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PH3-SM (PHY3032) Soft Matter Physics Lecture 3 Potential Energy in Condensed Matter and its Response to Mechanical Stress 18 October, 2011 See Jones’ Soft Condensed Matter, Chapt. 2; Ashcroft and Mermin, Ch. 20 In the previous lecture: Interaction Potentials: w = -Cr -n • If n <3, molecules interact with all others in the system of size, L. If n >3, molecules interact only with the “nearer” neighbours. • Gravity: negligible at the molecular level. W(r) = -Cr -1 • Coulombic: relevant for salts, ionic liquids and charged molecules. W(r) = -Cr -1 • van der Waals’ Interactions: three types; usually quite weak; cause attraction between ANY two molecules. W(r) = -Cr -6 • Covalent bonds: usually the strongest type of bond; directional forces - not described by a simple potential. • Hydrogen bonding: stronger than van der Waals bonds; charge attraction resulting from unshielded proton of H. Last Lecture: • Discussed polar molecules and dipole moments (Debye units) and described charge-dipole and dipole-dipole interactions. +Q - u - + + u u - + • Discussed polarisability of molecules (electronic and orientational) and described charge-nonpolar, polarnonpolar, and dispersive (London) interactions. +Q a a - u + • Summarised ways to measure polarisability. a a Summary Type of Interaction Charge-charge In vacuum: =1 Interaction Energy, w(r) Q 1Q 2 Coulombic 4 o r _ Qu cos Dipole-charge 4 o r 2 _Q u 2 2 6 ( 4 o ) kTr 2 4 _ u 1 u 2 f ( 1 , 2 , ) 2 2 4 o r Dipole-dipole 2 _ u1 u 2 2 3( 4 o ) kTr 2 Charge-nonpolar Keesom 2 2 ( 4 o ) r 2 4 _ u a (1 + 3 cos ) 2 2 ( 4 o ) r 2 _u a 2 ( 4 o ) r 2 Nonpolar-nonpolar 6 _Q a 2 Dipole-nonpolar 3 Dispersive 6 6 Debye w (r ) = _ 3 a o 2h 4 ( 4 o ) 2 r 6 Cohesive Energy • Def’n.: Energy, E, required to separate all molecules in the condensed phase or energy holding molecules in the condensed phase. • We found previously (Lecture 2, slide 17) for a single molecule, and with n>3: 4 C r E ( n 3 )s n3 We can write r = number of molecules per unit volume s -3, where s is the molecular diameter. So, for van der Waals’ interactions with n = 6: E 4 C r 3s 3 4 C 3s 6 1/2 to avoid double counting! • For one mole, Esubstance = (1/2)NAE • Esubstance = sum of heats of melting + vaporisation. • Predictions agree well with experiment! Boiling Point • At the boiling point, TB, for a liquid, the thermal energy of a monoatomic molecule (3 degrees of freedom), 3/2 kTB, 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 s, then the shortest centre-to-centre distance will likewise be s. • Thus the boiling point is approximately: TB w (s ) 3 k 2 2C 3ks 6 Comparison of Theory and Experiment London equation Non-polar Note that ao and C increase with s. (P a V 2 )(V b ) RT (Per mole, n = 1) E mole ~ N A 4 C 2 3s 6 TB = w (r ) 3 k 2 Evaluated at close contact where r = s. Additivity of Interactions Molecule Mol. Wt. H H C-C H H Ethane: C2H6 u (D) H TB(°C) Dispersive only H 30 0 -89 H C=O H Formaldehyde: CH2O Keesom + dispersive 30 2.3 -21 H H C-O-H H Methanol: CH3OH H-bonding + Keesom + dispersive 32 1.7 64 Lennard-Jones Potential • To describe the total interaction energy (and hence the force) between two molecules at a distance r, a pair potential is used. • The pair potential for isolated molecules that are affected only by van der Waals’ interactions can be described by a Lennard-Jones potential: w(r) = +B/r12 - C/r6 • 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. L-J Potential for Ar London Constant for Ar is calculated to be C = 4.5 x 10-78 Jm6 We can guess that B = 10-134 Jm12 Actual s 0.3 nm (Our guess for B is too large!) wmin -5 x 10-22 J Compare to: (3/2)kTB= 2 x 10-21 J (m) Ar boiling point: TB = 87 K (x 10-9 ) Intermolecular Force for Ar Pair Attractive force Very short-range force! Equ’m F = dw/dr s Repulsive force (m) r Comparison of Force and Potential Energy for Ar Pair F= dw/dr (m) (m) Weak Nano-scale Forces Can be Measured with an Atomic Force Microscope The AFM probe is exceedingly sharp so that – in principle - only a few atoms are at its tip! Sensitive to forces on the order of nano-Newtons. Tips for Scanning Probe Microscopy F The tip is on a cantilever, which typically has a spring constant on the order of k = 10 N/m. Radius of curvature ~ 10 nm Ideally, one of the atoms at the tip is slightly above the others. Modelled as a simple spring: F = kz where z is the deflection in the vertical direction. AFM tips from NT-MDT. See www.ntmdt.ru Tip/Sample Interactions: Function of Distance