A fictitious fragile liquid Z has fusion entropy at its melting point

PHYSICS · Multi-part · Fermi University Physics · Condensed Matter, Materials & Soft Matter · Crystal structure, bonding and diffraction · Lattice dynamics and thermal properties of solids

Solid State Physics, Ch Solids without a lattice: glasses and quasicrystals

Problem

A fictitious fragile liquid Z has fusion entropy ΔSm=20 J mol−1 K−1\Delta S_m=20\,\mathrm{J\,mol^{-1}\,K^{-1}} at its melting point Tm=300 KT_m=300\,\mathrm{K}. Its excess heat capacity divided by temperature, ΔCp/T\Delta C_p/T (with ΔCp=Cpliquid−Cpcrystal\Delta C_p=C_p^{\rm liquid}-C_p^{\rm crystal}), has been measured on cooling: ΔCp/T=0.10\Delta C_p/T=0.10, 0.120.12 and 0.14 J mol−1 K−20.14\,\mathrm{J\,mol^{-1}\,K^{-2}} at T=300T=300, 275275 and 250 K250\,\mathrm{K} respectively. Liquid Z has Tg=250 KT_g=250\,\mathrm{K}. A second, unrelated glass W has its heat capacity presented as C/TC/T against T2T^2 below 1 K1\,\mathrm{K} (the standard way of exposing the linear two-level term against the Debye T3T^3 term, since C=c1T+c3T3C=c_1T+c_3T^3 gives C/T=c1+c3T2C/T=c_1+c_3T^2): C/T=1.55×10−6 J g−1 K−2C/T=1.55\times10^{-6}\,\mathrm{J\,g^{-1}\,K^{-2}} at T=0.20 KT=0.20\,\mathrm{K} and 1.85×10−6 J g−1 K−21.85\times10^{-6}\,\mathrm{J\,g^{-1}\,K^{-2}} at T=0.40 KT=0.40\,\mathrm{K}. ((a)) Using ΔS(T)=ΔSm−∫TTm(ΔCp/T′) dT′\Delta S(T)=\Delta S_m-\int_T^{T_m}(\Delta C_p/T')\,dT' and the trapezoidal rule on the table for liquid Z, find ΔS\Delta S at Tg=250 KT_g=250\,\mathrm{K}. ((b)) Extrapolating linearly below TgT_g using the slope ΔCp/T\Delta C_p/T at TgT_g, estimate the temperature TKT_K at which ΔS\Delta S would extrapolate to zero for liquid Z. ((c)) For glass W, use the two tabulated points to find the intercept c1c_1 of C/TC/T plotted against T2T^2, and hence, via c1=π26kB2Pˉc_1=\frac{\pi^2}{6}k_B^2\bar P, the two-level density of states Pˉ\bar P per unit mass.

Browse the problem library