The MOLECULESof LIFEPhysical and Chemical Principles

Selected Solutions for Students

Prepared by James Fraser and Samuel Leachman

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer Garland Science

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 moleculesd. 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. 5/10b. 1/10c. 1/9d. 5/51e. 1/6

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.

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 Pascals triangle, with the tenth row filled in.

a. Fill in the values for the eleventh row. What simple rule can you use to fill in these values?

b. Using Pascals triangle, calculate how much more likely it is to get five heads and six tails in a series of 11 coin tosses than getting four heads and seven tails.

Answer:

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 Pascals triangle, the multiplicity for getting five heads is 462, and for four heads it is 330. So, the ratio of the probabilities is:

Entropy

Chapter 7

Problems

True/False and Multiple Choice

1 11 55 165 330 462 462 330 165 55 11

1 10 45 120 210 252 210 120 45 10 1

1

1 10 45 120 210 252 210 120 45 10 1

1 11 55 165 330 462 462 330 165 55 11

1 10 45 120 210 252 210 120 45 10 1

1

1 10 45 120 210 252 210 120 45 10 1

2 ChAPTEr 7: Entropy

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer Garland Science

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. (Hint: Use Equation 7.2.2.)

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!) = 308,897,820,000The 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 one-letter 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 = 1.46 104.

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.

22. Considering the following system at two time points, A 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?

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 Stirlings approximation.)

Answer:

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:

28. In our calculations, why do we work with the natural log of the multiplicity?

X

Y

Z

O

Problems, long answer, Q3 (entropy_28_v1)

X

YY

Z

O

O

O

462= 1.4

330

XB. X XXX

XA. X XXX

Problems, long answer, Q9 (entropy_31_v1)

W =Spaces!

Green!Yellow!Empty!=

10000!

2000!4000!4000!

S = kB ln10000!

2000!4000!4000!

= kB ln((10000 ln10000 10000) (2000 ln 2000 2000 + 4000 ln 4000 4000 + 4000 ln 4000 4000)

= 10549.2 kB

V2V1

w = nRT ln

= 293 K 8.31 JKmol1 ln (0.250)

= 3375.3 J

S = wT

=3375.3 J

293 K= 11.5 JK1

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.