h Physical contact between tip and surface Measuring Attractive Forces at the Nano-Scale Tip deflection B = “jump” to contact C 0 A = approach Force C A B C = contact E D Vertical position D = adhesion E = pull-off Measuring Force of Attraction to a Polymer Surface 200 Pushing on AFM probe tip Force /nN 100 trace retrace 0 -100 -200 Pulling on the AFM probe tip -300 -400 0 2 4 Distance/ m 6 8 Imaging with the AFM Tip The AFM tip is held at a constant distance from the surface - or a constant force is applied - as it scanned back and forth. Surface Force Apparatus Mica has an atomistically smooth surface. A piezoelectric moves the arm up by a known amount. Force on the mica is determined by measuring the distance between the mica and knowing the arm’s stiffness. www.fisica.unam.mx/liquids/tutorials/surface.html Distance between mica sheets is measured with interferometry. 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. We will analyse them as a very simple form of “soft matter.” In molecular crystals, the pair potential for neighbouring atoms (or molecules) is written as s w 4 r 12 s r 6 The molecular diameter in the gas state is s. Note that when r = s, then w = 0. is a bond energy (related to the London constant), such that w(r) = - when r is at the equilibrium spacing of r = ro. Lennard-Jones Potential for Molecular Pairs in a Crystal + w(r) - - s ro r 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. 12 s F 0 4 13 dr r dw 12 6s 7 r 6 We can solve this expression for r to find the pair’s equilibrium spacing, ro: 1 ro = 2 6s = 1 . 12 s To find the minimum energy in the potential, we can evaluate it when r = ro: s 1/ 6 W ( 2 s ) 4 1 2 6 s 12 s 1 2 6s 6 1 1 4 4 2 Variety of Atomic Spacings in Cubic Crystals 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. 6 nearest neighbours; 12 second nearest 8 nearest neighbours; 6 2nd nearest; 12 3rd nearest 12 nearest neighbours; 6 second nearest; 24 3rd nearest Image from: http://www.uccs.edu/~tchriste/courses/PHYS549/549lectures/figures/cubes.gif 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 Wtot = 1/2 S w(r) since r each atom in a pair “owns” only 1/2 of the interaction energy. • As shown already, the particular crystal structure (FCC, BCC, etc.) defines the distances of 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: A12 and A6 (where the 12 and 6 represent the two terms of the L-J potential.) • For FCC crystals, A12 = 12.13 and A6= 14.45. There are different values for BCC, SC, etc. Cohesive Energy of Atoms in a Molecular Crystal So, for a pair we write the interaction potential as: s w 4 r 12 s 6 r For each atom in a molecular crystal, however, we write that the cohesive energy is: W tot 12 6 s s 2 A12 A6 r r From the first derivative, we can find the equilibrium spacing 1 for an FCC crystal: 6 2 A12 ro s 1 . 09 s A6 We notice that the molecules are slightly closer together in a crystal compared to when they are in an isolated pair (ro=1.12 s). Cohesive Energy of Atoms in a Molecular Crystal We can evaluate Wtot(r) when r = ro to find for an FCC crystal: W TOT FCC - ε A6 2 - 8 .6 ε 2 A12 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, but the equivalent of 8.6 pairs of atoms (17.2) contribute to the energy!) Connection between the macroscopic and the atomic? st A F L The Young’s modulus, Y, relates tensile stress and strain: s Y = t t How does the interatomic force, F, relate to the macroscopic Y? Y t Model for Elastic Modulus of Molecular Crystals We can model the interatomic force using a spring with a spring constant, k. The force, F, to separate two atoms in the crystal is: F = k(r - ro). At equilibrium, r = ro and F = 0. ro ao ao F ro The tensile stress st is defined as a force acting per unit area, so that: st F A k ( r - ro ) ro 2 The tensile strain t is given as the change in length as a result of the r ro r stress: t ro ro Y can thus be expressed in terms of atomic interactions: k ( r - ro ) Y st t ro ( r - ro ) ro 2 k ro What is k? Elastic Modulus of Molecular Crystals + Wtot The Young’s modulus is sometimes known simply as the elastic modulus. F dW tot s ro -8.6 r + F = 0 when r = ro dr F k dF dr r rO - ro r Linear region around ro is approximated as: F = k(r - ro). Elastic Modulus of Molecular Crystals Force to separate atoms is the derivative of the potential: - 12 A12 s 12 6 A6s 6 F 2 13 7 dr r r dW So, taking the derivative again to find k: dF dr 2 13( 12 ) A12 s But we already know that: Re-arranging, we see that: r 7 ( 6 ) A 6s 12 14 2 A12 ro A6 r A6 s ro 2 