Liquids and Solutions Flashcards

Question

What are the assumptions in enthalpy of regular solution?

Answer 1

Random mixing Only nearest neighbour interactions, w Volume constant, and as H = U + PV, so ΔH = ΔU

Answer 2

Number of neighbours at a lattice site z=3, triangle z=4, square z=6, hexagonal

Answer 3

Δ_mH = (n₁ + n₂) Wx₁x₂ where W = N_Az[w₁₂ - (w₁₁+w₂₂)/2]

Answer 4

Δ_mG = Δ_mH - TΔ_mS Δ_mG = (n₁+n₂)Wx₁x₂ + RT[n₁lnx₁ + n₂lnx₂]

Answer 5

μ₁ = μ₁`*` + RTlnx₁ + wx₂² can relate to f₁ = exp[wx₂²/RT]

Answer 6

f₁ = exp[wx₂²/RT] ideal limit when w=0 so f₁ = 1

Answer 7

w₁₁ = w₁₂ = w₂₁ = w₂₂

Answer 8

P₁ = x₁P₁`*` exp[wx₂²/RT) = x₁P₁`*` f₁ Remember: for ideal gas and regular solution chemical eqm

Answer 9

Boundary of ideal gas and regular solution in eqm, μ_solution = μ_vapour μ₁`*` + RTlnx₁ + Wx₂² = μ₁^θ + RTln(P₁/P^θ) when x₁ = 1 and x₂ = 0, μ₁`*` = μ₁^θ + RTln(P₁/P^θ) then 1st - 2nd equation RTlnx₁ + Wx₂² = RTln(P₁/P₁`*`) rearrange to P₁ = x₁P₁`*` exp[Wx₂²/RT]

Answer 10

Starts with mix Finishes with two phases, neither are pure but just richer in one

Answer 11

g = wx₁x₂ + RT[x₁lnx₁ + x₂lnx₂] when high w then encourages demixing, when low then mixing when high T encourages mixing, when low T is demixing Wants to min Gibbs free energy

Answer 12

W = N_ARz[w₁₂ - (w₁₁+w₂₂)/2] if W +ve then w₁₁ and w₂₂ more -ve so more attractive than w₁₂

Answer 13

Critical temperature - T at beyond which there is a one-phase region

Answer 14

Double well needed for phase separation As T increases then one phase more likely

Answer 15

μ₁ = μ₁`*` - X₂(dg/dX₂) + g μ₂ = μ₂`*` - (1 - X₂)(dg/dX₂) + g

Answer 16

Minima gives coesting densities after separation (dg/dx₂^a) = [g(x₂^a) - g(x₂^b)]/(x₂^a-x₂^b)

Answer 17

(dg/dx₂)^a = (dg/dx₂)^b = 0 This is required for 2 phases to be present

Answer 18

(dg/dx₂)_Tc = (d²g/dx₂²)_Tc = 0 When this satisfied then at the Tc

Answer 19

Binodal - line on graph when (dg/dx₂)^a = (dg/dx₂)^b = 0 , gives 2 phases, dominates at lower T Spinodal - line on graph when (d²g/dx₂²)_Tc = 0 , gives mix, dominates at higher T

Answer 20

Treat as huge (slow and see-able) atoms Same stat thermo and similar phase behaviour

Answer 21

Polymer is a random walk of N statistical segments of length l Stat segments can include multiple momomers

Answer 22

Volume fraction formula for entropy of mixing of polymer solutions

Answer 23

N₁ molecules of type 1 (solvent) N₂ molecules of type 2 (polymer, with r stat segments per molecule N₂ Number of lattice sites, N = N₁ + rN₂

Answer 24

Equivalent to mole fractions but for space taken by solvent (1) or polymer (2) φ₁ = N₁/(N₁ + rN₂) φ₂ = rN₂/(N₁ + rN₂)

Answer 25

Ω_unmix = [rN₂!/N₂)(z/rN₂)]^(r-1)N2

Answer 26

Expected without corrections: Ω_mix = N!/(N₁! rN₂!) Must implement corrections: 1) random mix doesnt apply, so z/n for each segment other than first one (where z is coord) 2) Cannot switch chem segments unlike separate molecules, so multiply by rN₂!/N₂! Actual result Ω_mix = N!/(N₁! N₂!) (z/N)^(r-1)N2

Answer 27

Find by taking logs of Ω and using stirling's approx as for ideal Δ_mS = k_b[ln(N!/(N₁! N₂!) + (r-1)N₂ ln(rN₂/N)] = -R[n₁lnφ₁ + n₂lnφ₂]

