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 IDENTIFY Known Unknown [latex]∆H_{vap}[/latex] = 38.6 kJ/mol Vapor pressure at 65.0 °C ([latex]P_{2}[/latex]) Vapor pressure at 34.7 °C ([latex]P_{1}[/latex] = 100.0 mm Hg) [latex]T_{1} [/latex]= 34.7 °C [latex]T_{2}[/latex] = 65.0 °C STRATEGY Use the rearranged form of the Clausius–Clapeyron equation that enables us to calculate the vapor pressure of the liquid at any other temperature if we know the heat of vaporization and the vapor pressure at one temperature. Convert temperature to Kelvin and [latex]∆H_{vap}[/latex] to units of joules. [latex] ln(\frac{P_{1}}{P_{2}})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})[/latex]

Substitute all known values into the equation and solve for[latex]P_{2}[/latex] . [latex]ln(\frac{100.0 mm Hg}{P_{2}} )=\frac{38,600 J/mol}{ 8.314 J/mol .K}(\frac{1}{ 338.2}-\frac{1}{307.9})[/latex] [latex]ln(\frac{100.0 mm Hg}{P_{2}} )= -1.3509[/latex] [latex]\frac{100.0 mm Hg}{P_{2}}=e^{-1.3509}=0.2590[/latex] [latex]P_{2}=386 mm Hg[/latex] CHECK We would expect the vapor pressure to be larger at a higher temperature so the answer is physically reasonable. It is difficult to estimate the magnitude of the new vapor pressure due to the complexity of the calculation.

Question 11.1: calculating a Vapor pressure Using the clausius–clapeyron Eq...

Chemistry
General Chemistry: Atoms First [1015060]
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

IDENTIFY

Known	Unknown
$∆H_{vap}$ = 38.6 kJ/mol	Vapor pressure at 65.0 °C ( $P_{2}$ )
Vapor pressure at 34.7 °C ( $P_{1}$ = 100.0 mm Hg)
$T_{1}$ = 34.7 °C
$T_{2}$ = 65.0 °C

STRATEGY

Use the rearranged form of the Clausius–Clapeyron equation that enables us to calculate the vapor pressure of the liquid at any other temperature if we know the heat of vaporization and the vapor pressure at one temperature. Convert temperature to Kelvin and $∆H_{vap}$ to units of joules.

$ln(\frac{P_{1}}{P_{2}})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$

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Substitute all known values into the equation and solve for $P_{2}$ .

$ln(\frac{100.0 mm Hg}{P_{2}} )=\frac{38,600 J/mol}{ 8.314 J/mol .K}(\frac{1}{ 338.2}-\frac{1}{307.9})$

$ln(\frac{100.0 mm Hg}{P_{2}} )= -1.3509$

$\frac{100.0 mm Hg}{P_{2}}=e^{-1.3509}=0.2590$

$P_{2}=386 mm Hg$

CHECK

We would expect the vapor pressure to be larger at a higher temperature so the answer is physically reasonable. It is difficult to estimate the magnitude of the new vapor pressure due to the complexity of the calculation.