Question

The solid/liquid phase transition of water near 1 atm external pressure is an important feature of our daily lives. Some relevant data concerning the transition include the density of liquid water at 273.15 K (0.999 g cm^-3), the density of ice at 273.15 K (0.917 g cm^-3), and the enthalpy of fusion of water at 273.15 K (6.008 kJ mol^-1).

(a)         On a phase diagram, this transition appears as a line. Calculate the slope of this transition line near the triple point of water (in math terms, calculate dp/dT near T = 273.16 K, p = 0.006 atm.). Assume all densities and enthalpies are independent of external pressure.

Answer 1

Answer :- in the graph ,

● at point A- temperature =273.15 K and pressure =0.006 atm

●at point B- temperature = 278.15 and pressure = 0.05 atm

Therefore by using slope formula - slope = dp/dT

We get, slope = 0.05 - 0.006 / 278.15 - 273.15

= 0.044 / 5

= 0.0088