Answer 28

H_unmix = (z/2)N₁w₁₁ + (z/2)rN₂w₂₂ = (1/2)(N₁+rN₂)[w₁₁N₁/N + w₂rN₂/N] H_unmix = z/2 (N₁+rN₂)(w₁₁φ₁ + w₂φ₂)

Answer 29

Start with H_mix = (z/2)N₁w₁₁ϕ₁ + (z/2)N₁w₁₂ϕ₂ +(z/2)N₁w₁₁ϕ₁ + (z/2)N₁w₁₂ϕ₂ + (z/2)rN₂w₂₂ϕ₂ + (z/2)rN₂w₂₁ϕ₁ Simplifies down to: H_mix = (z/2)(N₁ + rN₂)[w₁₁ϕ₁² + w₂₂ϕ₂² + 2w₁₂ϕ₁ϕ₂]

Answer 30

Δ_mH = (n₁ + rn₂)wφ₁φ₂ = RT(n₁ + rn₂)χφ₁φ₂

Answer 31

Δ_mG = RT[n₁lnφ₁ + n₂lnφ₂ + (n₁+ rn₂)χφ₁φ₂

Answer 32

μ₁ = μ₁`*` + RT[lnφ₁ + (1-1/r)φ₂ + χφ₂²]

Answer 33

Follow same process as with regular, just replace values

Answer 34

Semi-permeable membrane which polymer cannot pass through, but solvent can Osmotic pressure is difference in pressure on either side

Answer 35

Π = P^β - P^α Where β contains the polymer

Answer 36

Π = P^β - P^α = (μ₁ - μ₁`*`)/V₁`*` then sub in: μ₁= μ₁`*` + RT[lnφ₁ + (1-1/r)φ₂ + χφ₂² gives: Π = RT[ (1/M₂)c₂ + (1/2 - χ)/(ρ₂²V₁`*`) c₂² + ..]

Answer 37

Π = RT[ (1/M₂)c₂ + (1/2 - χ)/(ρ₂²V₁`*`) c₂² + ..] Virial expansion for gases: P = RT[(1/V) + B(T)(1/V_mean²) + ...]

Answer 38

Π = RT[c₂/ M₂] This is from the first term of expansion

Answer 39

Π = RT[ (1/M₂)c₂ + (1/2 - χ)/(ρ₂²V₁`*`) c₂² + ..] 2nd osmotic virial coefficient, A₂ = (1/2 - χ)/(ρ₂²V₁`*`) Plot c₂ on x and Π/c₂ on y y-intercept = RT/M₂ gradient = RTA₂

Answer 40

Good - max interactions with solvent, polymer spread out Theta - ideal osmotic pressure Poor - max interactions with self, polymer folded in

Answer 41

Good solvent : χ < 1/2 and A₂ > 0, T = θ Theta: χ = 1/2 and A₂ = 0, T =θ Poor solvent: χ > 1/2 and A₂ < 0, T =θ

Answer 42

**R** = Σ **I** <**R**> = Σ <**I**> = 0 This is 0 due to the randomness of the polymer

Answer 43

Removes effect of -ve contribution to the mean <**R**²> = Nl² = Ml²/m where M is mol weight and m is segment mol weight

Answer 44

R_g is just useful experimental value < R_g² > = < **R**² > / 6 so R_g proportional to sqrt(M)

Answer 45

A_v+B_v- -> v₊ A^Z+ + v_- B^Z-

Answer 46

Long-range electrostatic interactions Presence of ionic atmosphere of counter-ions

Answer 47

-ve of ion surrounded by +ve ions and vice cersa

Answer 48

μ₊ = μ₊^θ + RTlnc₊ + RTlny₊ c_i is molarity, equivalent for mole fraction y_i is activity coefficient equivalent

Answer 49

μ = μ^θ + vRTlnc_+- + vRTlny_+-

Answer 50

Potential of an isolated charge: ψ(r) = (1/r)(e/4πε₀ε_r) in solution: ψ(r) = (exp-κr/r) (e/4πε₀ε_r) then taylor expansion: ψ(r) = (1/r - κ) (e/4πε₀ε_r) where κ = debyte length

Answer 51

κ^-1 is range of ionic atmosphere κ^-1 = Sqrt[ε₀ε_rk_BT/2e²n₀] n₀ = N_+/-/V κ^-1 α 1/Sqrt[c] where c is salt conc

Answer 52

ψatm = -eκ/4πε₀ε_r

Answer 53

w_el = ∫ψ_atm dq = ∫-eκ/4πε₀ε_r dq where ψ_atm is potential of ionic atmosphere w_el = -κe²/8πε₀ε_r

