1. van der Waals’ Interactions Refers to all interactions between polar or nonpolar molecules, varying as r -6. Includes Keesom, Debye and dispersive interactions. Values of interaction energy are usually only a few kT.

2. Summary Type of Interaction Interaction Energy, w(r) Charge-charge Coulombic Nonpolar-nonpolar Dispersive Charge-nonpolar Dipole-charge Dipole-dipole Keesom Dipole-nonpolar Debye In vacuum:  =1

3. Interaction between ions and polar molecules Interactions involving charged molecules (e.g. ions) tend to be stronger than polar-polar interactions. For freely-rotating dipoles with a moment of u interacting with molecules with a charge of Q we saw: One result of this interaction energy is the condensation of water (u = 1.85 D) caused by the presence of ions in the atmosphere. During a thunderstorm, ions are created that nucleate rain drops in thunderclouds (ionic nucleation).

4. Comparison of the Dependence of Interaction Potentials on r Not a comparison of the magnitudes of the energies! n = 1 n = 2 n = 3 n = 6 Coulombic van der Waals Dipole-dipole

5. Cohesive Energy Def’n.: Energy required to separate all molecules in the condensed phase or energy holding molecules in the condensed phase. In Lecture 1, we found that for single molecules with a potential w(r) = Cr -n, and with n>3: 1/2 to avoid double counting! For one mole, E substance = (1/2)N A E E substance = sum of heats of melting + vaporisation. Predictions agree well with experiment! with  = number of molecules per unit volume   -3, where  is the molecular diameter. So for dispersive interactions, n = 6 and C is the London constant: 

6. Boiling Point At the boiling point, T B, for a liquid, the thermal energy of a molecule, 3/2 kT B, will exactly equal the energy of attraction between molecules. Of course, the strongest attraction will be between the “nearest neighbours”, rather than pairs of molecules that are farther away. The interaction energy for van der Waals’ interactions is of the form, w(r) = -Cr -6. If molecules have a diameter of , then the shortest centre-to-centre distance will likewise be . Thus the boiling point is approximately:

7. Comparison of Theory and Experiment Evaluated at close contact where r = . Note that  o and C increase with . C can be found experimentally from deviations from the ideal gas law:

8. Additivity of Interactions Molecule Mol. Wt. u (D) T B (°C) Ethane: CH 3 CH 3 300-89 Formaldehyde: HCHO 30 2.3-21 Methanol: CH 3 OH 32 1.7 64 C=O H H C-O-H H H H C-C H H H H H H Dispersive only Keesom + dispersive H-bonding + Keesom + dispersive

9. Problem Set 1 1. Noble gases (e.g. Ar and Xe) condense to form crystals at very low temperatures. As the atoms do not undergo any chemical bonding, the crystals are held together by the London dispersion energy only. All noble gases form crystals having the face-centred cubic (FCC) structure and not the body-centred cubic (BCC) or simple cubic (SC) structures. Explain why the FCC structure is the most favourable in terms of energy, realising that the internal energy will be a minimum at the equilibrium spacing for a particular structure. Assume that the pairs have an interaction energy, u(r), described as where r is the centre-to-centre spacing between atoms. The so-called "lattice sums", A n, are given below for each of the three cubic lattices. SCBCCFCC A 6 8.4014.4512.25 A 12 6.2012.13 9.11 Then derive an expression for the maximum force required to move a pair of Ar atoms from their point of contact to an infinite separation. 2. (i) Starting with an expression for the Coulomb energy, derive an expression for the interaction energy between an ion of charge ze and a polar molecule at a distance of r from the ion. The dipole moment is tilted by an angle  with relation to r, as shown below. (ii) Evaluate your expression for a Mg 2+ ion (radius of 0.065 nm) dissolved in water (radius of 0.14 nm) when the water dipole is oriented normal to the ion and when the water and ion are at the point of contact. Express your answer in units of kT. Is it a significant value? (The dipole moment of water is 1.85 Debye.) r  ze

10. Molecular Crystals and Response of Condensed Matter to Mechanical Stress 3SCMP 2 February, 2006 Lecture 3 See Jones’ Soft Condensed Matter, Chapt. 2; Ashcroft and Mermin, Ch. 20

11. Lennard-Jones Potential The pair potential for isolated molecules affected by van der Waals’ interactions only can be described by a Lennard-Jones potential: w(r) = +B / r 12 - C / r 6 The -ve r -6 term is the attractive v.d.W. contribution The +ve r -12 term describes the hard-core repulsion stemming from the Pauli-exclusion. 12 is a mathematically-convenient exponent with no physical significance! The two terms are additive.

