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Data Provided: A formula sheet and table of physical constants is attached to this paper. Ancillary Material: 3 x Graph paper (linear) DEPARTMENT OF PHYSICS AND ASTRONOMY Autumn Semester (2018) THE PHYSICS OF SOFT CONDENSED MATTER 2 HOURS Instructions: Answer any 3 questions of your choice out of the 5 questions offered. There is no compulsory question. Questions are marked out of twenty. The breakdown on the right- hand side of the paper is meant as a guide to the marks that can be obtained from each part. Please clearly indicate the question numbers on which you would like to be examined on the front cover of the answer book. Cross through any work you do not wish to be examined.

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1. The table shows the temperature-dependent viscosity (h) of the glass-forming solvent o-terphenyl.

T [oC] h [Poise]

250 0.01 200 0.016 100 0.04 50 0.5 35 1.58 13 316 -2 158,500

-15 3.16*108 -20 7.94*1011

a.) Plot the data in appropriate form to decide if o-terphenyl forms a ‘strong’ or a ‘fragile’ glass. If it forms a strong glass, estimate the activation energy of liquid flow from your plot. If it forms a fragile glass, estimate the conventional glass temperature (the temperature where viscosity reaches 1013 Poise) from your plot, and give an estimate of the Vogel-Fulcher temperature in oC. b.) If the relaxation time of a polymer melt at the glass transition TG is 10 s by definition, how long is the relaxation time 8 oC below the glass transition? Hint: Polymers follow WLF ‘time-temperature superposition’,

log𝑎& = log𝜏(𝑇)𝜏(𝑇,)

= −𝐶/(𝑇 − 𝑇,)𝐶0 +(𝑇 − 𝑇,)

with C1 = 17.4 and C2 = 51.6 K.

c.) The ‘90/10’ time for the relaxation of a time-dependent elastic modulus E(t) is defined as the time it takes to relax from 90% of its initial value E(0) to 10% of E(0). An elastic modulus relaxes according to a ‘stretched exponential’ law,

𝐸(𝑡) = 𝐸(0)exp 8−9𝑡 𝜏: ;

<=

with an exponent b £ 1 within the exponential.

Calculate the ‘90/10’ time for the ‘stretched exponential’ relaxation, in multiples of t, for b = 0.3, 0.5, 0.7, 0.9, and 1.

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Question 1 continued

d.) Copy the molecules shown below into your answer book. Place an asterisk (*) at each chiral centre.

Warfarin

Vitamin C

Methadone Note: You will score a mark for each correctly placed asterisk, but lose one for a misplaced asterisk. So best avoid unsubstantiated guesses. There will be no overall negative score for Question 1d though.

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2. The structure of polystyrene (PS) is given by the chemical formula below:

a.) Calculate the molar mass of the PS repeat unit. Assume the molar masses of hydrogen and carbon are 1 g/mol and 12 g/mol exactly.

Below is a table of the radius-of-gyration Rg of monodisperse PS samples of different molar masses (Mw), determined by light scattering from dilute solutions in the solvent cyclohexane at 35 oC. Rg is a measure of the size of a dissolved polymer coil which is related to rms end-to-end distance, R, by Rg = 0.4825 R.

Mw [g/mol] Rg [nm]

2*104 4.3 4*104 6.5 105 10

4*105 20 106 29

4*106 58 107 95

2.5*107 150 5.8*107 225

b.) Plot the data in the table in an appropriate form to decide if cyclohexane at 35 oC is a good solvent, or a Q solvent, for PS. Briefly explain your reasoning in constructing this plot.

c.) From known chemical bond lengths and angles, it can be determined that the length a of the PS repeat unit is a = 2.33 Angstrom. Plot the data in the table above again in a different form than in b.) to find the characteristic ratio, C¥ , of PS. Briefly explain your reasoning in constructing this plot. Hint: In a polymer melt, R2 = C¥ N a2, where N is the number of repeat units in a chain.

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Question 2 continued

Polystyrene (PS) is miscible without phase separation with poly(para-phenylene oxide) (PPO) for all volume (or weight) fractions and forms amorphous (non-crystalline) blends. The diagram below shows the glass transition temperatures of such blends as a function of PS weight fraction, and a fit to the model of Gordon and Taylor.

d.) Explain how this diagram shows that the glass transition is not a phase transition between a solid and a liquid state.

