CHEM2524Chemical Physics

    Formula sheet

    Every equation from the lectures, grouped by module. Each one links back to the page that explains it.

    443 of 443 formulas

    01 Symmetry

    Models: simple pictures that still predict things

    1. Gauss's law (electric)

      ∇⋅E=ρ/ε0\nabla \cdot \mathbf{E} = \rho/\varepsilon_0

      E\mathbf{E} electric field, ρ\rho charge density, ε0\varepsilon_0 permittivity of free space

      Charge makes an electric field.

    2. Gauss's law (magnetic)

      ∇⋅B=0\nabla \cdot \mathbf{B} = 0

      B\mathbf{B} magnetic field

      There are no single magnetic charges.

    3. Faraday's law

      ∇×E=−∂B/∂t\nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t

      tt time

      A changing magnetic field makes an electric field.

    4. Ampère–Maxwell law

      ∇×B=μ0j+(1/c2) ∂E/∂t\nabla \times \mathbf{B} = \mu_0 \mathbf{j} + (1/c^2)\,\partial \mathbf{E}/\partial t

      μ0\mu_0 permeability of free space, j\mathbf{j} current density, cc speed of light

      Current (and a changing electric field) makes a magnetic field.

    5. Mass scaling

      m∝L3m \propto L^3

      mm mass, LL length scale

    6. Bone cross-section requirement

      A∝L3A \propto L^3

      AA cross-section area needed

      Bone area has to grow as fast as mass to carry the load.

    7. Uniform-scaling area

      A∝L2A \propto L^2

      AA area, LL length

      What the area actually does if you stretch every dimension equally.

    8. Molecular volume

      v∝L3v \propto L^3

      vv volume per molecule

    9. Area–volume scaling

      A∝v2/3A \propto v^{2/3}

      AA area, vv volume

    10. Surface tension scaling

      ΔHvap∝σ v2/3\Delta H_{vap} \propto \sigma\, v^{2/3}

      ΔHvap\Delta H_{vap} enthalpy of vaporisation, σ\sigma surface tension, vv volume per molecule

      Plot ΔHvap\Delta H_{vap} against σv2/3\sigma v^{2/3} and you get a straight line.

    11. de Broglie wavelength

      λ=h/p\lambda = h/p

      λ\lambda de Broglie wavelength, hh Planck's constant, pp momentum

      Compare it with the size of the system to choose classical or quantum.

    12. Planck's constant

      h=6.626×10−34 J sh = 6.626 \times 10^{-34}\ \text{J s}

      hh Planck's constant

    13. Momentum

      p=mvp = m v

      pp momentum, mm mass, vv velocity

    Why atoms hold together and how particles attract

    1. de Broglie relation

      p=h/λp = h/\lambda

      pp momentum, hh Planck's constant, λ\lambda wavelength

      Ties how much the electron wave is squeezed to how much energy it carries.

    2. Momentum

      p=mup = mu

      mm mass, uu speed

    3. Electron kinetic energy

      KE=12mu2=h2/(2mλ2)\text{KE} = \tfrac{1}{2} m u^2 = h^2/(2m\lambda^2)

      KE\text{KE} kinetic energy

      Squeeze the wave (small λ\lambda) and KE grows fast.

    4. Electron potential energy

      PE=−Ze2/(4πϵ0r)≈−Ze2/(4πϵ0λ)\text{PE} = -Ze^2/(4\pi\epsilon_0 r) \approx -Ze^2/(4\pi\epsilon_0\lambda)

      ZZ nuclear charge number, ee elementary charge, ϵ0\epsilon_0 permittivity, rr separation

      The slide takes r≈λr \approx \lambda.

    5. Total energy vs wavelength

      E=KE+PE∝1/λ2−1/λE = \text{KE} + \text{PE} \propto 1/\lambda^2 - 1/\lambda

      EE total energy, λ\lambda matter wavelength

    6. Energy minimum

      Emin at ⟨r⟩=1.5E_{min}\ \text{at}\ \langle r\rangle = 1.5

      ⟨r⟩\langle r\rangle mean radius in units of a0a_0

    7. Generic multipole potential

      V∝1/rn+m−1V \propto 1/r^{n+m-1}

      VV potential energy, rr separation, nn and mm equal 1 for a monopole, 2 for a dipole, 4 for a quadrupole

      Gives the falloff table: ion-ion 1/r1/r, ion-dipole 1/r21/r^2, dipole-dipole 1/r31/r^3, quadrupole-quadrupole 1/r71/r^7.

    8. Coulomb potential

      V(r)=−q1q2/(4πϵ0r)V(r) = -q_1 q_2/(4\pi\epsilon_0 r)

      q1,q2q_1, q_2 charges, rr separation

    9. Dipole–dipole potential

      V(r)=−μ1μ2/(2πϵ0r3)V(r) = -\mu_1\mu_2/(2\pi\epsilon_0 r^3)

      μ1,μ2\mu_1, \mu_2 dipole moments

    10. Induced dipole–induced dipole

      V(r)=−C6/r6≈−32 α1α2/r6⋅I1I2/(I1+I2)V(r) = -C_6/r^6 \approx -\tfrac{3}{2}\,\alpha_1\alpha_2/r^6 \cdot I_1I_2/(I_1+I_2)

      C6C_6 dispersion coefficient, α\alpha polarisabilities, II ionisation energies

    Symmetry operations and rotation axes

    1. Rotation operator

      Cn: rotation by 2π/n (=360∘/n)C_n:\ \text{rotation by}\ 2\pi/n\ (=360^\circ/n)

      nn order of the axis

      A C2C_2 axis means a half turn, a C3C_3 axis a third of a turn.

    2. Power identities

      Cn n≡E,C4 2≡C2C_n^{\,n} \equiv E, \qquad C_4^{\,2} \equiv C_2

      EE identity operation

      Doing the full set of turns brings you back to where you started. Two quarter turns make a half turn.

    Mirror planes, inversion and improper rotation

    1. Improper rotation

      Sn≡σh×Cn≡Cn×σhS_n \equiv \sigma_h \times C_n \equiv C_n \times \sigma_h

      SnS_n improper rotation axis of order nn, σh\sigma_h horizontal mirror plane, CnC_n rotation

      Turn, then flip in the plane across the axis. The order does not matter.

    2. Low-order improper rotations

      S1≡σ,S2≡iS_1 \equiv \sigma, \qquad S_2 \equiv i

      σ\sigma mirror plane, ii inversion centre

      So every operation is either a proper or an improper rotation.

    Point groups: sorting molecules by their symmetry

    1. Closure

      A,B∈G⇒A⋄B∈GA, B \in G \Rightarrow A \diamond B \in G

      GG point group, ⋄\diamond combining two symmetry operations

      Doing one operation then another is itself an operation in the group. Beyond the scope of the course, per the slide.

    2. Identity

      A⋄E=E⋄A=AA \diamond E = E \diamond A = A

      EE the identity operation

    3. Associativity

      (A⋄B)⋄C=A⋄(B⋄C)(A \diamond B)\diamond C = A \diamond (B \diamond C)

      A,B,CA, B, C symmetry operations in the group

    4. Inverse

      A⋄A−1=A−1⋄A=EA \diamond A^{-1} = A^{-1} \diamond A = E

      A−1A^{-1} the operation that undoes AA

    5. Commutativity (abelian group)

      A⋄B=B⋄AA \diamond B = B \diamond A

      Holds only if the group is abelian

    02 Crystals and diffraction

    Unit cells, crystal systems and Bravais lattices

    1. Position in a unit cell

      r=x a+y b+z c,0≤x,y,z≤1\mathbf{r} = x\,\mathbf{a} + y\,\mathbf{b} + z\,\mathbf{c}, \qquad 0 \le x,y,z \le 1

      r\mathbf{r} position vector, a,b,c\mathbf{a},\mathbf{b},\mathbf{c} unit cell vectors, x,y,zx,y,z fractional coordinates

      Say where an atom is, as a fraction of each cell edge.

    2. Position anywhere in the crystal

      r=(u+x) a+(v+y) b+(w+z) c\mathbf{r} = (u+x)\,\mathbf{a} + (v+y)\,\mathbf{b} + (w+z)\,\mathbf{c}

      u,v,wu,v,w integers (which cell), x,y,zx,y,z fractional coordinates (where in the cell)

    3. Translation of an atom

      (x,y,z)⇒(1+x, y, z),(x, 3+y, z)(x,y,z) \Rightarrow (1+x,\,y,\,z),\quad (x,\,3+y,\,z)

      x,y,zx,y,z fractional coordinates

      An atom here means a copy one cell over, or three cells over.

    4. Cell lengths and angles

      a=∣a∣,b=∣b∣,c=∣c∣;α,β,γ angles between the cell vectorsa = |\mathbf{a}|,\quad b = |\mathbf{b}|,\quad c = |\mathbf{c}|;\quad \alpha,\beta,\gamma\ \text{angles between the cell vectors}

      a,b,ca, b, c lengths of the cell vectors, α,β,γ\alpha, \beta, \gamma angles between the cell vectors

    Crystal symmetry: glides, screws and space groups

    1. Space group count

      32 point groups+14 Bravais lattices⇒230 space groups\text{32 point groups} + \text{14 Bravais lattices} \Rightarrow \text{230 space groups}

      Point group symmetry combined with translation in the 14 lattices

      Only some point group symmetries fit each lattice, which is why the product is not 32 times 14.

    2. Screw axis

      mn: translate nm of the cell, then rotate 360∘mm_n:\ \text{translate } \tfrac{n}{m}\text{ of the cell, then rotate } \tfrac{360^\circ}{m}

      mm order of the rotation, nn translation index

      Examples from the slide: 212_1 (½, 180°), 313_1 (⅓, 120°), 424_2 (²⁄₄, 90°), 636_3 (³⁄₆, 60°).

    3. Glide translations

      glide: 12 a, 12 b, 12 c,n: 12 along diagonal,d: 14 along face or diagonal\text{glide: } \tfrac12\,\mathbf{a},\ \tfrac12\,\mathbf{b},\ \tfrac12\,\mathbf{c}, \quad n:\ \tfrac12 \text{ along diagonal}, \quad d:\ \tfrac14 \text{ along face or diagonal}

      a,b,c\mathbf{a},\mathbf{b},\mathbf{c} cell edges; a glide is a reflection plus this translation

    Close packing and the holes between the atoms

    1. fcc vertex atoms

      8×18=18 \times \tfrac{1}{8} = 1

      8 vertex atoms, each shared by 8 cells

    2. fcc face atoms

      6×12=36 \times \tfrac{1}{2} = 3

      6 face atoms, each shared by 2 cells

    3. ccp atoms per cell

      1+3=41 + 3 = 4

      vertex atoms plus face atoms

      Four atoms in each fcc cell.

    4. fcc packing efficiency

      74%74\%

      fraction of space filled by spheres in fcc

    5. Octahedral hole radius ratio

      rh=0.414 rr_h = 0.414\,r

      rhr_h radius of the hole, rr radius of the packed spheres

    6. Tetrahedral hole radius ratio (Lecture 5 slide)

      rh=0.255 rr_h = 0.255\,r

      rhr_h radius of the hole, rr radius of the packed spheres

      Lecture 7 gives 0.225 for this ratio. The two slides disagree.

    7. Octahedral edge holes per cell

      12×14=312 \times \tfrac{1}{4} = 3

      12 edge holes, each shared by 4 cells

    8. Octahedral holes per cell

      3+1=43 + 1 = 4

      edge holes plus the centre hole

    9. Tetrahedral holes per cell

      88

      one per octahedron face, or per cube vertex

    10. Tetrahedral holes per ccp atom

      8/4=28 / 4 = 2

      8 holes per cell divided by 4 atoms per cell

    Ionic structures: rock salt, zinc blende, antifluorite, CsCl and rutile

    1. Na count in rock salt

      8×18+6×12=48 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4

      Na atoms per cell: 8 corners shared by 8 cells, 6 faces shared by 2 cells

    2. Cl count in rock salt

      12×14+1×1=412 \times \tfrac{1}{4} + 1 \times 1 = 4

      Cl atoms per cell: 12 edges shared by 4 cells, 1 at the centre

    3. Rock-salt formula

      Na4Cl4=NaCl\text{Na}_4\text{Cl}_4 = \text{NaCl}

      cell contents reduced to the empirical formula

    4. Zinc blende composition

      ZnS\text{ZnS}

      ccp S²⁻ with Zn²⁺ in half the tetrahedral holes

    5. Antifluorite composition

      Li2O\text{Li}_2\text{O}

      ccp O²⁻ with Li⁺ in all tetrahedral holes

    6. CsCl composition

      CsCl\text{CsCl}

      two interpenetrating simple cubic cells

    7. Rutile composition

      TiO2\text{TiO}_2

      distorted hcp O²⁻ with Ti⁴⁺ in half the octahedral holes

    Measuring crystal structure: Miller indices and Bragg diffraction

    1. Miller plane

      ha+kb+lch\mathbf{a} + k\mathbf{b} + l\mathbf{c}

      (h k l)(h\,k\,l) integers, a,b,c\mathbf{a},\mathbf{b},\mathbf{c} cell vectors

      The slide ties each index to an axis intercept of 1/h1/h, 1/k1/k, 1/l1/l.

    2. Bragg equation

      nλ=2dsin⁡θn\lambda = 2d\sin\theta

      nn integer order of reflection, λ\lambda X-ray wavelength, dd lattice plane spacing, θ\theta incidence angle

      Use it to get dd from a measured angle, or the angle from dd.

    3. Path difference

      2dsin⁡θ2d\sin\theta

      extra distance travelled by beam 2 compared with beam 1

      Constructive interference needs this to equal nλn\lambda.

    Ionic or covalent, ionic radii and radius ratios

    1. Charge neutrality, general form

      AmXpA_m X_p

      mm and pp are set by charge neutrality

      The total positive charge equals the total negative charge.

    2. Octahedral hole ratio

      roct/rsphere=2−1=0.414r_\text{oct}/r_\text{sphere} = \sqrt{2}-1 = 0.414

      rr radii of the octahedral hole and of the packed spheres

    3. Tetrahedral hole ratio

      rtet/rsphere=1sin⁡(109.47/2)−1=0.225r_\text{tet}/r_\text{sphere} = \dfrac{1}{\sin(109.47/2)}-1 = 0.225

      rr radii of the tetrahedral hole and of the packed spheres

      Lecture 5 gives 0.255 for the same ratio, which does not match.

