.e., for association).
(b) Which type of pair interaction (AB, AC, BC) will dominate the attraction?
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S SS SS S SSS
S
S
Unbound Bound
C C
CC
B
B A B
CC
C
C
C C
CC
B
B A B
CC
C
C
SS SS SS SS
S
S
S
S
S
S S
S
S
S
189
C C
CC
B
B A B
CC
C
C
S
S
S
S
C C
CC
B
B A
S
Unbound
Bound
S
S
S
S
S
S
S
S
S
S
B
CC
C
CS
Break 2AS 1BS 3CS
Make 6SS
Break 1BS 5CS
Make 6SS
Make 1AB 2AC 1BC 3CC
(a) K =
(qbound
qlockqkey
)exp
[1
kT(2wAS + 2wBS + 8wCS − wAB
−wAC − wBC − 3wCC − 6wss)]
(b) CS and SS interactions will dominate overall because there are the largest number ofthem. Among AB, AC, BC, it will just depend on which is largest.
190
9. Oil/water partitioning of drugs.
In the process of partitioning of a drug from oil into water, ∆s◦ = −50 cal (mol deg)−1 and∆h◦ = 0 at T = 300 K.
(a) What is ∆µ◦ at T = 300 K?
(b) What is the partition coefficient from oil to water, Kwateroil at T = 300 K?
(c) Assume that ∆s◦ and ∆h◦ are independent of temperature. Calculate Kw0 at T =
320 K.
(a) ∆µ0 = ∆h0 − T∆s0
= −(300 K)
(−50
cal
mol K
)
= 15.0 kcal/mol
(b) ∆µ0 = −RT lnKw0
Kw0 = exp
(−∆µ0
RT
)= exp
− 15kcal
mol(1.987 cal
mol K
)(300 K)
= 1.18× 10−11
(c) ∆µ0(320 K) = ∆h0 − T∆s0
= −(320)
(−50
cal
mol K
)
= 16.0 kcal/mol
Kw0 = exp
− 16kcal
mol(1.987 cal
mol K
)(320 K)
= 1.18× 10−11
Since ∆h◦ = 0, K doesn’t change with temperature, according to the van’t Hoff equation.
191
10. Another oil/water partitioning problem.
Assume that a drug molecule partitions from water to oil with a partition coefficients K1 =1000 at T1 = 300 K, and K2 = 1400 at T2 = 320 K.
(a) Calculate the free energy of transfer, ∆µ◦ at T1 = 300 K.
(b) Calculate the enthalpy of transfer, ∆h◦ (water to oil).
(c) Calculate the entropy of transfer, ∆s◦ (water to oil).
(a) ∆µotransfer = −RT lnK
= −(1.987cal
mol deg) (300 K) ln 1000
= −4.118kcal
mol
(b) Here we assume that ∆h◦ is independent of temperature. Then we use the van’t Hoffformula:
∆h◦ = −R(lnK2 − lnK1)(1T2− 1
T1
)
= −(1.987cal
mol deg)ln 1400− ln 1000(
1320 K
− 1300 K
) = 3.21kcal
mol
(c) ∆s◦ =∆h◦ −∆µ◦
T=
3.21 kcalmol
+ 4.12kcalmol
300 K= 24.4
cal
mol K
11. Oil/water partitioning of benzene.
You put the solute benzene into a mixture containing the two solvents oil and water. Youobserve the benzene concentration in water to be xw = 2.0 × 10−6 M and in oil to bexo = 5.08× 10−3 M.
(a) What is the partition coefficient Koilwater (from water into oil)?
(b) What is ∆µ◦ for the transfer of benzene from water into oil at T = 300 K?
192
(a) Kow =
xoxw
=5.08× 10−3M
2.0× 10−6M= 2540
(b) ∆µo = −RT lnKow = −
(1.987
cal
mol K
)(300 K) ln(2540)
= −4.673kcal
mol
12. Balancing osmotic pressures.
Consider a membrane-enclosed vesicle that contains within it a protein species A that cannotexchange across the membrane. This causes an osmotic flow of water into the cell. Couldyou reverse the osmotic flow by a sufficient concentration of a different nonexchangeableprotein species B in the external medium?
Yes; the osmotic flow could be reversed because at low concentrations each distinguishablespecies of nonexchangeable component reduces the chemical potential independently of everyother species.
13. Vapor pressure of large molecules.
Why do large molecules, such as polymers, proteins, and DNA, have very small vaporpressures?
Large molecules have low vapor pressure because the vapor pressure decreases exponentiallywith the energy of attraction of the molecule for its neighbors. A large molecule has manysuch attractive interactions. For example, if the vapor pressure for a“single-lattice-site-solute” is
pA = po exp[zwAA2RT
]
then for a chain of n A-type monomers, the vapor pressure will be
pchain = p′o exp
[n(z − 2)wAA
2RT
]
193
Since wAA < 0, if n is large, pchain will be small.
194
14. Osmosis in plants.
Consider the problem of how plants might lift water from ground level to their leaves.Assume that there is a semipermeable membrane at the roots, with pure water on theoutside, and an ideal solution inside a small cylindrical capillary inside the plant. Thesolute mole fraction inside the capillary is x = 0.001. The radius of the capillary is 10−2 cm.The gravitational potential energy that must be overcome is mgh, where m is the mass ofthe solution, and g is the gravitational constant, 980 cm s−2. The density of the solution= 1 g cm−3. What is the height of the solution at room temperature? Can osmotic pressureaccount for raising this water?
For the given ideal solution, the osmotic pressure is given by
∆µ = πv = kTxsolute
where v equals the molar volume of solvent (water).
This free energy equals the gravitational potential energy at equilibrium:
∆µ = mgh
But mgh = ρV gh and Vcylinder = πr2h so
∆µ = ρgπr2h2 = πv = kTxsolute.
Therefore
h2 =kTxsoluteρgπr2
h2 =(4.14× 10−14 g cm2
s2)(0.001)
(1 gcm3 )(980 cm
s2)(3.14)(10−4 cm2)
=⇒ h ≈ 10−8 cm
Therefore, although osmotic pressure does contribute to water transport inside of trees, otherfactors, such as transpiration of the water from the leaves, are much more important.
195
15. Polymerization in solution.
Using the lattice dimerization model (see the figure below), derive the equilibrium constantfor creating a chain of m monomers of type A in a solvent s.
S
The generalization of Equation (16.53) for the association of m monomers is
ln
[xAmxmA
]=(m
z
)χsA + ln
[qAAqmA
].
16. Ethane association in water
The association of ethane molecules in water is accompanied by an enthalpy of 2.5 kcal mol−1
at T = 25◦C. Calculate (∂ lnKassoc/∂T ) at this temperature. Does the ‘hydrophobic effect’get stronger or weaker as temperature is increased?
∆µoassoc = −RT lnKassoc =⇒ lnKassoc = − 1
RT[∆hoassoc − T∆soassoc]
assume ∆ho, ∆so are independent of T ; then
∂ lnKassoc
∂T=
∆hoassoc
RT 2=
2500 calmol
(1.987 calmol K
)(298 K)2= 0.0142 K−1
As T increases, Kassoc increases =⇒ hydrophobic effect gets stronger.
196
17. Freezing point depression by large molecules.
Freezing temperature depression is a useful way of measuring the molecular weights of somemolecules. Is it useful for proteins? One g of protein of molecular weight 100, 000 g mol−1
is in 1 g of water. Calculate the freezing temperature of the solution.
Freezing temperature depression may be calculated for ideal solutions by
To − T =RT 2
f
∆Hfusxsolute
where the mole fraction of solute is given by
xsolute ≈
(1 g solute
105g/mol
)
(1 g water18g/mol
) = 1.8× 10−4
Therefore
∆T =(1.987 cal/mol K)(273.16 K)2
(2350 cal/mol)(1.8× 10−4)
= 0.0114 K
The freezing temperature depression is too small to measure molecular weights of very largemolecules.
18. Partitioning into small droplets is opposed by interfacial tension.
When a solute molecule (s) partitions from water (A) into a small oil droplet (B), the dropletwill grow larger, creating a larger surface of contact between the droplet and the water. Thuspartitioning the solute into the droplet will be opposed by the droplet’s interfacial tensionwith water. Derive an expression to show how much the partition coefficient will be reducedas a function of the interfacial tension and radius of the droplet.
