Question

When a certain liquid (MM = 54.6 g mol-1) freezes at 9.62°C and 1 atm, its density changes from 0.781 g cm-3 to 0.803 g cm-3. Its enthalpy of fusion is 11.16 kJ mol-1. Find the freezing point of the liquid in °C at 79 MPa.

Answer 1

dP/dT = deltaS_fusion / deltaV_fusion

=> deltaP/deltaT = deltaS_fus/deltaV_fus

deltaS_fus = deltaH_fus/T

so, deltaP/deltaT = deltaH_fus/TdeltaV_fusion

delta T = T * deltaP * deltaV_fus / delta H_fus

deltaV_fus = MM(( 1/density₁ - 1/density₂)

= 54.6g/mol ( 1/0.781gcm^-3 - 1/0.803gcm^-3)

= 1.92 * 10^-6 m³/mol

delta T =[ 282.62K * (79*10⁶ Pa - 101325Pa) * 1.92*10^-6m³/mol ] / 11.16*10³ J/mol

delta T = 3.84K

Freezing point = T + deltaT = 282.62K+3.84K = 286.46K

Freezing point = 286.46 - 273 = 13.5⁰C