    Lattice enthalpy: the Born–Mayer equation and its trends

    1. Coulomb potential

      V(r)=−q1q24πε0rV(r) = -\dfrac{q_1 q_2}{4\pi\varepsilon_0 r}

      q1,q2q_1,q_2 charges, ε0\varepsilon_0 permittivity, rr separation

    2. Lattice reaction

      NaCl(s)→Na+(g)+Cl−(g),ΔH=ΔHL0\text{NaCl}(s) \rightarrow \text{Na}^+(g) + \text{Cl}^-(g), \qquad \Delta H = \Delta H^0_L

      ΔHL0\Delta H^0_L lattice enthalpy

      The energy to break the crystal into separate gas-phase ions.

    3. Born–Mayer equation

      ΔHL0=NA∣zAzB∣e24πε0d(1−d∗d)A\Delta H^0_L = \dfrac{N_A |z_A z_B| e^2}{4\pi\varepsilon_0 d}\left(1 - \dfrac{d^*}{d}\right)A

      NAN_A Avogadro's number, ee electronic charge, zA,zBz_A, z_B ionic charges, ε0\varepsilon_0 vacuum permittivity, d=rA+rBd = r_A + r_B separation, d∗d^* repulsion constant (0.345 Å), AA Madelung constant

      For one structure, only dd, zAz_A and zBz_B change.

    4. Madelung series

      V=−e2∣zAzB∣d(6−122+83−⋯ )V = -\dfrac{e^2|z_A z_B|}{d}\left(6 - \dfrac{12}{\sqrt{2}} + \dfrac{8}{\sqrt{3}} - \cdots\right)

      sum over successive coordination spheres of NaCl

    5. Madelung form

      V=−e2∣zAzB∣d⋅AV = -\dfrac{e^2|z_A z_B|}{d}\cdot A

      AA Madelung constant

    6. First three NaCl sphere terms

      V=−6e2∣zAzB∣d,V=12e2∣zAzB∣2 d,V=−8e2∣zAzB∣3 dV = -\dfrac{6e^2|z_A z_B|}{d}, \quad V = \dfrac{12e^2|z_A z_B|}{\sqrt{2}\,d}, \quad V = -\dfrac{8e^2|z_A z_B|}{\sqrt{3}\,d}

      six nearest neighbours attract, twelve next neighbours repel, eight further ones attract

    03 Solid-state materials

    Energy bands: how a pile of atoms becomes a metal, semiconductor or insulator

    1. Band-gap thermal condition

      Eg∼kTE_g \sim kT

      EgE_g band gap, kk Boltzmann constant, TT temperature

      For a material to be an intrinsic semiconductor the gap must be small enough that thermal energy, of order kTkT, can lift electrons across it. The slide writes it as a band gap of order kTkT.

    Conductors in one, two and three dimensions

    1. Krogmann salt oxidation

      K2[Pt(CN)4]⋅3H2O+0.15 X2→K2[Pt(CN)4]X0.3⋅3H2O\text{K}_2[\text{Pt(CN)}_4]\cdot 3\text{H}_2\text{O} + 0.15\,\text{X}_2 \rightarrow \text{K}_2[\text{Pt(CN)}_4]\text{X}_{0.3}\cdot 3\text{H}_2\text{O}

      X\text{X} is Cl or Br

      The halogen removes some electrons from the Pt chains. That makes the band partly empty, so the chain conducts.

    2. Electron count in C60

      4×60=2404 \times 60 = 240

      4 valence electrons per carbon atom, times 60 carbon atoms

      Use it to count the electrons that fill the molecular orbitals of C60.

    Superconductivity, BCS theory and Mott-Hubbard localisation

    1. BCS critical temperature

      kBTc=ℏω exp⁡ ⁣(−1V N(Ef))k_B T_c = \hbar\omega\,\exp\!\left(\frac{-1}{V\,N(E_f)}\right)

      kBk_B Boltzmann constant, TcT_c critical temperature, ℏω\hbar\omega Debye energy, VV strength of the attraction between electrons, N(Ef)N(E_f) density of states at the Fermi energy

      A larger N(Ef)N(E_f) or a stronger attraction VV raises TcT_c.

    2. Debye temperature

      ℏωkB\frac{\hbar\omega}{k_B}

      ℏ\hbar reduced Planck constant, ω\omega phonon frequency, kBk_B Boltzmann constant

      The slide describes it as the collective (acoustic) vibration of the lattice.

    Conductivity, carriers and how temperature changes them

    1. Conductivity and resistivity

      σ=1ρ\sigma = \frac{1}{\rho}

      σ\sigma conductivity (S cm−1^{-1}), ρ\rho resistivity

      Conductivity is how well a material carries current. Resistivity is how much it resists.

    2. Conductivity from carriers

      σ=neμ\sigma = n e \mu

      nn carrier concentration, ee electronic charge, μ\mu carrier mobility

      More carriers, or carriers that move more easily, mean higher conductivity.

    3. Carrier mobility

      μ=eτm∗\mu = \frac{e\tau}{m^*}

      ee electronic charge, τ\tau relaxation time between scattering events, m∗m^* effective mass

      Longer time between scattering gives a higher mobility.

    4. Drude conductivity

      σ=ne2τm∗\sigma = \frac{n e^2 \tau}{m^*}

      nn carrier concentration, ee electronic charge, τ\tau relaxation time, m∗m^* effective mass

      Put the mobility into the carrier equation and you get this. Use it to see why a metal loses conductivity when τ\tau falls on heating.

    5. Semiconductor carrier density

      n∝exp⁡{−Eg2kT}n \propto \exp\left\{-\frac{E_g}{2kT}\right\}

      EgE_g band gap, kk Boltzmann constant, TT temperature

      Heating a semiconductor puts many more carriers across the gap.

    Dielectrics, capacitors and the structure of barium titanate

    1. Capacitance (definition)

      C=QΔVC = \frac{Q}{\Delta V}

      CC capacitance, QQ charge on either plate, ΔV\Delta V potential difference between the plates

      Capacitance says how much charge the plates hold per volt.

    2. Parallel-plate capacitance in vacuum

      C=ε0AdC = \frac{\varepsilon_0 A}{d}

      ε0\varepsilon_0 permittivity of vacuum, AA plate area, dd plate separation

      Bigger plates or a smaller gap give more capacitance.

    3. Energy stored in a capacitor

      U=12C(ΔV)2U = \tfrac{1}{2} C (\Delta V)^2

      UU electric potential energy, CC capacitance, ΔV\Delta V potential difference

    4. Capacitance with a dielectric

      C=εrε0AdC = \varepsilon_r \varepsilon_0 \frac{A}{d}

      εr\varepsilon_r relative permittivity, ε0\varepsilon_0 permittivity of vacuum, AA plate area, dd plate separation

      Put in a dielectric and the capacitance goes up by the factor εr\varepsilon_r.

    Polarisation loops, lone pairs, piezoelectrics and second harmonic generation

    1. Polarisation

      P=ε0χeEP = \varepsilon_0 \chi_e E

      PP polarisation, ε0\varepsilon_0 permittivity of vacuum, χe\chi_e electric susceptibility, EE applied electric field

      For a paraelectric, polarisation grows in proportion to the field. The slope at the origin is ε0χe\varepsilon_0\chi_e.

    2. Electric susceptibility

      χe=εr−1\chi_e = \varepsilon_r - 1

      χe\chi_e electric susceptibility, εr\varepsilon_r relative permittivity

      Vacuum has εr=1\varepsilon_r = 1 so χe=0\chi_e = 0, meaning no polarisation response.

    Magnetic moments, diamagnetism and paramagnetism

    1. Magnetisation

      M=χH\mathbf{M} = \chi \mathbf{H}

      M\mathbf{M} magnetisation (magnetic moment per unit volume), χ\chi magnetic susceptibility, H\mathbf{H} applied magnetic field

      Susceptibility tells you how strongly the material responds to a field.

    2. Diamagnetic susceptibility

      χ<0,χ=const\chi < 0, \quad \chi = \text{const}

      χ\chi magnetic susceptibility, which is negative and does not change with temperature

      The sign tells you the material is repelled by a field.

    3. Curie law

      χ=CCurieT\chi = \frac{C_{\text{Curie}}}{T}

      χ\chi susceptibility, CCurieC_{\text{Curie}} Curie constant, TT temperature

      Use it for a paramagnet. Susceptibility falls as temperature rises.

    4. Total spin of high-spin Cr2+

      S=4(+12)=2S = 4\left(+\tfrac{1}{2}\right) = 2

      SS total spin quantum number, s=±12s = \pm\tfrac{1}{2} for each electron; four unpaired electrons

    5. Total spin of low-spin Cr2+

      S=3(+12)+(−12)=1S = 3\left(+\tfrac{1}{2}\right) + \left(-\tfrac{1}{2}\right) = 1

      SS total spin quantum number, s=±12s = \pm\tfrac{1}{2} for each electron; three up and one down

    6. Bohr magneton

      μB=9.27×10−24 A m2\mu_B = 9.27 \times 10^{-24}\ \text{A m}^2

      μB\mu_B Bohr magneton, the unit of an electron magnetic moment

    7. Nuclear magneton

      μN=5.05×10−27 A m2\mu_N = 5.05 \times 10^{-27}\ \text{A m}^2

      μN\mu_N nuclear magneton, the unit of a nuclear magnetic moment

    Magnetic order: ferro-, antiferro- and ferrimagnets and the exchange that makes them

    1. Curie-Weiss law

      χ=CT−θ\chi = \frac{C}{T - \theta}

      χ\chi susceptibility, CC Curie constant, TT temperature, θ\theta Weiss constant

      Use it above the ordering temperature of a ferromagnet or antiferromagnet. It reduces to Curie law when θ=0\theta = 0.

    2. Weiss constant of a ferromagnet

      θ>0,θ=TC\theta > 0, \quad \theta = T_C

      θ\theta Weiss constant, TCT_C Curie temperature

    3. Weiss constant of an antiferromagnet

      θ<0,θ=−TN\theta < 0, \quad \theta = -T_N

      θ\theta Weiss constant, TNT_N Néel temperature

      Extrapolating a straight line of χ−1\chi^{-1} against TT back to the axis hits −TN-T_N.

    04 Statistical mechanics

    Microstates, configurations and the Boltzmann distribution

    1. Harmonic oscillator levels

      En=hν(n+12)E_n = h\nu\left(n + \tfrac{1}{2}\right)

      hh Planck constant, ν\nu frequency, nn quantum number

    2. Harmonic oscillator levels, zero-point energy ignored

      En=hνnE_n = h\nu n

      hh Planck constant, ν\nu frequency, nn quantum number

      Evenly spaced rungs. Dropping the constant half does not change the gaps.

    3. Factorial

      n!=n(n−1)⋯2⋅1n! = n(n-1)\cdots 2\cdot 1

      nn an integer

    4. Weight of a configuration

      W(n0,n1,… )=N!n0! n1! n2!⋯W(n_0,n_1,\dots) = \frac{N!}{n_0!\,n_1!\,n_2!\cdots}

      WW number of microstates, NN total molecules, nin_i occupation of state ii

      Count the microstates behind a configuration.

    5. Worked weight

      W(2,0,0,1)=3!2! 0! 0! 1!=62⋅1⋅1⋅1=3W(2,0,0,1) = \frac{3!}{2!\,0!\,0!\,1!} = \frac{6}{2\cdot 1\cdot 1\cdot 1} = 3

      The blue configuration of three quanta on three oscillators.

    6. Stirling approximation

      ln⁡n!≈nln⁡n−n\ln n! \approx n\ln n - n

      nn a large integer

      Turns awkward factorials into something you can differentiate.

    7. ln W to maximise

      ln⁡W=Nln⁡N−∑iniln⁡ni\ln W = N\ln N - \sum_i n_i \ln n_i

      NN total molecules, nin_i occupation of state ii

    8. Fixed number of molecules

      ∑ini=N\sum_i n_i = N

      nin_i occupation of state ii, NN total molecules

    9. Fixed total energy

      ∑iεini=E\sum_i \varepsilon_i n_i = E

      εi\varepsilon_i energy of level ii, EE total energy

    10. Boltzmann distribution

      pi=niN=e−βεi∑ie−βεip_i = \frac{n_i}{N} = \frac{e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}

      pip_i probability of level ii, β\beta Lagrange multiplier (inverse temperature)

      Gives the share of molecules in each level at equilibrium.

    11. Partition function

      q=∑ie−βεiq = \sum_i e^{-\beta\varepsilon_i}

      εi\varepsilon_i energy of level ii, β\beta as above

    12. Population ratio

      ninj=e−β(εi−εj)\frac{n_i}{n_j} = e^{-\beta(\varepsilon_i - \varepsilon_j)}

      ni,njn_i, n_j populations of levels ii and jj

      Only the energy gap matters.

    13. Definition of temperature

      β=1kT\beta = \frac{1}{kT}

      kk Boltzmann constant, TT temperature

    14. Air in a room, all on one side (very rough)

      ≈e−1023\approx e^{-10^{23}}

      Chance that more than 51% of the air in a 40 m3^3 room is on the left

    Temperature, absolute zero and the zeroth law

    1. Expectation value

      ⟨O⟩=∑ipiOi\langle O\rangle = \sum_i p_i O_i

      OO an observable, pip_i probability of state ii, OiO_i its value in state ii

      A weighted average: each state counts as often as it occurs.

    2. Average energy per molecule

      ⟨ε⟩=∑ipiεi=∑iεie−βεi∑ie−βεi\langle\varepsilon\rangle = \sum_i p_i\varepsilon_i = \frac{\sum_i \varepsilon_i e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}

      ε\varepsilon energy, β=1/kT\beta = 1/kT

    3. Total energy of N molecules

      E=N⟨ε⟩=∑iniεi=N∑iεie−βεi∑ie−βεiE = N\langle\varepsilon\rangle = \sum_i n_i\varepsilon_i = N\frac{\sum_i \varepsilon_i e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}

      EE total energy, NN number of molecules, nin_i occupation of level ii

      For identical, independent molecules.

    4. Boltzmann distribution

      pi=niN=e−βεi∑ie−βεip_i = \frac{n_i}{N} = \frac{e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}

      pip_i probability of level ii, nin_i its occupation, NN total molecules

    5. Definition of temperature

      β=1kT\beta = \frac{1}{kT}

      kk Boltzmann constant = 1.380649×10−231.380649 \times 10^{-23} J K−1^{-1}, TT absolute temperature

      beta has units of inverse energy.

