Question

The normal freezing point of water is 0.0 oC. At this temperature the density of liquid water is 1.000 g/ml and the density of ice is 0.917 g /ml. The increase in enthalpy for the melting of ice at this temperature is 6010 J/mol. What is the freezing point of water at 200 atmospheres?

Answer 1

We will use Clausius-Clapeyron Equation, which is

Volume = Mass / Density

Volume ice = 18.02 / 0.917 = 19.65

Volume water = 18.02 / 1 = 18.02

volume change in melting = 18.02 - 19.65 = -1.63 cm^3 = -1.63 X 10^-6 m^3 / mole

Delta H fusion = 6010 J / mole

Delta P / Delta T = 6010 J / mole / 1.63 X 10^-6 m^3 / mole X 273.15 = 13.49 X 10^6 Pascal / K

P1 = 1 atm = 101325 pascal

P2 = 200 atm = 2.026 X 10^7 pascal

DeltaP = P2-P1 = (2.026 -0.010133 )X 10^7 = 2.016 X 10^7 pascals

delta T = 2.016 X 10^7 pascals / 13.49 X 10^6 Pascal / K = 1.494 K

TM(200atm) = 273.15 + 1.494 = 274.644 K