12. L-J Potential for Ar (boiling point = 87 K) London Constant, C = 4.5 x 10 -78 Jm 6 ; Guessing B = 10 -134 Jm 12 (m) w min  -5 x 10 -22 J Compare to: (3/2)kT B = 2 x 10 -21 J Actual  ~ 0.3 nm (Guess for B is too large!)

13. Intermolecular Force for Ar (boiling point = 87 K) F= dw / dr (m)

14. Intermolecular Force for Ar (boiling point = 87 K) F= dw / dr (m)

15. Weak Nano-scale Forces Can be Measured The AFM probe is exceedingly sharp so that only a few atoms are at its tip! Sensitive to forces on the order of nano-Newtons.

16. Measuring Attractive Forces at the Nano-Scale A = approach B = “jump” to contact C = contact D = adhesion E = pull-off Tip deflection  Force Vertical position A B C DE C

17. Latex Particle Packing T g = 20 °C

18. L-J Potential in Molecular Crystals Noble gases, such as Ar and Xe, form crystalline solids (called molecular crystals) that are held together solely by the dispersive energy. In molecular crystals, the pair potential for neighbouring atoms (or molecules) is written as The molecular diameter in the gas state is . Note that when r = , then w = 0.  is a bond energy, such that w(r) = -  when r is at the equilibrium spacing of r = r o. w(r) = 4  [( ) 12 -( ) 6 ]

19. L-J Potential in Molecular Crystals The minimum of the potential is found from the first derivative of the potential. Also corresponds to the point where F = 0. [] - We can solve this expression for r to find the equilibrium spacing, r o : ) [ [([( ) - ( ]] -- To find the minimum energy in the potential, we can evaluate it when r = r o :

20. Lennard-Jones Potential r w(r) + - -- roro 

21. Potential Energy of an Atom in a Molecular Crystal For each atom/molecule in a molecular crystal, we need to sum up the interaction energies between all pairs (assuming additivity of the potential energies). The total cohesive energy per atom is W = 1 / 2  r  w(r) since each atom in a pair “owns” only 1 / 2 of the interaction energy. The particular crystal structure (FCC, BCC, etc.) defines the distances between nearest neighbours, 2nd neighbours, 3rd neighbours, etc., and it defines the number of neighbours at each distance. This geometric information that is determined by the crystal structure can be described by constants known as the lattice sums: A 12 and A 6. For FCC crystals, A 12 = 12.13 and A 6 = 14.45.

22. Cohesive Energy of Atoms in a Molecular Crystal w(r) = 4  [( ) 12 -( ) 6 ] So, for a pair we write the interaction potential as: We notice that the molecules are slightly closer together in a crystal compared to when they are in an isolated pair (r o =1.12  ). From the first derivative, we can find the equilibrium spacing for an FCC crystal: () For each atom in a molecular crystal, however, we write that the cohesive energy is: W = 2  [ A 12 ( ) 12 - A 6 ( ) 6 ]

23. Cohesive Energy of Atoms in a Molecular Crystal We can evaluate W when r = r o to find for an FCC crystal: - - This expression represents the energy holding an atom/molecule within the molecular crystal. Its value is only 8.6 times the interaction energy for an isolated pair. This result demonstrates that the dispersive energy is operative over fairly short distances, so that most of the interaction energy is contributed by the nearest neighbours. In an FCC crystal, each atom has 12 nearest neighbours! W

24. Elastic Modulus of Molecular Crystals We can model the intermolecular force using a spring with a spring constant, k. The force, F, to separate two atoms in the crystal is: F = k(r - r o ). The tensile stress  t is defined as a force acting per unit area, so that:  - F roro The tensile strain  t is given as the change in length as a result of the stress: - oo oo

25. F A L The Young’s modulus, Y, relates tensile stress and strain: Connection between the atomic and the macroscopic Y tt tt Y can thus be expressed in terms of atomic interactions: - - What is k?

26. r W + - -8.6  roro  F = 0 when r = r o Elastic Modulus of Molecular Crystals r F + - roro

27. Force to separate atoms is the derivative of the potential : [] - So, taking the derivative again: [] - But we already know that: () So we see that: () We will therefore make a substitution for  when finding k.