    6. Population ratio

      ninj=e−β(εi−εj)\frac{n_i}{n_j} = e^{-\beta(\varepsilon_i-\varepsilon_j)}

      ni,njn_i, n_j populations of levels ii and jj

    7. Absolute-zero limit

      εi−ε0kT=∞(T=0)\frac{\varepsilon_i-\varepsilon_0}{kT} = \infty \quad (T = 0)

      ε0\varepsilon_0 ground-state energy

    8. Ground-state occupation at T = 0

      ni=n0e−(εi−ε0)/kT=n0e−∞=0(ni≠0)n_i = n_0 e^{-(\varepsilon_i-\varepsilon_0)/kT} = n_0 e^{-\infty} = 0 \quad (n_i \ne 0)

      n0n_0 ground-state occupation

    9. All molecules in the ground state

      n0=Nn_0 = N

      At absolute zero

    The partition function, degeneracy and characteristic temperature

    1. Partition function

      q=∑ie−βεiq = \sum_i e^{-\beta\varepsilon_i}

      εi\varepsilon_i energy of state ii, β=1/kT\beta = 1/kT

      Roughly the number of thermally accessible states per molecule.

    2. Occupation relative to the ground state

      e−βεi=nin0e^{-\beta\varepsilon_i} = \frac{n_i}{n_0}

      nin_i occupation of state ii, n0n_0 ground-state occupation

    3. Oscillator levels

      εi=ihν=iβ−1,e−βεi=e−i\varepsilon_i = ih\nu = i\beta^{-1}, \qquad e^{-\beta\varepsilon_i} = e^{-i}

      Case where hν=β−1h\nu = \beta^{-1}

    4. Oscillator partition function

      q=∑i=0∞e−i=1+e−1+e−2+⋯=11−e−1=1.58q = \sum_{i=0}^{\infty} e^{-i} = 1 + e^{-1} + e^{-2} + \cdots = \frac{1}{1-e^{-1}} = 1.58

      A geometric series, summed in closed form.

    5. Oscillator probabilities, Case 1

      p0=0.633,p1=e−11.58=0.233,p2=e−21.58=0.086p_0 = 0.633, \quad p_1 = \frac{e^{-1}}{1.58} = 0.233, \quad p_2 = \frac{e^{-2}}{1.58} = 0.086

      q=1.58q = 1.58

    6. Oscillator probabilities, Case 2 (q = 2.54)

      p0=0.394,p1=e−1/22.54=0.239,p2=e−12.54=0.145p_0 = 0.394, \quad p_1 = \frac{e^{-1/2}}{2.54} = 0.239, \quad p_2 = \frac{e^{-1}}{2.54} = 0.145

      Levels brought closer together, so qq is larger

    7. Energy from the partition function

      E=∑iεini=N∑iεie−βεiq=−Nqdqdβ=−Ndln⁡qdβ=NkT2dln⁡qdTE = \sum_i \varepsilon_i n_i = N\sum_i \varepsilon_i \frac{e^{-\beta\varepsilon_i}}{q} = -\frac{N}{q}\frac{dq}{d\beta} = -N\frac{d\ln q}{d\beta} = NkT^2\frac{d\ln q}{dT}

      EE total energy, NN molecules, kk Boltzmann constant, TT temperature

      Bulk quantities are derivatives of q.

    8. Derivative of q

      dqdβ=−∑iεie−βεi\frac{dq}{d\beta} = -\sum_i \varepsilon_i e^{-\beta\varepsilon_i}
    9. Partition function over levels

      q=∑εig(εi) e−βεiq = \sum_{\varepsilon_i} g(\varepsilon_i)\, e^{-\beta\varepsilon_i}

      g(εi)g(\varepsilon_i) degeneracy of the level with energy εi\varepsilon_i

      Sum over levels, with each level counted g times.

    10. Level population ratio

      n(εj)n(εi)=g(εj)g(εi) e−β(εj−εi)\frac{n(\varepsilon_j)}{n(\varepsilon_i)} = \frac{g(\varepsilon_j)}{g(\varepsilon_i)}\, e^{-\beta(\varepsilon_j-\varepsilon_i)}

      gg degeneracies of the two levels

    11. Characteristic temperature

      Θ=ΔE/k\Theta = \Delta E / k

      ΔE\Delta E typical energy spacing, kk Boltzmann constant

      Roughly the lowest temperature at which excited states start to fill.

    12. Population versus characteristic temperature

      njni∼e−ΔE/kT=e−Θ/T\frac{n_j}{n_i} \sim e^{-\Delta E/kT} = e^{-\Theta/T}

      nj,nin_j, n_i populations of an upper and a lower level

    13. Worked values for Theta = 300 K

      e−1=0.37 (300 K);e−0.1=0.90 (3000 K);e−10=4.5×10−5 (30 K)e^{-1} = 0.37\ (300\ \text{K}); \quad e^{-0.1} = 0.90\ (3000\ \text{K}); \quad e^{-10} = 4.5\times10^{-5}\ (30\ \text{K})

      Ratio of first excited state to ground state

    05 Thermodynamics

    The first law: internal energy, work and heat

    1. Internal energy (microscopic)

      U=E=∑iniεiU = E = \sum_i n_i\varepsilon_i

      nin_i occupation of microstate ii, εi\varepsilon_i its energy

      Total energy of the occupied states.

    2. First law

      ΔU=Q+W\Delta U = Q + W

      QQ heat entering the system, WW work done on the system

      The lecture convention. Work done ON the system is positive.

    3. First law, alternate sign convention

      ΔU=Q−W\Delta U = Q - W

      WW is the work done by the system

      Used by some books and Wikipedia. Not the lecture convention.

    4. Work

      dW=F⋅dxdW = \mathbf{F}\cdot d\mathbf{x}

      F\mathbf{F} generalised force, x\mathbf{x} generalised displacement

    5. Expansion work

      dW=−pext dVdW = -p_{\text{ext}}\,dV

      pextp_{\text{ext}} external pressure, dVdV change in volume

    6. Total expansion work

      W=−∫ViVfpext dVW = -\int_{V_i}^{V_f} p_{\text{ext}}\,dV

      ViV_i initial volume, VfV_f final volume

    7. Surface expansion work

      dw=γ dσdw = \gamma\,d\sigma

      γ\gamma surface tension, σ\sigma surface area

    8. Extension work

      dw=f dldw = f\,dl

      ff tension, ll length

    9. Electrical work

      dw=ϕ dQdw = \phi\,dQ

      ϕ\phi electric potential, QQ charge

    10. Energy balance

      ΔUtotal=ΔUsystem+ΔUsurroundings=0\Delta U_{\text{total}} = \Delta U_{\text{system}} + \Delta U_{\text{surroundings}} = 0
    11. Isolated-system relation

      ΔUsystem=−ΔUsurroundings\Delta U_{\text{system}} = -\Delta U_{\text{surroundings}}

      What the system gains, the surroundings lose.

    12. Microscopic differential of U

      dU=∑ini dεi+∑iεi dnidU = \sum_i n_i\,d\varepsilon_i + \sum_i \varepsilon_i\,dn_i

      First sum is work (levels shift), second sum is heat (populations change)

    The second law and entropy

    1. Second law, version 1

      ΔS≥0\Delta S \ge 0

      For an isolated system

    2. Dog-flea jump probability

      P+(NRover)=NRover2RP_+(N_{\text{Rover}}) = \frac{N_{\text{Rover}}}{2R}

      RR half the number of fleas, NRoverN_{\text{Rover}} fleas on Rover

      Chance that the next jump goes from Rover to Rex.

    3. Reverse-jump probability

      P−(NRover)=1−P+(NRover)P_-(N_{\text{Rover}}) = 1 - P_+(N_{\text{Rover}})
    4. Equilibrium in the dog-flea model

      NRover=NRex=RN_{\text{Rover}} = N_{\text{Rex}} = R
    5. Direction of change

      ΔW≥0;dln⁡W≥0;dln⁡W=0 at equilibrium\Delta W \ge 0; \qquad d\ln W \ge 0; \qquad d\ln W = 0\ \text{at equilibrium}

      WW number of microstates, at constant NN and EE

    6. Entropy

      S=kln⁡WS = k\ln W

      kk Boltzmann constant, WW number of microstates

    7. Second law for an isolated system

      dS≥0dS \ge 0
    8. Weight of a configuration

      W(n0,n1,… )=N!n0! n1!⋯W(n_0,n_1,\dots) = \frac{N!}{n_0!\,n_1!\cdots}
    9. Stirling form of ln W

      ln⁡W=Nln⁡N−∑iniln⁡ni\ln W = N\ln N - \sum_i n_i \ln n_i
    10. Total differential of ln W

      dln⁡W=−∑iln⁡ni dnid\ln W = -\sum_i \ln n_i\,dn_i

      With NN constant, so dN=0dN = 0

    11. Equilibrium population

      ni=n0e−βεi;ln⁡ni=ln⁡n0−βεin_i = n_0 e^{-\beta\varepsilon_i}; \qquad \ln n_i = \ln n_0 - \beta\varepsilon_i
    12. Expansion of the differential

      dln⁡W=−ln⁡n0∑idni+β∑iεi dnid\ln W = -\ln n_0 \sum_i dn_i + \beta\sum_i \varepsilon_i\,dn_i

      The first term vanishes because NN is constant

    13. Reversible heat relation

      dln⁡W=β dQrevd\ln W = \beta\,dQ_{\text{rev}}
    14. Thermodynamic entropy

      dS=dQrevTdS = \frac{dQ_{\text{rev}}}{T}

      QrevQ_{\text{rev}} heat in a reversible process, TT temperature

    15. Entropy change, reversible process

      ΔS=∫dQrevT≥0\Delta S = \int \frac{dQ_{\text{rev}}}{T} \ge 0

      Slide wording. See the aside about the inequality.

    16. State-function equality

      ΔSrev=S2−S1=ΔSirrev\Delta S_{\text{rev}} = S_2 - S_1 = \Delta S_{\text{irrev}}

      Entropy depends only on the state, not the path.

    17. Entropy from q (1)

      S=UT+Nkln⁡qS = \frac{U}{T} + Nk\ln q

      UU internal energy, qq partition function

    18. Entropy from q (2)

      S=NkT∂ln⁡q∂T+Nkln⁡qS = NkT\frac{\partial\ln q}{\partial T} + Nk\ln q
    19. Entropy from q (3)

      S=Nk∂∂T(Tln⁡q)S = Nk\frac{\partial}{\partial T}\left(T\ln q\right)

      The lecture says there is no need to memorise these.

    20. Additivity of entropy

      Wtotal=WxWy;ln⁡Wtotal=ln⁡Wx+ln⁡Wy;Stotal=Sx+SyW_{\text{total}} = W_x W_y; \quad \ln W_{\text{total}} = \ln W_x + \ln W_y; \quad S_{\text{total}} = S_x + S_y

      Two independent, isolated systems xx and yy

    Heat flow, the third law and residual entropy

    1. Total entropy on thermal contact

      dStotal=dSx+dSy≥0dS_{\text{total}} = dS_x + dS_y \ge 0

      xx and yy two systems in thermal contact, isolated from everything else

    2. Heat terms in the entropy sum

      dQxTx+dQyTy≥0\frac{dQ_x}{T_x} + \frac{dQ_y}{T_y} \ge 0
    3. Combined form

      dQx(1Tx−1Ty)≥0dQ_x\left(\frac{1}{T_x} - \frac{1}{T_y}\right) \ge 0
    4. Direction of heat flow

      Tx<Ty ⇒ dQx>0T_x < T_y \ \Rightarrow\ dQ_x > 0

      Heat flows into xx when xx is the colder system

      Heat flows spontaneously from hot to cold.

    5. Third law, perfect crystal

      S(0)=kln⁡(1)=0S(0) = k\ln(1) = 0

      One microstate at T=0T = 0 for a perfect crystal of a pure substance

    6. Residual entropy of CO

      Sresidual=Nkln⁡2S_{\text{residual}} = Nk\ln 2

      NN molecules, each in one of two orientations

    Open systems, free energy and spontaneity

    1. Total entropy with a bath

      dStotal=dS+dSbath≥0;dSbath=−dQTdS_{\text{total}} = dS + dS_{\text{bath}} \ge 0; \qquad dS_{\text{bath}} = -\frac{dQ}{T}

      SS system entropy, SbathS_{\text{bath}} bath entropy, QQ heat entering the system

    2. Clausius form of the second law

      dS≥dQTdS \ge \frac{dQ}{T}

      SS system entropy, QQ heat

    3. Energy balance

      dQ=dU−dWdQ = dU - dW
    4. Combined inequality

      dU−T dS≤dWdU - T\,dS \le dW
    5. Helmholtz free energy

      A≡U−TSA \equiv U - TS

      AA Helmholtz free energy, UU internal energy, TT temperature, SS entropy

      A is for Arbeit, German for work.

    6. Helmholtz change at constant T

      dA=dU−T dSdA = dU - T\,dS
    7. Work bound

      dA≤dWdA \le dW

      At constant T, A is the maximum work you can extract.

    8. Spontaneity at constant T and V

      dA≤0dA \le 0

      No pV work and no non-pV work

    9. Internal energy differential

      dU=T dS−p dVdU = T\,dS - p\,dV

      No non-expansion work

    10. Constant-volume heat

      dU=dQrevdU = dQ_{\text{rev}}

      At constant VV

    11. Derivative identity

      (dUdS)V=T\left(\frac{dU}{dS}\right)_V = T
    12. Enthalpy

      H≡U+pVH \equiv U + pV

      HH enthalpy, pp pressure, VV volume

    13. Enthalpy differential

      dH=d(U+pV)=T dS−p dV+p dV+V dp=T dS+V dpdH = d(U+pV) = T\,dS - p\,dV + p\,dV + V\,dp = T\,dS + V\,dp
    14. Constant-pressure heat

      dH=dQrevdH = dQ_{\text{rev}}

      At constant pp and no non-pV work

      Exothermic and endothermic usually refer to delta H.

    15. Gibbs free energy

      G≡U−TS+pV=H−TS=A+pVG \equiv U - TS + pV = H - TS = A + pV

      GG Gibbs free energy

    16. Gibbs bound

      dG≤dW+p dV+V dpdG \le dW + p\,dV + V\,dp
    17. Constant T and p

      dG≤dW+p dV=dWnon-expansiondG \le dW + p\,dV = dW_{\text{non-expansion}}
    18. Spontaneity at constant T and p

      ΔG≤0\Delta G \le 0

      No non-pV work

    19. Spontaneity summary

      dS≥0 (adiabatic, isochoric);dA≤0 (isothermal, isochoric);dH≤0 (adiabatic, isobaric);dG≤0 (isothermal, isobaric)dS \ge 0\ (\text{adiabatic, isochoric}); \quad dA \le 0\ (\text{isothermal, isochoric}); \quad dH \le 0\ (\text{adiabatic, isobaric}); \quad dG \le 0\ (\text{isothermal, isobaric})
    20. Gibbs from the partition function

      G=−NkTln⁡qNG = -NkT\ln\frac{q}{N}

      NN molecules, qq partition function

    The ideal gas from statistical mechanics

    1. Partition function of two systems

      qtotal=q1q2q_{\text{total}} = q_1 q_2

      qq molecular partition function of each system

      The lecture sets showing this as homework.