Begin with the Gibbs free energy function, since the process is at constant T and p, butrecognize that the droplet surface area a increases if a solute is added, so you need to include
197
the conjugate pair γa. Therefore G(T, p,N, a) and
dG = −SdT + V dp+ µA(s) dNSA + µB(s) dNSB + γda,
where NSB is the number of solute (s) molecules in the droplet B and NSA is the number ofsolute molecules in the external water phase. The constraint is NSA +NSB = constant, so theequilibrium condition (at constant T and p) is
[µB(s)− µA(s)] dNSB + γda = 0. (a)
Now changes in area a are not independent of changes in NSB because adding a soluteincreases the volume of the droplet. To find the relationship between dNSB and da first notethat the increased volume, dV , of the droplet is
dV = vdNSB (b)
where v is the volume of a solute molecule. If the droplet is a sphere before and after theuptake of solute, then the area is A = 4πr2 and the volume is V = (4/3)πr3, where r is theradius. So we have
dA = 8πrdr
and
dV = 4πr2dr
so
dA =2
rdr (c)
Substituting (b) and (c) into (a) gives
[µB(s)− µA(s) +
2γv
r
]dNSB = 0.
198
Using the lattice model for solute at infinite dilution in both phases gives
lnKBA = ln
(xSBxSA
)= − (χSB − χSA)− 2γv
r.
Here are some implications of this expression. (a) As r ∞, the partition coefficientapproaches the bulk value. (2) As r, the droplet radius, becomes small, solute is increasinglyexpelled from the droplet. (3) As the solute radius increases or the droplet interfacial tensionwith water increases, solute is expelled.
19. Alternative description of Henry’s Law
Show that an alternative way to express Henry’s law of gas solubility is to say that thevolume of gas that dissolves in a fixed volume of solution is independent of pressure at agiven temperature.
Henry’s Law of solubility: pa = kaxa. pa is the vapor pressure of the gas, ka is Henry’s Lawconstant for the gas in the particular solvent, and xa is the mole fraction of gas in solution.
The essence of this problem is in the conversion from mole fraction of gas to volume fraction:
φa = vol fraction =Vgas
Vsolution≈ Vgas
Vsolvent
xa = φa
(moles
Vgas
molesV
solvent
)
Assume the gas is ideal. Then
moles gas
vol. gas=n
V=
paRT
For the solvent
moles solvent
vol. solvent=
1
v
199
where v is the molar volume of the solvent. Therefore
xa = φa
( paRT1v
)= φa
(pav
RT
)
Substitute into Henry’s Law:
pa = kaφa
(pav
RT
)
so φa =RT
kav,
i.e., the volume fraction of gas dissolved is independent of pressure at constant temperature.
20. Benzene transfer into water.
At T = 25◦C, benzene dissolves in water up to a mole fraction of 4.07× 10−4 (its solubilitylimit).
(a) Compute ∆µ◦ for transferring benzene to water.
(b) Compute χbenzene,water.
(c) Write an expression for the temperature dependence of ∆µ◦ as a function of ∆h◦, ∆s◦,the molar enthalpy and entropy at 25◦ C, and ∆Cp.
(d) Using the expression you wrote for (c), calculate ∆µ◦ for transferring benzene to waterat T = 100◦ C, if ∆h◦ = 2 kJ mol−1, ∆s◦ = −58 J mol K−1, and ∆Cp = 225 J mol K−1.
(a) ∆µ◦ = −kT ln x = −(
1.987cal
mol K
)(298 K) ln(4.07× 10−4)
= −4.62kcal
mol
(b) χAB = − ln x = − ln(4.07× 10−4)
= 7.81
200
(c) Assuming that ∆h◦, ∆s◦, and ∆Cp are all temperature-independent, then
∆µ◦ = ∆h− T∆s
= ∆h◦ − T∆s◦ + ∆Cp(T − 298 K)− T∆Cp ln(
T
298 K
)
(d) ∆µ◦ =
(2
kJ
mol
)+ (373 K)
(58
J
mol K
)
+(225
J
mol K
)(75 K)− (373 K)
(225
J
mol K
)ln(
373
298
)
= (2000 + 21, 634 + 16, 875− 18, 840)J
mol
= 21.7kJ
mol
21. Raoult’s Law for methanol-water mixtures.
201
The figure below plots the vapor pressure, in mm Hg, versus composition for methanol ina binary solution with water at 300 K. Raoult’s Law behavior is indicated by the straightline.
(a) Using the lattice model, calculate the contact energy w for methanol. To simplify,neglect the internal degrees of freedom for methanol, and assume a lattice with z = 6.
(b) From the graph, estimate the value of the exchange parameter, χ, for the methanol-water mixture.
(c) Using the lattice model expression for the activity coefficient, show that pmeth ap-proaches P ∗meth (the equilibrium, vapor pressure above pure methanol) as with thesame slope as Raoult’s law. That is, the plot for Pmeth should resemble the figureabove, and not the figure shown below.
(a) From the graph, it is clear that for pure methanol (χmeth = 1), the equilibrium vaporpressure is 140 mm Hg. This can be used with the equation from the text just after eq.15.3:
P ∗ = P ointe
zw2kT , where eq. 10.49 gives P o
int = kT (2πmkTh2 )3/2
Note that in calculating the internal pressure term you do not have to include theinternal degrees of freedom; but you still need to include the translational degrees of
202
freedom for the P oint term. Solving this for w:
w = 2kTzln( P ∗
P oint) = 2kT
zln(P
∗kT
( h2
2πmkT)3/2)
w = 2·1.38·10−23 ·3006
ln(140760·1.013·105
1.38·10−23·300(6.6262·10−68
2·3.14·32·1.78·10−27·1.38·10−23 ·300)3/2)
w = −2.427 · 10−20 Jmolecule
= −14.6 kJmol
(b) At Xmeth = 0.2, for example, Pmeth = 100 mm Hg, while Raoult’s Law givesPmeth = P ∗methXmeth = 28 mm Hg. The activity coefficient, on the solvent convention, γ,is Pmeth
P ∗meth
Xmeth= 100
28= 3.75.
This can be related to the exchange parameter, χ, by γ = eχ(1−Xmeth)2
=⇒ χ = lnγ(1−Xmeth)2
= ln3.570.64
= 1.99
(c) According to the lattice model, we have Pmeth = γXmethP∗meth
=⇒ dPmethdXmeth
= dγdXmeth
XmethP∗meth + γ(dXmeth
dXmeth)P ∗meth
dPmethdXmeth
= −(1−Xmeth)eχ(1−Xmeth)2XmethP
∗meth + eχ(1−Xmeth)2P ∗meth
and in the limit as Xmeth → 1, dPmethdXmeth
→ 0 + P ∗meth = P ∗methThis is the same slope as that for Raoult’s Law line.
22. Vapor Pressure of Water.
(a) At a vapor pressure of p = 0.03 atm, and a temperature T = 25◦C, what is the densityof water vapor? Assume an ideal gas.
(b) What is the corresponding density of liquid water under the same conditions?
(c) If the vapor pressure of water is 0.03 atm at T = 25◦ C, and 1 atm at T = 100◦ C,what is the pairwise interaction energy wAA, if each water molecule has 4 nearestneighbors?
(a) PV = nRT
ρ = nV
= PRT
= 0.03atm
8.206·10−5m3atmK·mol (300K)
≈ 1.22molm3
(b) pBpoB,int
= xBezwBB2kT
xB = pBpoB,int
e−zwBB2kT , where pB = poB,inte
zwBB2kT
=⇒ xB =poB,intpoB,int
e−zwBB2kT e
zwBB2kT = 1
(c) p1ApoA,int
= e4wAA2kT1 , p2A
poA,int
= e4wAA2kT2
203
Solving for poA,int, and equating,=⇒ p1A
e
4wAA2kT1
= p2A
e
4wAA2kT2
p1Ap2A
= e
4wAA2k( 1
T1− 1T2
)
=⇒ wAA =kln(
p1Ap2A
)
2( 1T1− 1T2
)
23. Osmotic pressure of sucrose.
What is the osmotic pressure of a 0.1 molar solution of sucrose in water at T = 300 K?
π ≈ cART = 0.1molL
(8.314 JK·mol)(300K)(103 L
m3 ) = 249.42 · 103 Jm3 ≈ 2.5 · 105 J
m3
204
Chapter 19Chemical kinetics
1. Isotope substitution experiments can detect hydrogen tunneling.
Isotope substitution experiments can sometimes be used to determine whether hydrogensare cleared from molecules through mechanisms that involve tunnelling. To test this, twoisotopes are substituted for hydrogen: (1) deuterium (D, mass = 2) is substituted and theratio of rate coefficients kH/kD is measured, and (2) tritium (T, mass = 3) is substitutedand kH/kT is measured.
(a) Using the isotope substitution model in this chapter, show that
kHkT
=
(kHkD
)α.
(b) compute the numerical value of α.
You have
kH
kD
= exp
[−hνH
2kT
(1√2− 1
)]
Similarly
kH
kT= exp
[−hνH
2kT
(1√3− 1
)]
221
sokH
kT=
(kH
kD
)α
where α =
(1√3− 1
)
(1√2− 1
) =−0.4226
−0.2929= 1.443
2. Relating stability to activation barriers.
Using the Evans–Polanyi model, with r1 = 5, r2 = 15, m1 = 1, and m2 = −2:
(a) Compute the activation barriers Ea for three systems having product stabilities ∆G =−2 kcal mol−1, −5 kcal mol−1, and −8 kcal mol−1.
