The MOLECULES
of LIFE
Physical and Chemical Principles
Selected Solutions for Students
Prepared by James Fraser and Samuel Leachman
Chapter 7
Entropy
Problems
True/False and Multiple Choice
2. Two sets of molecules are mixed into a system at time
zero. After 10 seconds, equilibrium has been reached. At
what time was the entropy of the system maximal?
a. 10 seconds (equilibrium)
b. 5 seconds (half the time it takes to reach
equilibrium)
c. immediately upon mixing of the two sets of
molecules
d. prior to mixing of the two sets of molecules
4. On a 10-sided die, with a side for each number from 1 to
10, the probability of rolling a 5 is:
a.
b.
c.
d.
e.
5/10
1/10
1/9
5/51
1/6
Answer: greater
10. The log of the multiplicity of the system is an
__________________ property of the system.
Answer: additive/extensive
12. A drop of dye is added to a container of solvent. When
equilibrium is reached the concentration of dye will be
____________ within the container.
Answer: uniform
Quantitative/Essay
14. Shown below is a portion of Pascal’s triangle, with the
tenth row filled in.
1
6. A system is divided into two halves separated by a
removable divider. Initially, with the divider in place,
the left half has only red molecules and the right
half has only blue molecules. The divider is removed
and equilibrium is reached. Which of the following
statements correctly characterizes the
system?
a. It contains only blue molecules on the left side.
b. It contains only red molecules on the left side.
c. It contains a mixture of red and blue molecules
throughout the system.
d. It contains only blue molecules at the bottom of the
system.
e. It has an equal number of red and blue molecules
on each side.
Fill in the Blank
8. The work done in a near-equilibrium expansion of an
ideal gas is _____________ than for a nonequilibrium
expansion.
10
45
120 210 252 210 120 45
10
1
a. 1Fill 10
in the
45values
120 for
210the
252eleventh
210 120row.
45What
10 simple
1
rule can you use to fill in these values?
b. Using Pascal’s triangle, calculate how much more
likely
and 210
six tails
a series
1 it is
10to get
45 five
120 heads
210 252
120 in45
10 of
1 11
coin tosses than getting four heads and seven tails.
1
11
Answer:
1
1
55
10
11
165 330 462 462 330 165
45
55
120 210 252 210 120
55
45
165 330 462 462 330 165
11
10
55
1
1
11
1
a. Simple rule: The numbers at the edges are always
1. All the other numbers are the sums of the two
numbers diagonally above.
b. According to Pascal’s triangle, the multiplicity for
getting five heads is 462, and for four heads it is 330. So,
the ratio of the probabilities is:
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
2
Chapter 7: Entropy
462
330
= 1.4
16. Calculate the entropy of the system depicted below.
There are four types of molecules (that is, X, O, Y, and Z)
that can be arranged in any way in the available boxes.
Problems,
long answer, Q3 (entropy_28_v1)
(Hint: Use Equation 7.2.2.)
X
Y
Z
O
O
X
Y
Y
O
Z
O
Answer:
The multiplicity of the system (W) is given by
M!/(X!O!Y!Z!(M – X – O – Y – Z)!).
W = 25!/(2!3!4!2!14!) = 308,897,820,000
The entropy is given by kB lnW = kB ln(308,897,820,000)
= 26.46 kB.
18. A bag contains 20 coins, each marked by the oneletter code for an amino acid. What is the probability of
drawing each of the three large amino acids (Y, W, and
F) once, in any order, without returning the coins to the
bag after each draw? What is the probability if coins are
returned to the bag after each draw?
What is the probability of drawing the three letters
out in exactly the order of hydrophobicity (a unique
ordering) without returning any coins to the bag?
Answer:
Large amino acids:
Drawing each without replacement = 3/20 × 2/19 ×
1/18 = 0.00088.
With replacement = 3/20 × 2/20 × 1/20 = 0.00075.
Hydrophobicity:
Probability of drawing the first amino acid = 1/20,
second = 1/19, third = 1/18. 1/20 × 1/19 × 1/18 =
1/6840 = 1.46 × 10–4.
20. System A has a multiplicity of 15, whereas System B has
a multiplicity of 12. What is the total entropy of Systems
A and B?
Answer:
The entropies of the two parts will be additive.
Therefore we can calculate them separately.
SA = kB ln(15) = 2.71 kB; SB = kB ln(12) = 2.48 kB.
Total entropy = 5.19 kB.
A.
X
X
B.
X
X
X
X
Answer:
At time point A:
The multiplicity of the left side is: M!/(X!(M – X)!) =
4!/3! = 4.
The multiplicity of the right side is: M!/(X!(M – X)!) =
6!/(2!4!) = 15.
The total multiplicity = mulitpicityright × multiplicityleft
= 4 × 15 = 60.
At time point B:
The multiplicity of the left side is: M!/(X!(M – X)!) =
5!/(3!2!) = 10.
The multiplicity of the right side is: M!/(X!(M – X)!) =
5!/(2!3!) = 10.
The total multiplicity = mulitpicityright × multiplicityleft
= 10 × 10 = 100.
The multiplicity of time point B is higher. Since the
multiplicity of time point B is higher, it is closer to
equilibrium than time point A.
24. A system consists of molecules that convert between
two colors (green and yellow). There are spaces for
10,000 molecules, but there are only 6000 molecules
in the system. The system starts in a state with 2000
green molecules and 4000 yellow molecules. What
is the entropy of the system? (Hint: Use Stirling’s
approximation.)
Answer:
W=
Spaces!
Green!Yellow!Empty!
S = kB ln
=
10000!
2000!4000!4000!
10000!
2000!4000!4000!
= kB ln((10000 ln10000 – 10000) – (2000 ln 2000 – 2000 +
4000 ln 4000 – 4000 + 4000 ln 4000 – 4000)
= 10549.2 kB
26. How much work is done in compressing one mole of an
ideal gas from a starting volume of 1 L to a final volume
of 250 mL at a constant temperature of 293 K? What is
the change in entropy? Assume that the process occurs
near-equilibrium (reversibly).
Answer:
V2
V1
= –293 K × 8.31 J•K•mol–1 × ln (0.250)
w = –nRT ln
22. Considering the following system at two time points, A
Problems, long answer, Q9 (entropy_31_v1)
and B.
The system is divided by a movable partition. The
molecules (X) cannot move across the partition, but
they can move between the grid boxes on the same
side of the partition. At which time point does the
system have the higher multiplicity? At which time point
is the system closer to equilibrium?
X
X
X
X
= 3375.3 J
–w –3375.3 J
= 11.5 J•K–1
∆S =
=
T
293 K
28. In our calculations, why do we work with the natural log
of the multiplicity?
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
PROBLEMS and solutions — EVEN NUMBERS ONLY
3
Answer:
ln(W) is an additive property of the system, and is more
convenient to work with than W because its numerical
value remains manageable as the number of molecules
in the system increases.
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science