    2. Distinguishable components

      Q=qNQ = q^N

      QQ total partition function, NN number of components

    3. Indistinguishable components

      Q=qNN!Q = \frac{q^N}{N!}

      N!N! factorial

      Use this for molecules in a gas.

    4. 1D box energy levels

      εi=i2h28mL2\varepsilon_i = \frac{i^2h^2}{8mL^2}

      ii quantum number, mm mass, LL box length, hh Planck constant

    5. 3D box energy levels

      εix,iy,iz=h28mL2(ix2+iy2+iz2)\varepsilon_{i_x,i_y,i_z} = \frac{h^2}{8mL^2}\left(i_x^2+i_y^2+i_z^2\right)

      ix,iy,izi_x, i_y, i_z quantum numbers

    6. 1D molecular partition function

      q=∑i=1∞e−βεiq = \sum_{i=1}^{\infty} e^{-\beta\varepsilon_i}

      β=1/kT\beta = 1/kT

    7. Translational temperature scale

      Θtrans∼h28mL2k∼10−18 K\Theta_{\text{trans}} \sim \frac{h^2}{8mL^2k} \sim 10^{-18}\ \text{K}

      kk Boltzmann constant. Value for helium in a 1 m3^3 box

      T is far above this, so sums can be replaced by integrals.

    8. Sum replaced by integral

      q≈∫0∞e−βi2h2/8mL2 diq \approx \int_0^\infty e^{-\beta i^2h^2/8mL^2}\,di

      ii treated as continuous

    9. 1D translational partition function

      q=(2πmh2β)1/2Lq = \left(\frac{2\pi m}{h^2\beta}\right)^{1/2} L
    10. 3D translational partition function

      qtrans=qxqyqz=(2πmh2β)3/2Vq_{\text{trans}} = q_x q_y q_z = \left(\frac{2\pi m}{h^2\beta}\right)^{3/2} V

      VV volume

    11. Thermal wavelength

      Λ=(h2β2πm)1/2\Lambda = \left(\frac{h^2\beta}{2\pi m}\right)^{1/2}

      Λ\Lambda thermal wavelength

    12. Translational partition function via the thermal wavelength

      qtrans=VΛ3q_{\text{trans}} = \frac{V}{\Lambda^3}
    13. Typical q_trans for 1 m^3 of gas

      qtrans=1030q_{\text{trans}} = 10^{30}
    14. Energy from the partition function

      E=NkT2 dln⁡qdTE = NkT^2\,\frac{d\ln q}{dT}

      EE energy, NN number of molecules

    15. Translational energy

      Etrans=32NkT=32nRTE_{\text{trans}} = \frac{3}{2}NkT = \frac{3}{2}nRT

      nn moles, RR gas constant

      U depends only on T for an ideal gas.

    16. Work on an expanding gas

      dW=−p dVdW = -p\,dV

      pp pressure, VV volume

    17. Internal energy change

      dU=∑ini dεi+∑iεi dnidU = \sum_i n_i\,d\varepsilon_i + \sum_i \varepsilon_i\,dn_i

      nin_i occupation of level ii. First sum is work, second is heat

    18. Pressure from level energies

      p=−∑inidεidVp = -\sum_i n_i \frac{d\varepsilon_i}{dV}
    19. Level energy scaling with volume

      ε=aV−2/3\varepsilon = aV^{-2/3}

      aa is constant with respect to VV

    20. Derivative of level energy

      dεidV=−23εiV−1\frac{d\varepsilon_i}{dV} = -\frac{2}{3}\varepsilon_i V^{-1}
    21. Ideal gas law from microscopic principles

      p=NkTV=nRTVp = \frac{NkT}{V} = \frac{nRT}{V}
    22. Helmholtz energy differential

      dA=−S dT−p dVdA = -S\,dT - p\,dV

      AA Helmholtz energy, SS entropy

    23. Pressure from Helmholtz energy

      p=−(∂A∂V)Tp = -\left(\frac{\partial A}{\partial V}\right)_T
    24. Helmholtz energy from the partition function

      A=−kTln⁡Q=−kTln⁡qNN!A = -kT\ln Q = -kT\ln\frac{q^N}{N!}
    25. Pressure from the partition function

      p=NkT(∂ln⁡q∂V)T=NkTVp = NkT\left(\frac{\partial \ln q}{\partial V}\right)_T = \frac{NkT}{V}
    26. Heat capacity

      C=dQdTC = \frac{dQ}{dT}

      CC heat capacity

      Heat needed to raise the temperature by one unit.

    27. Monatomic ideal gas, constant volume

      CV=32Nk=32nRC_V = \frac{3}{2}Nk = \frac{3}{2}nR

      CVC_V heat capacity at constant volume

    28. Monatomic ideal gas, constant pressure

      CP=52NkC_P = \frac{5}{2}Nk

      CPC_P heat capacity at constant pressure

      The lecture sets this as homework to show.

    29. Ideal gas law

      pV=NkTpV = NkT

      Still true for ideal polyatomic gases.

    Vibrations and the Einstein solid

    1. Harmonic oscillator levels

      En=hν(n+12)E_n = h\nu\left(n + \tfrac{1}{2}\right)

      ν\nu frequency, nn quantum number

    2. Harmonic oscillator levels, zero-point energy ignored

      En=hνnE_n = h\nu n
    3. Vibrational characteristic temperature

      Θvib=hνk\Theta_{\text{vib}} = \frac{h\nu}{k}

      Typical values: Ar(s) 40 K, Pb(s) 60 K, SO2(g) stretch 2000 K, H2O(g) stretch 5000 K.

    4. Vibrational partition function, series

      qvib=∑n=0∞e−βnhν=1+e−Θvib/T+(e−Θvib/T)2+⋯q_{\text{vib}} = \sum_{n=0}^{\infty} e^{-\beta nh\nu} = 1 + e^{-\Theta_{\text{vib}}/T} + \left(e^{-\Theta_{\text{vib}}/T}\right)^2 + \cdots
    5. Geometric series identity

      ∑n=0∞xn=11−x(∣x∣<1)\sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \quad (|x| < 1)
    6. Vibrational partition function, closed form

      qvib=11−e−Θvib/Tq_{\text{vib}} = \frac{1}{1 - e^{-\Theta_{\text{vib}}/T}}
    7. Vibrational energy

      E=NkT2dln⁡qdT=NkΘvibexp⁡(Θvib/T)−1E = NkT^2\frac{d\ln q}{dT} = \frac{Nk\Theta_{\text{vib}}}{\exp(\Theta_{\text{vib}}/T)-1}

      Energy above the zero-point energy

    8. Vibrational heat capacity

      CV=(dEdT)V=Nk(ΘvibT)2exp⁡(Θvib/T)(exp⁡(Θvib/T)−1)2C_V = \left(\frac{dE}{dT}\right)_V = Nk\left(\frac{\Theta_{\text{vib}}}{T}\right)^2\frac{\exp(\Theta_{\text{vib}}/T)}{\left(\exp(\Theta_{\text{vib}}/T)-1\right)^2}
    9. Vibrational heat capacity limit

      CVvib→RC_V^{\text{vib}} \to R

      At high T, for one vibration per molecule. A diatomic gas goes from 5R/2 towards 7R/2.

    10. Dulong-Petit limit

      3R=24.9 J K−1 mol−13R = 24.9\ \text{J K}^{-1}\,\text{mol}^{-1}

      RR gas constant

      High-temperature molar heat capacity of a simple solid.

    06 Defects and phase transitions

    Defects: why every real crystal has some, and how many

    1. Number of arrangements of defects

      W=N!nd! (N−nd)!W = \frac{N!}{n_d!\,(N-n_d)!}

      WW number of arrangements, NN number of possible sites, ndn_d number of defects

      Counts the ways to choose which sites hold a defect. More ways means more entropy.

    2. Entropy of the defect distribution

      ΔS=kln⁡W≈k(Nln⁡N−(N−nd)ln⁡(N−nd)−ndln⁡nd)\Delta S = k\ln W \approx k\left(N\ln N - (N-n_d)\ln(N-n_d) - n_d\ln n_d\right)

      ΔS\Delta S entropy change, kk Boltzmann constant

      The second form uses Stirling's approximation.

    3. Gibbs energy of defect formation

      ΔG=ndΔHd−kT(Nln⁡N−(N−nd)ln⁡(N−nd)−ndln⁡nd)\Delta G = n_d\Delta H_d - kT\left(N\ln N - (N-n_d)\ln(N-n_d) - n_d\ln n_d\right)

      ΔHd\Delta H_d enthalpy to make one defect, TT temperature

      Energy cost rises in a straight line with ndn_d. The entropy term curves and wins at small ndn_d.

    4. Equilibrium condition

      dΔGdnd=0\frac{d\Delta G}{dn_d} = 0

      ΔG\Delta G Gibbs energy change, ndn_d number of defects

      Equilibrium is the bottom of the G curve, where the slope is zero.

    5. Equilibrium condition, expanded

      ΔHd−kT(ln⁡(N−nd)−ln⁡nd)=0\Delta H_d - kT\left(\ln(N-n_d) - \ln n_d\right) = 0

      What you get after differentiating the Gibbs energy.

    6. Equilibrium ratio

      ndN−nd=exp⁡(−ΔHdkT)\frac{n_d}{N-n_d} = \exp\left(-\frac{\Delta H_d}{kT}\right)

      Defects over non-defects.

    7. Defect fraction

      ndN=exp⁡(−ΔHdkT)\frac{n_d}{N} = \exp\left(-\frac{\Delta H_d}{kT}\right)

      nd/Nn_d/N fraction of sites that are defects

      Use it when N≫ndN \gg n_d. A Boltzmann factor: costly defects are rare and heat makes more of them.

    Ionic conductivity: ions hop through gaps, and doping adds more gaps

    1. Arrhenius conductivity

      σ=σ0e−Ea/kT\sigma = \sigma_0 e^{-E_a/kT}

      σ\sigma conductivity, σ0\sigma_0 pre-exponential factor, EaE_a activation energy, kk Boltzmann constant, TT temperature

      Ion hopping is thermally activated. A plot of ln⁡σ\ln\sigma against 1/T1/T is a straight line of slope −Ea/k-E_a/k.

    2. Doped NaCl composition

      Na1−2xMnxVxCl\mathrm{Na}_{1-2x}\mathrm{Mn}_x\mathrm{V}_x\mathrm{Cl}

      xx fraction of Mn, V\mathrm{V} cation vacancy

      For every Mn cation added there is one cation vacancy.

    3. Li-ion anode reaction

      LiC6⟶Li++6C+e−\mathrm{LiC_6} \longrightarrow \mathrm{Li^+} + 6\mathrm{C} + e^-

      Lithium leaves the graphite and gives up an electron.

    4. Li-ion cathode reaction

      Li++CoO2+e−⟶LiCoO2\mathrm{Li^+} + \mathrm{CoO_2} + e^- \longrightarrow \mathrm{LiCoO_2}

      Lithium ion and electron arrive at the cobalt oxide.

    5. SOFC anode reaction

      2H2+2O2−⟶2H2O+4e−2\mathrm{H_2} + 2\mathrm{O^{2-}} \longrightarrow 2\mathrm{H_2O} + 4e^-

      Hydrogen is oxidised by oxide ions that crossed the electrolyte.

    6. SOFC cathode reaction

      O2+4e−⟶2O2−\mathrm{O_2} + 4e^- \longrightarrow 2\mathrm{O^{2-}}

      Oxygen picks up electrons and becomes oxide ions.

    Phase transitions need interactions, and only sharpen with many particles

    1. Partition function, non-interacting particles

      Q=qN (distinguishable)Q=qNN! (indistinguishable)Q = q^N \ \text{(distinguishable)} \qquad Q = \frac{q^N}{N!} \ \text{(indistinguishable)}

      qq molecular partition function, NN number of particles

      Works only because each particle can be treated alone.

    2. Extensive quantities of n copies

      En=nE1Sn=nS1Gn=nG1E_n = nE_1 \qquad S_n = nS_1 \qquad G_n = nG_1

      subscript 1 is a single molecule, nn number of copies

      For non-interacting particles, totals are just multiples of one particle.

    3. Molecular partition function

      q=∑i=1Ne−βεiq = \sum_{i=1}^{N} e^{-\beta\varepsilon_i}

      εi\varepsilon_i energy of state ii, β=1/kT\beta = 1/kT

    4. Canonical partition function

      Q=∑j=1N~e−βEjQ = \sum_{j=1}^{\tilde N} e^{-\beta E_j}

      EjE_j total energy of system state jj, N~\tilde N number of system states

      Use this when particles interact. The sum runs over states of the whole system.

    5. DNA zipper partition function

      Q=∑s=0N−1g(s)e−βE(s)=∑s=0N−1gse−βsε=∑s=0N−1xs=1−xN1−xQ = \sum_{s=0}^{N-1} g(s)e^{-\beta E(s)} = \sum_{s=0}^{N-1} g^s e^{-\beta s\varepsilon} = \sum_{s=0}^{N-1} x^s = \frac{1-x^N}{1-x}

      ss number of open links, gg orientational degeneracy per open link, ε\varepsilon energy per open link, NN number of links

      A geometric series, so it sums to a closed form.

    6. Definition of x

      x=ge−βεx = g e^{-\beta\varepsilon}

      Combines the entropy gain gg and the energy cost ε\varepsilon of opening one link.

    7. Expectation value, sum form

      ⟨s⟩=∑s=0N−1s ps=∑s=0N−1s gse−βsεQ=∑s=0N−1s xsQ\langle s\rangle = \sum_{s=0}^{N-1} s\,p_s = \sum_{s=0}^{N-1} \frac{s\,g^s e^{-\beta s\varepsilon}}{Q} = \sum_{s=0}^{N-1} \frac{s\,x^s}{Q}

      psp_s probability of ss open links

    8. Derivative identity

      sxs=x ddxxss x^s = x\,\frac{d}{dx} x^s

      A trick to bring down the power of xx.

    9. Expectation value, derivative form

      ⟨s⟩=1Q xddxQ\langle s\rangle = \frac{1}{Q}\,x\frac{d}{dx}Q
    10. Expected number of open links

      ⟨s⟩=NxNxN−1−xx−1\langle s\rangle = \frac{Nx^N}{x^N-1} - \frac{x}{x-1}
    11. Fraction unzipped

      ⟨s⟩N=xNxN−1−xN(x−1)\frac{\langle s\rangle}{N} = \frac{x^N}{x^N-1} - \frac{x}{N(x-1)}

      Plot it against xx to see the transition sharpen as NN grows.

