Answer :
Final answer:
The entropy change (ΔS) for the vaporization of one mole of ethanol at its boiling point can be calculated using the Clausius-Clapeyron equation. Substituting the given values into the equation, the entropy change is approximately 133.6 J/mol·K.
Explanation:
The entropy change (ΔS) for the vaporization of one mole of ethanol at its boiling point can be calculated using the Clausius-Clapeyron equation:
ln(P₂/P₁) = -ΔHvap/R * (1/T₂ - 1/T₁)
where P₁ and P₂ are the vapor pressures of ethanol at two different temperatures (T₁ and T₂), ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. Rearranging the equation gives:
ΔS = (-R * ln(P₂/P₁)) / (1/T₂ - 1/T₁)
Substituting the given values (P₁ = 5.95 kPa, P₂ = 53.3 kPa, T₁ = 20.0 °C = 293 K, T₂ = 63.5 °C = 336.5 K) into the equation, we can calculate the entropy change:
ΔS = (-8.314 kJ/mol·K * ln(53.3/5.95)) / (1/336.5 - 1/293)
ΔS ≈ 133.6 J/mol·K
