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Question 7.1: The equilibrium vapor pressure of a single component liquid ...

The equilibrium vapor pressure of a single component liquid at 300 K was found to be 0.2 bar. The heat of vaporization of the liquid is 40 kJ/mol. Assume the vapor to be an ideal gas, its density to be much less than that of the liquid. Estimate the temperature at which the equilibrium vapor pressure is doubled.

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The logarithmic form of (7.45) – the solution of the Clausius– Clapeyron equation – can be used in the following form:

P_{2}=P_{1} e^{-\frac{\Delta_{\text {vap }} h}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)} .                (7.45)

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

Solution of the equation inserting P_{1}/P_{2} = 2 yields 313.55 K at which the vapor pressure is doubled.

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