    12. Critical temperature of unzipping

      Tc=εkln⁡gT_c = \frac{\varepsilon}{k\ln g}

      TcT_c critical temperature, kk Boltzmann constant

      Where x=1x = 1.

    13. Partition function, continuous in T

      Q=∑s=0N−1gse−sε/kTQ = \sum_{s=0}^{N-1} g^s e^{-s\varepsilon/kT}

      A finite sum of smooth terms is smooth. So no jump can appear unless N→∞N \to \infty.

    Simulations: Monte Carlo, molecular dynamics and the Ising magnet

    1. Thermodynamic average by importance sampling

      ⟨X⟩=∑j=1N~trialsXj e−βEj∑j=1N~trialse−βEj\langle X\rangle = \frac{\sum_{j=1}^{\tilde N_{\text{trials}}} X_j\, e^{-\beta E_j}}{\sum_{j=1}^{\tilde N_{\text{trials}}} e^{-\beta E_j}}

      XjX_j value of XX in trial configuration jj, N~trials\tilde N_{\text{trials}} number of trials, EjE_j energy of configuration jj

      If configurations are generated with probability e−βEe^{-\beta E}, the weights are already in the sampling, so you just average XjX_j.

    2. Metropolis acceptance probability for an uphill move

      exp⁡(−βΔE)\exp(-\beta\Delta E)

      ΔE\Delta E energy change of the move, β=1/kT\beta = 1/kT

      Moves that lower the energy are always accepted. Moves that raise it are accepted with this probability.

    3. Ising energy

      E=−B∑iσi−J∑⟨ij⟩σiσjE = -B\sum_i\sigma_i - J\sum_{\langle ij\rangle}\sigma_i\sigma_j

      σi=±1\sigma_i = \pm 1 spin, BB applied field, JJ nearest-neighbour coupling, ⟨ij⟩\langle ij\rangle nearest-neighbour pairs

      Aligned neighbours lower the energy when JJ is positive.

    4. Net magnetisation

      m=∑iσim = \sum_i \sigma_i

      mm magnetisation, σi\sigma_i spin at site ii

    5. Magnetic phase behaviour

      m∼0 (T≫Tc)∣m∣∼1 (T≪Tc)m \sim 0 \ (T \gg T_c) \qquad |m| \sim 1 \ (T \ll T_c)

      TcT_c critical temperature

      Hot magnet: disordered. Cold magnet: aligned.

    6. Newton's law for molecular dynamics

      Fi=miai=−∂U∂ri\mathbf{F}_i = m_i\mathbf{a}_i = -\frac{\partial U}{\partial \mathbf{r}_i}

      Fi\mathbf{F}_i force on atom ii, mim_i mass, ai\mathbf{a}_i acceleration, UU potential energy, ri\mathbf{r}_i position

      Force is minus the slope of the potential energy.

    Mean-field theory: replace the crowd with its average and a transition appears

    1. Ising energy

      E=−B∑iσi−J∑⟨ij⟩σiσjE = -B\sum_i\sigma_i - J\sum_{\langle ij\rangle}\sigma_i\sigma_j

      σi=±1\sigma_i = \pm 1 spin, BB applied field, JJ nearest-neighbour coupling

    2. Net magnetisation

      m=∑iσim = \sum_i\sigma_i

      mm magnetisation, σi\sigma_i spin at site ii

    3. Non-interacting Ising energy

      E=−B∑iσiE = -B\sum_i\sigma_i

      No JJ term, so spins are independent.

    4. Single-spin partition function

      q=eβB+e−βBq = e^{\beta B} + e^{-\beta B}

      β=1/kT\beta = 1/kT

      Two states, up and down.

    5. Spin probabilities

      p↑=eβBeβB+e−βBp↓=e−βBeβB+e−βBp_\uparrow = \frac{e^{\beta B}}{e^{\beta B}+e^{-\beta B}} \qquad p_\downarrow = \frac{e^{-\beta B}}{e^{\beta B}+e^{-\beta B}}

      p↑p_\uparrow probability a spin is up, p↓p_\downarrow probability it is down

    6. Non-interacting magnetisation

      m=Ntanh⁡(βB)m = N\tanh(\beta B)

      N↑N_\uparrow and N↓N_\downarrow numbers of up and down spins, m=N↑−N↓m = N_\uparrow - N_\downarrow

    7. Mean-field energy

      E=−B∑iσi−J∑iσi⟨σ⟩dE = -B\sum_i\sigma_i - J\sum_i\sigma_i\langle\sigma\rangle d

      ⟨σ⟩\langle\sigma\rangle average spin, dd number of nearest neighbours

      Each neighbour spin is replaced by the average spin.

    8. Mean-field energy, single effective field

      E=−(B+J⟨σ⟩d)∑iσiE = -\left(B + J\langle\sigma\rangle d\right)\sum_i\sigma_i

      Looks like the non-interacting model with a bigger field.

    9. Mean-field energy at B = 0

      E=−JmdN∑iσiE = -\frac{Jmd}{N}\sum_i\sigma_i

      Uses ⟨σ⟩=m/N\langle\sigma\rangle = m/N.

    10. Mean-field self-consistency

      mN=tanh⁡(βJd mN)\frac{m}{N} = \tanh\left(\beta J d\,\frac{m}{N}\right)

      mm is on both sides, so solve it graphically.

    11. Reduced form

      kTJd x=tanh⁡(x)x=βJd mN\frac{kT}{Jd}\,x = \tanh(x) \qquad x = \frac{\beta J d\, m}{N}

      A straight line of slope kT/JdkT/Jd meets the tanh curve.

    12. Critical temperature (mean field)

      Tc=JdkT_c = \frac{Jd}{k}

      TcT_c Curie temperature

    The van der Waals gas: two corrections to the ideal gas that predict liquids

    1. Interaction-averaged potential per molecule

      Upot,mol=−a NVU_{\text{pot,mol}} = -a\,\frac{N}{V}

      aa attraction constant, N/VN/V number density

      Mean-field: a molecule feels an energy proportional to how crowded its surroundings are.

    2. Total attractive potential energy

      Upot=−a N2VU_{\text{pot}} = -a\,\frac{N^2}{V}

      Multiply by NN molecules.

    3. Pressure from internal energy

      p=−(∂U∂V)Sp = -\left(\frac{\partial U}{\partial V}\right)_S

      From the first law dU=T dS−p dVdU = T\,dS - p\,dV at constant entropy.

    4. Attractive pressure

      ppot=−aN2V2p_{\text{pot}} = -\frac{aN^2}{V^2}

      Attraction lowers the pressure.

    5. van der Waals equation of state

      p=NkTV−Nb−aN2V2p = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2}

      bb excluded volume per molecule, aa attraction constant

      Excluded volume raises the pressure, attraction lowers it.

    6. van der Waals equation, product form

      (p+aN2V2)(V−Nb)=NkT\left(p + \frac{aN^2}{V^2}\right)(V - Nb) = NkT

      Same equation. Looks like pV=NkTpV = NkT with pp and VV corrected.

    7. Gibbs identity

      dG=−S dT+V dP+μ dNdG = -S\,dT + V\,dP + \mu\,dN

      μ\mu chemical potential

      Appendix, eq. 5.53.

    8. Gibbs derivative

      (∂G∂V)N,T=V(∂P∂V)N,T\left(\frac{\partial G}{\partial V}\right)_{N,T} = V\left(\frac{\partial P}{\partial V}\right)_{N,T}

      Appendix, eq. 5.54. At fixed NN and TT, dG=V dPdG = V\,dP.

    9. Gibbs derivative from the vdW equation

      (∂G∂V)N,T=−NkTV(V−Nb)2+2aN2V2\left(\frac{\partial G}{\partial V}\right)_{N,T} = -\frac{NkTV}{(V-Nb)^2} + \frac{2aN^2}{V^2}

      Appendix, eq. 5.55.

    10. Gibbs energy of a vdW fluid

      G=−NkTln⁡(V−Nb)+(NkT)(Nb)V−Nb−2aN2V+c(T)G = -NkT\ln(V-Nb) + \frac{(NkT)(Nb)}{V-Nb} - \frac{2aN^2}{V} + c(T)

      c(T)c(T) integration constant that depends on TT

      Appendix, eq. 5.56. Lets you plot GG at fixed TT.

    Chemical potential decides which phase wins, and phase diagrams map the winners

    1. Total differential of G

      dG=(∂G∂T)p,nidT+(∂G∂p)T,nidp+∑i(∂G∂ni)T,pdnidG = \left(\frac{\partial G}{\partial T}\right)_{p,n_i} dT + \left(\frac{\partial G}{\partial p}\right)_{T,n_i} dp + \sum_i \left(\frac{\partial G}{\partial n_i}\right)_{T,p} dn_i

      GG Gibbs energy, TT temperature, pp pressure, nin_i moles of species ii

      The slide labels the three slopes as −S-S, VV and μi\mu_i.

    2. Chemical potential as molar Gibbs energy

      μi=ΔGm,i\mu_i = \Delta G_{m,i}

      μi\mu_i chemical potential, ΔGm,i\Delta G_{m,i} molar Gibbs energy of ii

      Gibbs energy added per mole of ii added.

    3. Gibbs energy change at constant T and p

      dG=∑iμi dnidG = \sum_i \mu_i\, dn_i

      μi\mu_i chemical potential, nin_i amount of ii

    4. Transport criterion between regions I and II

      dG=(μiII−μiI)dni<0dG = \left(\mu_i^{II} - \mu_i^{I}\right) dn_i < 0

      μiI\mu_i^{I}, μiII\mu_i^{II} chemical potentials in regions I and II

      Species ii moves from I to II only if μiI\mu_i^{I} is higher than μiII\mu_i^{II}.

    5. Gibbs energy differential

      dG=V dp−S dTdG = V\,dp - S\,dT

      VV volume, SS entropy

      Compare the sign of SS here with the slide on S=(∂G/∂T)pS = (\partial G/\partial T)_p in Lecture 23.

    6. Molar chemical potential differential

      dμ=Vm dp−Sm dTd\mu = V_m\,dp - S_m\,dT

      VmV_m molar volume, SmS_m molar entropy

    7. Slope of mu with T

      (∂μ∂T)p=−Sm<0\left(\frac{\partial \mu}{\partial T}\right)_p = -S_m < 0

      SmS_m molar entropy

      Why every mu against T curve slopes down.

    8. Slope of mu with p

      (∂μ∂p)T=Vm>0\left(\frac{\partial \mu}{\partial p}\right)_T = V_m > 0

      VmV_m molar volume

      Why raising pressure lifts every curve.

    9. Coexistence condition

      μα(p,T)=μβ(p,T)\mu_\alpha(p,T) = \mu_\beta(p,T)

      α\alpha, β\beta two phases

      On a boundary the two phases have equal chemical potential.

    10. Coexistence differential

      dμα=dμβd\mu_\alpha = d\mu_\beta

      Stay on the boundary: the two potentials change together.

    11. Expanded coexistence

      −Sm,α dT+Vm,α dp=−Sm,β dT+Vm,β dp-S_{m,\alpha}\,dT + V_{m,\alpha}\,dp = -S_{m,\beta}\,dT + V_{m,\beta}\,dp
    12. Rearranged coexistence

      (Vm,β−Vm,α)dp=(Sm,β−Sm,α)dT\left(V_{m,\beta} - V_{m,\alpha}\right) dp = \left(S_{m,\beta} - S_{m,\alpha}\right) dT
    13. Clausius-Clapeyron

      dpdT=ΔSm,trΔVm,tr\frac{dp}{dT} = \frac{\Delta S_{m,tr}}{\Delta V_{m,tr}}

      ΔSm,tr\Delta S_{m,tr}, ΔVm,tr\Delta V_{m,tr} molar entropy and volume change on the transition

      The slope of a phase boundary on a pT diagram.

    07 Heat capacity, kinetics and glasses

    Heat capacity and calorimetry: how much heat it takes, and how transitions show up

    1. Heat capacity

      C=dQdTC = \frac{dQ}{dT}

      CC heat capacity (J K−1^{-1}), QQ heat, TT temperature

      Heat in, per kelvin of warming.

    2. Specific heat capacity

      Cs=CmC_s = \frac{C}{m}

      CsC_s specific heat capacity, mm mass (the slide says per gram)

      Use it to compare materials of different size.

    3. Molar heat capacity

      Cm=CnC_m = \frac{C}{n}

      CmC_m molar heat capacity, nn amount in mol

    4. Entropy from G

      S=−(∂G∂T)pS = -\left(\frac{\partial G}{\partial T}\right)_p

      SS entropy, GG Gibbs energy, TT temperature, pp pressure

      The Lecture 23 slide prints this without the minus sign, S=(∂G/∂T)pS = (\partial G/\partial T)_p. The standard relation has the minus sign. It follows from dG=V dp−S dTdG = V\,dp - S\,dT, and Lecture 22 uses it as (∂μ/∂T)p=−Sm(\partial\mu/\partial T)_p = -S_m.

    5. Volume from G

      V=(∂G∂p)TV = \left(\frac{\partial G}{\partial p}\right)_T

      VV volume, pp pressure

    6. Heat capacity at constant p from G

      Cp=(dQdT)p=T(dSdT)p=−T(d2GdT2)pC_p = \left(\frac{dQ}{dT}\right)_p = T\left(\frac{dS}{dT}\right)_p = -T\left(\frac{d^2G}{dT^2}\right)_p

      CpC_p heat capacity at constant pp, QQ heat

      The slide prints the last form as +T(d2G/dT2)p+T(d^2G/dT^2)_p, carrying over the missing minus sign in SS. With the standard sign of SS it is −T(d2G/dT2)p-T(d^2G/dT^2)_p. The point for the exam is that CpC_p is a second derivative of GG.

    7. DSC relation

      Heat flowT scan rate=dQ/dtdT/dt=C\frac{\text{Heat flow}}{T\ \text{scan rate}} = \frac{dQ/dt}{dT/dt} = C

      tt time, dT/dtdT/dt scan rate

      At a constant scan rate, the thermogram is a plot of heat capacity.

    Nucleation and growth: a new phase must first beat a surface-energy barrier

    1. Gibbs energy change on nucleation

      ΔG=Gf−Gi=−V(GVL−GVS)+Aγ=−V ΔGV+Aγ\Delta G = G_f - G_i = -V\left(G_V^{L} - G_V^{S}\right) + A\gamma = -V\,\Delta G_V + A\gamma

      VV volume of the nucleus, AA interfacial area, γ\gamma interfacial energy (always positive), ΔGV\Delta G_V volumetric driving force

      Gain from becoming solid (negative term) against the cost of making a surface (positive term).