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3. The phase behavior of a binary mixture of two liquids is described by the free energy

of mixing expression derived from the lattice model,

∆?@AB

= 𝛷 ln𝛷 + (1 − 𝛷)ln(1 − 𝛷) + 𝜒(𝑇)𝛷(1 − 𝛷)

with F the volume fraction of one of the liquids, and an interaction parameter c of the form 𝜒(𝑇) = −𝐴 +𝐵 𝑇: with positive constants A and B. Assume c is independent of F. We observe cloud points at 64 oC for F = 0.3 and 76 oC for F = 0.4.

a.) What will be the cloud point temperatures at F = 0.6 and F = 0.7?

b.) Calculate the values of constants A and B.

c.) What is the critical temperature (in oC) of this mixture? Give justified and/or worked answers. Hint: In your calculations, carry as many digits as you can. Only final results should be rounded reasonably.

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Question 3 continued

Here is the phase diagram for mixtures of two other organic molecules. There never is a phase separation in the liquid phase.

a.) b.) c.) d.) e.) f.) g.) h.) i.) j.) k.) l.) m.) n.) o.) p.) q.) r.)

d.) How are the two molecules in this mixture related to each other? Justify your answer.

Note: Stating the obvious ‘They have the same melting point’ is not a sufficient answer.

Liquid

Solid

Solid +

Liquid

Solid +

Liquid

F

Eutectic

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4. The dihedral angle f describes the angle of rotation of two parts of a molecule around a carbon-carbon single bond. A typical example is n-butane, which serves as a model for isotactic polypropylene (iPP). It is a good approximation to assume f can only be one of three angles, called ‘trans’ or ‘anti’ (a), 180o; ‘gauche+’ (g+), 60o; or ‘gauche-’ (g-), 300o.

a.) Assuming all three dihedral angles are equally probable, what would be the loss of entropy (in units of kB) when an n-butane molecule would be forced to adopt one particular dihedral angle, rather than being allowed to adopt any of three randomly? In reality, different dihedral angles lead to different potential energies and therefore are not equally probable. This is described by the rotational potential V(f), as shown below for n-butane. g+ and g- have the same V(f), a has lower V(f) than g+ and g-.

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3.8 kJ/mole kJ/mole

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Question 4 continued

b.) At T = 150 oC, a typical crystallisation temperature of iPP, what are the probabilities pa, pg+, pg- that the dihedral angle is a, g+, or g- in n-butane? What would be the loss of entropy (in units of kB) when an n-butane molecule would be forced to adopt one particular dihedral angle, rather than being allowed to adopt any of three with their respective probabilities? Hint: The entropy for a system in contact with a thermal reservoir is 𝑆 = −𝑘K ∑ 𝑝NlnN 𝑝N where the pi are the probabilities of the states accessible to the system. The thickness d of crystalline lamellae in iPP depends on crystallisation temperature, TC, as shown in the table below:

TC [oC] d [nm]

150 17.2 140 13.3 130 11.1 120 9.3 110 8.1 100 7.1 90 6.3 80 5.8

(Data from Macromolecules 33, 14, 5204 (2000))

The equilibrium melting point of iPP, TM(¥), is 184 oC. c.) Polymer crystal lamella thickness depends on undercooling DT = TM(¥) - TC by an expression of the form 𝑑 = 𝐴 +𝐵 ∆𝑇: . Plot the data for iPP lamella thickness in a form appropriate to this expression and extract parameters A and B, with their units. d.) Briefly explain the relationship between the results from Question 4b, and 4c.

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5. The structure of the molecule abbreviated as ‘5CB’ is shown below.

5CB displays a transition from a crystalline to a nematic phase at 22.5 oC (cn transition), and a transition from the nematic to the isotropic liquid phase at 35 oC (ni transition). The latent heats associated with these transitions are DHcn = 56.1 J/g and DHni = 1.76 J/g. a.) Calculate the molar mass of 5CB, assuming atomic masses of 1 g/mol for hydrogen (H), 12 g/mol for carbon (C), and 14 g/mol for nitrogen (N).

b.) Calculate the change of entropy per 5CB molecule, in units of kB, at the cn and the ni transitions. In an ensemble of 10 rod-shaped molecules, 6 are aligned with their long axis exactly along the z-axis, 3 make an angle of 30o with the z-axis, and one makes an angle of 60o with the z-axis.

c.) Calculate the nematic order parameter, S, for this ensemble.