(b) Plot the three points Ea versus ∆G, to show how the activation barrier is related tostability.
(a) Use Eqn. 19.52:
Ea =(
m1
m1 −m2
)[∆G−m2(r2 − r1)]
=(
1
3
)[∆G + 2(10)] =
∆G+ 20
3
so for ∆G = −2, Ea = 6
∆G = −5, Ea = 5
∆G = −8, Ea = 4
(b) This shows that the activation barrier decreases linearly with increasing stability of theproduct: the more stable the product, the faster the reaction.
3. Reduced masses.
Equations (19.40) and (19.41) give the reduced masses for C–H and C–D bonds as µCH ≈ mH
and µCD ≈ 2mH. The approximation is based on assuming the mass of carbon is muchgreater than of H or D. Give the more correct values of these reduced masses if you don’tmake this assumption.
222
µCH =mCmH
mC +mH=
12
13mH = 0.923mH
µCD =mCmD
mC +mD
=12
14(2mH) = 1.714mH
4. Classical collision theory.
According to the kinetic theory of gases, the reaction rate k2 of a sphere of radius rA withanother sphere of radius rB is
k2 = πR2
(8kT
πµAB
)1/2
e−∆ε‡0/kT ,
where R = rA + rB is the distance of closest approach, µAB is the reduced mass of the twospheres, and ∆ε‡0 is the activation energy. Derive this from transition state theory.
The transition state is an AB dimer having mass mA +mB and a rotational partitionfunction with moment of inertia I = µABR
2. Using µAB = (mAmB)/(mA +mB) andtransition state theory gives
k2 = q‡vqrqt
=
(kT
h
)(8π2µABR
2kT
h2
)[2π(mA +mB)kT h−2
(2πmAkT h−2)(2πmBkT h−2)
]3/2
e−∆ε
‡◦
kT
=
(8π2µABR
2(kT )2
h3
)[h2
2πkT
]3/2 [1
µAB
]3/2
e−∆ε
‡◦
kT
= πR2
(8kT
πµAB
)1/2
e−∆ε
‡◦
kT
223
5. The pressure dependence of rate constants.
(a) Show that the pressure dependence of the rate constant k for the reaction
Akf−→ B
is proportional to an activation volume v‡,
(∂ ln kf∂p
)= −(v‡ − vA)
kT.
(b) Show that the expression in (a) is consistent with Equation (12.45),K = kf/kr, where kr is the rate of the reverse reaction.
(a) Here kf = K‡ k‡, where K‡ is like an equilibrium constant between the reactant andtransition state, and where k‡ depends only on T and not on pressure. Then fromEquation (12.45) we get
∂ ln kf∂p
=∂ lnK‡
∂p+∂ ln k‡
∂p= −v
‡ − vAkT
+ 0
where v‡ is the volume of the transition state.
(b) Since K =kfkr
,
∂ lnK
∂p=∂ ln kf∂p
− ∂ ln kr∂p
= −v‡ − vAkT
+v‡ − vBkT
= −vB − vAkT
which is what we get in Eqn.12.45.
6. Relating the Arrhenius and activated state parameters.
Derive the relationship of the activation parameter ∆H ‡ in Equation (19.19) to the Arrhe-nius activation energy Ea in Equation (19.4) for a gas-phase reaction.
224
For the forward reaction, you have the Arrhenius Equation (19.15)
d ln kfdT
=EakT 2
(1)
Using the Gibbs–Helmholtz Eqn (13.43) gives
d lnK‡
dT=
∆U ‡
kT 2(2)
(Since K‡
has partition functions that depend on V , we are using the (T, V,N) ensemble, sothe right side is ∆U ‡ rather than ∆H‡.) Substituting thermodynamic relationships and theideal gas law into the equation gives
d lnK‡
dT=
∆H‡ − p∆V ‡kT 2
=∆H‡ −∆n‡kT
kT 2(3)
where ∆V ‡ is the activation volume and ∆n‡ is the stoichiometric change during the
reaction. Taking the derivative of Equation (19.28), kf =(kTh
)K‡, gives
d ln kfdT
=1
T+d lnK
‡
dT(4)
Substitute Equations (1) and (3) into Equation (4) to get
EakT 2
=1
T+
∆H‡ −∆n‡kT
kT 2
=⇒ Ea = ∆H‡ + kT (1−∆n‡)
225
7. Enzymes accelerate chemical reactions.
The figure below shows an Arrhenius plot for the uncatalyzed reaction of 1-methyloroticacid (OMP).
(a) Estimate ∆H‡ from the figure.
(b) Estimate ∆S‡ at T = 300 K.
(c) At T = 25◦ C, the enzyme OMP decarboxylase accelerates this reaction 1.4×1017-fold.How fast is the catalyzed reaction at 25◦ C?
(d) What is the binding constant of the enzyme to the transition state of the reaction atT = 300 K?
10−20
1
10−10
0.0018 0.0026 0.0034
200 C 100 C 25 C
T −1 (K−1)
Rate Constant (s−1)
Source: A Radzicka and R Wolfenden, Science 267, 90–93 (1995).
(a) Since
k2
k1=
(kT2/h
kT1/h
)exp
[−∆H‡
k
(1
T2− 1
T1
)]
you have
lnk2
k1= ln
(T2
T1
)− ∆H‡
k
(1
T2− 1
T1
)
Take 2 points, such as (T1, k1) = (560 K, 1 s−1), (T2, k2) = (294 K, 1× 10−16 s−1). Then
∆H‡ =k[ln(T2
T1
)− ln
(k2k1
)]
1T2− 1
T1
226
=(1.987 cal mol−1K−1)
[ln(
294560
)− ln (1× 10−16)
]
(0.0034− 0.0018) K−1
= 45 kcal mol−1
(b) To get ∆S‡, choose a point, say (T, k) = (300 K, 1× 10−16 s−1), and use the expansion
ln k2 = ln
(kT
h
)− ∆H‡
kT+
∆S‡
k
and rearrange to get
∆S‡ = k
[ln
(k2
kTh
)+
∆H‡
kT
]
= (1.987 cal mol−1 K−1)
[ln
(1× 10−16 s−1
6.25× 1012 s−1
)+
45, 000 cal mol−1
600 cal mol−1
]
= 17 cal mol−1 K−1
(c) (1.4× 1017)(1× 10−16 s−1) = 14 s−1
(d) The binding constant is simply the rate enhancement, 1.4× 1017. This is one of thelargest catalytic rate enhancements known.
8. Rate increase with temperature.
A rule-of-thumb used to be that chemical reaction rates would roughly double for a ten-degree increase in temperature, say from T1 = 300 K to T2 = 310 K. For what activationenergy Ea would this be exactly correct?
Using k2/k1 = 2 gives
Ea =R ln 2(1T1− 1
T2
) =(1.987 cal mol−1 K−1) ln 2
(3.33× 10−3 − 3.226× 10−3)K−1= 12.8 kcal mol−1 = 53.5 kJ/mol
This rule of thumb is not particularly useful in general.
227
9. Catalytic rate enhancement.
The reaction rate of an uncatalyzed reaction is k0 = 103 M−1 s−1 at T = 300 K. A catalystC binds to the transition state with free energy ∆G = −5 kcal mol−1. What is the rate kcof the catalyzed reaction at T = 300 K?
kck0
= K = exp−∆G/kT = exp[
5000 cal mol−1
(2 cal mol−1 K−1)(300K)
]= 4160
Therefore kc = 4160k0 = 4.16× 106 m−1s−1.
10. Negative activation energies.
Activation energy barriers often indicate a process that is rate-limited by the need to breaka critical bond, or achieve some particular strained state. In contrast, some processesinvolve negative activation energies—their rates decrease with increasing temperature (seethe figures below. Table 247 shows some Arrhenius parameters for the formation of duplexnucleic acids from single strands of various nucleotide sequences. The second figure showsan example of the kind of data from which the parameters in the table were derived. Whichmolecules have negative activation energies?
