Question

The following paragraphs should be calculated assuming that the molar volume of Ethanol in its gaseous phase is far larger than its molar volume in either the solid or liquid phases.

Moreover, one should assume that the enthalpy involved in the transition between phases does not change with temperature.

Ethanol boils at 73.30 degrees Celsius at atmospheric pressure. The heat of vaporization, ΔΗ_vapis 39.3 kJ/mol. The melting point at atmospheric pressure is -114.10 degrees Celsius. The triple point is at pressure 0.00043 Pa and temperature -123.00 degrees Celsius. The density of liquid Ethanol at 25.0 degrees Celsius is 0.783 gr/cm³.

1. The heat of fusionΔΗ_meltis 4.9 kJ/mol. Calculate the change in volume brought about by a transition from the liquid phase to the solid phase at 300 K. Assume that the coexistence curve solid-liquid is a straight line with a constant slope along its entire trajectory. Is the result logical?
2. Draw the phase diagram, including the points referred to in the question.

Answer 1

Phase Diagram is given below, in the attachment,

From Clausius-Clapeyron equation,

Slope is change in pressure divided by change in temperature.

At triple point temperature = 150.15 K, pressure = 0.00043 Pa and at normal melting point temperature = 159.05 K, pressure = 10⁵ Pa.

So,

For melting,

T=300 K

According to the given data, at 25 ^oC or 298 K, 1 mol of EtOH (46 g) has a volume (46 / 0.783) cm³ = 58.748 cm³.

Our experimental temperature is only 2 K higher. So, at this temperature even if at very high pressure solid EtOH exists, volume of that would be negative. That is unacceptable. So, the answer is not at all logical. The assumption of linear slope is thus, wrong.