Hint: P2(cosq) = ½ (3 cos2q - 1) d.) If the birefringence of the same rod-shaped molecules perfectly aligned along the z-axis is Dn = 0.35, what birefringence does the ensemble described in 5 c.) display? What angle (known as ‘magic’ angle) would a rod have to make to the z-axis so it makes no contribution to birefringence, despite its innate anisotropy?

-Question 5 is continued on the next page-

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Question 5 continued A homologous series of amphiphile molecules is a set of molecules with the same hydrophilic head group, but hydrophobic ‘tails’ of different lengths, given by the number of saturated carbons, n, in the tail. The table below shows the critical micelle concentration (CMC, in millimoles/litre, mM) for the homologous series of sodium-n-alkyl-1-sulfates, all at 40 oC, and all forming spherical micelles.

n CMC [mM] 8 136

10 33.7 12 8.7 14 2.21 16 0.52

(Data from P Mukerjee, K J Mysels, ‘Critical Micelle Concentrations of Aqueous Surfactant Systems’, NSRDS – NBS 36, Washington 1971)

e.) Propose a relationship for CMC as a function of n, plot the data in the form suggested by this relationship to confirm it, and evaluate the parameters of the function CMC(n) for this homologous series.

Hint: Guess first how DE, the difference in energy between a single amphiphile molecule dissolved in water, and an amphiphile in a micelle with ‘optimum’ number of molecules, depends on n. Then use CMC ~ Fc = exp(-DE/kBT).

- END OF EXAMINATION PAPER -

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PHYSICAL CONSTANTS &MATHEMATICAL FORMULAE

Physical Constants

electron charge e = 1.60×10−19 Celectron mass me = 9.11×10−31 kg = 0.511MeV c−2proton mass mp = 1.673×10−27 kg = 938.3MeV c−2neutron mass mn = 1.675×10−27 kg = 939.6MeV c−2Planck’s constant h = 6.63×10−34 J sDirac’s constant (~ = h/2π) ~ = 1.05×10−34 J sBoltzmann’s constant kB = 1.38×10−23 J K−1 = 8.62×10−5 eVK−1speed of light in free space c = 299 792 458 ms−1 ≈ 3.00×108 ms−1permittivity of free space ε0 = 8.85×10−12 Fm−1permeability of free space µ0 = 4π×10−7 Hm−1Avogadro’s constant NA = 6.02×1023 mol−1gas constant R = 8.314 Jmol−1K−1ideal gas volume (STP) V0 = 22.4 l mol−1gravitational constant G = 6.67×10−11 Nm2 kg−2Rydberg constant R∞ = 1.10×107 m−1Rydberg energy of hydrogen RH = 13.6 eVBohr radius a0 = 0.529×10−10 mBohr magneton µB = 9.27×10−24 J T−1fine structure constant α ≈ 1/137Wien displacement law constant b = 2.898×10−3 mKStefan’s constant σ = 5.67×10−8 Wm−2K−4radiation density constant a = 7.55×10−16 Jm−3 K−4mass of the Sun M� = 1.99×1030 kgradius of the Sun R� = 6.96×108 mluminosity of the Sun L� = 3.85×1026 Wmass of the Earth M⊕ = 6.0×1024 kgradius of the Earth R⊕ = 6.4×106 m

Conversion Factors1 u (atomic mass unit) = 1.66×10−27 kg = 931.5MeV c−2 1 Å (angstrom) = 10−10 m1 astronomical unit = 1.50×1011 m 1 g (gravity) = 9.81 ms−21 eV = 1.60×10−19 J 1 parsec = 3.08×1016 m1 atmosphere = 1.01×105 Pa 1 year = 3.16×107 s