Relaxation kinetics of oligonucleotides (21◦C to 23◦C). Source: CR Cantor and PRSchimmel, Biophysical Chemistry, Volume 3, WH Freeman, San Francisco, 1980.
k1 Ea
Ordered form (M−1s−1) (kcal mol−1)A9 · U9 5.3× 105 −8A10 · U10 6.2× 105 −14A11 · U11 5.0× 105 −12A14 · U14 7.2× 105 −17.5A4U4 · A4U4 1.0× 106 −6A5U5 · A5U5 1.8× 106 −4A6U6 · A6U6 1.5× 106 −3A7U7 · A7U7 8.0× 105 +5A2GCU2 · A2GCU2 1.6× 106 +3A3GCU3 · A3GCU3 7.5× 105 +7A4GCU4 · A4GCU4 1.3× 105 +8
228
k−1
k1
+ →→
0.0033 0.0034 0.0035 0.0036
T − 1 (K−1)
k1 (M−1s−1)
Ea = −14 kcal mol−1
1×106
2×106
6×105
8×105
4×105
2×105
4×106
Source: CR Cantor and PR Schimmel, Biophysical Chemistry, Volume 3, WH Freeman, SanFrancisco, 1980.
Notice that there are 3 possible thermal signatures: positive activation energy (where ratesincrease with temperature), negative activation energy (where rates decrease withtemperature), and entropy-dominated processes in which the rate doesn’t change much withtemperature. All the sequences having only A and U nucleotides appear to have negativeactivation energy barriers. It’s not understood why. None of the cases shown areentropy-dominated.
11. Drug dissociation from DNA.
229
Drug A is an uncharged molecule containing a three-ring aromatic chromophore linked tothree amino acids: valine, N-methyl valine and proline. This drug binds to double-strandedDNA, with the planar ring system intercalating between base pairs and the peptide portionlying in the minor groove.
(a) The aqueous solubility of Drug A decreases with increasing temperature. What is thesign of ∆H◦ for the dissolution process?
(b) This drug is generally poorly soluble in water, i.e. ∆G◦ > 0 for dissolution. What isthe sign of ∆S◦ for this process?
(c) Give a brief explanation for your answer to part b in terms of solvent properties.
(d) The figure below shows data for the temperature dependence of the rate constant fordissociation of Drug A from duplex DNA. Use this plot to estimate the activationenergy, Ea, and the entropy of activation, ∆S‡. Comment on the role of solvent indetermining the value for ∆S‡.
(a) For the process Drug (solid) → Drug (aqueous), the vant Hoff equation gives:∂lnK∂T
= ∆Ho
RT 2
And since K goes down when T goes up, ∆Ho must be negative.
(b) ∆Go = ∆Ho − T∆So. Since ∆Ho < 0, then ∆So must be < 0.
(c) This result is unusual in that dissolution (change from solid to solution phase) usuallyentails an increase in S. The observed decrease in S must be due to the hydrophobiceffect, that is the drug is lipophilic and thus when dissolved in water, the waterundergoes an ordering which actually lowers S. This decrease in S more thancompensates the increase in S due to the change in state of the drug.
230
(d) The Arrhenius equation for rate constant k is:
k = Ae−EaRT =⇒ lnk = lnA− Ea
RT
Thus, Ea = −R·(slope of the given plot)Determine the slope by picking two points on the line, such aslnk = −2 at 1
T= 0.0032 and lnk = −4 at 1
T= 0.0033
=⇒ slope = (−2−(−4))0.0032−0.0033
= −20000
Ea = −8.314(−2000) = 166 kJmol
∆S‡ is related to the pre-exponential factor A by the equation:
A = κkThe
1+∆S‡R =⇒ ∆S‡
R= ln( hA
κkT)− 1
Set κ, the transmission factor, to 1, and determine the intercept, lnA, of the plot above.y = mx + b, use point (x, y) = (0.0032,−2), m = −20000=⇒ b = y −mx = −2 + 20000 · 0.0032 = 62 = lnAA = 8.438 · 1026 =⇒ ∆S‡ = 262 J
K
This represents an increase in S, which is expected when a complex of drug with DNAstarts to fall apart. Working against this increase in S is a decrease in S due to partialordering of water around the lipophilic drug as it starts to come off the DNA, andhence be exposed to water. However, the net balance appears to be in favor theincrease in S due to the loosening of the complex.
12. Series vs. parallel reactions.
231
(a) You have a reaction in which A converts to B through 2 steps in series:
Ak1−→ I
k2−→ B,
where the temperature dependencies of the steps are given by the Arrhenius law:
k1 = a exp(−E1/(RT )) and
k2 = a exp(−E2/(RT )).
Derive an expression for the total rate ktot, from A to B, as a function of a, E1, E2,and T. (Hint: the time from A to B is the sum of times, A to I and I to B).
(b) Now consider instead 2 steps in parallel:
Ak1−→ B, A
k2−→ B.
Using the same Arrhenius equations above for the individual steps, derive the tem-perature dependence for k in this case. (Hint: now, the rates add.)
(a) Since the times add,
ttot = t1 + t21ktot
= 1k1
+ 1k2
ktot = k1k2k1+k2
ktot = Ae−E1+E2
RT
e− E1RT +e
− E2RT
(b) Since the rates add,
ktot = k1 + k2 = A(e−E1RT + e−
E2RT )
13. A cricket thermometer.
232
The frequency that crickets chirp increase with temperature. You can calculate the tem-perature in degrees Celsius by adding 4 to the number of chirps you hear from a cricket in8 seconds. In this way, you can use a cricket as a thermometer.
(a) What four temperatures would you have measured if the number of chirps in 8 secondswere 16, 21, 26, and 31?
(b) From this data, make an Arrhenius plot and calculate the activation energy inkcal/mol. This will tell you about a rate-limiting biochemical reaction that underliescricket chirping.
(c) You observe that crickets outside a fast-food restaurant chirp faster than their cousinsin the wild. The fast-food crickets chirp 30 times in 8 seconds at 25◦ C and 40 timesin 8 seconds at 35◦C. By how much did the fast-food additive reduce the activationbarrier for cricket chirping?
(d) If the equilibrium enthalpy for the chirping reaction, ∆h◦, is 10 kcal/mol, what is theactivation energy for the reverse chirp reaction for the normal crickets?
(a) T (oC) = (# chirps in 8 seconds) +4
(b) Plot lnK vs. 1T→ lnK = −3971.8( 1
T) + 14.27, where K is measured in chirps/sec
slope = m = −EaR
=⇒ Ea = −mR
233
Ea = −(−3971.8)(8.314) = 33 kJmol
= 7.89kcalmol
(c) Fast-food crickets: lnK = −2640.5( 1T) + 10.18
Ea = −(−2640.5)(8.314) = 22 kJmol
= 75.75kcalmol
∆Ea = 7.89− 5.25 = 2.64kcalmol
(d) Reaction Diagram:
Er = Ea + ∆ho = 7.89 + 10 = 17.89kcalmol
14. Onset of Huntington’s disease.
Huntingtons is an example of a polyglutamine disease. The severity of the disease correlateswith the degree of polyglutamine fibril formation. The age of onset decreases exponentiallywith increasing length of the polyglutamine chains (see red curve in the figure below takenfrom Gusella and Macdonald (2000), Nat. Review Neuroscience, 1:109-115). Fibrils formquickly after an initial nucleation event, the binding of two polyglutamine chains. Assumedimeratization is the rate limiting step in the onset of the disease. Our aim is to interpretthis data with a simple dimerization model.
234
(a) Use the lattice model below. Define wss, wsg, and wgg to be the solvent-solvent, solvent-glutamine, and glutamine-glutamine interaction energies. If each glutamine chain isof length L, fill in the table below for the pairwise interactions that are broken andformed when two chains dimerize.
ggggg g
ggggg g
gggggg
gggggg
s s s s s s
s s s s s s
s s s s s s
s s s s s s
s s s s s s
s s s s s s
s s s s s s
s s s s s s
K
wss wsg wggBrokenFormed
(b) Write an expression for the dimerization equilibrium constant in terms of the interac-tion energies.
(c) Use the expression from answering (b) to explain the exponential form of the givenfigure.
(d) Estimate the polymer-solvent interaction energy, χsg, from the given figure.
235
(a) # Broken = # Formed
(b) Kdim = Keq ∝ e−∆µkT , where ∆µ = Lwss + Lwgg − 2Lwsg
Kdim = ( qdimerqmonomer
)e−Lwss+Lwgg−2Lwsg
kT
Kdim = ( qdimerqmonomer
)e−L(wss+wgg−2wsg)
kT
(c) Assume the rate of onset of the disease = Kdim[dimer][free monomer]
rate of onset ∝ 1/(time of onset)
rate of onset of [dimer] ∝ Keq ∝ e−L(wss+wgg−2wsg)
kT =⇒ time of onset ∝ 1
e−L(wss+wgg−2wsg)
kT
=⇒ time of onset ∝ e2L(wss+wgg−2wsg)
kT , since z = 2.=⇒ time of onset ∝ e−Lχsg , χsg = z
kT(wsg − wss+wgg
2)
(d) Plot ln(age of onset) vs. L, or pick two values:time1 = 60, L1 = 40time2 = 5, L2 = 100time2 − time1 = e−(L2−L1)χsg
ln( time2time1
) = −(L2 − L1)χsgln( 5
60) = −(100− 40)χsg
=⇒ χsg ≈ 0.04, G and S like each other less than they like themselves.
