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PROTEIN PHYSICS LECTURE 5

PROTEIN PHYSICS LECTURE 5. Energy: E (enthalpy: H=E+PV) PV=N k B T for gases, PV << N k B T for solids & liquids Entropy: S = k B • ln(#STATES) If AB=A+B  E AB = E A +E B ; V AB = V A +V B ; S AB = S A +S B

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PROTEIN PHYSICS LECTURE 5

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  1. PROTEIN PHYSICS LECTURE 5

  2. Energy: E (enthalpy: H=E+PV) PV=NkBT for gases, PV << NkBTfor solids & liquids Entropy: S = kB•ln(#STATES) If AB=A+B  EAB= EA+EB; VAB= VA+VB; SAB= SA+SB while #AB= #A• #B Free energy: G = H-TS  F = E-TS ------- G: used when P=constfor solids & liquids Chemical potential:   G(1) = G/N Probability ~ exp(-G/kBT)  max G  min

  3. G = H-TS; dG = (dH-TdS) - SdT • At T=const (i.e., at dT=0): • at equilibrium, G = min, i.e., dH-TdS=0. • Thus: • T = dH/dS • S = -dG/dT • H = G+TS = G-T(dG/dT) • Probability ~ exp(-G/kBT) • kB= 2cal/(61023) = 2cal/mol = R • kB  300o = R  300o = 0.6 kcal/mol

  4. int: “Free energy of interactions” Chemical potential: m G(1) = Gint - T•kBln(V(1))  Gint + T•kBln[C] EQUILIBRIUM for transition of molecule 1 from A to B: GA(1) = GB(1) chemical potentials in A and B are equal GintAB = GintB – GintA GintAB= kBT•ln([CinA]/[CinB]) ===================================================

  5. Experiment: GintAB= kBT•ln([C1inA]/[C1inB]) SintAB = -d(GintAB)/dT HintAB = GintAB +TSintAB [C] of C6H12 in H2O: 50 times less than in gas; 100000 times less than in liquid C6H12 C6H12 T=2980K=250C

  6. -2/3 Loss: S (usual case) -2/3 +1/3 H-bond: directed Loss: LARGE E (rare case)

  7. High heat capacity d(H)/dT: Melting of “iceberg”

  8. 20-25 cal/mol per Å2 of accessible non-polar surface

  9. ______ large effect _______ small ______ large

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