Polar Coordinates

x = r cos θ y = r sin θ dA = r dr dθ

∇2 =1

r

∂

∂r

(r∂

∂r

)+

1

r2∂2

∂θ2

Spherical Coordinates

x = r sin θ cosφ y = r sin θ sinφ z = r cos θ dV = r2 sin θ dr dθ dφ

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

Calculusf(x) f ′(x) f(x) f ′(x)

xn nxn−1 tanx sec2 x

ex ex sin−1(xa

)1√

a2−x2

lnx = loge x1x

cos−1(xa

)− 1√

a2−x2

sinx cosx tan−1(xa

)a

a2+x2

cosx − sinx sinh−1(xa

)1√

x2+a2

coshx sinhx cosh−1(xa

)1√

x2−a2

sinhx coshx tanh−1(xa

)a

a2−x2

cosecx −cosecx cotx uv u′v + uv′

secx secx tanx u/v u′v−uv′v2

Definite Integrals∫ ∞0

xne−ax dx =n!

an+1(n ≥ 0 and a > 0)

∫ +∞

−∞e−ax

2 dx =

√π

a∫ +∞

−∞x2e−ax

2 dx =1

2

√π

a3

Integration by Parts:∫ b

a

u(x)dv(x)dx

dx = u(x)v(x)∣∣∣ba−∫ b

a

du(x)dx

v(x) dx

Series Expansions

Taylor series: f(x) = f(a) +(x− a)

1!f ′(a) +

(x− a)2

2!f ′′(a) +

(x− a)3

3!f ′′′(a) + · · ·

Binomial expansion: (x+ y)n =n∑k=0

(n

k

)xn−kyk and

(n

k

)=

n!

(n− k)!k!

(1 + x)n = 1 + nx+n(n− 1)

2!x2 + · · · (|x| < 1)

ex = 1+x+x2

2!+x3

3!+ · · · , sinx = x− x

3

3!+x5

5!−· · · and cosx = 1− x

2

2!+x4

4!−· · ·

ln(1 + x) = loge(1 + x) = x− x2

2+x3

3− · · · (|x| < 1)

Geometric series:n∑k=0

rk =1− rn+1

1− r

Stirling’s formula: logeN ! = N logeN −N or lnN ! = N lnN −N

Trigonometry

sin(a± b) = sin a cos b± cos a sin b

cos(a± b) = cos a cos b∓ sin a sin b

tan(a± b) = tan a± tan b

1∓ tan a tan b

sin 2a = 2 sin a cos a

cos 2a = cos2 a− sin2 a = 2 cos2 a− 1 = 1− 2 sin2 a

sin a+ sin b = 2 sin 12(a+ b) cos 1

2(a− b)

sin a− sin b = 2 cos 12(a+ b) sin 1

2(a− b)

cos a+ cos b = 2 cos 12(a+ b) cos 1

2(a− b)

cos a− cos b = −2 sin 12(a+ b) sin 1

2(a− b)

eiθ = cos θ + i sin θ

cos θ =1

2

(eiθ + e−iθ

)and sin θ =

1

2i(eiθ − e−iθ

)cosh θ =

1

2

(eθ + e−θ

)and sinh θ =

1

2

(eθ − e−θ

)Spherical geometry:

sin a

sinA=

sin b

sinB=

sin c

sinCand cos a = cos b cos c+sin b sin c cosA

Vector Calculus

A ·B = AxBx + AyBy + AzBz = AjBj

A×B = (AyBz − AzBy) i+ (AzBx − AxBz) j+ (AxBy − AyBx) k = εijkAjBk

A×(B×C) = (A ·C)B− (A ·B)C

A · (B×C) = B · (C×A) = C · (A×B)

gradφ = ∇φ = ∂jφ =∂φ

∂xi+

∂φ

∂yj+

∂φ

∂zk

divA = ∇ ·A = ∂jAj =∂Ax∂x

+∂Ay∂y

+∂Az∂z

curlA = ∇×A = εijk∂jAk =

(∂Az∂y− ∂Ay

∂z

)i+

(∂Ax∂z− ∂Az

∂x

)j+

(∂Ay∂x− ∂Ax

∂y

)k

∇ · ∇φ = ∇2φ =∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2

∇×(∇φ) = 0 and ∇ · (∇×A) = 0

∇×(∇×A) = ∇(∇ ·A)−∇2A