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 [latex]P_{vap}[/latex] 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: [latex]C = ln P_{1} + \frac{ΔH_{vap}}{RT_{1}} = ln P_{2} + \frac{ΔH_{vap}}{RT_{2}}[/latex] This equation can be rearranged to solve for the desired quantity, In [latex]P_{2}[/latex]: [latex]ln P_{2} = ln P_{1} + \left(\frac{ΔH_{vap}}{R}\right)\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)[/latex] where [latex]P_{1}[/latex] = 100.0 mm Hg and ln [latex]P_{1}[/latex] = 4.6052, [latex]ΔH_{vap}[/latex] = 38.6 kJ/mol, R = 8.3145 J/((K · mol), [latex]T_{2}[/latex] = 338.2 K (65.0 °C), and [latex]T_{1}[/latex] = 307.9 K (34.7 °C).

ln [latex]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) [/latex] ln [latex]P_{2}[/latex] = 4.6052 + 1.3509 = 5.9561 [latex]P_{2}[/latex] = antiln (5.9561) = 386.1 mm Hg Antilogarithms are reviewed in Appendix A.2.

Question 10.5: CALCULATING A VAPOR PRESSURE USING THE CLAUSIUS–CLAPEYRON EQ...

Chemistry

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).

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