Under an increase of pressure of atmosphere, the volume of `1m^(3)` of ice is decreased to `0.986m^(3)`. Calculate the fall in the freezing point of water.

Explanation

`dP=1.013xx10^(5)=1.013xx10^(5) Nm^(-2)`
`V_(2)-V_(2)=-(1-0.986)=-0.014 m^(3)`
`T=273K,L=3.36xx10^(5)J kg^(-1)`
We know from Classius-Clapeyron equation,
`(dP)/(dT)=(L)/(T(V_(2)V_(1)))`
or `dT=(T(V_(2)-V_(1))dP)/(L)=(273xx(-0.014)(1.0130xx10^(3)))/(3.36xx10^(3))=-11.52K`