236
Chapter 22Electrochemical Equilibria
1. A charged protein.
Model a protein as a sphere with a radius of 20 A and a charge of 20e in water at 25◦ C andDwater = 78.54. Assume the sphere is uniformly charged on its surface.
(a) In units of kT , what is the potential at a distance 30 A from the protein surface?
(b) What is the electrostatic free energy of the charge distribution on the protein inkcal mol−1?
(a) Outside of the sphere of charge, we can consider it to be a point charge at the sphere’scenter. Additionally, since the point is far away compared to the diameter of thesphere, we can neglect the perturbation of the dielectric by the sphere and assume auniform dielectric of 78.54. ψ is therefore given by
eψ
kT= C q
Dr· ekT
=(8.988× 109 kg m3 C−2 s−2)(20)(1.602× 10−19 C)2
(78.54)[(20 + 30)× 10−10 m](1.381× 10−23 J K−1)(298 K)· J
kg m2 s−2
≈ 2.85
(b) The free energy of charging a hollow sphere with uniform surface charge is given byEquation (22.57):
∆Gel =Cq2
2Da
255
=8.988× 109 kg m3
C2s2· (20)2(1.602× 10−19 C)2
(2)(78.54)(20× 10−10 m)· J
kg m2 s−2
·6.022× 1023
mol· 1 kcal
4184 J
≈ 42.3 kcal mol−1
2. pKa’s in a protein.
Suppose that the protein in problem 1 has an aspartic acid residue and a lysine residue atits surface. In bulk water the pKa of the aspartic acid side group is 3.9, and the pKa of thelysine NH2 side group is 10.8. What are the pKa’s of these groups in a protein if it has anet charge of +20e?
Use Equation (22.37):
Asp : pKapp = 3.9− 0.4343 z lBa
= 3.9− 0.4343(20)(7.13 A)
20 A
= 0.80
Lys : pKapp = 10.8− 0.4343(20)(7.13 A)
20 A
= 7.70
3. Acid dissociation near a protein.
Now put the protein of problems 1 and 2 in a 0.01 M aqueous solution of acetic acid (HAc).In bulk solution the acetic acid has a pKa = 4.0. It is dissociated to give ion concentrations[H+]bulk and [Ac−]bulk.
(a) What are the concentrations of H+ and Ac− ions at a location x near the proteinwhere the potential is ψ(x)?
(b) What is the pKa of the acetic acid near the protein at x? Explain the differencebetween the situation of this acid and the situation of the aspartic acid in problem 2.
256
(a) Since both ions can freely diffuse, use Equation (22.33):
[H+
]x
=[H+
]bulk
e−eψ(x)kT
[Ac−
]x
=[Ac−
]bulk
e+eψ(x)kT
(b) As with uncharged compounds, the dissociation constant Ka is determined byshort-range binding interactions and by local concentrations. So we have
Ka(x) =[H+]x [Ac−]x
[HAc]x
=[H+]bulk e−
eψ(x)kT [Ac−]bulk e+
eψ(x)kT
[HAc]bulk
([HAc] same throughout since uncharged, so unaffected by ψ(x))
=[H+]bulk [Ac−]bulk
[HAc]bulk
= Ka(∞)
For the aspartic acid at the protein surface, only the H+ has a Boltzmann distribution,because the acid ion cannot diffuse freely. In fact, the ratio [acid−]/[acid] depends onψ(x) only through [H+], not directly.
4. Acids and bases near a charged surface.
On a negatively charged surface:
(a) Are basic groups ionized more or less than in bulk water?
(b) Are acid groups ionized more or less than in bulk water?
(a) More. Bases will pick up protons and become net positively charged. Near a negativelycharged surface, this charge–charge interaction becomes much more favorable.
(b) Less. Acids lose protons to become negatively charged. Near a negatively chargedsurface, this ionized state becomes more unfavorable.
257
5. Free energy of charging a sphere.
A sphere with radius a in a medium with dielectric constant D is uniformly filled with acharge of volume density ρ. Derive the electrical free energy of this sphere in two differentways:
(a) Derive the potential field and then charging the sphere from charge density 0 to ρ.
(b) Add fully charged shells of thickness dr from r = 0 to r = a.
(a) A shell of radius R < a, and thickness dR contributes to the potential inside the shelland outside the shell:
Equation (21.45): dψin =ρ · 4πR2dR
4πε0DR=
ρ
ε0DRdR r < R
Equation (21.41): dψout =ρ · 4πR2dR
4πε0Dr=
ρ
ε0D
R2
rdR r > R
Total potential at r < a:
ψ =∫ R=r
R=0dψout +
∫ R=a
R=rdψin
=ρ
ε0D
[∫ R=r
R=0
R2
rdR +
∫ R=a
R=rRdR
]
=ρ
ε0D
(R3
3r|rR=0 +
R2
2|aR=r
)
=ρ
ε0D
(1
2a2 − 1
6r2)
So ∆Gel =1
2
∫ρψdV
=1
2
∫ a
r=0ρ · ρ
ε0D
(1
2a2 − 1
6r2)· 4πr2dr
=2πρ2
ε0D
∫ a
r=0
(1
2a2r2 − 1
6r4)
=4π
15
ρ2a5
ε0D
258
(b) Sphere with radius r has surface potential ψr as if all charge for ≤ r was at center:
ψr = ρ · 43πr3 1
4πε0Dr
=ρr2
3ε0D
Adding a shell with thickness dr costs
d(∆Gel) = ψr · ρ · 4πr2dr
Total work: ∆Gel =∫ r=a
r=0d(∆Gel)
=∫ r=a
r=0
ρr2
3ε0D· ρ · 4πr2dr
=4πρ2
3ε0D
∫ r=a
r=0r4dr
=4π
15
ρ2a5
ε0D
6. Burying a charge in a protein.
As an estimate for the free energy of burying a charged amino acid such as aspartic orglutamic acid in protein folding, compute the free energy of transferring an ion of radius3 A and charge +1 from water to oil. Assume that water has a dielectric constant Dw = 80,and oil has Do = 2.
Use the Born energy expression,
∆G =Cq2
2a
(1
D2− 1
D1
)
in terms of the Bjerrum length in vacuum
∆G
kT=
lB2a
(1
D2− 1
D1
)
259
=560 A
(2)(3 A)
(1
2− 1
80
)
≈ 45.5
For T = 300 K,
∆G = 45.5 · 1.987 cal
K mol· 300 K ≈ 27.1 kcal mol−1
7. A solvated charge in a protein.
The Born energy in Equation (22.55) allows you to estimate the free energy cost for acharged group to be deep inside a fully folded protein (instead of at the protein surfacein contact with water). Estimate the free energy cost for a charged group to stay inside apartly folded protein. Consider the ionic group as a sphere with radius 2 A and total surfacecharge e (see the figure below). This group is first surrounded by a shell of bulk water, 4 Athick. This in turn is surrounded by a shell of protein, 14 A thick with dielectric constantDP = 4, which is in contact with bulk water at 25◦ C. Derive the electrostatic free energyof the ionic group
(a) in water,
(b) in the partly folded protein, and
(c) as in (b) but with the water shell replaced by protein, DP = 4, to simulate thecompletely folded protein.
14 Å 4 Å
2 Å
DP = 4
Dw
Dw
260
(a) Use Equation (22.57):
∆Gw =Cq2
2Dw a=
(7.13 A)
(2)(2 A)· kT = 1.78kT = 1.06 kcal mol−1
(b) To get the charge inside the protein, we (1) discharge the sphere of part (a), (2)transfer into the hydrophobic region, (3) transfer from the hydrophobic region into thewater cavity, and (4) recharge the sphere. So
∆Gpf =Cq2
2Dw a− Cq2
2Dw a+ 0 + 0 +
Cq2
2Dw a= ∆Gw,
the same as part (a)
(c) Use Equation (22.57) again, with D = 4:
∆Gf =Cq2
2Dp a=
(7.13 A)
(2)(2 A)· 78.54
4· kT = 35kT = 20.7 kcal mol−1
This is a much more unfavorable situation.
8. Oil/water interfacial potential.
Consider an uncharged oil/water interface. On the aqueous side, the proximity of the oilphase biases the orientation of the water molecules in an unknown way. Calculate theresulting potential across the boundary layer for the maximum bias, the complete line-up ofthe first layer of water dipoles perpendicular to the interface. Treat this layer as a parallelplate capacitor, with one water molecule occupying 10 A2 of the interfacial area. The dipolemoment is µ = 1.85 Debye (1 Debye = 3.336 × 10−30 C m) per water molecule. Take twovalues of the dielectric constant D between the capacitor plates:
(a) D = 2, as for oil, and
(b) D = 80, as for bulk water.
