Respuesta :
Explanation:
According to Clausius-Claperyon equation,
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
The given data is as follows.
[tex]T_{1} = 63.5^{o}C[/tex] = (63.5 + 273) K
= 336.6 K
[tex]T_{2} = 78^{o}C[/tex] = (78 + 273) K
= 351 K
[tex]P_{1}[/tex] = 1 atm, [tex]P_{2}[/tex] = ?
Putting the given values into the above equation as follows.
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
[tex]ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}][/tex]
[tex]\Delta H[/tex] = [tex]\frac{0.559}{1.466 \times 10^{-4}} J/mol[/tex]
= [tex]0.38131 \times 10^{4} J/mol[/tex]
= 3813.1 J/mol
Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.