A 12 8 6 1 6 s 1 6 We will therefore make a substitution for s when finding k. Elastic Modulus of Molecular Crystals 13( 12 ) A12 s 12 7 ( 6 ) A6s 6 dF 2 14 8 dr r r with A6 s ro 2 A 12 1 6 To find k, we now need to evaluate dF/dr when r = ro. Combining the constants to create new constants, C1 and C2, and setting r = ro, we can write: k 2 C 1 ro ro 14 12 6 C 2 ro C1 - C 2 2 ( ) 8 2 ro ro Finally, we find the Young’s modulus to be: Y k ro 2 (C1 - C 2 ) ro 3 As ro3 can be considered an atomic volume, we see that the modulus can be considered an energy density, directly related to the pair interaction energy. Response of Condensed Matter to Shear Stress A F y A ss = F A How does soft matter respond 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) Elastic Response of Hookean Solids The shear strain gs is given by the angle (in units of radians). ss = A x y A F F A gs = ~ x y The shear strain gs is linearly related to the shear stress by the shear ss modulus, G: gs = G No time-dependence in the response to stress. Strain is instantaneous and constant over time. Viscous Response of Newtonian Liquids The top plane moves at a constant velocity, v, in response to a shear stress: F ss = A x A F v y v = A x t There is a velocity gradient (v/y) normal to the area. The viscosity h relates the shear stress, ss, to the velocity gradient. s s= h v y =h x t y h has S.I. units of Pa s. The shear strain increases by a constant amount over a time interval, allowing g us to define a strain rate: -1 g t Units of s The viscosity can thus be seen to relate the shear stress to the shear rate: x x g ss h h h h g t y y t t Hookean Solids vs. Newtonian Liquids Hookean Solids: Newtonian Liquids: s s Gg ss h dg h g dt Many substances, i.e. “structured liquids”, display both types of behaviour, depending on the time scale. At short time scales, the response is solid-like. At longer time scales, the response is liquid like. This type of response is called “viscoelastic”. Examples of viscoelastic systems include colloidal dispersions and melted polymers. The simple Maxwell model assumes that the elastic and viscous responses act in series and are additive. Example of Viscoelasticity Video: Viscoelastic gel Maxwell Model of Viscoelasticity Spring: Elastic element Dashpot: ss Viscous element The model says that the elastic and viscous elements act in series, such that the shear stress, ss, is the same for both of them. The elastic shear strain, ge, and the viscous shear strain, gv, are additive: γ γe γv As ss is the same for both elements, we have: dg v s s GM g e h dt For a constant applied shear stress, the total strain can be ss ss written as: g (t ) t GM h We define a Maxwell relaxation time: t = h/GM Stress Relaxation after a Step Strain g Constant shear strain applied time ss Hookean viscoelastic solid s s g G eq viscoelastic liquid Newtonian t time If viscoelastic, then the stress relaxes over time as molecules re-arrange. Stress Relaxation after a Step Strain For a viscoelastic liquid, the shear stress relaxation in the Maxwell model is described as: s s (t ) G M ge -t t t is the relaxation time For typical solids, t is exceedingly large: t 1012 s, such that there is no observed relaxation – just a Hookean response. For melted polymers, however, t 1 s. From here, we can define a time-dependent stress relaxation t modulus: s s (t ) t G (t ) g GM e For a viscoelastic solid, the equilibrium shear modulus (non-zero) is defined as: G eq lim( t ) G ( t ) Response of Soft Matter to a Constant Shear Stress Apply a constant shear stress, ss, and observe the time dependence of the shear strain, g(t) In this experiment, we find the creep compliance: J(t) = 1/G(t)= g(t)/ss. s s Recalling that: g ( t ) s s t GM h 1 t We see in the Maxwell model that: J (t ) GM η Response of Soft Matter to a Constant Shear Stress J (t ) g s (t ) ss (Pa-1) Slope: d Jeq ( g dt s s ) g ss 1 h Viscous response (strain increases linearly over time) Elastic response (provides initial and recoverable strain) t Steady-state compliance = Jeq= 1/Geq In the Maxwell model, Geq = h/t so, t = hJeq and Jeq = t/h Viscoelastic Response after Stress is Removed J (t ) g s (t ) ss Jeq (Pa-1) Elastic strain is recovered Jeq Constant stress, ss, is applied Stress removed Viscous flow is permanent t Viscosity of Soft Matter Often Depends on the Shear Rate h ss Newtonian: h (simple liquids like water) g s h ss Shear thinning or thickening: g s h g s g s An Example of Shear Thickening Future lectures will explain how polymers and colloids respond to shear stress. Video: shear thickening 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 s u ( r ) 2 A12 r 12 s A6 , r 6 where r is the centre-to-centre spacing between atoms. The so-called "lattice sums", An, are given below for each of the three cubic lattices. SC BCC FCC A6 8.40 12.25 14.45 A12 6.20 9.11 12.13 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. ze r (ii) Evaluate your expression for a Mg2+ 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.) 