Question 10.5: Calculating a Vapor Pressure Using the Clausius–Clapeyron Eq......
Calculating a Vapor Pressure Using the Clausius–Clapeyron
Equation The vapor pressure of ethanol at 34.7 °C is 100.0 mm Hg, and the heat of vaporization of ethanol is 38.6 kJ/mol. What is the vapor pressure of ethanol in millimeters of mercury at 65.0 °C?
STRATEGY
There are several ways to do this problem. One way is to use the vapor pressure at T = 307.9 K (34.7 °C) to find a value for C in the Clausius–Clapeyron equation. You could then use that value to solve for ln P_{vap} at T = 338.2 K (65.0 °C).
Alternatively, because C is a constant, its value is the same at any two pressures and temperatures. That is:
C=\ln P_1+\frac{\Delta H_{\text {vap }}}{R T_1}=\ln P_2+\frac{\Delta H_{\text {vap }}}{R T_2}Tis equation can be rearranged to solve for the desired quantity, ln P_2:
\ln P_2=\ln P_1+\left(\frac{\Delta H_{\text {vap }}}{R}\right)\left(\frac{1}{T_1}-\frac{1}{T_2}\right)where P_1 = 100.0 mm Hg and ln P_1 = 4.6052, ΔH_{vap} = 38.6 kJ/mol, R = 8.3145 J/(K ° mol), T_2 = 338.2 K (65.0 °C), and T_1 = 307.9 K (34.7 °C).
Antilogarithms are reviewed in Appendix A.2.