    2. Sphere volume

      V=43πr3V = \frac{4}{3}\pi r^3

      rr radius

    3. Sphere area

      A=4πr2A = 4\pi r^2
    4. Delta G for a spherical nucleus

      ΔG=−43πr3 ΔGV+4πr2γ\Delta G = -\frac{4}{3}\pi r^3\,\Delta G_V + 4\pi r^2\gamma

      The r3r^3 term wins at large rr, the r2r^2 term at small rr.

    5. Critical radius

      r∗=2γΔGVr^* = \frac{2\gamma}{\Delta G_V}

      r∗r^* critical nucleus radius

      Where dΔG/dr=0d\Delta G/dr = 0.

    6. Nucleation barrier

      ΔG∗=16πγ33 ΔGV2\Delta G^* = \frac{16\pi\gamma^3}{3\,\Delta G_V^2}

      ΔG∗\Delta G^* height of the barrier

    7. Volumetric driving force

      ΔGV=GVL−GVS\Delta G_V = G_V^{L} - G_V^{S}

      GVLG_V^L, GVSG_V^S Gibbs energies of liquid and solid (the slide says molar)

    8. Driving force at the melting point

      ΔGV=ΔHVm−Tm ΔSVm=0\Delta G_V = \Delta H_V^{m} - T_m\,\Delta S_V^{m} = 0

      ΔHVm\Delta H_V^m, ΔSVm\Delta S_V^m enthalpy and entropy of transition, TmT_m melting temperature

      At TmT_m liquid and solid are at equilibrium, so there is no driving force.

    9. Entropy of transition

      ΔSVm=ΔHVmTm\Delta S_V^{m} = \frac{\Delta H_V^{m}}{T_m}
    10. Driving force near the melting point

      ΔGV≈ΔHVm ΔTTm\Delta G_V \approx \frac{\Delta H_V^{m}\,\Delta T}{T_m}

      ΔT=Tm−T∗\Delta T = T_m - T^* undercooling

      Assumes HH and SS do not depend on TT for small undercooling.

    11. Critical radius against undercooling

      r∗=2γTmΔHVm 1ΔTr^* = \frac{2\gamma T_m}{\Delta H_V^{m}}\,\frac{1}{\Delta T}
    12. Barrier against undercooling

      ΔG∗=16πγ3Tm23(ΔHVm)2 1ΔT2\Delta G^* = \frac{16\pi\gamma^3 T_m^2}{3\left(\Delta H_V^{m}\right)^2}\,\frac{1}{\Delta T^2}
    13. Nucleation rate (Arrhenius)

      N˙∼vdexp⁡(−ΔG∗kBT)∼exp⁡(−QdkBT)exp⁡(−AΔT2)\dot{N} \sim v_d \exp\left(-\frac{\Delta G^*}{k_B T}\right) \sim \exp\left(-\frac{Q_d}{k_B T}\right)\exp\left(-\frac{A}{\Delta T^2}\right)

      vdv_d attempt frequency, kBk_B Boltzmann constant, QdQ_d diffusion-like activation energy, AA constant

      First exponential: atomic mobility. Second: the barrier.

    08 Soft matter and liquids

    Soft matter and the kT yardstick

    1. Boltzmann distribution

      pi∝e−εi/kTp_i \propto e^{-\varepsilon_i/kT}

      pip_i probability of state ii, εi\varepsilon_i energy of the state (J), kk Boltzmann constant (1.38×10−231.38\times10^{-23} J K−1^{-1}), TT temperature (K)

      Use it to see how likely a state is. Higher energy means a lower chance.

    2. Population ratio

      NiNj=pipj=e−(εi−εj)/kT\frac{N_i}{N_j} = \frac{p_i}{p_j} = e^{-(\varepsilon_i-\varepsilon_j)/kT}

      NiN_i, NjN_j numbers of molecules in states ii and jj, other symbols as above

      Use it to compare two states. A gap much bigger than kTkT means very few molecules reach the upper state.

    Self-organisation and mesophases between liquid and crystal

    1. Director

      nˉ\bar{n}

      nˉ\bar{n} average direction of the particle axis

      A symbol rather than an equation. It says which way the rods point on average.

    The order parameter and the hard-particle model of nematics

    1. Order parameter

      S=12⟨3cos⁡2θ−1⟩S = \frac{1}{2}\left\langle 3\cos^2\theta - 1\right\rangle

      θ\theta angle between the molecular long axis and the director nˉ\bar{n}, ⟨⟩\langle\rangle average over all molecules

      S is 1 for perfect alignment and 0 for isotropic. It sums up a messy crowd of angles in one number.

    2. Reduced energy unit

      e0=kBTe_0 = k_B T

      kBk_B Boltzmann constant, TT temperature

      Energy measured in units of the thermal energy.

    3. Reduced length unit

      l0=ρ−1/3l_0 = \rho^{-1/3}

      ρ\rho number density

      The typical spacing between particles.

    4. Reduced time unit

      t0=ρ−1/3mkBTt_0 = \rho^{-1/3}\sqrt{\frac{m}{k_B T}}

      ρ\rho density, mm particle mass (the slide does not define mm), kBk_B Boltzmann constant, TT temperature

      Roughly the time to travel one spacing at thermal speed.

    5. Gibbs free energy

      G=H−TS=U+PV−TSG = H - TS = U + PV - TS

      HH enthalpy, UU internal energy, PP pressure, VV volume, TT temperature, SS entropy. For hard particles U=0U = 0

      Minimise it for a system at fixed pressure and temperature.

    6. Helmholtz free energy

      F=U−TSF = U - TS

      UU internal energy, TT temperature, SS entropy

      Minimise it for a system at fixed volume.

    Excluded volume, entropy and Onsager theory of the nematic

    1. Translational entropy

      S=kBln⁡WS = k_B \ln W

      kBk_B Boltzmann constant, WW roughly the product of the accessible volumes for each particle

      More room for each particle means a bigger WW and more entropy.

    2. Excluded volume, isotropic rods

      Vexcl∼DL2V_{\text{excl}} \sim D L^2

      DD rod diameter, LL rod length

      A scaling, not an exact equation. Much larger than the volume of one rod.

    3. Excluded volume, aligned rods

      Vexcl∼D2LV_{\text{excl}} \sim D^2 L

      DD rod diameter, LL rod length

      A scaling. About 4 times the volume of one rod.

    4. Onsager isotropic limit

      concentration<4.2DL2(isotropic stable)\text{concentration} < \frac{4.2}{D L^2}\quad\text{(isotropic stable)}

      DD rod diameter, LL rod length

      Below this the isotropic phase is the stable one.

    5. Onsager nematic limit

      concentration>7.0DL2(nematic stable)\text{concentration} > \frac{7.0}{D L^2}\quad\text{(nematic stable)}

      DD rod diameter, LL rod length

      Above this the nematic is stable. Between 4.2 and 7.0 the phases coexist.

    6. Volume of a rod

      Vrod=πD2L4V_{\text{rod}} = \frac{\pi D^2 L}{4}

      DD rod diameter, LL rod length

      Used to turn concentrations into a packing fraction.

    7. Reduced concentration

      c=14πL2Dρc = \frac{1}{4}\pi L^2 D \rho

      LL length, DD diameter, ρ\rho number density

      From the caption of the slide 16 theory versus simulation plot.

    Random walks and diffusion: Fick's two laws

    1. Fick's first law

      Jx=−DAdcdxJ_x = -D A \frac{dc}{dx}

      JxJ_x diffusion flux (mol s−1^{-1}), AA surface area, dc/dxdc/dx concentration gradient, DD diffusion coefficient (not defined on the slide)

      Use it to get how fast material crosses a surface when you know the gradient.

    2. Fick's second law

      dcdt=Dd2cdx2\frac{dc}{dt} = D \frac{d^2 c}{dx^2}

      cc concentration, tt time, xx position, DD diffusion coefficient

      Use it to get how the concentration at one point changes as time passes.

    09 Chemical kinetics

    Rate laws, half-lives and diffusion-controlled reactions

    1. First-order rate law

      d[A]dt=−k[A]\frac{d[A]}{dt} = -k[A]

      [A][A] concentration of A, kk rate coefficient, tt time

      The rate of loss is proportional to how much A is left.

    2. First-order integrated rate law

      [A]=[A]0exp⁡(−kt)ln⁡[A]=−kt+ln⁡[A]0[A] = [A]_0 \exp(-kt) \qquad \ln[A] = -kt + \ln[A]_0

      [A]0[A]_0 initial concentration

      Plot ln⁡[A]\ln[A] against tt and the gradient is −k-k.

    3. Integration step for first order

      ∫d[A][A]=∫−k dt\int \frac{d[A]}{[A]} = \int -k\,dt

      [A][A] concentration, kk rate coefficient, tt time

      The step that turns the differential law into the integrated one.

    4. Hydrogen-bromine type rate law

      d[HBr]dt=ka[H2][Br2]3/2[Br2]+kb[HBr]\frac{d[\mathrm{HBr}]}{dt} = \frac{k_a[\mathrm{H}_2][\mathrm{Br}_2]^{3/2}}{[\mathrm{Br}_2] + k_b[\mathrm{HBr}]}

      kak_a, kbk_b rate coefficients

      An example of a differential rate law that is not a simple order.

    5. Zeroth-order law, integrated form and half-life

      d[A]dt=−k[A]=[A]0−ktt1/2=[A]02k\frac{d[A]}{dt} = -k \qquad [A] = [A]_0 - kt \qquad t_{1/2} = \frac{[A]_0}{2k}

      kk rate coefficient, [A]0[A]_0 initial concentration

      Test by plotting [A][A] against tt.

    6. Second-order law, integrated form and half-life

      d[A]dt=−k[A]21[A]=1[A]0+ktt1/2=1k[A]0\frac{d[A]}{dt} = -k[A]^2 \qquad \frac{1}{[A]} = \frac{1}{[A]_0} + kt \qquad t_{1/2} = \frac{1}{k[A]_0}

      kk rate coefficient, [A]0[A]_0 initial concentration

      Test by plotting 1/[A]1/[A] against tt.

    7. First-order half-life

      t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k}

      kk rate coefficient

      It does not depend on the starting concentration.

    8. Einstein-Smoluchowski time

      time=x22D\text{time} = \frac{x^2}{2D}

      xx distance, DD diffusion coefficient

      The average time to diffuse a distance xx.

    9. Exchange time

      τ=(2r)22D=2r2D\tau = \frac{(2r)^2}{2D} = \frac{2r^2}{D}

      rr molecular radius, DD diffusion coefficient

      The time for a molecule to swap places with a neighbour.

    10. Encounter probability

      P∼12 ngVgV=12 ng4π3r3VP \sim 12\,\frac{n_g V_g}{V} = 12\,\frac{n_g \frac{4\pi}{3} r^3}{V}

      ngn_g number of green molecules, VgV_g volume of one green molecule, VV volume, rr radius

      About 12 neighbours in 3-d, times the fraction of sites that are green.

    11. Successful collisions, one radius

      Z=noΔtτP=nongV 4π 2r D ΔtZ = n_o \frac{\Delta t}{\tau} P = \frac{n_o n_g}{V}\, 4\pi\, 2r\, D\, \Delta t

      non_o, ngn_g numbers of orange and green molecules, Δt\Delta t time interval, τ\tau exchange time, PP encounter probability

      Exchanges in Δt\Delta t times the chance each exchange meets a green molecule.

    12. Successful collisions, two radii

      Z=nongV 4π (ro+rg)(Do+Dg) ΔtZ = \frac{n_o n_g}{V}\, 4\pi\,(r_o + r_g)(D_o + D_g)\,\Delta t

      ror_o, rgr_g radii, DoD_o, DgD_g diffusion coefficients

      Lets both partners diffuse and have different sizes.

    13. Diffusion-controlled rate law

      dcodt=−k cocg\frac{dc_o}{dt} = -k\,c_o c_g

      coc_o, cgc_g concentrations of orange and green

      A second-order differential rate law.

    14. Diffusion-controlled rate coefficient

      k=4π (ro+rg)(Do+Dg) NAk = 4\pi\,(r_o + r_g)(D_o + D_g)\,N_A

      ro+rgr_o + r_g distance at which reactants touch, Do+DgD_o + D_g sum of diffusion coefficients, NAN_A Avogadro constant

      Bigger partners that diffuse faster give a bigger kk.

    15. Arrhenius equation (introduced here)

      k=Aexp⁡(−Ea/RT)k = A\exp(-E_a/RT)

      AA pre-exponential factor, EaE_a activation energy, RR gas constant, TT temperature

      Slows the diffusion-controlled rate by a temperature-sensitive factor. Covered in the next topic.

    Activation barriers, transition state theory and catalysis

    1. Arrhenius equation

      k=Aexp⁡(−Ea/RT)k = A\exp(-E_a/RT)

      AA pre-exponential factor, EaE_a activation energy, RR gas constant, TT temperature

      Use it to see how a rate coefficient changes with temperature.

    2. Hydrogen exchange reaction

      Ha+Hb-Hc→Ha-Hb+Hc\mathrm{H_a + H_b\text{-}H_c \rightarrow H_a\text{-}H_b + H_c}

      A hydrogen atom attacks a hydrogen molecule and swaps places. A chemical equation, not a rate law.

      The simplest bimolecular reaction, studied by Eyring and Polanyi in 1931.

    3. Transition-state equilibrium constant

      K‡=exp⁡(−ΔG‡/RT)=exp⁡(−ΔH‡/RT)exp⁡(ΔS‡/R)K^\ddagger = \exp(-\Delta G^\ddagger/RT) = \exp(-\Delta H^\ddagger/RT)\exp(\Delta S^\ddagger/R)

      ΔG‡\Delta G^\ddagger Gibbs free energy of activation, ΔH‡\Delta H^\ddagger enthalpy part, ΔS‡\Delta S^\ddagger entropy part, RR gas constant, TT temperature

      A smaller barrier or a bigger entropy of activation puts more molecules at the transition state.

    4. Rate coefficient for leaving the transition state

      kp=kBThk_p = \frac{k_B T}{h}

      kBk_B Boltzmann constant, TT temperature, hh Planck constant

      A very fast rate for the dynamics of moving off the transition state.

    5. TST rate constant

      kTST=kpK‡2=kBT2hexp⁡(ΔS‡/R)exp⁡(−ΔH‡/RT)k_{\mathrm{TST}} = \frac{k_p K^\ddagger}{2} = \frac{k_B T}{2h}\exp(\Delta S^\ddagger/R)\exp(-\Delta H^\ddagger/RT)

      kpk_p rate coefficient for leaving the transition state, K‡K^\ddagger transition-state equilibrium constant

      Combine the equilibrium with the escape rate.