28. Elastic Modulus of Molecular Crystals To find k, we now need to evaluate dF/dr when r = r o. [ ] - Combining the constants to create new constants, C 1 and C 2, and setting r = r o, we can write: [] - - Finally, we find the Young’s modulus to be: - As r o 3 can be considered an atomic volume, we see that the modulus can be considered an energy density, directly related to the pair interaction energy.

29. Bulk Modulus of Molecular Crystals We recall the thermodynamic identity: dU = TdS - PdV The definition of the bulk modulus, B, is: - This identity tells us that: - If we neglect the kinetic energy in a crystal, then U  W. So B can be written as: -- After writing V in terms of , and differentiating W, we obtain for an FCC molecular crystal:

30. Theory of Molecular Crystals Compares Well with Experiments w(r o )

31. Response of Condensed Matter to Shear Stress When exposed to a shear stress, the response of condensed matter can fall between two extremes: Hookean (solid-like) or Newtonian (liquid-like) How does soft matter respond to shear stress? A A y F

32. Elastic Response of Hookean Solids No time-dependence in the response to stress. Strain is instantaneous and constant over time. The shear strain  s is given by the angle  (in units of radians). The shear strain  s is linearly related to the shear stress by the shear modulus, G: A A y F  xx

33. Viscous Response of Newtonian Liquids A A y F xx There is a velocity gradient (v/y) normal to the area. The viscosity  relates the shear stress,  s, to the velocity gradient. The viscosity can thus be seen to relate the shear stress to the shear rate: The top plane moves at a constant velocity, v, in response to a shear stress: v  has S.I. units of Pa s. The shear strain increases by a constant amount over a time interval, allowing us to define a strain rate: Units of s -1

34. Hookean Solids vs. Newtonian Liquids Hookean Solids: Newtonian Liquids: Many substances, i.e. “structured liquids”, display both type of behaviour, depending on the time scale. Examples include colloidal dispersions and melted polymers. This type of response is called “viscoelastic”.

35. Response of Soft Matter to a Constant Shear Stress: Viscoelasticity When a constant stress is applied, the molecules initially bear the stress. Over time, they can re-arrange and flow to relieve the stress: The shear strain, and hence the shear modulus, both change over time:  s (t) and t Elastic response Viscous response (strain is constant over time) (strain increases over time)

36. Response of Soft Matter to a Constant Shear Stress: Viscoelasticity t Slope: We see that 1/G o  ( 1 /  )    is the “relaxation time” Hence, viscosity can be approximated as   G o 

37. Example of Viscoelasticity

38. Physical Meaning of the Relaxation Time time  Constant strain applied  Stress relaxes over time as molecules re-arrange time Stress relaxation:

39. Typical Relaxation Times For solids,  is exceedingly large:   10 12 s For simple liquids,  is very small:   10 -12 s For soft matter,  takes intermediate values. For instance, for melted polymers,   1 s.

40. Viscosity Sometimes Depends on the Shear Rate Newtonian: ss   Shear thinning or thickening:  ss 

41. An Example of Shear Thickening Future lectures will explain how polymers and colloids respond to shear stress.

42. Problem Set 2 1. Calculate the energy required to separate two atoms from their equilibrium spacing r o to a very large distance apart. Then calculate the maximum force required to separate the atoms. The pair potential is given as w(r) = - A/r 6 + B/r 12, where A = 10 -77 Jm 6 and B = 10 -134 Jm 12. Finally, provide a rough estimate of the modulus of a solid composed of these atoms. 2. The latent heat of vaporisation of water is given as 40.7 kJ mole -1. The temperature dependence of the viscosity of water  is given in the table below. (i) Does the viscosity follow the expectations of an Arrhenius relationship with a reasonable activation energy? (ii) The shear modulus G of ice at 0  C is 2.5 x 10 9 Pa. Assume that this modulus is comparable to the instantaneous shear modulus of water G o and estimate the characteristic frequency of vibration for water,. Temp (  C)01020304050  (10 -4 Pa s)17.9313.0710.027.986.535.47 Temp (  C)60708090100  (10 -4 Pa s) 4.674.043.543.152.82 3. In poly(styrene) the relaxation time for configurational rearrangements  follows a Vogel-Fulcher law given as  =  o exp(B/T-T o ), where B = 710  C and T o = 50  C. In an experiment with an effective timescale of  exp = 1000 s, the glass transition temperature T g of poly(styrene) is found to be 101.4  C. If you carry out a second experiment with  exp = 10 5 s, what value of T g would be obtained?