Question 7.2: The triple point of methanol is at T =175.610 K and p =0.000...
The triple point of methanol is at T =175.610 K and p =0.0002 Pa.
Sketch the phase diagram at the vicinity of the triple point, using the following data: \Delta _{vap} h = 35.2 kJ/mol , \Delta _{fus} h = 3.173 kJ/mol , \rho _{l} = 0.7918 g/cm³ and \rho _{s} = 0.7802 g/cm³ . Suppose the shape of the coexistence curves to be linear within the range of the sketch and the vapor as an ideal gas.
The slope of each line is given by the Clapeyron equation (7.38) as
\frac{dP}{dT}= \frac{\Delta ^{\beta }_{\alpha }h }{T \Delta ^{\beta }_{\alpha }v}.The missing heat of sublimation can be calculated based on the state function property of enthalpy as \Delta _{sub} h =\Delta _{vap} h + \Delta _{fus} h. Molar volumes of solid and liquid methanol can be calculated from the densities as v=M /ρ, and the missing molar volume from the ideal gas equation of state. Using these quantities, the slopes of the three coexistence curves are the following:
v/l : 27.45 μPa/K ; v/s : 29.93 μPa/K ; l/s : -30.07 kPa/K.
Using these quantities as the slopes of linear portions of coexistence curves, the sketch looks like the one below. Note that the solid/liquid coexistence line looks completely vertical at a scale (shown) where the vapor/liquid and vapor/solid lines are not looking completely horizontal. However, the solid/liquid coexistence line is tilted to the left – similarly to that of water.