3. Show that 1 kJ mole-1 = 0.4 kT per molecule at 300 K. Problem Set 2 1. Calculate the energy required to separate two atoms from their equilibrium spacing ro 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/r6 + B/r12, where A = 10-77 Jm6 and B = 10-134 Jm12. 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 h 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 109 Pa. Assume that this modulus is comparable to the instantaneous shear modulus of water Go and estimate the characteristic frequency of vibration for water, . Temp (C) h (10-4 Pa s) 0 17.93 10 13.07 20 10.02 30 7.98 40 6.53 Temp (C) h (10-4 Pa s) 60 4.67 70 4.04 80 3.54 90 3.15 100 2.82 50 5.47 3. In poly(styrene) the relaxation time for configurational rearrangements t follows a Vogel-Fulcher law given as t = to exp(B/T-To), where B = 710 C and To = 50 C. In an experiment with an effective timescale of texp = 1000 s, the glass transition temperature Tg of poly(styrene) is found to be 101.4 C. If you carry out a second experiment with texp = 105 s, what value of Tg would be obtained? Typical Relaxation Times For solids, t is exceedingly large: t 1012 s For simple liquids, t is very small: t 10 -12 s For soft matter,t takes intermediate values. For instance, for melted polymers, t 1 s. Slime movie Relaxation Time in a Creep Experiment •In a “creep” experiment, a constant stress is applied for a fixed period of time and then released. •The strain is recorded while the stress is applied and after it is released. •The instantaneous (elastic) response is modeled as a spring giving an elastic modulus, Y. •The time-dependent (viscous flow) response is modeled as a “dashpot” giving a viscosity, h. Stress, Strain, s Y h 0 Time 0 Time Response of Soft Matter to a Constant Shear Stress: Viscoelasticity 1 G (t ) = g (t ) ss Slope: 1 d Go ( gs dt s s )= g s ss t is the “relaxation time” We see that 1/Go (1/h)t t Hence, viscosity can be approximated as h Got t = 1 h Modelling of Creep and Relaxation Maxwell model: Spring and dashpot in series s Instantaneous elastic deformation is followed by flow. Elastic deformation is recovered but not the flow. s Y s h t Voigt-Kelvin model: Spring and dashpot in parallel s Anelastic deformation is followed by recovery with an exponential dependence on t = h/Y, which is called the relaxation time. Deformation: s Yt s t 1 exp 1 exp Y h Y t Total Creep and Recovery Constant stress applied Displacement Anelastic Elastic Flow Spring and dashpot acting both in series Time and in parallel. Strain is described by the sum of the Maxwell and Voigt-Kelvin models. Creation of a New Surface Leads to a “Thermodynamic” Adhesion Force F Surface area increases when tip is removed. R G is the surface tension (energy) of the tip and the surface assumed here to be equal. Work of adhesion: W = G dA ∫ W L GLSA L G dA W = GLVA + GSVA - GLSA GLVA Work per unit area, W: S W GLV+ GSV - GSL GSVA GLV S When liquid (L) and solid (S) are separated, two new interfaces with the vapour (V) are created. Work per unit area, W: W GLV (1 cos ) Young-Dupré Equation GSL GSV GSL GLV cos GSV GSL GLV cos W GLV GLV cos GSV The Capillary Force Pressure is required to bend a surface with a surface tension, G P = G[ F 1 1 r1 r2 ]≈ G r1 F = PA ≈ G r1 Max. Capillary Force: 2 Rd F 4RG cos With G = 0.072 N/m for water and R = 10 nm, F is on the order of 10-8 -10-9 N! Force for Polymer Deformation 20 Indent Pull 10 Force/ nN 0 -10 -20 -30 -40 -50 0 1 2 3 4 5 6 Distance/ m 7 8 9 Hookean Solids vs. Newtonian Liquids Hookean Solids: Newtonian Liquids: σ Gγ σ η dγ η γ dt Many substances, i.e. “structured liquids”, display both types of behaviour, depending on the time scale: solid-like on short timescales and liquid-like on longer time-scales. Examples include colloidal dispersions and melted polymers. This type of response is called “viscoelastic”. The simplest model of viscoelasticity is the Maxwell model, which assumes that the viscous and elastic elements act in a series. Under a constant shear stress, the shear strains are additive.