A liquid has a molar heat of vaporization of 22.7 kJ/mol.Its normal boiling point is 459 K.What is the vapor pressure,in mmHg,at 70 °C?

Question

Quizplus · Accepted Answer

The Clausius-Clapeyron equation can be used to calculate the vapor pressure at a new temperature given the heat of vaporization and the normal boiling point. The equation is ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1), where ΔHvap is the molar heat of vaporization, R is the gas constant (8.314 J/mol·K), T1 is the normal boiling point in Kelvin, and T2 is the new temperature in Kelvin. Converting 70°C to Kelvin gives 343K. Since the normal boiling point is 459K, where the vapor pressure equals 760 mmHg (standard atmospheric pressure), we can solve for P2, the vapor pressure at 343K. The calculation shows that the vapor pressure at 70°C is closest to 102 mmHg, making option A correct.

A Liquid Has a Molar Heat of Vaporization of 22