(If water is perfectly oriented, a situation called dielectric saturation, D = 2 is more likely.)
Let water be a dipole of charges +q and −q separated by distance d. Then the charge density
of the “plate” surface is σ = q/10 A2. So we can calculate the potential across this boundary
261
“capacitor” as
∆ψ =σd
ε0D=
( q
10 A2 )d
ε0D=
µ
(10 A2)ε0D
(a) For D=2
∆ψ =(1.85 D)
(3.336× 10−30 C m
D
)
(10× 10−20 m2)(8.85× 10−12 C2
J m
)(2)
= 3.48 V
(b) For D=80
∆ψ =(1.85 D)
(3.336× 10−30 Cm
D
)
(10× 10−20 m2)(8.85× 10−12 C2
J m
)(80)
= 87.2 mV
Even an uncharged oil/water interface can have a significant interfacial potential.
9. Small electrostatic potentials.
For a monovalent ion at T = 300 K, what is the value of ψ such that eψ = kT ?
ψ =kT
e=
(1.38× 10−23 J K−1)(300 K)
1.602× 10−19 C
= 25.8 mV
262
10. Sodium potential in a frog muscle.
Inside a frog muscle cell, the sodium concentration is [Na]in = 9.2 mM. Outside, the sodiumconcentration is [Na]out = 120 mM.
(a) Compute the sodium potential ∆ψ across the membrane, at T = 300 K.
(b) Which side of the membrane has the more positive potential, the inside or the outside?
(a) Use Equation (22.25):
∆ψ = ψin − ψout =kT
eln
[Na+]out
[Na+]in
= (25.8 mV) ln120 mM
9.2 mM= 66 mV
(b) The inside, since ∆ψ > 0 =⇒ ψin is more positive.
11. Membrane pores.
A neutral protein ‘carrier’ may help an ion to transfer into and across a lipid membrane.
(a) What is the electrostatic free energy change when a monovalent ion is transferred fromwater at 25◦ C to a hydrocarbon solvent with dielectric constant Dhc = 2? The radiusof the ion is 2 A.
(b) Now wrap the ion in a neutral protein to produce a spherical complex with radiusb = 15 A. What is the electrostatic free energy of transfer from water to hydrocarbonof the ion–protein complex?
For both parts, use Equation (22.63)
(a) ∆Gel =Ce2
2r
(1
D(hc)− 1
D(water)
)=
lB2r
(1
D(hc)− 1
D(water)
)· kT
=560 A
2(2 A)
(1
2− 1
78.54
)(1.987 kcal mol−1 K−1)(298.15 K)
263
= 40.4 kcal mol−1
highly unfavorable.
(b) ∆Gel =560 A
2(15 A)
(1
2− 1
78.54
)(1.987 kcal mol−1 K−1)(298.15 K)
= 5.39 kcal mol−1
much less unfavorable.
12. Solvating a protein.
A spherical protein has a valency of z = −5, and a radius of a = 10 A. The change inelectrostatic free energy, ∆Gel, when you transfer the protein from vacuum (D = 1) towater (D ≈ 80) is a part of the solvation free energy. Compute ∆Gel.
We use here radius a = 10 A. Equation (22.65) gives us
∆Gel =Cq2
2a
(1
D2
− 1
D1
)=
25 lB2a
(1
D2
− 1
D1
)· kT
=25(560 A)
2(10 A)
(1
80− 1
)(1.987 kcal mol−1 K−1)(298.15 K)
= −409.5 kcal mol−1
Solvation is extremely favorable.
13. Burying an ion pair in oil.
What is the free energy cost of transferring a monovalent anion of radius a = 2 A and amonovalent cation of the same radius a from vacuum into oil (D = 2) at an ion-pairedseparation of 2a?
Calculate this transfer by breaking into 3 steps: (1) separate the two ions to infinite distance,(2) transfer each from vacuum into oil, (3) bring the two ions back together. Combine the
264
energy changes for each step, to get:
∆Gel =Ce2
2a(1)+ 2 · Ce
2
2a
(1
2− 1
)− Ce2
2a(2)
=Ce2
2a(1)(1− 1− 1
2) = −Ce
2
4a
= −(560 A)(1.987 kcal mol−1)(298.15 K)
4(2 A)
= −41.5 kcal mol−1
The transfer from vacuum is favorable.
14. Temperature dependent dielectric constants.
Show that Coulombic interactions between charges in liquids can have an entropic compo-nent if the dielectric constant, D, depends on temperature.
Equation 20.66 of MDF shows how to get the enthalpic component for dielectric constantsthat change with temperature; there’s also an entropic part.
15. Ligand-protein electrostatic interactions.
265
Consider a ligand with charge of +1e and a spherical protein of +4e and radius of 20 Ainwater (D = 80).
(a) What is the work of bringing the ligand from r =∞ to the surface of the protein?
(b) Now assume there is a hollow cavity of radius a = 5 Ainside the protein. The proteincavity is filled with water, and there is a net charge of +4e on the shell surroundingthe cavity. What is the work of transferring the ligand from the bulk water solutionfar away from the protein to the center of the water filled cavity?
(c) Now consider the ligand to be an uncharged carboxylic acid with pKa = 3.0. If youbring this ligand into the water-filled protein cavity as described in part (b), what willbe its apparent pKa?
(a) Work to bring ligand from r =∞ to r = 20A
Work = ψa, ψr = eQDr
= 14πε0
4e80(20·10−10m)
ψr = 4(1.6·10−19C)
4π(8.89·10−19 CV ·m )(80)(20·10−10m)
ψr = 0.3596VWork = ψq = 0.03596(e) = 5.755 · 10−21J
(b) Potential inside charged spherical shell:
ψin = CqDa
, where a = 14r
So ψin = 4ψr, W = ψrq = 2.3 · 10−20J
(c) pKapp(rL = 0) = pKa(∞)− 0.4343eψ(r<a)kT
pKapp(rL = 0) = 3− 0.4343(1.6·10−19C)(4·0.03596V )
(1.38·10−23 JK
)(298K)
pKapp(rL = 0) = 3− 2.43 = 0.57The RCOOH group is more acidic inside a cavity with a positively charged shell.
266
Chapter 23Salt Ions Shield Charged Objects
1. The potential around colloidal spheres.
What is the dimensionless surface potential Φ = eψ/kT at a distance of 50 A from
(a) A colloidal sphere with a radius of 20 A and charge 20e in pure water?
(b) The same sphere in 0.1 M NaCl at 25◦C?
(c) What are the potentials in volts?
(a) φ =eψ
kT=CqeDrkT
=q
e
lBr
= 20 · 7.13 A
50 A
= 2.85
(b) Use Equation (23.22), with 1/κ = 9.62 A
φ =eψ
kT=
CqeDrkT (1 + κa)
e−κ(r−a)
=20(7.13 A)
(50 A)(1 + 20 A
9.62 A)e− 50 A−20 A
9.62 A
267
= 0.041
(c)kT
e=
(1.38× 10−23 J K−1)(298.15 K)
(1.602× 10−19 C)= 25.7 mV
(a) ψ = φ · kTe
= 2.85 · 25.7 mV = 73.2 mV
(b) ψ = 0.041 · 25.7 mV = 1.05 mV
2. The potential near a protein in salt solution.
Consider a protein sphere with a radius of 18 A, and charge Q = −10 e, in an aqueoussolution of 0.05 M NaCl at 25◦C. Consider the small ions as point charges and use theDebye–Huckel linear approximation of the Poisson–Boltzmann equation.
(a) What is the dimensionless surface potential eψa/kT of the protein?
(b) What are the concentrations of Na+ and Cl− ions at the surface of the protein?
(c) What are the concentrations of Na+ and Cl− ions at a distance of 3 A from the proteinsurface?
From Table 23.1, 1/κ = 13.6 A in 0.05 M salt solution.
(a)eψakT
=Cqe
DkTa(1 + κa)
=(−10)(7.13 A)
(18 A)(1 + 18 A
13.6 A)
= −1.705
(b) Use Equations (23.1) and (23.2) to get the concentrations at the protein face:
[Na+] = 0.05e1.705 = 275 mM
[Cl−] = 0.05e−1.705 = 9.1 mM
268
(c) Following Equation (23.18), the potential at r = 21Ais
eψ
kT=
eψakT
a
re−κ(r−a) = −1.705 · 18 A
21 Ae− 3 A
13.6 A = −1.172
Therefore
[Na+] = 0.05e1.172 = 161 mM
[Cl−] = 0.05e−1.172 = 15.5 mM
3. Surface potentials and Debye lengths.
You have a uniformly charged sphere with radius a = 50 A in a 0.02 M NaCl solution. Ata distance of 30 A from the surface of the sphere the potential ψ = 20 mV.
