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 in 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{ΔH_{vap}}{RT_{1}} = ln P_{2} + \frac{ΔH_{vap}}{RT_{2}}This equation can be rearranged to solve for the desired quantity, In P_{2}:
ln P_{2} = ln P_{1} + \left(\frac{ΔH_{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).
ln P_{2} = 4.6052 +\left(\frac{38,600 \frac{J}{mol}}{8.3145 \frac{J}{K · mol}}\right) \left( \frac{1}{307.9 K} – \frac{1}{338.2 K}\right)
ln P_{2} = 4.6052 + 1.3509 = 5.9561
P_{2} = antiln (5.9561) = 386.1 mm Hg
Antilogarithms are reviewed in Appendix A.2.