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The normal boiling point of ethanol is 78.4 oC. Its enthalpy of vaporization is 38.6 kJ/mol. Estimate the vapor pressure of ethanol at 26.3 oC.

Sagot :

Answer: The vapor pressure of ethanol at [tex]26.3^{o}C[/tex] is 238.3 torr.

Explanation:

Given: [tex]\Delta H_{vap}[/tex] = 38.6 kJ/mol

[tex]T_{1} = 26.3^{o}C = (26.3 + 273) K = 299.3 K[/tex]

[tex]T_{2} = 78.4^{o}C = (78.4 + 273) K = 351.4 K[/tex]

Formula used to calculate the vapor pressure of ethanol is as follows.

[tex]ln\frac{P_{2}}{P_{1}} = \frac{\Delta H_{vap}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]\\[/tex]

Substitute the values into above formula as follows.

[tex]ln\frac{P_{2}}{P_{1}} = \frac{\Delta H_{vap}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]\\ \\ln \frac{760 torr}{P_{1}} = \frac{38600 J}{8.314 J/mol K}[\frac{1}{299.3} - \frac{1}{351.4}]\\\frac{760}{P_{1}} = 3.18\\P_{1} = 238.3 torr[/tex]

Thus, we can conclude that the vapor pressure of ethanol at [tex]26.3^{o}C[/tex] is 238.3 torr.

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