Question

When a certain liquid (MM = 49.9 g mol-1) freezes at -8.51°C and 1 atm, its density changes from 0.787 g cm-3 to 0.808 g cm-3. Its enthalpy of fusion is 6.27 kJ mol-1. Find the freezing point of the liquid in °C at 52 MPa.

Answer 1

ans)

from above data that

The Clausius-Clapeyron equation for solid-liquid phase transition is given by

dP/dT = ΔH/(T.ΔV)

where ΔH is the enthalpy of fusion; T = temperature of fusion and ΔV = change in molar volume due to the phase transition.

The left hand side can be approximated as ΔP/ΔT; therefore,

ΔP/ΔT = ΔH/(T.ΔV)

===> ΔT = T.ΔV.ΔP/ΔH where we will write T = T_fus = melting point of the liquid at 1 atm pressure = -8.51⁰C = [273 + (-8.51)] K = 264.49 K.

Therefore,

T₂ – T₁ = T_fus*ΔV*(P₂ – P₁)/ΔH …..(1)

Given P₁ = 1 atm, P₂ = 52 MPa = (52 MPa)*(9.869 atm/1 MPa) = 513.188 atm.

Calculate ΔV = V_s – V_l where V_s (49.9 g/mol)/(0.808 g/cm³)*(1 L/1000 cm³) = 0.062 L/mol.

V_l = (49.9 g/mol)/(0.787 g/cm³)*(1 L/1000 cm³) = 0.063 L/mol.

Therefore, ΔV = (0.062 L/mol) – (0.063 L/mol) = -0.001 L/mol.

Also note that ΔH = 6.27 kJ/mol = (6.27 kJ/mol)*(1000 J/1 kJ)*(1 L-atm/101.32 J) = 61.88 L-atm/mol

Now plug in values in (1) and write

T₂ – T₁ = (264.49 K)*(-0.001 L/mol)*(513.188– 1)atm/(61.88 L-atm/mol) = -2.1892 K

Therefore, T₂ = T₁ – 2.1892 K = 264.49 K – 2.1892 K = 262.30 K = (262.30 – 273)⁰C = -10.7⁰C (ans).

The freezing point of the liquid is -10.7 ⁰C at 52 MPa pressure