Answer 54

RTlny = N_Aw_el = -κe²/8πε₀ε_rk_BT

Answer 55

logy_+- = - A Sqrt[c]

Answer 56

I = 1/2 Σ c_iz_i² where c is concentration and z is ionic charge

Answer 57

logy_+- = -z₊z_-A Sqrt[I] where I is an i

Answer 58

I = 1/2 Σ c_iz_i² I = (1/2) (c) (1`*`1² + 1`*`1²) = c

Answer 59

log c = (1/2)logK_s + A Sqrt[I] Sol increases if inert electrolyte conc increases

Answer 60

Adding more inert salt decreases work to bring in more ions, so more soluble As decreases debye length

Answer 61

Force per unit length which resists an external force Minmises surface area

Answer 62

γ = surface tension σ = interface

Answer 63

Work against surface tension which wants to min surface area F = 2γl, where l is length and 2 as bottom and top surface of film dw = Fdx = 2γldx = γdσ

Answer 64

Bulk dominates for large systems (length³) Surface dominates for smaller systems (length²)

Answer 65

Molecules attracted to surrounded in bulk Costs energy to move a molecule from bulk to surface as loses attraction so it costs work to create a surface This is surface tension

Answer 66

Small bubble goes into large This is because pressure in smaller bubble is larger

Answer 67

ΔP = 2γ/R where R is radius of droplet when isothermal enlargement

Answer 68

P_inside > P_outside so inside can act against force preventing it from forming

Answer 69

outside force (expansion work then expansion work against outside P0) = inside force (work to increase bubble size) (P₀+ΔP)dV - P₀dV = γdσ σ = 4πR² dσ = 8πRdR ΔP (4πR²dr) = γ8πRdR ΔP = 2γ/R

Answer 70

Vapour pressure of a liquid droplet

Answer 71

P_r = P₀ exp[2γv_I/rRT] where P_r is vap pressure of a liquid droplet with radius r, with respect to P₀

Answer 72

Small droplet has large lapace pressure so large (2γ/r) P_r >> P₀

Answer 73

Large droplet has small lapace pressure so small (2γ/r) P_r ~ P₀

Answer 74

Eqm between liquid droplet and gas 1) μ_liq(P₀₎) = μ_gas (P₀) 2) μ_liq(P_r + 2γ/r) = μ_gas(P_r) then do 2-1 and then integrate use molar volume liquid (v_l) = molar volume gas (v_g) = RT/P Then RTln(Pr/P0) >> vl(Pr-P0) so approx to give (2γv_l)/r = RTln(Pr/P0) rearrange to give P_r = P₀ exp[2γv_l/rRT)

Answer 75

Smallest droplet size possible in a pressure P r`*` = 2γv_l/RTln(P/P0)

Answer 76

When smaller vap pressure (Pr) than required due to bigger size of a droplet then is more likely to nucleate

Answer 77

Relation between adsorption and surface tension

Answer 78

Approx to conc profile Gibbs dividing surface has no colume No excess or depletion of solvent molecules

Answer 79

Dividing surface volume, V^σ = 0 V_total = V^α + V^β

Answer 80

No excess/depletion of solcent molecules (no molecules in interface) n₁^σ n_1,total = n₁^α + n₁^β

Answer 81

n_i^σ = n_i,total - (n₁^α + n₁^β) XS of this amount of this component actually present in system over that present in a reference system

Answer 82

n₁^α = V^αc_i^α n₁^β = V^βc_i^β These come from integral beneath conc profile

Answer 83

Over and underestimates the curve by the same amount so interface is 0

Answer 84

Positive - surfactants wants to be at interface more than bulk Negative - surfactant does not want to be at the surface

Answer 85

Positive - surfactants wants to be at interface more than bulk Negative - surfactant does not want to be at the surface

Answer 86

Γ_i¹ = n_i^σ/σ this is for solvent - component 1

Answer 87

dA^σ = -S^σdT + Σμ_idn_i^σ + γdσ

Answer 88

dγ = -(S^σ/σ)dT - Σ (n_i^σ/σ) dμ Where the sum is from i to c S^σdT + Σμ_idn_i^σ + γdσ

Answer 89

dγ = - Σ Γ_i¹ dμ_i

Answer 90

Sub in μ2 = μ2^θ + RTlnc₂ Into following equation: dγ = - Σ Γ_i¹ dμ_i Gives dγ = - RTΓ₂¹ dlnc₂ Surface tension therefore a function of conc c2