    6. TST in Arrhenius form

      kTST=Aexp⁡(−Ea/RT)A=kBT2hexp⁡(ΔS‡/R)k_{\mathrm{TST}} = A\exp(-E_a/RT) \qquad A = \frac{k_B T}{2h}\exp(\Delta S^\ddagger/R)

      EaE_a is often written for ΔH‡\Delta H^\ddagger (activation energy)

      Shows that TST gives back the Arrhenius temperature dependence.

    Elementary steps, mechanisms and detailed balance

    1. First-order rate law

      d[A]dt=−k[A]\frac{d[A]}{dt} = -k[A]

      [A][A] concentration of A, kk rate coefficient, tt time

      Also the rate law for a unimolecular elementary step.

    2. First-order integrated rate law

      [A]=[A]0exp⁡(−kt)[A] = [A]_0 \exp(-kt)

      [A]0[A]_0 initial concentration

      The concentration decays exponentially.

    3. Integration step

      ∫d[A][A]=∫−k dt\int \frac{d[A]}{[A]} = \int -k\,dt

      [A][A] concentration, kk rate coefficient, tt time

      The step used to get the integrated law from the differential one.

    4. Hydrogen-bromine rate law

      d[HBr]dt=ka[H2][Br2]3/2[Br2]+kb[HBr]\frac{d[\mathrm{HBr}]}{dt} = \frac{k_a[\mathrm{H}_2][\mathrm{Br}_2]^{3/2}}{[\mathrm{Br}_2] + k_b[\mathrm{HBr}]}

      kak_a, kbk_b rate coefficients

      The observed law for H2_2 + Br2_2 giving 2HBr. It is not a simple order, so it must come from a mechanism.

    5. Unimolecular elementary step

      d[A]dt=−k[A]\frac{d[A]}{dt} = -k[A]

      [A][A] concentration, kk rate coefficient

      One molecule reacts, so the rate is first order in A.

    6. Bimolecular elementary step

      d[A]dt=−k[A][B]\frac{d[A]}{dt} = -k[A][B]

      [A][A], [B][B] concentrations, kk rate coefficient

      Two molecules must meet, so the rate is first order in each.

    7. Steady-state condition

      d[NO]dt=0\frac{d[\mathrm{NO}]}{dt} = 0

      [NO][\mathrm{NO}] concentration of the intermediate in the lecture example

      A reactive intermediate stays at an unchanging concentration.

    8. Detailed balance, step i

      k1[NO]eq2=k−1[N2O2]eqk_1[\mathrm{NO}]^2_{eq} = k_{-1}[\mathrm{N_2O_2}]_{eq}

      k1k_1, k−1k_{-1} forward and backward rate coefficients, subscript eq means at equilibrium

      Forward and backward rates are equal for this step.

    9. Detailed balance, step ii

      k2[N2O2]eq[H2]eq=k−2[N2O]eq[H2O]eqk_2[\mathrm{N_2O_2}]_{eq}[\mathrm{H_2}]_{eq} = k_{-2}[\mathrm{N_2O}]_{eq}[\mathrm{H_2O}]_{eq}

      k2k_2, k−2k_{-2} forward and backward rate coefficients

      Forward and backward rates are equal for this step.

    10. Detailed balance, step iii

      k3[N2O]eq[H2]eq=k−3[N2]eq[H2O]eqk_3[\mathrm{N_2O}]_{eq}[\mathrm{H_2}]_{eq} = k_{-3}[\mathrm{N_2}]_{eq}[\mathrm{H_2O}]_{eq}

      k3k_3, k−3k_{-3} forward and backward rate coefficients

      Forward and backward rates are equal for this step.

    11. Overall equilibrium constant from rate coefficients

      k1k2k3k−1k−2k−3=[N2]eq[H2O]eq2[NO]eq2[H2]eq2=K\frac{k_1k_2k_3}{k_{-1}k_{-2}k_{-3}} = \frac{[\mathrm{N_2}]_{eq}[\mathrm{H_2O}]^2_{eq}}{[\mathrm{NO}]^2_{eq}[\mathrm{H_2}]^2_{eq}} = K

      KK equilibrium constant of the overall reaction

      The equilibrium constant is the ratio of the products of forward and backward rate coefficients.

    10 Surfaces and interfaces

    Surfaces cost energy: surface tension, work of cohesion and why drops are round

    1. Surface tension as a free-energy derivative

      γ=(∂G∂A)T,V\gamma = \left(\frac{\partial G}{\partial A}\right)_{T,V}

      γ\gamma surface tension, GG Gibbs free energy, AA surface area, TT temperature, VV volume

      It says how much free energy you pay for each extra bit of area.

    2. Work to create surface

      work=γ⋅ΔA\text{work} = \gamma \cdot \Delta A

      ΔA\Delta A new surface area created

      Bigger new area or bigger γ\gamma means more work.

    3. Force-per-length form

      work=γ⋅l⋅dx=F⋅dx⇒γ=F/l\text{work} = \gamma \cdot l \cdot dx = F \cdot dx \quad\Rightarrow\quad \gamma = F/l

      ll length of the movable rod, dxdx distance it moves, FF force

      Use it to read γ\gamma as a sideways pull along the surface.

    4. Work of cohesion, two materials

      WAB=γAV+γBV−γABW_{AB} = \gamma_{AV} + \gamma_{BV} - \gamma_{AB}

      WW work per unit area, γAV\gamma_{AV} and γBV\gamma_{BV} surface tensions against vapour, γAB\gamma_{AB} interfacial tension between A and B

      Two new surfaces are made, one old interface is lost.

    5. Work of cohesion, one material

      WAA=2γAVW_{AA} = 2\gamma_{AV}

      γAV\gamma_{AV} surface tension of A against vapour

      Splitting a block of A into two makes two identical new surfaces.

    6. Area of a subdivided cube

      A=6×d2=6 cm2=6⋅10−4 m2A=1018×600 nm2=600 m2A = 6 \times d^2 = 6\ \text{cm}^2 = 6\cdot10^{-4}\ \text{m}^2 \qquad A = 10^{18} \times 600\ \text{nm}^2 = 600\ \text{m}^2

      dd side of the cube

      The lecture example: 1 cm cube against a million-fold smaller cubes.

    7. Units of surface tension

      Jm2=N mm2=Nm\frac{J}{m^2} = \frac{N\,m}{m^2} = \frac{N}{m}

      Energy per area and force per length are the same unit.

    Wetting: spreading, contact angle, Young's equation and measuring surface tension

    1. Spreading parameter

      S=γSV−(γSL+γ)S = \gamma_{SV} - (\gamma_{SL} + \gamma)

      SS spreading parameter, γSV\gamma_{SV} solid-vapour surface energy, γSL\gamma_{SL} solid-liquid interfacial tension, γ\gamma liquid-vapour surface tension

      Negative means the drop stays a drop. Positive means it spreads into a film.

    2. Young's equation

      γSV=γSL+γcos⁡θE\gamma_{SV} = \gamma_{SL} + \gamma\cos\theta_E

      θE\theta_E equilibrium contact angle

      Use it to get the contact angle from the three surface energies.

    3. Young-Dupre (intermediate step)

      S=γSL+γcos⁡θE−(γSL+γ)=γ(cos⁡θE−1)S = \gamma_{SL} + \gamma\cos\theta_E - (\gamma_{SL} + \gamma) = \gamma(\cos\theta_E - 1)

      Substitute Young's equation for γSV\gamma_{SV} in SS.

    4. Young-Dupre equation

      S=γ(cos⁡θE−1)S = \gamma(\cos\theta_E - 1)

      S depends only on the liquid and the angle. Since cos is at most 1, S is at most 0 for a drop with an angle.

    5. Contact angle hysteresis

      Δθ=θA−θR\Delta\theta = \theta_A - \theta_R

      θA\theta_A advancing contact angle, θR\theta_R receding contact angle

      A measure of how much the contact line sticks.

    6. Wilhelmy surface tension force

      surface tension force=p γcos⁡θ\text{surface tension force} = p\,\gamma\cos\theta

      pp perimeter of the contact line, γ\gamma surface tension, θ\theta contact angle of the liquid

      The surface part of the force on a partly dipped plate or fibre.

    7. Capillary rise

      H=2γcos⁡θEρgRH = \frac{2\gamma\cos\theta_E}{\rho g R}

      HH height of liquid in the capillary, ρ\rho liquid density, gg gravitational acceleration (about 9.8 m/s2^2), RR capillary radius

      Narrow tube and high surface tension give a high rise.

    Friction and adhesion: why surfaces grip and how glues hold

    1. Adhesional friction force

      Fad=A sF_{ad} = A\,s

      AA actual area of contact, ss adhesive shear strength

      Friction from breaking the bonds at the real contact spots.

    2. Real contact area (plastic deformation)

      A=Wτ0A = \frac{W}{\tau_0}

      WW normal load, τ0\tau_0 yield strength of the interface

      More load gives more real contact.

    3. Friction coefficient

      μ=FW\mu = \frac{F}{W}

      μ\mu friction coefficient (no units), FF friction force, WW applied load

      A bigger μ\mu means a stickier pair of surfaces.

    4. Work of adhesion

      WAB=γAV+γBV−γABW_{AB} = \gamma_{AV} + \gamma_{BV} - \gamma_{AB}

      WABW_{AB} work of adhesion per unit area, γAV\gamma_{AV}, γBV\gamma_{BV} and γAB\gamma_{AB} interfacial energies

      The ideal, reversible upper bound for the strength of a joint.

    Adsorption onto solids: physisorption, chemisorption, thin films and self-assembled monolayers

    1. Depletion adsorption

      nads=([X]initial−[X]eq)Vn_{ads} = \left([X]_{initial} - [X]_{eq}\right)V

      nadsn_{ads} amount adsorbed, [X]initial[X]_{initial} and [X]eq[X]_{eq} solute concentration before and after, VV solution volume

      What the surface took is what the solution lost.

    2. Disulfide adsorption on gold

      RS-SR+Aun0→RS−Au+⋅Aun0RS\text{-}SR + Au^0_n \rightarrow RS^- Au^+ \cdot Au^0_n

      RR alkyl group, Aun0Au^0_n gold surface atoms

      Probably oxidative addition of the S-S bond to the gold. The arrows are not in the extracted text and are written here in the usual way.

    3. Thiol adsorption on gold

      R-S-H+Aun0→R-S−Au+⋅Aun0+12H2R\text{-}S\text{-}H + Au^0_n \rightarrow R\text{-}S^- Au^+ \cdot Au^0_n + \tfrac{1}{2}H_2

      RR alkyl group

      A thiol loses its hydrogen as 12H2\tfrac12 H_2 as it binds.

    4. Octadecyltrichlorosilane (OTS)

      Cl3-Si-(CH2)17-CH3\mathrm{Cl_3\text{-}Si\text{-}(CH_2)_{17}\text{-}CH_3}

      The head group reacts with surface silanols and the long chain sticks out.

    Langmuir and BET isotherms: how much gas a surface holds, and how big the surface is

    1. Langmuir equilibrium

      X+S0⇌S1X + S_0 \rightleftharpoons S_1

      XX adsorbing species, S0S_0 bare surface sites, S1S_1 occupied sites

      Adsorption treated as one simple equilibrium.

    2. Langmuir coverage, solute

      θ=nadsnmono=K1[X]1+K1[X]\theta = \frac{n_{ads}}{n_{mono}} = \frac{K_1[X]}{1+K_1[X]}

      θ\theta fractional coverage, nadsn_{ads} amount adsorbed, nmonon_{mono} amount for a monolayer, K1K_1 equilibrium constant, [X][X] solute concentration

      Use it for adsorption from solution.

    3. Langmuir coverage, gas

      θ=nadsnmono=K1p1+K1p\theta = \frac{n_{ads}}{n_{mono}} = \frac{K_1 p}{1+K_1 p}

      pp gas pressure

      Low pp: θ\theta rises almost linearly. High pp: θ\theta levels off at 1.

    4. Linearised Langmuir, solute

      [X]nads=[X]nmono+1K1nmono\frac{[X]}{n_{ads}} = \frac{[X]}{n_{mono}} + \frac{1}{K_1 n_{mono}}

      A straight line in [X][X]. Slope 1/nmono1/n_{mono}, intercept 1/(K1nmono)1/(K_1 n_{mono}).

    5. Linearised Langmuir, gas

      pnads=pnmono+1K1nmono\frac{p}{n_{ads}} = \frac{p}{n_{mono}} + \frac{1}{K_1 n_{mono}}

      Plot p/nadsp/n_{ads} against pp to get nmonon_{mono} and K1K_1.

    6. Worked example: monolayer area of N2

      5.0×10−3 g28 g mol−1=0.178×10−3 mol;0.178×10−3 mol×6.022×1023=1.08×1020;1.08×1020×0.162×10−18 m2=17.4 m2\frac{5.0\times10^{-3}\ \text{g}}{28\ \text{g mol}^{-1}} = 0.178\times10^{-3}\ \text{mol};\quad 0.178\times10^{-3}\ \text{mol}\times6.022\times10^{23} = 1.08\times10^{20};\quad 1.08\times10^{20}\times0.162\times10^{-18}\ \text{m}^2 = 17.4\ \text{m}^2

      Mass of monolayer to moles, to molecules, to area.

    7. Free energy of condensation

      ΔG∘=−RTln⁡K=ΔH∘−TΔS∘\Delta G^\circ = -RT\ln K = \Delta H^\circ - T\Delta S^\circ

      KK equilibrium constant for adsorbing into layers 2, 3, 4..., RR gas constant, TT temperature

      The BET constant KK is the same one that describes condensation.

    8. BET isotherm

      θ=nadsnmono=cP(1−P)(1+(c−1)P)\theta = \frac{n_{ads}}{n_{mono}} = \frac{cP}{(1-P)\left(1 + (c-1)P\right)}

      P=p/p0P = p/p^0 relative pressure, c=K1/Kc = K_1/K BET parameter

      θ\theta may exceed 1 because layers can pile up.

    9. Linearised BET (moles)

      pnads(p0−p)=1c nmono+c−1c nmono pp0\frac{p}{n_{ads}\left(p^0 - p\right)} = \frac{1}{c\,n_{mono}} + \frac{c-1}{c\,n_{mono}}\,\frac{p}{p^0}

      Form y=b+mxy = b + mx: y=p/[nads(p0−p)]y = p/[n_{ads}(p^0-p)], x=p/p0x = p/p^0, intercept b=1/(c nmono)b = 1/(c\,n_{mono}), slope m=(c−1)/(c nmono)m = (c-1)/(c\,n_{mono})

      A straight line gives both cc and nmonon_{mono}.