(a) What is the Debye length 1/κ in the solution?
(b) What is the surface potential ψa of the sphere? (Assume that the potential field inthe solution around the sphere can be derived from the linear Poisson–Boltzmannequation.)
(c) What is the charge Q on the sphere?
(d) Sketch the potential as a function of distance from the sphere.
(a) Table 23.1 tells us that 1/κ = 21.5 A for a 0.02 M solution
(b) Rearrange Equation (23.18) to get
ψa = ψ(r)r
aeκ(r−a) = 20 mV · 80 A
50 A· e
30 A
21.5 A
= 129.16 mV
(c) The number of elementary charges e on the sphere follow from Equation (23.20):
Q
e=ψae
4πε0Da(1 + κa)
269
It is convenient to work with the dimensionless potential:
eψakT
=ψa
25.8 mV=
129.16 mV
25.8 mV= 5.0
ThenQ
e=
eψakT· 4πε0DkT
e2a(1 + κa) =
eψakT· l−1B · a(1 + κa)
= 5 · 7.13 A · 50 A(1 +50 A
21.5 A) = 116.6
4. The potential near a plane.
You have a uniformly charged flat plate in contact with a 0.02 M NaCl solution at 25◦ C.At a distance of 3 nm from the plate, the potential is 30 mV.
(a) What is the Debye length 1/κ in the solution?
(b) What is the surface potential of the charged plane in mV, and in units of kT/e? Usethe linear Poisson–Boltzmann equation.
(c) If you had used the nonlinear Poisson–Boltzmann equation, would you find the surfacepotential to be larger or smaller than you found under (b)? Why?
(d) Use the surface potential from (b) to find the surface charge density σ of the plane.
(a) From Table 23.1, we get 1/κ = 21.5 Ain a 0.02 M salt solution
(b) Rearrange Equqtion (23.10) to get
ψ0 = ψ(x)eκx = (30 mV)e30 A
21.5 A
= 121.1 mV
φ0 =eψ0
kT=
121.1 mV
25.7 mV
= 4.71
270
(c) Larger, because the curvature of the (non-linearized) potential-distance curve isgreater, while the curves for the two P-B equations coincide at large distances (bothhave ψ → 0).
(d) From Equation (23.13)
σ
e=
Dε0κψ0
e=kTDε0φ0κ
e2
=κφ0
4πlB=
4.69
(21.5 A)(4π)(7.13 A)
= 0.00245 A−2
or σ = (0.00245 A−2)e = (0.00245× 1020 m−2)(1.602× 10−19 C)
= 0.039 C m−2.
5. A charged protein near a chromatographic surface.
Consider a spherical protein in water with charge −q (q > 0) at a distance R0 from a planarion chromatography column surface with net positive charge, shown in the figure below.
(a) Draw the field lines approximately. Include arrows to indicate the field direction.
(b) Draw the equipotential contours.
(c) If the solvent water were replaced by a water/methanol mixture with a lower dielectricconstant, would it weaken or strengthen the attraction between the protein and theplanar surface?
R0 −q
271
(a)
−q
equipotential lines (b)
field lines (a)
+
+
+
+
+
(b) Equipotential lines are ⊥ to field lines.
(c) It would strengthen the attraction.
6. The potential of a membrane.
Consider a phospholipid bilayer membrane consisting of 90% uncharged lipid (zwitterionicphosphatidylcholine) and 10% acid lipid (singly charged phosphatidylserine or phosphatidyl-glycerol). Assume 68 A2 surface area per lipid head group. The membrane is in contact withan aqueous solution of NaCl concentration cNaCl at 25◦C. Calculate the surface potential ofthe membrane for cNaCl = 0.05 and 0.1 mol−1.
Inside the membrane there is no space charge: (d2ψ)/(dx2) = 0. This gives ψ = C1x+C2. Atx = −∞ ψ must be finite, so we choose C1 = 0; therefore ψ = C2. On the aqueous side wesolve the linear PB equation (Equation (23.13)) which yields
ψ0 =σ
κε0D.
272
Converting to dimensionless quantities,
eψ0
kT=σ
e
4πlBκ
where the Bjerrium length is lB = 7.13 Aand σ/e = 0.1/68 A2. In 0.05 M NaCl, 1/κ = 13.6 A;therefore
eψ0
kT=
0.1
68 A24π(7.13 A)(13.6 A) = 1.79
In 0.1 M NaCl, 1/κ = 9.62 A, so eψ0/kT = 1.27.
7. Binding to a membrane.
What is the electrostatic free energy of binding to the membrane in problem 6 of a trivalentpositive ion such as spermidine (a biologically active polyamine), assuming that
(a) binding occurs at the membrane surface, and
(b) owing to steric factors, the charges of the bound spermidine stay in the water 5 Adistant from the membrane surface?
In 0.05 M NaCl the electrostatic free energy of binding is in case (a)
∆Gb = 3eψ0 = 3(1.79 kT ) = 5.37 kT per bound molecule, or
∆Gb = 5.37RT per mole = 3.18 kcal/mol
In case (b)
ψ = ψ0e−κx =
(1.79
kT
e
)e−
513.6 = 1.24
kT
e
so ∆Gb = 3eψ = 3(1.24kT
e) = 3.72 kT per bound molecule
= 3.72RT per mole = 2.20 kcal/mol.
For 0.1 M NaCl, use the answers from problem 6 and the Debye length 1/κ = 9.62 Ato get
273
(a) ∆Gb = 2.26 kcal/mol and (b) ∆Gb = 1.34 kcal/mol.
8. Electrostatic potential near a protein.
A protein in aqueous solution with 0.1M monovalent salt has 9 positive and 22 negativecharges at 25◦ C. It is modelled as a sphere with radius 20 A and uniform surface charge.What is the potential in units of kT/e
(a) at the surface of the protein, and
(b) at a distance 10 A from the protein surface into the solution?
Use (i) Coulomb’s equation (no shielding by small ions) and (ii) the linear Poisson–Boltzmann equation.
(a) (i) Use Equation (21.45)
eψakT
=ze
4πε0Da
e
kT=lBz
a=
(7.13 A)(−13)
20 A= −4.63
(ii) Equation (23.20) for ψa is the same as case (ai) but for the extra factor1 + κa = 1 + (20/9.62) = 3.079:
eψakT
=−4.63
3.079= −1.51
(b) (i) Use Equation (21.35)
eψ
kT=eψakT
a
a + 10 A= −4.63
20 A
30 A= −3.09
(ii) We use the surface potential of (bi) with Equation (23.17)
eψ
kT=eψakT
a
a + 10 Ae−κ(r−a) = −3.09e
− 10 A
9.62 A = −0.356
9. Debye–Huckel model.
Apply the Debye–Huckel theory to a 0.01M monovalent salt solution at 25◦C. Treat theions as hard spheres with a radius of 2 A. Assume D = 80. What is the change in chemicalpotential in cal mol−1 due to the ion interactions? What is the activity coefficient of theions?
274
Apply Equation (23.29):
∆ψ = 2kT ln γ = − e2
4πε0D
κ
1 + κb
= − (1.90× 10−19 C)2
4π(8.85× 10−12 C2
J m
)(80)
(30.4× 10−10 m)−1
1 + 2(2 A)
30.4 A
= −1.18× 10−21 J
molecule
= (−1.18× 10−21 J)(6.022× 1023 mol−1)
(1 cal
4.184 J
)= −169.8
cal
mol
ln γ =∆ψ
2kT=
−1.18× 10−21 J
2(1.38× 10−23 JK
)(298.15 K)= −0.1434
=⇒ γ = 0.866
10. Electrostatic potential around a line charge.
A line charge with density λ and with length ` is in a salt solution with Debye length 1/κ.Use Equation (23.24) to find an integral expression for the potential in the plane bisectingthe line charge.
(Similar to problem 20.4(a)) Choose the z coordinate along the line charge, from − l2
to + l2.
A piece dz at point z has charge λdz, at distance√r2 + z2 from P . The contribution of this
charge to the potential ψ(r) at P is
dψ(r) =λdze−κ
√r2+z2
4πε0D√r2 + z2
The total potential at P is then
ψ(r) =∫ z=+ l
2
z=− l2
dψ(r) =λ
4πε0D
∫ l2
− l2
e−κ√r2+z2
√r2 + z2
dz
Let y =√κ2(z2 + r2), then
275
=λ
4πε0D
∫ κ
√r2+ l2
4
−κ√r2+ l2
4
e−y
ydy
The restriction is that the answer is valid only for low potential, like Equation (23.21)
11. Electrostatic potential near a vesicle.
A simple model for a spherical vesicle is a spherical shell with radius a, charge density σ,permeated by a salt solution with Debye length 1/κ. Use Equation (23.24) to derive thepotential outside the shell at distance r from the center.
