Solutions practice exam - PRACTICE EXAM FOR FINAL EXAM AT 22 JUNE 2022 NOTES AND INSTRUCTIONS - Studeersnel (2024)

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PRACTICE EXAM FOR FINAL EXAM

AT 22 JUNE 2022

NOTES AND INSTRUCTIONS

• Support all your answers with short arguments, or as much as is needed.

• Make sure to be systematic in calculations and derivations. Intermediate steps are expected and rewarded, unless the question explicitly states that you can skip the derivation.

• Some questions may be more difficult than others. In case you would get stuck on one, you can skip to the next. Most questions do not depend on the previous answers.

• You can use all your notes and materials, but let it not distract from what you know. By all your training, your knowledge and intuition may guide you well.

• Working together is not allowed.

• Answers can given with any type of preferred materials (pencil / pen / ... on any type of paper) as long as the answers are clear and legible, and preferably pre- sented in an organised way.

• For the exam you can use two full hours (13:45h - 15:45h) plus 20 minutes for scan- ning and submission. Please report technical issues via email (j.w@tudelft) or the regular zoom meeting room: tudelft.zoom/j/98394956411?pwd=T0MrZlRHblNvVnplR01jZmdzR2ltQT also for other issues and questions.

• The moral values of science are in your hands. Guarding them well, and contem- plating them, will help supporting the quality of science and education, and will strengthen your own confidence and abilities.

1

PART I

A drunk person is lost on a 3-dimensional Cartesian grid. Starting at the origin, the per-son is taking steps with a step size 1, in the positive or negative x, y, and z-direction withequal probability (there are 6 different possibilities, so each direction has a probability 1 6 ). Each second, the person takes one step. (a) Calculate the mean after 3 hours. The mean x position after 1 step is

〈x〉 = −1 1 6

+ 1 1

6

= 0 (1)

The same for y and z, so after N steps it is still 0. (The probability of going left isthe same as going right, so it should be zero)

(b) Calculate the standard deviation after 3 hours The standard deviation in x after 1 step is not zero.

〈x 2 〉 = (−1) 2

1

6

+ 12

1

6

=

1

3

(2)

so σ 1 (x) =

√

〈x 2 〉 − 〈x〉 2 =

1

p 3

(3)

The same for the standard deviation in the y and z-direction. So after N steps

σN (x) = σN (y) = σN (z) =

√

N

3

(4)

After three hours, N = 10800, so σx (N ) = 60.

(c) What is the probability that the person will get home, at position (x, y, z) = (500, 0, 0)? Using the central limit theorem

P (x, y, z) = P (x)P (y)P (z) ∝ e−

x 2 +y 2 +z 2 2 σN = e− 3(x

2 +y 2 +z2) 2 N (5)

Substituting the values x = 500, y = 0, and z = 0 and N = 60 yields a fantasticallysmall number, so the probability is just zero (not mathematically zero, but for allother reasons it is).

(d) Estimate the time needed to get home, and compare this to the time needed for a sober person. The question is not asking for a specific time, but an estimation. One could ar- gue that ’home’ needs to be within the standard deviation to have a reasonable probability of getting there. In that case one would have

σN =

√

N

3

≈ 500 (6)

(a) How do the particles distribute in this system? According to the Boltzmann distribution

ρ(r ) ∝ e−βar

2 = e−βV (r ) (14)

(b) A hom*ogeneous distribution has the highest entropy. Explain why your previous answer is not in violation with the second law of thermodynamics. Because the system is not closed, but in thermal contact with a reservoir. The state of the system is connected to the entropy of the reservoir, and it is such that the total entropy will tend to a maximum (not the entropy of the system).

(c) What is the average kinetic energy of the particles? Fro any 3-dimensional system 〈 N ∑i = 0

p 2 i2 m

〉

=

3

2

N kBT (15)

PART III

(a) What is the first law of thermodynamics? ...

(b) Given that d F = −Sd T − pdV + μd N , find the differential form for the grand po- tential Φ(μ,V, T ), d Φ = ..., using a Legendre transform.