    10. Linearised BET (volume)

      pVads(p0−p)=1c Vmono+c−1c Vmono pp0\frac{p}{V_{ads}\left(p^0 - p\right)} = \frac{1}{c\,V_{mono}} + \frac{c-1}{c\,V_{mono}}\,\frac{p}{p^0}

      The same line written with gas volume instead of moles.

    11. Ideal gas equation

      pV=nRTpV = nRT

      To first approximation, converts between volume VV and moles nn.

    12. Worked example: BET surface area of silica

      c−1c nmono=319 mol−1; 1c nmono=3.35 mol−1; c=319×13.35+1=96.2; nmono=196.2×3.35=3.10×10−3 mol; Area=3.10×10−3×0.162×10−18×6.02×1023=302 m2\frac{c-1}{c\,n_{mono}} = 319\ \text{mol}^{-1};\ \frac{1}{c\,n_{mono}} = 3.35\ \text{mol}^{-1};\ c = 319\times\frac{1}{3.35}+1 = 96.2;\ n_{mono} = \frac{1}{96.2\times3.35} = 3.10\times10^{-3}\ \text{mol};\ \text{Area} = 3.10\times10^{-3}\times0.162\times10^{-18}\times6.02\times10^{23} = 302\ \text{m}^2

      From a fitted slope and intercept to the surface area of the sample.

    The Gibbs adsorption isotherm: surface excess, surfactants and the cmc

    1. Surface excess

      Γi=niσA\Gamma_i = \frac{n_i^{\sigma}}{A}

      niσn_i^{\sigma} extra moles of ii at the interface, AA interface area; units mol m−2^{-2}

      How much extra solute sits on each square metre of surface.

    2. Gibbs isotherm (ideal solution)

      Γi=−1RT dγdln⁡ci=−ciRT dγdci\Gamma_i = \frac{-1}{RT}\,\frac{d\gamma}{d\ln c_i} = \frac{-c_i}{RT}\,\frac{d\gamma}{dc_i}

      γ\gamma surface tension, cic_i concentration of ii, RR gas constant, TT temperature

      Read Γ\Gamma off the slope of γ\gamma against ln⁡c\ln c.

    3. Gibbs isotherm (ionic surfactant)

      Γi=−ci+2RT dγdci+\Gamma_i = \frac{-c_i^{+}}{2RT}\,\frac{d\gamma}{dc_i^{+}}

      ci+c_i^{+} concentration of the surfactant ion

      The 2 appears because two ions adsorb and both chemical potentials change.

    4. Electroneutral interface

      ΓNa+=ΓCl−\Gamma_{Na^+} = \Gamma_{Cl^-}

      The interface must stay electrically neutral.

    5. Butanol surface tension fit

      Y=42.5+30.6 exp⁡ ⁣(−X139)Y = 42.5 + 30.6\,\exp\!\left(\frac{-X}{139}\right)

      YY surface tension (mN m−1^{-1}), XX butanol concentration (mol m−3^{-3}), at 298 K

      A fitted curve you can differentiate to get Γ\Gamma.

    6. Logarithm conversion

      log⁡10x=ln⁡x2.303\log_{10} x = \frac{\ln x}{2.303}

      Lets you convert between plots against log⁡C\log C and ln⁡c\ln c.

    7. DDA+ surface excess, no salt

      ΓDDA+=−12RT dγdln⁡[DDA+]\Gamma_{DDA^+} = \frac{-1}{2RT}\,\frac{d\gamma}{d\ln[DDA^+]}

      Printed with −1/(2RT)-1/(2RT) on slide 15, because two ions adsorb.

    8. Gibbs isotherm with NaCl present

      −dγ=ΓDDA+RT dln⁡[DDA+]+ΓNa+RT dln⁡[Na+]+ΓCl−RT dln⁡[Cl−]-d\gamma = \Gamma_{DDA^+}RT\,d\ln[DDA^+] + \Gamma_{Na^+}RT\,d\ln[Na^+] + \Gamma_{Cl^-}RT\,d\ln[Cl^-]

      Each species adds a term. With lots of salt the last two are negligible.

    9. DDA+ surface excess, with 0.20 M NaCl

      ΓDDA+=−1RT dγdln⁡[DDA+]\Gamma_{DDA^+} = \frac{-1}{RT}\,\frac{d\gamma}{d\ln[DDA^+]}

      Printed with −1/(RT)-1/(RT) on slide 16, because only the surfactant ion changes.

    10. Area per molecule

      A=1NA ΓA = \frac{1}{N_A\,\Gamma}

      NAN_A Avogadro constant, Γ\Gamma surface excess of the surfactant ion only; units m2^2 per molecule

      For a full monolayer, the average space each molecule takes.

    11. Micelle condition

      Γ=0;dγ=0\Gamma = 0; \quad d\gamma = 0

      Micelles are not surface active, so above the cmc the surface tension does not change.

    11 Metal-organic frameworks

    Thermal expansion and why cyanide frameworks can shrink when heated

    1. Volumetric coefficient of thermal expansion

      αV=1V(dVdT)\alpha_V = \frac{1}{V}\left(\frac{dV}{dT}\right)

      αV\alpha_V volumetric expansion coefficient, VV volume, TT temperature

      Fractional change in volume for each kelvin.

    2. Linear coefficient of thermal expansion

      αl=1l(dldT)\alpha_l = \frac{1}{l}\left(\frac{dl}{dT}\right)

      αl\alpha_l linear expansion coefficient, ll length

      Fractional change in length for each kelvin. Negative means the material shrinks on heating.

    3. Zn(CN)2

      α=−16.5×10−6 K−1\alpha = -16.5 \times 10^{-6}\,\mathrm{K^{-1}}

      α\alpha thermal expansion coefficient

      A clear negative value: the lattice contracts as it warms.

    4. Cd(CN)2 with CCl4 clathrate guest

      α=+10.0(2)×10−6 K−1\alpha = +10.0(2) \times 10^{-6}\,\mathrm{K^{-1}}

      GG guest molecule in the cavity

      With the guest present the framework expands normally.

    5. Desolvated Cd(CN)2 on cooling

      α=−33.4(5)×10−6 K−1\alpha = -33.4(5) \times 10^{-6}\,\mathrm{K^{-1}}

      With the guest gone the framework shows strong NTE.

    6. N(CH3)4CuZn(CN)4

      α≈−2.5, +0.2, +6.0 MK−1 at about 250, 300, 370 K\alpha \approx -2.5,\ +0.2,\ +6.0\ \mathrm{MK^{-1}}\ \text{at about 250, 300, 370 K}

      MK−1^{-1} means 10−610^{-6} K−1^{-1}

      Values read off labels on the lower panel of the slide. Close to zero over a wide range.

    7. ZnPt(CN)6 with water, below room temperature

      α=+2×10−6 K−1\alpha = +2 \times 10^{-6}\,\mathrm{K^{-1}}

      Near zero expansion while water fills the cubic holes.

    8. ZnPt(CN)6 framework NTE

      α=−3×10−6 K−1\alpha = -3 \times 10^{-6}\,\mathrm{K^{-1}}

      After water is removed, transverse vibrations give NTE.

    Tuning NTE: how the metal ion and crystal field stabilisation set bond strength

    1. NTE trend across M

      αMn>αFe>αCo>αNi<αCu<αZn≪αCd\alpha_{Mn} > \alpha_{Fe} > \alpha_{Co} > \alpha_{Ni} < \alpha_{Cu} < \alpha_{Zn} \ll \alpha_{Cd}

      αM\alpha_M NTE coefficient for metal M in MII^{II}PtIV^{IV}(CN)6_6

      Ni sits at the trough. Read it as: the stronger the M-N bond, the smaller the NTE.

    2. CFSE, Mn(II)

      CFSE=0\mathrm{CFSE} = 0

      d5^5 high spin

      Weak bonds.

    3. CFSE, Fe(II)

      CFSE=−0.4 Δo\mathrm{CFSE} = -0.4\,\Delta_o

      Δo\Delta_o octahedral splitting

      Partially weak bonds.

    4. CFSE, Co(II)

      CFSE=−0.8 Δo\mathrm{CFSE} = -0.8\,\Delta_o

      Partially strong bonds.

    5. CFSE, Ni(II)

      CFSE=−1.2 Δo\mathrm{CFSE} = -1.2\,\Delta_o

      Largest CFSE, strongest bonds.

    6. CFSE, Cu(II)

      CFSE=−0.6 Δo\mathrm{CFSE} = -0.6\,\Delta_o

      Intermediate.

    7. CFSE, Zn(II)

      CFSE=0\mathrm{CFSE} = 0

      Weak bonds.

    8. e_g and t_2g levels

      eg=+0.6 Δo,t2g=−0.4 Δoe_g = +0.6\,\Delta_o, \qquad t_{2g} = -0.4\,\Delta_o

      Δo\Delta_o octahedral splitting energy

      The labelled levels used to count CFSE.

    MOFs as sponges for gas: hydrogen, carbon dioxide and methane

    1. Fractional sorption (Langmuir)

      θ=KP1+KP\theta = \dfrac{KP}{1+KP}

      θ\theta fractional sorption, KK Langmuir constant, PP pressure

      The same shape as the gas Langmuir isotherm in the adsorption lecture.

    2. Langmuir constant

      K=eΔS∘/R e−ΔH∘/RTP∘K = \dfrac{e^{\Delta S^\circ/R}\,e^{-\Delta H^\circ/RT}}{P^\circ}

      ΔS∘\Delta S^\circ standard entropy change of adsorption, ΔH∘\Delta H^\circ standard enthalpy change, RR gas constant, TT temperature, P∘P^\circ standard pressure

      A more negative ΔH∘\Delta H^\circ makes KK bigger and the uptake easier.

    3. Entropy assumption

      ΔS∘≅−8R\Delta S^\circ \cong -8R

      The value used in the back-of-envelope calculation.

    4. Optimum interaction energy

      ΔHopt∘=TΔS∘+RT2ln⁡ ⁣(P1P2P∘2)=−13.6 kJ mol−1\Delta H^\circ_{opt} = T\Delta S^\circ + \dfrac{RT}{2}\ln\!\left(\dfrac{P_1 P_2}{P^{\circ 2}}\right) = -13.6\ \mathrm{kJ\,mol^{-1}}

      P1,P2P_1, P_2 practical pressure limits (1.5 and 100 bar), P∘P^\circ standard pressure, T=298T = 298 K

      The binding enthalpy that best balances filling at high pressure with emptying at low pressure.

    5. MOF-5 alpha-site binding (quantum, 0 K)

      ΔHads=−6.4 kJ mol−1\Delta H_{ads} = -6.4\ \mathrm{kJ\,mol^{-1}}

      Cup-shaped site of three Zn atoms.

    6. MOF-5 binding (experiment, 77 K)

      ΔHads=−3.8 to −4.8 kJ mol−1\Delta H_{ads} = -3.8\ \text{to}\ -4.8\ \mathrm{kJ\,mol^{-1}}

      Weaker than the 0 K simulation.

    7. Physisorption target for H2

      ΔH∼14 kJ mol−1 (ideal);ΔH∼4 to 6 kJ mol−1 (typical)\Delta H \sim 14\ \mathrm{kJ\,mol^{-1}}\ \text{(ideal)};\quad \Delta H \sim 4\text{ to }6\ \mathrm{kJ\,mol^{-1}}\ \text{(typical)}

      Typical MOFs bind H2 too weakly at room temperature.

    8. Ideal surface area

      >4000 m2 g−1> 4000\ \mathrm{m^2\,g^{-1}}

      Uptake scales with surface area.

    9. HKUST-1 CH4 binding enthalpies

      ΔH=−21.8 (windows); −9.4 (metal only); ∼−15 (with CH4-CH4) kJ mol−1\Delta H = -21.8\ \text{(windows)};\ -9.4\ \text{(metal only)};\ \sim -15\ \text{(with CH}_4\text{-CH}_4\text{)}\ \mathrm{kJ\,mol^{-1}}

      Three binding situations in the same material.

    10. HKUST-1 CH4 uptake

      230 cm3(STP) cm−3 at 35 bar;270 cm3(STP) cm−3 at 65 bar230\ \mathrm{cm^3(STP)\,cm^{-3}}\ \text{at 35 bar};\quad 270\ \mathrm{cm^3(STP)\,cm^{-3}}\ \text{at 65 bar}

      Total uptake, meeting the volumetric target set by DOE.

    11. Zero-coverage Q_st for CO2 (amine grafting)

      −Qst: 21 (non-grafted), 90 (en), 96 (pip), 82 (men), 96 (mmen), 46 (dimen) kJ mol−1-Q_{st}:\ 21\ (\text{non-grafted}),\ 90\ (\mathrm{en}),\ 96\ (\mathrm{pip}),\ 82\ (\mathrm{men}),\ 96\ (\mathrm{mmen}),\ 46\ (\mathrm{dimen})\ \mathrm{kJ\,mol^{-1}}

      A jump from 21 to 82 to 96 shows physisorption turning to chemisorption.

    12. Bare-metal-site distance label

      2.39 A˚2.39\ \mathrm{\AA}

      Label on the slide for the bare metal site. The slide does not name the two atoms it spans.

    13. H2 loading in HKUST-1 cages

      48 D2 (cage B);32 D2 (cage A)48\ \mathrm{D_2}\ (\text{cage B});\quad 32\ \mathrm{D_2}\ (\text{cage A})

      Neutron diffraction counts in the large cages once the bare metal sites are full.

    MOFs that act as magnets, wires, proton conductors and spin switches

    1. Fe(II) high-spin configuration

      (t2g)4(eg)2(Δ<P)(t_{2g})^4(e_g)^2 \quad (\Delta < P)

      Δ\Delta ligand field splitting, PP spin-pairing energy

      Weak ligand field: electrons spread out over both sets of orbitals.

    2. Fe(II) low-spin configuration

      (t2g)6(Δ>P)(t_{2g})^6 \quad (\Delta > P)

      Δ\Delta ligand field splitting, PP spin-pairing energy

      Strong ligand field: all six electrons pair up in t2gt_{2g}.

    3. Proton conductivity of a sulfonated MOF

      σ=1.27×10−1 S cm−1 at 100% RH and 80 °C\sigma = 1.27 \times 10^{-1}\ \mathrm{S\,cm^{-1}}\ \text{at 100\% RH and 80 °C}

      σ\sigma proton conductivity, RH relative humidity

      A very high value for a proton conductor.

    4. Spin states of Cu(II) and the radical ligand

      S=12 (each)S = \tfrac{1}{2}\ \text{(each)}

      Cu(II) is d9^9 with one unpaired electron. The ligand is a stable radical.

    5. ReO3 conduction band

      Re6+: 5d1 (t2g1)Re^{6+}:\ 5d^1\ (t_{2g}^1)

      The one 5d electron delocalises through the band, so ReO3_3 is metallic.