Question

# The normal freezing point of water is 0.0 oC. At this temperature the density of liquid...

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?

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

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