Answer: - 38.1 kJ/mol
Explanation:
ln(P1/P2) = (-ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively,
ΔHvap is the molar heat of vaporization,
R is the ideal gas constant (8.314 J/(mol·K)), and
ln represents the natural logarithm.
P1 = 40.0 mm Hg
T1 = 326 K
P2 = 100.0 mm Hg
T2 = 343 K
P1 = 40.0 mm Hg = 40.0/760 atm ≈ 0.0526 atm
T1 = 326 K
P2 = 100.0 mm Hg = 100.0/760 atm ≈ 0.1316 atm
T2 = 343 K
ln(0.0526/0.1316) = (-ΔHvap/8.314) * (1/326 - 1/343)
n(0.4) = (-ΔHvap/8.314) * (0.00306748 - 0.00291545)
ln(0.4) = (-ΔHvap/8.314) * 0.00015203
ΔHvap = -(8.314 * ln(0.4))/0.00015203
ΔHvap ≈ -(8.314 * ln(0.4))/0.00015203 ≈ - 38,051.4 J/mol ≈ - 38.1 kJ/mol
