Question

Calculate the value of the heat of vaporization of ice at -1°C . The values of the pressure
of water vapor are 3.88 mm Hg at -2°C and 4.58 mm Hg at 0°C,
(a) 5.10 x10^ J/mol
(b) 1x10^ J/mol
(c) 9x109 J/mol
(d) 3.3x10^ J/mol​

Answer 1

by using the Clausius-Clapeyron equation

$\frac{dp}{dt}$ =Δs/Δv≈$\frac{L}{TVg} =\frac{PL}{T(PVg)}=\frac{PL}{RT^{2} }$

leading to in P= $\frac{-L}{RT}+CONSTANT$

$\frac{-P1}{P2}=-\frac{-L}{R}(\frac{1}{T1}-\frac{1}{T2} )$

$\frac{L}{R} =\frac{P1}{P2}/(\frac{1}{T1}-\frac{1}{T2}) =6135.55$

OR $5.10*10^ J/mol$