Question

Using the Clapeyron equation find the pressure at which water freezes at -10 degrees celsius. The experimental value is 1104 bar. Briefly explain why the answer you found is greatly in error.

ΔH_fus​(H₂O)=6005j/mol   ρ(ice)=0.914g/cm³ ρ(liquid,water)= 1.00 g/m³​ all at 0 degrees C. Molecular weight of H₂​O is 18.01 g/mol

Answer 1

Apply Clapeyron equation

T = 273 - 10 = 263 K

L = 6005 J/mol

del v = change in specific volume = 914^-1 -1000^-1 = 9.4 x 10^-5 m³/ kg

det t = 0-(-10) = 10 K

Putting all values

dP = 2.42 x 10⁶ bar

P at 0^oC = 0.006 bar

P at -10^oC = 2.42 x 10⁶ bar

which is much higher than experimental value is 1104 bar.

The reason for error is Clapeyron equation is applicable for closed systems and during phase change where pressure and temperature are nearly constant.

1. In present case we are calculating pressure at -10^oC which is far from 0^oC where phase transition takes place

2. System won't be a closed system