Answer to Question #142579 in Physical Chemistry for NimoRenold

Question

The enthalpy of fusion of ice is 6.01 kJ mol^-1, and ice shrinks by 1.70 cm³ per mole, when it melts at 1 atm pressure. What would be the melting point of ice at 1cde atm (in my case, 1354 atm)? Assume 1 atm = 101325 Pa.

Answer 1

Clausius-Clapeyron equation is:

dP/dT = $\Delta$H/T$\Delta$V

$\Delta$H = 6010 J/mole

$\Delta$V = -1.70*10^-6 m³/mole

Then dT/T = dP*$\Delta$V/$\Delta$H

ln(T2/T1) = (P2-P1)*$\Delta$V/$\Delta$H

normal melting point is T1 = 273.15 K, P1 = 1 atm = 101325 Pa

At raised pressure P2 = 1354 atm the melting temperature T2 is:

T2 = T1*exp{(P2-P1)*$\Delta$V/$\Delta$H} = 273.15*exp{-1353*101325*1.70*10^-6/6010} = 262.76 K

Melting point will be 262.76 K