Φ = F −

∂F

∂N N

= F − μN (16)sod Φ = d F − d μN − μd N = −Sd T − pdV − μd N (17)

(c) Describe the μ,V, T -ensemble and give an example. This is also called an open system, a system that can exchange particles and heat with an environment. For example, electrolytes in a hydrogel, or a template where particles can bind from solution.

(d) Argue that the grand potential can be written as Φ(μ,V, T ) = p(μ, T )V , with p(μ, T ) the pressure. (so, Φ/V represents the pressure of the system) μ and T are intensive, and V and Φ are extensive, so

Φ(μ, λV, T ) = λΦ(μ,V, T ) (18)

this implies that Φ has to be linearly dependent on V , and the proportionality fac-tor has the dimension of pressure, and is, in fact, the (osmotic) pressure of thesystem Φ(μ,V, T ) = p(μ, T )V (19)

Figure 1: The free energy density of a fluid at different temperatures

PART IV

The free energy density f of a certain fluid is plotted at different temperatures, in figure

1. The units of f and ρ are dimensionless, for convenience.

(a) Estimate the critical temperature. About 45oC. The lower temperatures have regions with negative curvature, so for those temperatures the system can phase separate (into two densities, that can be found by using this clever trick, the common tangent construction). At 45oC the curvature seems to be zero at 0 but it is nowhere negative. That is the signature of a critical point

(b) Estimate the critical density. About 0.

(c) Name two ways in which a phase transition can occur, and indicate in the graph where you would expect those phenomena to happen. Nucleation and growth, and spinodal decomposition. See set 6 for more explana- tion

(d) Describe what would happen if one would gradually increase the density from ρ = 0 to ρ = 1 at T = 35 oC The density would initially be hom*ogeneous, and it would increase up to about 0. After that point the system could in principle fall apart into two phases, but it would need to do that via nucleation and growth (which requires an initial energy

For a positive ion it should be high in the channel, and lower outside, and for anegative ion it should be low (negative) in the channel, and closer to zero outside.

(c) Sketch what would happen to the potential energy of an ion in the center of the channel if the salt concentration would increase, and give arguments. If the salt concentration increases, the screening length decreases (electric fields get screened over a shorter range), so the ions in the center of the channel would feel the boundaries less. The potential energy for a positive ion would drop, and for a negative ion would be less negative

(d) Argue why this channel could block the current of positive ions, if the conditions are right. If the potential energy is high in the channel, it requires work to get the ions through the channel. This would be the case at low salt concentrations. At high salt concen- trations, the energy barrier becomes lower, and it will be easier to pass a current. (actually for two reasons: there are more ions, and the barrier is lower)

(e) Why do the conditions have to be right? In other words, are there conditions under which the channel would not block the ions? See previous answer

[this question was not selected for the final exam for several reasons - but it seemsvery useful for the preparation nonetheless]

PART IV

A template with N binding sites is in contact with a solution of particles that can bind tothese sites with energy ε. The chemical potential of the solution is μ and the temperatureis T.

(a) What is the energy of the system for a configuration of n particles?

En = nε (20)

(b) What is the entropy for a configuration of n particles?

S(n) = kB ln Ω(n) = kB ln

(

N

n

)

= ... (21)

(you can use Stirling here, see set 3)

(c) When is the entropy maximal? And when is the energy minimal? Can both be achieved at the same time? See set 3 for maximal entropy. The minimal energy is obtained when all the sites are occupied, but that configuration has the lowest entropy. So no, it is not possible to achieve both. (a compromise is found in exercise 3)

(d) What is the (grand) partition function of the template?

see 5 and notes during lecture 5 (under slide shows on Brightspace)

(e) Calculate the average number of particles binding to the template for a given μ and T. Same...

and the grand canonical partition function

Z =

∑

n

e−βEn +βμNn (23)

=

∑

N

Ω(N )e−βE (N )+βμN

with Ω(N ) the number of states containing N particles.

〈N 〉 = kBT

∂

∂μ ln Z

In three dimensions 〈 p 22 m

〉

=

3

2

kBT

and under some conditions 〈 p 22 m

〉

= 〈V 〉 =

3

2

kBT

if the potential energy V is a quadratic function. This is the so-called equipartition the-orem.