Question

determine the change in the freezing point of ice with increasing pressure. the molar volume of water is 18.02 cm^3 mol^-1 and the molar volume of ice is 19.63 cm^3 mol^-1 at 273.15K. The molar heat of fusion is 6.009E3 J/mol.

Answer 1

From the Clapeyron equation

dP/dT = Hf /(Tf x V)

dT/dP = (Tf x V) / (Hf)

V = change in molar volume = molar volume of water - molar volume of ice

= 18.02 - 19.63

= - 1.61 cm3/mol x (1L/1000cm3)

= - 0.00161 L/mol

Freezing point of water Tf = 273.15 K

molar heat of fusion Hf = 6.009 x 10^3 J/mol

dT/dP = (Tf x V) / (Hf)

= (273.15 K) x (-0.00161 L/mol) / (6.009 x 10^3 J/mol)

= - (7.32 x 10^-5 K-L/J) x (101.33 J/L-atm)

dT/dP = - 7.42 x 10^-3 K/atm

Change in freezing point with increasing pressure from P1 to P2

Where P2 > P1

dT = - 7.42 x 10^-3 (P2 - P1) K