Answer:
286 J/K
Explanation:
The molar Gibbs free energy for the vaporization (ΔGvap) is:
ΔGvap = ΔHvap - T.ΔSvap
where,
ΔHvap: molar enthalpy of vaporization
T: absolute temperature
ΔSvap: molar entropy of the vaporization
When T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.
ΔHvap - Tb . ΔSvap = 0
ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.mol
When 1.35 mol of methanol vaporizes, the change in the entropy is:
[tex]1.35mol.\frac{212J}{K.mol} =286 J/K[/tex]
