Question

consider the substance, ethylene glycol. The vapor pressure at 100 C is 14.9 mmHg. The enthalpy of vaporization is 58.9 kJ/mol. what is the vapor pressure of ethylene glycol at 125 C?

Answer 1

T1 = 100.0 oC
=(100.0 + 273)K
= 373.0 K
T2 = 125.0 oC
=(125.0 + 273)K
= 398.0 K
P1 = 14.9 mmHg
ΔH = 58.9 KJ/mol
= 58900.0 J/mol

use:
ln(P2/P1) = (ΔH/R)*(1/T1 - 1/T2)
ln(P2/14.9) = (58900.0/8.314)*(1/373.0 - 1/398.0)
ln(P2/14.9) = 7084*(1.684*10^-4)
P2 = 49.1 mmHg
Answer: 49.1 mmHg