(a) Derive and discuss the limit of ψ(r) for κ→ 0.
(b) For the situation in (a), derive the potential inside the spherical shell. What is thelimit for κ→ 0?
(c) From the results of (a) and (b), derive the expression for the limit κa → ∞ andcompare these with the result of Example 23.3.
(a) Use the same set-up at with problem 7 in Chapter 21, with total charge on the sphereq = 4πσa2, and introducing the Debye term, so:
dψ(r) =(4πσa2)
8πε0aDr
(e−κR
)dR =
σa
2ε0Dr
(e−κR
)dR
Integrate: ψ(r) =σa
2ε0Dr
∫ r+a
R=r−ae−κRdR
=σa
2ε0Dκr
(e−κ(r−a) − e−κ(r+a)
).
For κ→ 0, remembering that e−x ≈ 1− x for small x:
ψ(r) ≈ σa
2ε0Dκr[(1− κ(r − a))− (1− κ(r + a))]
=σa2
2ε0Dr=
q
4πε0Dr
as expected from Coulomb’s Law.
276
(b) Use the setup for finding the inner potential equation derived for problem 21.7, withthe same changes used in part (a):
ψ(r) =σa
2ε0Dr
∫ a+r
R=a−re−κRdR
=σa
2ε0Dκr
(e−κ(a−r) − e−κ(a+r)
)
So for κ→ 0 we have e−κ(a−r) − e−κ(a+r) ≈ 2κr so ψ(r)→ σaε0D
= q4πε0Da
as found withCoulomb’s law.
(c) With κa→∞ we have e−κ(a+r) → 0, ar→ 1 and r − a = δ. So outside the charged shell
we have
ψδ =σa
2ε0Dκr
(e−κ(r−a) − e−κ(r+a)
)
=σ
2ε0Dκe−κδ δ > 0
Inside the shell:
ψδ =σa
2ε0Dκr
(e−κ(a−r) − e−κ(a+r)
)
=σ
2ε0Dκe+κδ δ < 0
We have for κa→∞ two flat double layers, one inside and one outside the shell, as inexample 23.4.
277
12. Debye lengths.
In the Debye–Huckel theory of monovalent salt solutions there is a characteristic lengthquantity κ, defined by κ2 = (2e2n∞)/(ε0DkT ), where n∞ is the salt concentration.
(a) Express κ in terms of the Bjerrum length `B.
(b) For water at room temperature, `B = 7.13 A. Compute the Debye length 1/κ in A fora solution of n∞ = 1 mol L−1.
(a) κ2 = (2)(4π)(n0) ·e2
4πε0DkT= 8π n0 lB
(b) κ2 = 8π(1 mol L−1)(7.13× 10−10 m)
=⇒ 1
κ= 3.04 A
13. Ion binding to a sphere.
Compute the free energy of bringing a divalent ion into contact with a spherical particle ofradius a = 14 A, from far away. The ion has valence z = +2 and the particle has a valenceof Z = +20. For water at room temperature, compute the free energy
(a) in a solution having monovalent salt concentration 0.1 M;
(b) for a solution having monovalent salt concentration 0.01 M.
(a) Use the Debye–Huckel model to get ψ(a), the electrostatic potential at the surface ofthe charged sphere (r = a), with ∆G = qψ(a):
∆G = qψ(a) =CQq
Dwa(1 + κa)
To simplify, kT =Ce2
DwlBw
so∆G
kT=
Zz
(1 + κa)
(lBwa
)
278
Since Q = Ze and q = ze, where Z = +20 and z = +2 are the valences, then
∆G
kT=
(20)(2)(1 + 14 A
9.62 A
)(
7.13 A
14 A
)= 8.30 so ∆G = 4.92
kcal
mol
(b)∆G
kT=
(20)(2)(1 + 14 A
30.4 A
)(
7.13 A
14 A
)= 13.95 so ∆G = 8.26
kcal
mol
14. Polymer on charged surface.
A polymer with a single positive charge at one end is attached to a uniformly charged sheetin a 0.1 M NaCl solution at the other end. It can take on the conformations shown. Thedistances, d, between model monomer beads is 2 A.
The electric potential at bead 4 due to the shielded sheet is: ψ(x) = σκε0D
e−κx,
where σ = 4× 10−4C/m2, D = 80, and κ = 10 A.Beads 1 and 4 have a favorable contact energy of ε = −2kBT , where T = 300 K. If the beadsare not adjacent to each other, there is no contact energy.
279
(a) Compute the energy and density of states for each energy level of the above ensemble.
(b) What is the partition function for this polymer? Express your answers in terms of theenergies computed above.
(c) Based on your partition function, what is the fraction of polymer in each state?
(d) Now assume that the system is moved to a lower dielectric medium, with D = 20,but still with 0.1 M NaCl. What is the fraction of polymer in the linear state whenimmersed in the new solvent? Explain why this fraction increased or decreased fromthe previous system.
(a) The electric potential at bead 4 due to the shielded sheet is: ψ(x) = σκε0D
e−κx, so theelectrostatic energy at distance x from the sheet is U(x) = q · ψ(x).
Linear polymer:D(s) = 1U(x) = eσ
κε0De−κ·4d
U(x) =(1.6·10−19C)(4·10−4 C
m2 )(10−9m)
(8.85·10−12 CV ·m )(80)
e−8A
10A
= 9.04 · 10−23CV · e−0.8 = 4.06 · 10−23JBent polymer:D(s) = 2U(x) = eσ
κε0De−κ·3d
U(x) = 9.04 · 10−23CV · e−6A
10A = 4.96 · 10−23J
(b) Q = e−U(4d)kBT + 2e
−U(3d)kBT + 2e
−U(2d)kBT
−2
= e−(4.06·10−23J)
4.14·10−21 + 2e−(4.96·10−23J)
4.14·10−21 + 2e−(6.06·10−23J)
4.14·10−21 −2
Q = 0.9902 + 1.9762 + 14.5634 = 17.5298
(c) Fraction of polymer in each state:
f(linear) = e−U(4d)kBT
Q= 0.9902
17.5298= 0.0565
f(bent) = 2e−U(3d)kBT
Q= 1.9762
17.5298= 0.1128
f(U-shaped) = 2e−U(2d)kBT
−2
Q= 14.5634
17.5298= 0.8308
(d) The dielectric constant changes from 80 to 20, so Dnew = Dold4
280
κ is proportional to D−12 , so κnew = 2 · κold = 0.2A = 2 · 109 1
m= 1
5·10−10m
New energy levels:Linear polymer:U(x) = eσ
2κε0(D4
)e−2κ·4d
U(x) =(1.6·10−19C)(4·10−4 C
m2 )(5·10−10m)
(8.85·10−12 CV ·m )(20)
e−8A
20A
= 1.81 · 10−22CV · e−0.4 = 1.21 · 10−22JBent polymer:U(x) = eσ
2κε0(D4
)e−2κ·3d
U(x) = 1.81 · 10−22J · e−6A
20A = 1.34 · 10−22JU-shaped polymer:U(x) = eσ
2κε0(D4
)e−2κ·2d − 2kBT
U(x) = 1.81 · 10−22J · e−4A
20A − 8.28 · 10−21J = −8.13 · 10−21J
Q = e−U(4d)kBT + 2e
−U(3d)kBT + 2e
−U(2d)kBT
−2
= e−(1.21·10−22J)
4.14·10−21 + 2e−(1.34·10−22J)
4.14·10−21 + 2e−(1.48·10−22J)
4.14·10−21 −2
= 0.9712 + 1.9363 + 14.2523 = 17.1598The new fraction of polymer in the linear state:
f(linear) = e−U(4d,Dnew)
kBT
Qnew= 0.9712
17.1598= 0.0566
The fraction of polymer in the linear state has increased slightly. There are twocompeting dielectric effects: (1) the decreased dielectric increases electrostatic repulsionbetween the surface and the charge and (2) charge shielding increases with decreaseddielectric (since the Debye length depends on the dielectric), decreasing theelectrostatic repulsion. Besides these effects nearly balancing each other out, theelectrostatic energy is overpowered by the contact energy.
15. Salting a Protein.
A particular protein has a large net charge at low pH. Adding a salt stabilizes the folded orthe unfolded state?
Salt stabilizes the folded state by shielding the charge.
16. Debye length.
What is the Debye length for a 0.01 M solution of magnesium sulfate (MgSO4), at T =300 K?
281
κ2 = 2(ze)2n∞NDε0RT
= 2 · 4π(1.386 · 10−4 Jmmol
)(0.01molL
)(103 Lm3 )(
6.022·1023moleculesmol
80(8.314 JK·mol )(300K)
)
κ2 = 0.001078 · 1020 1m2
1κ
= 3.046 · 10−9 = 3.046nm
282